Advanced Quantum Physics Ben Simons
Contents 1 Wave mechanics and the Schr¨ odinger equation 1.1 Historical foundations of quantum physics . . . . . 1.1.1 Black-body radiation . . . . . . . . . . . . . 1.1.2 Photoelectric effect . . . . . . . . . . . . . . 1.1.3 Compton Scattering . . . . . . . . . . . . . 1.1.4 Atomic spectra . . . . . . . . . . . . . . . . 1.2 Wave mechanics . . . . . . . . . . . . . . . . . . . 1.2.1 Maxwell’s wave equation . . . . . . . . . . . 1.2.2 Schr¨odinger’s equation . . . . . . . . . . . . 1.2.3 Time-independent Schr¨odinger equation . . 1.2.4 Particle flux and conservation of probability
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2 Quantum mechanics in one dimension 2.1 Wave mechanics of unbound particles . . . . . . 2.1.1 Free particle . . . . . . . . . . . . . . . . 2.1.2 Potential step . . . . . . . . . . . . . . . . 2.1.3 Potential barrier . . . . . . . . . . . . . . 2.1.4 The rectangular potential well . . . . . . 2.2 Wave mechanics of bound particles . . . . . . . . 2.2.1 The rectangular potential well (continued) 2.2.2 The δ-function potential well . . . . . . . 2.2.3 Info: The δ-function model of a crystal .
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10 10 10 12 13 15 15 15 16 17
3 Operator methods in quantum mechanics 3.1 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Time-evolution operator . . . . . . . . . . . . . . . 3.1.2 Uncertainty principle for non-commuting operators 3.1.3 Time-evolution of expectation values . . . . . . . . 3.2 Symmetry in quantum mechanics . . . . . . . . . . . . . . 3.2.1 Observables as generators of transformations . . . 3.2.2 Consequences of symmetries: multiplets . . . . . . 3.3 The Heisenberg Picture . . . . . . . . . . . . . . . . . . . 3.4 Quantum harmonic oscillator . . . . . . . . . . . . . . . . 3.5 Postulates of quantum theory . . . . . . . . . . . . . . . .
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19 20 21 22 23 24 24 26 27 27 31
4 Quantum mechanics in more than one-dimension 4.1 Rigid diatomic molecule . . . . . . . . . . . . . . . . . . 4.2 Angular momentum . . . . . . . . . . . . . . . . . . . . 4.2.1 Commutation relations . . . . . . . . . . . . . . . 4.2.2 Eigenvalues of angular momentum . . . . . . . . 4.2.3 Representation of the angular momentum states 4.3 The central potential . . . . . . . . . . . . . . . . . . . . 4.4 Atomic hydrogen . . . . . . . . . . . . . . . . . . . . . .
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CONTENTS
vi
5 Motion in a magnetic field 5.1 Classical mechanics of a particle in a field . . . . 5.2 Quantum mechanics of a particle in a field . . . . 5.3 Atomic hydrogen: Normal Zeeman effect . . . . . 5.4 Gauge invariance and the Aharonov-Bohm effect 5.5 Free electrons in a magnetic field: Landau levels
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6 Spin 6.1 Spinors, spin pperators, Pauli matrices . . . . . . . 6.2 Relating the spinor to the spin direction . . . . . . 6.3 Spin precession in a magnetic field . . . . . . . . . 6.3.1 Paramagnetic Resonance . . . . . . . . . . 6.4 Addition of angular momenta . . . . . . . . . . . . 6.4.1 Addition of two spin 1/2 degrees of freedom 6.4.2 Addition of angular momentum and spin . 6.4.3 Addition of two angular momenta J = 1 . .
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7 Approximation methods for stationary states 7.1 Time-independent perturbation theory . . . . . . . . . 7.1.1 The Perturbation Series . . . . . . . . . . . . . 7.1.2 First order perturbation theory . . . . . . . . . 7.1.3 Second order perturbation theory . . . . . . . . 7.2 Degenerate perturbation theory . . . . . . . . . . . . . 7.3 Variational method . . . . . . . . . . . . . . . . . . . . 7.4 Wentzel, Kramers and Brillouin (WKB) method . . . 7.4.1 Semi-classical approximation to leading order . 7.4.2 Next to leading order correction . . . . . . . . 7.4.3 Connection formulae, boundary conditions and zation rules . . . . . . . . . . . . . . . . . . . .
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8 Identical Particles 8.1 Quantum statistics . . . . . . . . . . . . . 8.2 Space and spin wavefunctions . . . . . . . 8.3 Physical consequences of particle statistics 8.4 Ideal quantum gases . . . . . . . . . . . . 8.4.1 Non-interacting Fermi gas . . . . . 8.4.2 Ideal Bose gas . . . . . . . . . . .
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9 Atomic structure 9.1 The “real” hydrogen atom . . . . . . . . . . . . . . 9.1.1 Relativistic correction to the kinetic energy 9.1.2 Spin-orbit coupling . . . . . . . . . . . . . . 9.1.3 Darwin term . . . . . . . . . . . . . . . . . 9.1.4 Lamb shift . . . . . . . . . . . . . . . . . . 9.1.5 Hyperfine structure . . . . . . . . . . . . . . 9.2 Multi-electron atoms . . . . . . . . . . . . . . . . . 9.2.1 Central field approximation . . . . . . . . . 9.3 Coupling schemes . . . . . . . . . . . . . . . . . . . 9.3.1 LS coupling scheme . . . . . . . . . . . . . 9.3.2 jj coupling scheme . . . . . . . . . . . . . . 9.4 Atomic spectra . . . . . . . . . . . . . . . . . . . . 9.4.1 Single electron atoms . . . . . . . . . . . . 9.4.2 Helium and alkali earths . . . . . . . . . . . 9.4.3 Multi-electron atoms . . . . . . . . . . . . .
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89 90 90 91 93 93 95 96 97 102 102 105 106 107 108 109
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CONTENTS 9.5
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Zeeman effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 9.5.1 Single-electron atoms . . . . . . . . . . . . . . . . . . . 109 9.5.2 Multi-electron atoms . . . . . . . . . . . . . . . . . . . . 110
10 From molecules to solids 10.1 The H+ 2 ion . . . . . . . . . . 10.2 The H2 Molecule . . . . . . . 10.3 From molecules to solids . . . 10.4 Molecular spectra . . . . . . . 10.4.1 Molecular rotation . . 10.4.2 Vibrational transitions
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112 113 116 118 122 124 125
11 Field theory: from phonons to photons 11.1 Quantization of the classical atomic chain . . . . . . . . 11.1.1 Classical chain . . . . . . . . . . . . . . . . . . . 11.1.2 Quantum chain . . . . . . . . . . . . . . . . . . . 11.2 Quantum electrodynamics . . . . . . . . . . . . . . . . . 11.2.1 Classical theory of the electromagnetic field . . . 11.2.2 Quantum field theory of the electromagnetic field
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127 127 127 130 134 134 137
12 Time-dependent perturbation theory 12.1 Time-dependent potentials: general formalism . . . . . . . . . . 12.2 Time-dependent perturbation theory . . . . . . . . . . . . . . . 12.3 “Sudden” perturbation . . . . . . . . . . . . . . . . . . . . . . . 12.3.1 Harmonic perturbations: Fermi’s Golden Rule . . . . . 12.3.2 Info: Harmonic perturbations: second-order transitions
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13 Radiative transitions 13.1 Coupling of matter to the electromagnetic field 13.1.1 Spontaneous emission . . . . . . . . . . 13.1.2 Absorption and stimulated emission . . 13.2 Selection Rules . . . . . . . . . . . . . . . . . . 13.3 Lasers . . . . . . . . . . . . . . . . . . . . . . . 13.3.1 Operating principles of a laser . . . . . 13.3.2 Gain mechanism . . . . . . . . . . . . . 13.4 Driven harmonic oscillator . . . . . . . . . . . .
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14 Scattering theory 14.1 Basics . . . . . . . . . . . . . . . . . . 14.2 Method of partial waves . . . . . . . . 14.3 The Born approximation . . . . . . . . 14.4 Info: Scattering of identical particles 14.5 Scattering by an atomic lattice . . . .
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15 Relativistic Quantum Mechanics 15.1 Klein-Gordon equation . . . . . . . . . . . . 15.2 Dirac Equation . . . . . . . . . . . . . . . . 15.2.1 Density and Current . . . . . . . . . 15.2.2 Relativistic Covariance . . . . . . . . 15.2.3 Angular momentum and spin . . . . 15.2.4 Parity . . . . . . . . . . . . . . . . . 15.3 Free Particle Solution of the Dirac Equation 15.3.1 Klein paradox: anti-particles . . . . 15.4 Quantization of relativistic fields . . . . . . Advanced Quantum Physics
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15.4.1 Info: Scalar field: Klein-Gordon equation 15.4.2 Info: Charged Scalar Field . . . . . . . . 15.4.3 Info: Dirac Field . . . . . . . . . . . . . 15.5 The low energy limit of the Dirac equation . . . 16 Problem sets 16.1 Problem Set 16.2 Problem Set 16.3 Problem Set 16.4 Problem Set
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i
Preface Quantum mechanics underpins a variety of broad subject areas within physics and the physical sciences from high energy particle physics, solid state and atomic physics through to chemistry. As such, the subject resides at the core of every physics programme. By building upon the conceptual foundations introduced in the IB Quantum Physics course, the aim of Part II Advanced Quantum Physics is to develop further conceptual insights and technical fluency in the subject. Although 24 lectures is a long course to prepare(!), it is still insufficient to cover all of the topics that different physicists will describe as “core”. For example, some will argue that concepts of quantum computation should already be included as an established component of the core. Simiilarly, in the field of quantum optics, some will say that a detailed knowledge of atomic and molecular spectroscopy is key. In the field of solid state physics, the concept of second quantization in many-body physics is also considered central. In all of these cases, we will be able to touch only the surface of the subject. However, the material included in this course has been chosen to cover the key conceptual foundations that provide access to these more advanced subjects, the majority of which will be covered in subequent optional courses in Part II and Part III. In the following, we list an approximate “lecture by lecture” synopsis of the different topics treated in this course. Those topics marked by a † will be covered if time permits. 1 Foundations of quantum physics: Overview of course structure and organization; brief revision of historical background: from wave mechanics to the Schr¨odinger equation. 2 Quantum mechanics in one dimension: Wave mechanics of unbound particles; potential step; potential barrier and quantum tunneling; bound states; rectangular well; δ-function potential well; † KronigPenney model of a crystal. 3 Operator methods in quantum mechanics: Operator methods; uncertainty principle for non-commuting operators; Ehrenfest theorem and the time-dependence of operators; symmetry in quantum mechanics; Heisenberg representation; postulates of quantum theory; quantum harmonic oscillator. 4 Quantum mechanics in more than one dimension: Rigid diatomic molecule; angular momentum; commutation relations; raising and lowering operators; representation of angular momentum states. 5 Quantum mechanics in more than one dimension: Central potential; atomic hydrogen; radial wavefunction. 6 Motion of charged particle in an electromagnetic field: Classical mechanics of a particle in a field; quantum mechanics of particle in a field; atomic hydrogen – normal Zeeman effect; † diamagnetic hydrogen and quantum chaos; gauge invariance and the Aharonov-Bohm effect; free electrons in a magnetic field – Landau levels. 7-8 Quantum mechanical spin: History and the Stern-Gerlach experiment; spinors, spin operators and Pauli matrices; relating the spinor to spin direction; spin precession in a magnetic field; parametric resonance; addition of angular momenta. Advanced Quantum Physics
ii 9 Time-independent perturbation theory: Perturbation series; first and second order expansion; degenerate perturbation theory; Stark effect; nearly free electron model. 10 Variational and † WKB method: Ground state energy and eigenfunctions; application to helium; excited states; † Wentzel-Kramers-Brillouin method. 11 Identical particles: Particle indistinguishability and quantum statistics; space and spin wavefunctions; consequences of particle statistics; ideal quantum gases; degeneracy pressure in neutron stars; Bose-Einstein condensation in ultracold atomic gases. 12-13 Atomic structure: Relativistic corrections; spin-orbit coupling; Darwin structure; Lamb shift; hyperfine structure; Multi-electron atoms; Helium; Hartree approximation and beyond; Hund’s rule; periodic table; coupling schemes LS and jj; atomic spectra; Zeeman effect. 14-15 Molecular structure: Born-Oppenheimer approximation; H2 +ion; H2 molecule; ionic and covalent bonding; molecular spectra; rotation; nuclear statistics; vibrational transitions. 16 Field theory of atomic chain: From particles to fields: classical field theory of the harmonic atomic chain; quantization of the atomic chain; phonons. 17 Quantum electrodynamics: Classical theory of the electromagnetic field; theory of waveguide; quantization of the electromagnetic field and photons. 18 Time-independent perturbation theory: Time-evolution operator; Rabi oscillations in two level systems; time-dependent potentials – general formalism; perturbation theory; sudden approximation; harmonic perturbations and Fermi’s Golden rule; second order transitions. 19 Radiative transitions: Light-matter interaction; spontaneous emission; absorption and stimulated emission; Einstein’s A and B coefficents; dipole approximation; selection rules; lasers. 20-21 Scattering theory I: Basics; elastic and inelastic scattering; method of particle waves; Born approximation; scattering of identical particles. 22-24 Relativistic quantum mechanics: History; Klein-Gordon equation; Dirac equation; relativistic covariance and spin; free relativistic particles and the Klein paradox; antiparticles and the positron; Coupling to EM field: gauge invariance, minimal coupling and the connection to nonrelativistic quantum mechanics; † field quantization.
Handout To accompany the course, a substantial handout has been prepared.1 In some cases, the handout contains material (usually listed as Info blocks) that goes beyond the scope of the lectures. Needless to say, the examination will be 1 I should note that, in preparing the handout, I have made use of some web-based material – particularly the excellent lecture notes by Fowler at Virginia – and notes prepared by David Ward in previous generations of the course. I have also included links to useful material on the course webpage.
Advanced Quantum Physics
iii limited to material that is covered in lectures and not the handout. Since this handout is substantially new, it is inevitable that there will be some typographical errors – some of them may even be important... I would be most grateful if you could e-mail the errors that you find to bds10@cam. I will try to maintain a “corrected” set of notes on the web. The overheads used in lectures can also be recovered from the course web2 site along with other relevant and useful material.
Problem Sets The problem sets are a vital and integral part of the course providing the means to reinforce key ideas as well as practice techniques. Problems indicated by a † symbol are regarded as challenging. Throughout these notes, I have included a number of simpler exercises which may be completed “along the way”, and aim to reinforce some of the ideas developed in the text.
Books As a core of every undergraduate and graduate physics programme, there is a wealth of excellent textbooks on the subject. Choosing ones that suit is a subjective exercise. Apart from the handout, I am not aware of a text that addresses all of the material covered in this course: Most are of course more dense and far-reaching, and others are simply more advanced or imbalanced towards specialist topics. At the same time, I would not recommend relying solely on the handout. Apart from the range of additional examples they offer, the textbooks will often provide a more erudite and engaging discussion of the material. Although there are too many texts on the subject to discuss every one, I have included below some of the books that I believe to be particularly useful.
2
http://www.tcm.phy.cam.ac.uk/~bds10/aqp.html
Advanced Quantum Physics
Bibliography [1] B. H. Bransden and C. J. Joachain, Quantum Mechanics, (2nd edition, Pearson, 2000). This is a classic text which covers core elements of advanced quantum mechanics. It is particularly strong in the area of atomic physics, but weaker on many-particle physics. [2] S. Gasiorowicz, Quantum Physics, (2nd edn. Wiley 1996, 3rd edition, Wiley, 2003). This is an excellent textbook that covers material at approximately the right level for the course. However, the published text (as opposed to the supplementary material available online) omits some topics which are addressed in this course. [3] H. Haken and H. C. Wolf, The Physics of Atoms and Quanta, (6th edn Springer, 2000). This is a more advanced text which addresses many aspects of atomic physics and quantum optics in a readable manner. [4] K. Konishi and G. Paffuti, Quantum Mechanics: A New Introduction, (OUP, 2009). This is a new text which includes some entertaining new topics within an old field. It also has a useful set of mathematica-based examples if you can get hold of the disk! [5] L. D. Landau and L. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, Volume 3, (Butterworth-Heinemann, 3rd edition, 1981). This is a rich and classic text which covers the many of the core topics in this course in most cases at a level that goes well-beyond our target. [6] F. Schwabl, Quantum Mechanics, (Springer, 4th edition, 2007). This book provides a clear exposition of the core topics addressed at the same level as the current text. [7] R. Shankar, Principles of Quantum Mechanics, (Springer; 2nd edition, 1994). This is a very interesting and idiosyncratic text that I particularly like. It is not comprehensive enough to cover every topic in this course. But where there is overlap, it provides an excellent and interesting commentary.
Advanced Quantum Physics
Chapter 1
Wave mechanics and the Schr¨ odinger equation William Thomson, 1st Baron Kelvin 1824-1907
Although this lecture course will assume a familiarity with the basic concepts of wave mechanics, to introduce more advanced topics in quantum theory, it makes sense to begin with a concise review of the foundations of the subject. In particular, in the first chapter of the course, we will begin with a brief discussion of the historical challenges that led to the development of quantum theory almost a century ago. The formulation of a consistent theory of statistical mechanics, electrodynamics and special relativity during the latter half of the 19th century and the early part of the 20th century had been a triumph of “unification”. However, the undoubted success of these theories gave an impression that physics was a mature, complete, and predictive science. Nowehere was confidence expressed more clearly than in the famous quote made at the time by Lord Kelvin: There is nothing new to be discovered in physics now. All that remains is more and more precise measurement. However, there were a number of seemingly unrelated and unsettling problems that challenged the prevailing theories.
1.1 1.1.1
Historical foundations of quantum physics Black-body radiation
In 1860, Gustav Kirchhoff introduced the concept of a “black body”, an object that absorbs all electromagnetic radiation that falls upon it – none passes through and none is reflected. Since no light is reflected or transmitted, the object appears black when it is cold. However, above absolute zero, a black body emits thermal radiation with a spectrum that depends on temperature. To determine the spectrum of radiated energy, it is helpful to think of a black body as a thermal cavity at a temperature, T . The energy radiated by the cavity can be estimated by considering the resonant modes. In three-dimensions, the number of modes, per unit frequency per unit volume is given by 8πν 2 N (ν)dν = 3 dν , c where, as usual, c is the speed of light.1 1 If we take the cavity to have dimension L3 , the modes of the cavity involve wave numbers k = πn/L where n = (nx , ny , nz ) denote the vector of integers nx = 0, 1, 2, · · · ∞, etc. The corresponding frequency of each mode is given by ν = c|k|/2π, where c is the velocity of light. The number of modes (per unit volume) having frequencies between ν and ν + dν is
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Kelvin was educated at Glasgow and Cambridge. He became professor of natural philosophy at Glasgow in 1846. From 1846 to 1851 Kelvin edited the “Cambridge and Dublin Mathematical Journal,” to which he contributed several important papers. Some of his chief discoveries are announced in the “Secular Coating of the Earth,” and the Bakerian lecture, the “Electrodynamic Qualities of Metals.” He invented the quadrant, portable, and absolute electrometers, and other scientific instruments. In January 1892, he was raised to the peerage as Lord Kelvin.
Gustav Robert Kirchhoff 18241887 A German physicist who contributed to the fundamental understanding of electrical circuits, spectroscopy, and the emission of black-body radiation by heated objects. He coined the term “black body” radiation in 1862, and two sets of independent concepts in both circuit theory and thermal emission are named “Kirchhoff’s laws” after him.
John William Strutt, 3rd Baron Rayleigh OM (1842-1919) An English physicist who, with William Ramsay, discovered the element argon, an achievement for which he earned the Nobel Prize in 1904. He also discovered the phenomenon of Rayleigh scattering, explaining why the sky is blue, and predicted the existence of surface waves known as Rayleigh waves.
1.1. HISTORICAL FOUNDATIONS OF QUANTUM PHYSICS
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Figure 1.1: The COBE (Cos-
mic Background Explorer) satellite made careful measurements of the shape of the spectrum of the emission from the cosmic microwave background. As one can see, the behaviour at low-frequencies (longwavelengths) conforms well with the predicted Rayleigh-Jeans law translating to a temperature of 2.728K. However, at high frequencies, there is a departure from the predicted ν 2 dependence.
The amount of radiation emitted in a given frequency range should be proportional to the number of modes in that range. Within the framework of classical statistical mechanics, each of these modes have an equal chance of being excited, and the average energy in each mode is kB T (equipartition), where kB is the Boltzmann constant. The corresponding energy density is therefore given by the Rayleigh-Jeans law, ρ(ν, T ) =
8πν 2 kB T . c3
This result predicts that ρ(ν, T ) increases without bound at high frequencies, ν — the untraviolet (UV) catastrophe. However, such behaviour stood in contradiction with experiment which revealed that the short-wavelength dependence was quite benign (see, e.g., Fig. 1.1). To resolve difficulties presented by the UV catastrophe, Planck hypothesized that, for each mode ν, energy is quantized in units of hν, where h denotes the Planck constant. In this case, the energy of each mode is given by2 !∞ −nhν/kB T hν n=0 nhνe !ε(ν)" = ! = hν/k T , ∞ −nhν/kB T B e e −1 n=0
leading to the anticipated suppression of high frequency modes. From this result one obtains the celebrated Planck radiation formula, ρ(ν, T ) =
8πν 2 c3
!ε(ν)" =
8πhν 3
1
c3
ehν/kB T − 1
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(1.1)
This result conforms with experiment (Fig. 1.1), and converges on the RayleighJeans law at low frequencies, hν/kB T → 0. Planck’s result suggests that electromagnetic energy is quantized: light of wavelength λ = c/ν is made up of quanta each of which has energy hν. The equipartion law fails for oscillation modes with high frequencies, hν % kB T . A 2
)dk therefore given by N (ν)dν = L13 × 2 × 18 × (4πk , where the factor of 2 accounts for the two (π/L)3 2 polarizations, the factor (4πk )dk is the volume of the shell from k to k + dk in reciprocal space, the factor of 1/8 accounts for the fact that only positive wavenumbers are involved in the closed cavity, and the factor of (π/L)3 denotes the volume of phase space occupied by each mode. Rearranging the equation, and noting that dk = 2πdν/c, we obtain the relation in the text. P∞ −βnhν 2 If we define the partition function, Z = , where β = 1/kBP T , #E$ = n=0 e ∞ n −∂β ln Z. Making use of the formula for the sum of a geometric progression, = n=0 r 1/(1 − r), we obtain the relation.
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Max Karl Ernst Ludwig Planck 1858-1947 German physicist whose work provided the bridge between classical and modern physics. Around 1900 Planck developed a theory explain the nature of black-body radiation. He proposed that energy is emitted and absorbed in discrete packets or “quanta,” and that these had a definite size – Planck’s constant. Planck’s finding didn’t get much attention until the idea was furthered by Albert Einstein in 1905 and Niels Bohr in 1913. Planck won the Nobel Prize in 1918, and, together with Einstein’s theory of relativity, his quantum theory changed the field of physics.
Albert Einstein 1879-1955 was a Germanborn theoretical physicist. He is best known for his theory of relativity and specifically mass-energy equivalence, expressed by the equation E = mc2 . Einstein received the 1921 Nobel Prize in Physics “for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect.”
1.1. HISTORICAL FOUNDATIONS OF QUANTUM PHYSICS
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Figure 1.2:
Measurements of the photoelectric effect taken by Robert Millikan showing the variation of the stopping voltage, eV , with variation of the frequency of incident light. Figure reproduced from Robert A. Millikan’s Nobel Lecture, The Electron and the Light-Quanta from the Experimental Point of View, The Nobel Foundation, 1923.
quantum theory for the specific heat of matter, which takes into account the quantization of lattice vibrational modes, was subsequently given by Debye and Einstein.
1.1.2
Photoelectric effect
We turn now to the second ground-breaking experiment in the development of quantum theory. When a metallic surface is exposed to electromagnetic radiation, above a certain threshold frequency, the light is absorbed and electrons are emitted (see figure, right). In 1902, Philipp Eduard Anton von Lenard observed that the energy of individual emitted electrons increases with the frequency of the light. This was at odds with Maxwell’s wave theory of light, which predicted that the electron energy would be proportional to the intensity of the radiation. In 1905, Einstein resolved this paradox by describing light as composed of discrete quanta (photons), rather than continuous waves. Based upon Planck’s theory of black-body radiation, Einstein theorized that the energy in each quantum of light was proportional to the frequency. A photon above a threshold energy, the “work function” W of the metal, has the required energy to eject a single electron, creating the observed effect. In particular, Einstein’s theory was able to predict that the maximum kinetic energy of electrons emitted by the radiation should vary as k.e.max = hν − W . Later, in 1916, Millikan was able to measure the maximum kinetic energy of the emitted electrons using an evacuated glass chamber. The kinetic energy of the photoelectrons were found by measuring the potential energy of the electric field, eV , needed to stop them. As well as confirming the linear dependence of the kinetic energy on frequency (see Fig. 1.2), by making use of his estimate for the electron charge, e, established from his oil drop experiment in 1913, he was able to determine Planck’s constant to a precision of around 0.5%. This discovery led to the quantum revolution in physics and earned Einstein the Nobel Prize in 1921.
1.1.3
Compton Scattering
In 1923, Compton investigated the scattering of high energy X-rays and γ-ray from electrons in a carbon target. By measuring the spectrum of radiation at different angles relative to the incident beam, he found two scattering peaks. The first peak occurred at a wavelength which matched that of the incident beam, while the second varied with angle. Within the framework of a purely classical theory of the scattering of electromagnetic radiation from a charged Advanced Quantum Physics
Robert Andrews Millikan 18681953 He received his doctorate from Columbia University and taught physics at the University of Chicago (1896-1921) and the California Institute of Technology (from 1921). To measure electric charge, he devised the Millikan oil-drop experiment. He verified Albert Einstein’s photoelectric equation and obtained a precise value for the Planck constant. He was awarded the 1923 Nobel Prize in physics.
Arthur Holly Compton 18921962: An American physicist, he shared the 1927 Nobel Prize in Physics with C. T. R. Wilson for his discovery of the Compton effect. In addition to his work on X rays he made valuable studies of cosmic rays and contributed to the development of the atomic bomb.
1.1. HISTORICAL FOUNDATIONS OF QUANTUM PHYSICS
4
particle – Thomson scattering – the wavelength of a low-intensity beam should remain unchanged.3 Compton’s observation demonstrated that light cannot be explained purely as a classical wave phenomenon. Light must behave as if it consists of particles in order to explain the low-intensity Compton scattering. If one assumes that the radiation is comprised of photons that have a well defined momentum as h well as energy, p = hν c = λ , the shift in wavelength can be understood: The interaction between electrons and high energy photons (ca. keV) results in the electron being given part of the energy (making it recoil), and a photon with the remaining energy being emitted in a different direction from the original, so that the overall momentum of the system is conserved. By taking into account both conservation of energy and momentum of the system, the Compton scattering formula describing the shift in the wavelength as function of scattering angle θ can be derived,4 ∆λ = λ# − λ =
h (1 − cos θ) . me c
The constant of proportionality h/me c = 0.002426 nm, the Compton wavelength, characterizes the scale of scattering. Moreover, as h → 0, one finds that ∆λ → 0 leading to the classical prediction.
1.1.4
Figure 1.3:
Variation of the wavelength of X rays scattered from a Carbon target. A. H. Compton, Phys. Rev. 21, 483; 22, 409 (1923)
Atomic spectra
The discovery by Rutherford that the atom was comprised of a small positively charged nucleus surrounded by a diffuse cloud of electrons lead naturally to the consideration of a planetary model of the atom. However, a classical theory of electrodynamics would predict that an accelerating charge would radiate energy leading to the eventual collapse of the electron into the nucleus. Moreover, as the electron spirals inwards, the emission would gradually increase in frequency leading to a broad continuous spectra. Yet, detailed studies of electrical discharges in low-pressure gases revealed that atoms emit light at discrete frequencies. The clue to resolving these puzzling observations lay in the discrete nature of atomic spectra. For the hydrogen atom, light emitted when the atom is thermally excited has a particular pattern: Balmer had discovered in 1885 that the emitted wavelengths follow the empirical law, λ = λ0 (1/4 − 1/n2 ) where n = 3, 4, 5, · · · and λ0 = 3645.6˚ A (see Fig. 1.4). Neils Bohr realized that these discrete vaues of the wavelength reflected the emission of individual photons having energy equal to the energy difference between two allowed orbits of the electron circling the nucleus (the proton), En − Em = hν, leading to the conclusion that the allowed energy levels must be quantized and varying H as En = − hcR , where RH = 109678 cm−1 denotes the Rydberg constant. n2 3
Classically, light of sufficient intensity for the electric field to accelerate a charged particle to a relativistic speed will cause radiation-pressure recoil and an associated Doppler shift of the scattered light. But the effect would become arbitrarily small at sufficiently low light intensities regardless of wavelength. 4 If we assume that the total energy and momentum are conserved in the scattering of a photon (γ) from an initially stationary target electron (e), we have Eγ + Ee = Eγ ! + Ee! and pγ = pγ ! + pe! . Here Eγ = hν and Ee = me c2 denote p the energy of the photon and electron before the collision, while Eγ ! = hν # and Ee! = (pe! c)2 + (mc2 )2 denote the energies after. From the equation for energy conservation, one obtains (pe! c)2 = (h(ν − ν # ) + me c2 )2 − (me c2 )2 . From the equation for momentum conservation, one obtains p2e! = p2γ + p2γ ! − 2|pγ ||pγ ! | cos θ. Then, noting that Eγ = pγ c, a rearrangement of these equations obtains the Compton scattering formula.
Advanced Quantum Physics
Ernest Rutherford, 1st Baron Rutherford of Nelson, 18711937 was a New Zealand born British chemist and Physicist who became known as the father of nuclear physics. He discovered that atoms have a small charged nucleus, and thereby pioneered the Rutherford model (or planetary model of the atom, through his discovery of Rutherford scattering. He was awarded the Nobel Prize in Chemistry in 1908. He is widely credited as splitting the atom in 1917 and leading the first experiment to “split the nucleus” in a controlled manner by two students under his direction, John Cockcroft and Ernest Walton in 1932.
1.1. HISTORICAL FOUNDATIONS OF QUANTUM PHYSICS
5
Figure 1.4: Schematic describ-
ing various transitions (and an image with the corresponding visible spectral lines) of atomic hydrogen.
How could the quantum hν restricting allowed radiation energies also restrict the allowed electron orbits? In 1913 Bohr proposed that the angular momentum of an electron in one of these orbits was quantized in units of Planck’s constant, L = me vr = n!, 2
!=
h . 2π
(1.2)
2
e )2 2πch3m . As a result, one finds that RH = ( 4π$ 0
( Exercise. Starting with the Bohr’s planetary model for atomic hydrogen, find how the quantization condition (1.2) restricts the radius of the allowed (circular) orbits. Determine the allowed energy levels and obtain the expression for the Rydberg constant above. But why should only certain angular momenta be allowed for the circling electron? A heuristic explanation was provided by de Broglie: just as the constituents of light waves (photons) are seen through Compton scattering to act like particles (of definite energy and momentum), so particles such as electrons may exhibit wave-like properties. For photons, we have seen that the relationship between wavelength and momentum is p = h/λ. de Broglie hypothesized that the inverse was true: for particles with a momentum p, the wavelength is λ=
h , p
i.e. p = !k ,
(1.3)
where k denotes the wavevector of the particle. Applied to the electron in the atom, this result suggested that the allowed circular orbits are standing waves, from which Bohr’s angular momentum quantization follows. The de Broglie hypothesis found quantitative support in an experiment by Davisson and Germer, and independently by G. P. Thomson in 1927. Their studies of electron diffraction from a crystalline array of Nickel atoms (Fig. 1.5) confirmed that the diffraction angles depend on the incident energy (and therefore momentum). This completes the summary of the pivotal conceptual insights that paved the way towards the development of quantum mechanics. Advanced Quantum Physics
Niels Henrik David Bohr 18851962 A Danish physicist who made fundamental contributions to the understanding atomic structure and quantum mechanics, for which he received the Nobel Prize in Physics in 1922. Bohr mentored and collaborated with many of the top physicists of the century at his institute in Copenhagen. He was also part of the team of physicists working on the Manhattan Project. Bohr married Margrethe Norlund in 1912, and one of their sons, Aage Niels Bohr, grew up to be an important physicist who, like his father, received the Nobel prize, in 1975.
Louis Victor Pierre Raymond, 7th duc de Broglie 1892-1987 A French physicist, de Broglie had a mind of a theoretician rather than one of an experimenter or engineer. de Broglie’s 1924 doctoral thesis Recherches sur la th´ eorie des quanta introduced his theory of electron waves. This included the particle-wave property duality theory of matter, based on the work of Einstein and Planck. He won the Nobel Prize in Physics in 1929 for his discovery of the wave nature of electrons, known as the de Broglie hypothesis or m´ ecanique ondulatoire.
1.2. WAVE MECHANICS
6
Figure 1.5: In 1927, Davisson and Germer bombarded a single crystal of nickel
with a beam of electrons, and observed several beams of scattered electrons that were almost as well defined as the incident beam. The phenomenological similarities with X-ray diffraction were striking, and showed that a wavelength could indeed be associated with the electrons. The first figure shows the intensity of electron scattering against co-latitude angle for various bombarding voltages. The second figure shows the intensity of electron scattering vs. azimuth angle - 54V, co-latitude 50. Figures taken taken from C. Davisson and L. H. Germer, Reflection of electrons by a crystal of nickel, Nature 119, 558 (1927).
1.2
Wave mechanics
de Broglie’s doctoral thesis, defended at the end of 1924, created a lot of excitement in European physics circles. Shortly after it was published in the Autumn of 1925, Pieter Debye, a theorist in Zurich, suggested to Erwin Schr¨odinger that he give a seminar on de Broglie’s work. Schr¨odinger gave a polished presentation, but at the end, Debye remarked that he considered the whole theory rather childish: Why should a wave confine itself to a circle in space? It wasnt as if the circle was a waving circular string; real waves in space diffracted and diffused; in fact they obeyed three-dimensional wave equations, and that was what was needed. This was a direct challenge to Schr¨odinger, who spent some weeks in the Swiss mountains working on the problem, and constructing his equation. There is no rigorous derivation of Schr¨odinger’s equation from previously established theory, but it can be made very plausible by thinking about the connection between light waves and photons, and constructing an analogous structure for de Broglie’s waves and electrons (and, of course, other particles).
1.2.1
Maxwell’s wave equation
For a monochromatic wave in vacuum, with no currents or charges present, Maxwell’s wave equation, ∇2 E −
1 ¨ E = 0, c2
(1.4)
admits the plane wave solution, E = E0 ei(k·r−ωt) , with the linear dispersion relation ω = c|k| and c the velocity of light. Here, (and throughout the text) ¨ ≡ ∂ 2 E. We know from the photoelectric effect we adopt the convention, E t and Compton scattering that the photon energy and momentum are related to the frequency and wavelength of light through the relations E = hν = !ω, p = λh = !k. The wave equation tells us that ω = c|k| and hence E = c|p|. If we think of ei(k·r−ωt) as describing a particle (photon) it would be more natural to write the plane wave in terms of the energy and momentum of the Advanced Quantum Physics
James Clerk Maxwell 1831-1879 was a Scottish theoretical physicist and mathematician. His greatest achievement was the development of classical electromagnetic theory, synthesizing all previous unrelated observations of electricity, magnetism and optics into a consistent theory. Maxwell’s equations demonstrated that electricity, magnetism and light are all manifestations of electromagnetic field.
1.2. WAVE MECHANICS
7
particle as E0 ei(p·r−Et)/!. Then, one may see that the wave equation applied to the plane wave describing particle propagation yields the familiar energymomentum relationship, E 2 = (cp)2 for a massless relativistic particle. This discussion suggests how one might extend the wave equation from the photon (with zero rest mass) to a particle with rest mass m0 . We require a wave equation that, when it operates on a plane wave, yields the relativistic energy-momentum invariant, E 2 = (cp)2 + m20 c4 . Writing the plane wave function φ(r, t) = Aei(p·r−Et)/!, where A is a constant, we can recover the energy-momentum invariant by adding a constant mass term to the wave operator, $ % " # (cp)2 − E 2 + m20 c4 i(p·r−Et)/! ∂t2 m20 c2 2 i(p·r−Et)/! ∇ − 2 − 2 e =− e = 0. c ! (!c)2 This wave equation is called the Klein-Gordon equation and correctly describes the propagation of relativistic particles of mass m0 . However, its form is seems inappropriate for non-relativistic particles, like the electron in hydrogen. Continuing along the same lines, let us assume that a non-relativistic electron in free space is also described by a plane wave of the form Ψ(x, t) = Aei(p·r−Et)/!. We need to construct an operator which, when applied to this wave function, just gives us the ordinary non-relativistic energy-momentum p2 . The factor of p2 can obviously be recovered from two relation, E = 2m derivatives with respect to r, but the only way we can get E is by having a single differentiation with respect to time, i.e. i!∂t Ψ(r, t) = −
!2 2 ∇ Ψ(r, t) . 2m
This is Schr¨odinger’s equation for a free non-relativistic particle. One remarkable feature of this equation is the factor of i which shows that the wavefunction is complex. How, then, does the presence of a spatially varying scalar potential effect the propagation of a de Broglie wave? This question was considered by Sommerfeld in an attempt to generalize the rather restrictive conditions in Bohr’s model of the atom. Since the electron orbit was established by an inversesquare force law, just like the planets around the Sun, Sommerfeld couldn’t understand why Bohr’s atom had only circular orbits as opposed to Keplerlike elliptical orbits. (Recall that all of the observed spectral lines of hydrogen were accounted for by energy differences between circular orbits.) de Broglie’s analysis of the allowed circular orbits can be formulated by assuming that, at some instant, the spatial variation of the wavefunction on going around the orbit includes a phase term of the form eipq/!, where here the parameter q measures the spatial distance around the orbit. Now, for an acceptable wavefunction, the total phase change on going around the orbit must be 2πn, where n is integer. For the usual Bohr circular orbit, where p = |p| is constant, this leads to quantization of the angular momentum L = pr = n!. Sommerfeld considered a general Keplerian elliptical orbit. Assuming that the de Broglie relation p = h/λ still holds, the wavelength must vary as the particle moves around the orbit, being shortest where the particle travels fastest, at its closest approach to the nucleus. Nevertheless, the phase change on moving a short distance ∆q should still be p∆q/!. Requiring the wavefunction to link up smoothly on going once around the orbit gives the Bohr-Sommerfeld Advanced Quantum Physics
Arnold Johannes Wilhelm Sommerfeld 1868-1951 A German theoretical physicist who pioneered developments in atomic and quantum physics, and also educated and groomed a large number of students for the new era of theoretical physics. He introduced the fine-structure constant into quantum mechanics.
1.2. WAVE MECHANICS
8
quantization condition &
p dq = nh ,
(1.5)
' where denotes the line integral around a closed orbit. Thus only certain elliptical orbits are allowed. The mathematics is non-trivial, but it turns out that every allowed elliptical orbit has the same energy as one of the allowed circular orbits. That is why Bohr’s theory gave the correct energy levels. This analysis suggests that, in a varying potential, the wavelength changes in concert with the momentum. ( Exercise. As a challenging exercise, try to prove Sommerfeld’s result for the elliptical orbit.
1.2.2
Schr¨ odinger’s equation
Following Sommerfeld’s considerations, let us then consider a particle moving in one spatial dimension subject to a “roller coaster-like” potential. How do we expect the wavefunction to behave? As discussed above, we would expect the wavelength to be shortest where the potential is lowest, in the minima, because that’s where the particle is going the fastest. Our task then is to construct a wave equation which leads naturally to the relation following from p2 (classical) energy conservation, E = 2m +V (x). In contrast to the free particle case discussed above, the relevant wavefunction here will no longer be a simple plane wave, since the wavelength (determined through the momentum via the de Broglie relation) varies with the potential. However, at a given position x, the momentum is determined by the “local wavelength”. The appropriate wave equation is the one-dimensional Schr¨odinger equation, i!∂t Ψ(x, t) = −
!2 ∂x2 Ψ(x, t) + V (x)Ψ(x, t) , 2m
(1.6)
with the generalization to three-dimensions leading to the Laplacian operator in place of ∂x2 (cf. Maxwell’s equation). So far, the validity of this equation rests on plausibility arguments and hand-waving. Why should anyone believe that it really describes an electron wave? Schr¨odinger’s test of his equation was the hydrogen atom. He looked for Bohr’s “stationary states”: states in which the electron was localized somewhere near the proton, and having a definite energy. The time dependence would be the same as for a plane wave of definite energy, e−iEt/!; the spatial dependence would be a time-independent function decreasing rapidly at large distances from the proton. From the solution of the stationary wave equation for the Coulomb potential, he was able to deduce the allowed values of energy and momentum. These values were exactly the same as those obtained by Bohr (except that the lowest allowed state in the “new” theory had zero angular momentum): impressive evidence that the new theory was correct.
1.2.3
Time-independent Schr¨ odinger equation
As with all second order linear differential equations, if the potential V (x, t) = V (x) is time-independent, the time-dependence of the wavefunction can be
Advanced Quantum Physics
Erwin Rudolf Josef Alexander Schrdinger 1887-1961 was an Austrian theoretical physicist who achieved fame for his contributions to quantum mechanics, especially the Schr¨ odinger equation, for which he received the Nobel Prize in 1933. In 1935, after extensive correspondence with personal friend Albert Einstein, he proposed the Schr¨ odinger’s cat thought experiment.
1.2. WAVE MECHANICS
9
separated from the spatial dependence. Setting Ψ(x, t) = T (t)ψ(x), and separating the variables, the Schr¨odinger equation takes the form, ) ( 2 2 ∂x ψ(x) + V (x)ψ(x) − !2m i!∂t T (t) = = const. = E . ψ(x) T (t) Since we have a function of only x set equal to a function of only t, they both must equal a constant. In the equation above, we call the constant E (with some knowledge of the outcome). We now have an equation in t set equal to a constant, i!∂t T (t) = ET (t), which has a simple general solution, T (t) = Ce−iEt/!, where C is some constant. The corresponding equation in x is then given by the stationary, or time-independent Schr¨ odinger equation, −
!2 ∂x2 ψ(x) + V (x)ψ(x) = Eψ(x) . 2m
The full time-dependent solution is given by Ψ(x, t) = e−iEt/!ψ(x) with definite energy, E. Their probability density |Ψ(x, t)|2 = |ψ(x)|2 is constant in time – hence they are called stationary states! The operator ˆ = − ! ∂x + V (x) H 2m 2 2
defines the Hamiltonian and the stationary wave equation can be written as ˆ ˆ the eigenfunction equation, Hψ(x) = Eψ(x), i.e. ψ(x) is an eigenstate of H with eigenvalue E.
1.2.4
Particle flux and conservation of probability
In analogy to the Poynting vector for the electromagnetic field, we may want to know the probability current. For example, for a free particle system, the probability density is uniform over all space, but there is a net flow along the direction of momentum. We can derive an equation showing conservation of probability by differentiating the probability density, P (x, t) = |ψ(x, t)|2 , and using the Schr¨odinger equation, ∂t P (x, t)+∂x j(x, t) = 0. This translates to the usual conservation equation if j(x, t) is identified as the probability current, i! ∗ [ψ ∂x ψ − ψ∂x ψ ∗ ] . (1.7) 2m *b *b If we integrate over some interval in x, a ∂t P (x, t)dx = − a ∂x j(x, t)dx it *b follows that ∂t a P (x, t)dx = j(x = a, t) − j(x = b, t), i.e. the rate of change of probability is equal to the net flux entering the interval. Extending this analysis to three space dimensions, we obtain the continuity equation, ∂t P (r, t) + ∇ · j(r, t) = 0, from which follows the particle flux, j(x, t) = −
j(r, t) = −
i! ∗ [ψ (r, t)∇ψ(r, t) − ψ(r, t)∇ψ ∗ (r, t)] . 2m
(1.8)
This completes are survey of the foundations and development of quantum theory. In due course, it will be necessary to develop some more formal mathematical aspects of the quantum theory. However, before doing, it is useful to acquire some intuition for the properties of the Schr¨odinger equation. Therefore, in the next chapter, we will explore the quantum mechanics of bound and unbound particles in a one-dimensional system turning to discuss more theoretical aspects of the quantum formulation in the following chapter. Advanced Quantum Physics
Chapter 2
Quantum mechanics in one dimension Following the rules of quantum mechanics, we have seen that the state of a quantum particle, subject to a scalar potential V (r), is described by the time-dependent Schr¨odinger equation, ˆ i!∂t Ψ(r, t) = HΨ(r, t) ,
(2.1)
ˆ = − ! ∇ + V (r) denotes the Hamiltonian. To explore its properwhere H 2m ties, we will first review some simple and, hopefully, familiar applications of the equation to one-dimensional systems. In addressing the one-dimensional geometry, we will divide our consideration between potentials, V (x), which leave the particle free (i.e. unbound), and those that bind the particle to some region of space. 2
2.1 2.1.1
2
Wave mechanics of unbound particles Free particle
In the absence of an external potential, the time-dependent Schr¨odinger equation (2.1) describes the propagation of travelling waves. In one dimension, the corresponding complex wavefunction has the form Ψ(x, t) = A ei(kx−ωt) , k where A is the amplitude, and E(k) = !ω(k) = !2m represents the free particle energy dispersion for a non-relativistic particle of mass, m, and wavevector k = 2π/λ with λ the wavelength. Each wavefunction describes a plane wave in which the particle has definite energy E(k) and, in accordance with the de Broglie relation, momentum p = !k = h/λ. The energy spectrum of a freely-moving particle is therefore continuous, extending from zero to infinity and, apart from the spatially constant state k = 0, has a two-fold degeneracy corresponding to right and left moving particles. For an infinite system, it makes no sense to fix the amplitude A by the normalization of the total probability. Instead, it is useful to fix the flux associated with the wavefunction. Making use of Eq. (1.7) for the particle current, the plane wave is associated with a constant (time-independent) flux, 2 2
j(x, t) = −
i! !k p (Ψ∗ ∂x Ψ − c.c.) = |A|2 = |A|2 . 2m m m
Advanced Quantum Physics
2.1. WAVE MECHANICS OF UNBOUND PARTICLES For a given value of the!flux j, the amplitude is given, up to an arbitrary constant phase, by A = mj/!k. To prepare a wave packet which is localized to a region of space, we must superpose components of different wave number. In an open system, this may be achieved using a Fourier expansion. For any function,1 ψ(x), we have the Fourier decomposition,2 " ∞ 1 ψ(k) eikx dk , ψ(x) = √ 2π −∞ where the coefficients are defined by the inverse transform, " ∞ 1 ψ(x) e−ikx dx . ψ(k) = √ 2π −∞ The normalization of ψ(k) #follows automatically from the normalization of #∞ ∗ ∞ ψ(x), −∞ ψ (k)ψ(k)dk = −∞ ψ ∗ (x)ψ(x)dx = 1, and both can represent probability amplitudes. Applied to a wavefunction, ψ(x) can be understood as a wave packet made up of contributions involving definite momentum states, eikx , with amplitude set by the Fourier coefficient ψ(k). The probability for a particle to be found in a region of width dx around some value of x is given by |ψ(x)|2 dx. Similarly, the probability for a particle to have wave number k in a region of width dk around some value of k is given by |ψ(k)|2 dk. (Remember that p = !k so the momentum distribution is very closely related. Here, for economy of notation, we work with k.) The Fourier transform of a normalized Gaussian wave packet, ψ(k) = 1/4 e−α(k−k0 )2 , is also a Gaussian (exercise), ) ( 2α π ψ(x) =
$
1 2πα
%1/4
x2
eik0 x e− 4α .
From these representations, we can see that it is possible to represent a single particle, localized in real space as a superposition of plane wave states localized in Fourier space. But note that, while we have achieved our goal of finding localized wave packets, this has been at the expense of having some non-zero width in x and in k. For the Gaussian wave packet, we can straightforwardly obtain the width (as measured by the root mean square – RMS) of the probability distribution, √ ∆x = (#(x − #x$)2 $)1/2 ≡ (#x2 $ − #x$2 $)1/2 = α, and ∆k = √14α . We can again see that, as we vary the width in k-space, the width in x-space varies to keep the following product constant, ∆x∆k = 12 . If we translate from the wavevector into momentum p = !k, then ∆p = !∆k and ∆p ∆x =
! . 2
If we consider the width of the distribution as a measure of the “uncertainty”, we will prove in section (3.1.2) that the Gaussian wave packet provides the minimum uncertainty. This result shows that we cannot know the position of a particle and its momentum at the same time. If we try to localize a particle to a very small region of space, its momentum becomes uncertain. If we try to 1 More precisely, we can make such an expansion providing we meet some rather weak conditions of smoothness and differentiability of ψ(x) – conditions met naturally by problems which derive from physical systems! 2 Here we will adopt an ecomony of notation using the same symbol ψ to denote the wavefunction and its Fourier coefficients. Their identity will be disclosed by their argument and context.
Advanced Quantum Physics
11
2.1. WAVE MECHANICS OF UNBOUND PARTICLES make a particle with a definite momentum, its probability distribution spreads out over space. With this introduction, we now turn to consider the interaction of a particle with a non-uniform potential background. For non-confining potentials, such systems fall into the class of scattering problems: For a beam of particles incident on a non-uniform potential, what fraction of the particles are transmitted and what fraction are reflected? In the one-dimensional system, the classical counterpart of this problem is trivial: For particle energies which exceed the maximum potential, all particles are eventually transmitted, while for energies which are lower, all particles are reflected. In quantum mechanics, the situation is richer: For a generic potential of finite extent and height, some particles are always reflected and some are always transmitted. Later, in chapter 14, we will consider the general problem of scattering from a localized potential in arbitrary dimension. But for now, we will focus on the one-dimensional system, where many of the key concepts can be formulated.
2.1.2
Potential step
As we have seen, for a time-independent potential, the wavefunction can be factorized as Ψ(x, t) = e−iEt/!ψ(x), where ψ(x) is obtained from the stationary form of the Schr¨odinger equation, & 2 2 ' ! ∂x + V (x) ψ(x) = Eψ(x) , − 2m and E denotes the energy of the particle. As |Ψ(x, t)|2 represents a probablility density, it must be everywhere finite. As a result, we can deduce that the wavefunction, ψ(x), is also finite. Moreover, since E and V (x) are presumed finite, so must be ∂x2 ψ(x). The latter condition implies that ' both ψ(x) and ∂x ψ(x) must be continuous functions of x, even if V has a discontinuity. Consider then the influence of a potential step (see figure) on the propagation of a beam of particles. Specifically, let us assume that a beam of particles with kinetic energy, E, moving from left to right are incident upon a potential step of height V0 at position x = 0. If the beam has unit amplitude, the reflected and transmitted (complex) amplitudes are set by r and t. The corresponding wavefunction is given by ψ< (x) = eik< x + re−ik< x x < 0 ψ> (x) = teik> x x>0 (
(
0) and k> = 2m(E−V . Applying the continuity conditions where k< = !2 on ψ and ∂x ψ at the step (x = 0), one obtains the relations 1 + r = t and ik< (1 − r) = ik> t leading to the reflection and transmission amplitudes,
2mE !2
r=
k< − k> , k< + k>
t=
2k< . k< + k>
The reflectivity, R, and transmittivity, T , are defined by the ratios, R=
reflected flux , incident flux
Advanced Quantum Physics
T =
transmitted flux . incident flux
12
2.1. WAVE MECHANICS OF UNBOUND PARTICLES With the incident, reflected, and transmitted fluxes given by |A|2 !km< , |Ar|2 !km< , and |At|2 !km> respectively, one obtains ) ) ) ) ) k< − k> )2 ) 2k< )2 k> 4k< k> 2 k> ) = |r|2 , ) ) R = )) T = ) k< + k> ) k< = |t| k< = (k< + k> )2 . k< + k> ) From these results one can confirm that the total flux is, as expected, conserved in the scattering process, i.e. R + T = 1.
' Exercise. While E −V0 remains positive, show that the beam is able to propagate across the potential step (see figure). Show that the fraction of the beam that is reflected depends on the relative height of the step while the phase depends on the sign of V0 . In particular, show that for V0 > 0, the reflected beam remains in phase with the incident beam, while for V0 < 0 it is reversed. Finally, when E − V0 < 0, show that the beam is unable to propagate to the right (R = 1). Instead show that there is an evanescent decay of the wavefunction into the barrier region with a decay ! length set by 2π !2 /2m(V0 − E). If V0 → ∞, show that the system forms a standing wave pattern.
2.1.3
Potential barrier
Having dealt with the potential step, we now turn to consider the problem of a beam of particles incident upon a square potential barrier of height V0 (presumed positive for now) and width a. As mentioned above, this geometry is particularly important as it includes the simplest example of a scattering phenomenon in which a beam of particles is “deflected” by a local potential. Moreover, this one-dimensional geometry also provides a platform to explore a phenomenon peculiar to quantum mechanics – quantum tunneling. For these reasons, we will treat this problem fully and with some care. Since the barrier is localized to a region of size a, the incident and(trans-
mitted wavefunctions have the same functional form, eik1 x , where k1 = 2mE , !2 and differ only in their complex amplitude, i.e. after the encounter with the barrier, the transmitted wavefunction undergoes only a change of amplitude (some particles are reflected from the barrier, even when the energy of the incident beam, E, is in excess of V0 ) and a phase shift. To deterimine the relative change in amplitude and phase, we can parameterise the wavefunction as ψ1 (x) = eik1 x + re−ik1 x x≤0 ψ2 (x) = Aeik2 x + Be−ik2 x 0 ≤ x ≤ a ψ3 (x) = teik1 x a≤x
( 0) where k2 = 2m(E−V . Here, as with the step, r denotes the reflected ampli!2 tude and t the transmitted. Applying the continuity conditions on the wavefunction, ψ, and its derivative, ∂x ψ, at the barrier interfaces at x = 0 and x = a, one obtains * * 1+r =A+B k1 (1 − r) = k2 (A − B) , . Aeik2 a + Be−ik2 a = teik1 a k2 (Aeik2 a − Be−ik2 a ) = k1 teik1 a Together, these four equations specify the four unknowns, r, t, A and B. Solving, one obtains (exercise) t=
2k1 k2 e−ik1 a , 2k1 k2 cos(k2 a) − i(k12 + k22 ) sin(k2 a)
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translating to a transmissivity of T = |t|2 =
1+
1 4
+
1 k1 k2
−
k2 k1
,2
, sin2 (k2 a)
and the reflectivity, R = 1 − T . As a consistency check, we can see that, when V0 = 0, k2 = k1 and t = 1, as expected. Moreover, T is restricted to the interval from 0 to 1 as required. So, for barrier heights in the range E > V0 > 0, the transmittivity T shows an oscillatory behaviour with k2 reaching unity when k2 a = nπ with n integer. At these values, there is a conspiracy of interference effects which eliminate altogether the reflected component of the wave leading to perfect transmission. Such a situation arises when the width of the barrier is perfectly matched to an integer or half-integer number of wavelengths inside the barrier. When the energy of the incident particles falls below the energy of the barrier, 0 < E < V0 , a classical beam would be completely reflected. However, in the quantum system, particles are able to tunnel through the barrier region and escape leading to a non-zero transmission coefficient. In this regime, k2 = iκ2 becomes pure imaginary leading to an evanescent decay of the wavefunction under the barrier and a suppression, but not extinction, of transmission probability, T = |t|2 =
1+
1 4
+
1 k1 κ2
+
κ2 k1
,2
Transmission probability of a √ finite potential barrier for 2mV0 a/! = 7. Dashed: classical result. Solid line: quantum mechanics.
. sinh2 κ2 a
For κ2 a ) 1 (the weak tunneling limit), the transmittivity takes the form T *
16k12 κ22 −2κ2 a e . (k12 + κ22 )2
Finally, on a cautionary note, while the phenomenon of quantum mechanical tunneling is well-established, it is difficult to access in a convincing experimental manner. Although a classical particle with energy E < V0 is unable to penetrate the barrier region, in a physical setting, one is usually concerned with a thermal distribution of particles. In such cases, thermal activation may lead to transmission over a barrier. Such processes often overwhelm any contribution from true quantum mechanical tunneling. ' Info. Scanning tunneling microscopy (STM) is a powerful technique for viewing surfaces at the atomic level. Its development in the early eighties earned its inventors, Gerd Binnig and Heinrich Rohrer (at IBM Z¨ urich), the Nobel Prize in Physics in 1986. STM probes the density of states of a material using the tunneling current. In its normal operation, a lateral resolution of 0.1 nm and a depth resolution of 0.01 nm is typical for STM. The STM can be used not only in ultra-high vacuum, but also in air and various other liquid or gas ambients, and at temperatures ranging from near zero kelvin to a few hundred degrees Celsius. The STM is based on the concept of quantum tunnelling (see Fig. 2.1). When a conducting tip is brought in proximity to a metallic or semiconducting surface, a bias between the two can allow electrons to tunnel through the vacuum between them. For low voltages, this tunneling current is a function of the local density of states at the Fermi level, EF , of the sample.3 Variations in current as the probe passes over the surface are translated into an image. STM can be a challenging technique, as it requires extremely clean surfaces and sharp tips. 3 Although the meaning of the Fermi level will be address in more detail in chapter 8, we mention here that it represents the energy level to which the electron states in a metal are fully-occupied.
Advanced Quantum Physics
Real part of the wavefunction for E/V0 = 0.6 (top), E/V0 = 1.6 (middle), and E/V0 = 1 + π 2 /2 (bottom), where mV0 a2 /!2 = 1. In the first case, the system shows tunneling behaviour, while in the third case, k2 a = π and the system shows resonant transmission.
STM image showing two point defects adorning the copper (111) surface. The point defects (possibly impurity atoms) scatter the surface state electrons resulting in circular standing wave patterns.
2.2. WAVE MECHANICS OF BOUND PARTICLES
Figure 2.1: Principle of scanning tunneling microscopy: Applying a negative sample
voltage yields electron tunneling from occupied states at the surface into unoccupied states of the tip. Keeping the tunneling current constant while scanning the tip over the surface, the tip height follows a contour of constant local density of states.
2.1.4
The rectangular potential well
Finally, if we consider scattering from a potential well (i.e. with V0 < 0), while E > 0, we can apply the results of the previous section to find a continuum of unbound states with the corresponding resonance behaviour. However, in addition to these unbound states, for E < 0 we have the opportunity to find bound states of the potential. It is to this general problem that we now turn. ' Exercise. Explore the phase dependence of the transmission coefficient in this regime. Consider what happens to the phase as resonances (bound states) of the potential are crossed.
2.2
Wave mechanics of bound particles
In the case of unbound particles, we have seen that the spectrum of states is continuous. However, for bound particles, the wavefunctions satisfying the Schr¨odinger equation have only particular quantized energies. In the onedimensional system, we will find that all binding potentials are capable of hosting a bound state, a feature particular to the low dimensional system.
2.2.1
The rectangular potential well (continued)
As a starting point, let us consider a rectangular potential well similar to that discussed above. To make use of symmetry considerations, it is helpful to reposition the potential setting x ≤ −a 0 V (x) = −V0 −a ≤ x ≤ a , 0 a≤x
where the potential depth V0 is assumed positive. In this case, we will look for bound state solutions with energies lying in the range −V0 < E < 0. ˆ Pˆ ] = 0 Since the Hamiltonian is invariant under parity transformation, [H, ˆ ˆ (where P ψ(x) = ψ(−x)), the eigenstates of the Hamiltonian H must also be
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eigenstates of parity, i.e. we expect the eigenfunctions to separate into those symmetric and those antisymmetric under parity.4 For E < 0 (bound states), the wavefunction outside the well region must have the form
with κ = the form
(
ψ(x < −a) = Ceκx ,
ψ(x > a) = De−κx ,
− 2mE while in the central well region, the general solution is of !2 ψ(−a < x < a) = A cos(kx) + B sin(kx) ,
( 0) where k = 2m(E+V . Once again we have four equations in four unknowns. !2 The calculation shows that either A or B must be zero for a solution. This means that the states separate into solutions with even or odd parity. For the even states, the solution of the equations leads to the quantization condition, κ = tan(ka)k, while for the odd states, we find κ = − cot(ka)k. These are transcendental equations, and must be solved numerically. The 2 0a − (ka)2 )1/2 with ka tan(ka) for the even figure (right) compares κa = ( 2mV !2 states and to −ka cot(ka) for the odd states. Where the curves intersect, we have an allowed energy. From the structure of these equations, it is evident that an even state solution can always be found for arbitrarily small values of the binding potential V0 while, for odd states, bound states appear only at a critical value of the coupling strength. The wider and deeper the well, the more solutions are generated. ' Exercise. Determine the pressure exerted on the walls of a rectangular potential well by a particle inside. For a hint on how to proceed, see the discussion on degeneracy pressure on page 85.
2.2.2
The δ-function potential well
Let us now consider perhaps the simplest binding potential, the δ-function, V (x) = −aV0 δ(x). Here the parameter ‘a’ denotes some microscopic length scale introduced to make the product aδ(x) dimensionless.5 For a state to be bound, its energy must be negative. Moreover, the form of the potential demands that the wavefunction is symmetric under parity, x → −x. (A wavefunction which was antisymmetric must have ψ(0) = 0 and so could not be influenced by the δ-function potential.) We therefore look for a solution of the form * κx e x<0 ψ(x) = A , −κx e x>0
! where κ = −2mE/!2 . With this choice, the wavefunction remains everywhere continuous including at the potential, x = 0. Integrating the stationary form of the Schr¨odinger equation across an infinitesimal interval that spans the region of the δ-funciton potential, we find that ∂x ψ|+% − ∂x ψ|−% = − 4 5
2maV0 ψ(0) . !2
Later, in section 3.2, we will discuss the role of symmetries in quantum mechanics. Note that the dimenions of δ(x) are [Length−1 ].
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0a Comparison of κa = ( 2mV − !2 2 1/2 (ka) ) with ka tan(ka) and 2 0a −ka cot(ka) for 2mV = 14. !2 For this potential, there are a total of three solutions labelled A, B and C. Note that, from the geometry of the curves, there is always a bound state no matter how small is the potential V0 .
2.2. WAVE MECHANICS OF BOUND PARTICLES From this result, we obtain that κ = maV0 /!2 , leading to the bound state energy E=−
ma2 V02 . 2!2
Indeed, the solution is unique. An attractive δ-function potential hosts only one bound state. ' Exercise. Explore the bound state properties of the “molecular” binding potential V (x) = −aV0 [δ(x + d) + δ(x − d)]. Show that it consists of two bound states, one bonding (nodeless) and one antibonding (single node). How does the energy of the latter compare with two isolated δ-function potential wells?
2.2.3
Info: The δ-function model of a crystal
Finally, as our last example of a one-dimensional quantum system, let us consider a particle moving in a periodic potential. The Kronig-Penney model provides a caricature of a (one-dimensional) crystalline lattice potential. The potential created by the ions is approximated as an infinite array of potential wells defined by a set of repulsive δ-function potentials, V (x) = aV0
∞ 0
n=−∞
δ(x − na) .
Since the potential is repulsive, it is evident that all states have energy E > 0. This potential has a new symmetry; a translation by the lattice spacing a leaves the protential unchanged, V (x + a) = V (x). The probability density must therefore exhibit the same translational symmetry, |ψ(x + a)|2 = |ψ(x)|2 , which means that, under translation, the wavefunction differs by at most a phase, ψ(x + a) = eiφ ψ(x). In the region from (n − 1)a < x < na, the general solution of the Schr¨odinger equation is plane wave like and can be written in the form, ψn (x) = An sin[k(x − na)] + Bn cos[k(x − na)] , ! where k = 2mE/!2 and, following the constraint on translational invariance, An+1 = eiφ An and Bn+1 = eiφ Bn . By applying the boundary conditions, one can derive a constraint on k similar to the quantized energies for bound states considered above. Consider the boundary conditions at position x = na. Continuity of the wavefunction, ψn |x=na = ψn+1 |x=na , translates to the condition, Bn = An+1 sin(−ka) + Bn+1 cos(−ka) or Bn+1 =
Bn + An+1 sin(ka) . cos(ka)
0 Similarly, the discontinuity in the first derivative, ∂x ψn+1 |x=na −∂x ψn |na = 2maV !2 ψn (na), 2maV0 leads to the condition, k [An+1 cos(ka) + Bn+1 sin(ka) − An ] = !2 Bn . Substituting the expression for Bn+1 and rearranging, one obtains
An+1 =
2maV0 Bn cos(ka) − Bn sin(ka) + An cos(ka) . !2 k
Similarly, replacing the expression for An+1 in that for Bn+1 , one obtains the parallel equation, Bn+1 =
2maV0 Bn sin(ka) + Bn cos(ka) + An sin(ka) . !2 k
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With these two eqations, and the relations An+1 = eiφ An and Bn+1 = eiφ Bn , we obtain the quantization condition,6 cos φ = cos(ka) +
maV0 sin(ka) . !2 k
√ As !k = 2mE, this result relates the allowed values of energy to the real parameter, φ. Since cos φ can only take values between −1 and 1, there are a sequence of allowed bands of energy with energy gaps separating these bands (see Fig. 2.2). Such behaviour is characteristic of the spectrum of periodic lattices: In the periodic system, the wavefunctions – known as Bloch states – are indexed by a “quasi”momentum index k, and a band index n where each Bloch band is separated by an energy gap within which there are no allowed states. In a metal, electrons (fermions) populate the energy states starting with the lowest energy up to some energy scale known as the Fermi energy. For a partially-filled band, low-lying excitations associated with the continuum of states allow electrons to be accelerated by a weak electric field. In a band insulator, all states are filled up to an energy gap. In this case, a small electric field is unable to excite electrons across the energy gap – hence the system remains insulating.
' Exercise. In the Kronig-Penney model above, we took the potential to be repulsive. Consider what happens if the potential is attractive when we also have to consider the fate of the states that were bound for the single δ-function potential. In this case, you will find that the methodology and conclusions mirror the results of the repulsive potential: all states remain extended and the continuum of states exhibits a sequence of band gaps controlled by similar sets of equations.
6
Eliminating An and Bn from the equations, a sequence of cancellations obtains „ « 2maV0 e2iφ − eiφ sin(ka) + 2 cos(ka) + 1 = 0 . 2 ! k
Then multiplying by e−iφ , we obtain the expression for cos φ.
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Figure 2.2: Solid line shows the
variation of cos φ with ka over a range from −1 to 1 for V0 = 2 and ma2 !2 = 1. The blue line shows 0.01×E = (!k)2 /2m. The shaded region represents values of k and energy for which there is no solution.
Chapter 3
Operator methods in quantum mechanics While the wave mechanical formulation has proved successful in describing the quantum mechanics of bound and unbound particles, some properties can not be represented through a wave-like description. For example, the electron spin degree of freedom does not translate to the action of a gradient operator. It is therefore useful to reformulate quantum mechanics in a framework that involves only operators. Before discussing properties of operators, it is helpful to introduce a further simplification of notation. One advantage of the operator algebra is that it ˆ = pˆ2 , does not rely upon a particular basis. For example, when one writes H 2m where the hat denotes an operator, we can equally represent the momentum operator in the spatial coordinate basis, when it is described by the differential operator, pˆ = −i!∂x , or in the momentum basis, when it is just a number pˆ = p. Similarly, it would be useful to work with a basis for the wavefunction which is coordinate independent. Such a representation was developed by Dirac early in the formulation of quantum mechanics. In the parlons of mathematics, square integrable functions (such as wavefunctions) are said form a vector space, much like the familiar three-dimensional vector spaces. In the Dirac notation, a state vector or wavefunction, ψ, is represented as a “ket”, |ψ". Just as we can express any three-dimensional vector in terms of the basis vectors, r = xˆ e1 + yˆ e2 + zˆ e3 , so we can expand any wavefunction as a superposition of basis state vectors, |ψ" = λ1 |ψ1 " + λ2 |ψ2 " + · · · . Alongside the ket, we can define the “bra”, #ψ|. Together, the bra and ket define the scalar product ! ∞ #φ|ψ" ≡ dx φ∗ (x)ψ(x) , −∞
from which follows the identity, #φ|ψ"∗ = #ψ|φ". In this formulation, the real space representation of the wavefunction is recovered from the inner product ψ(x) = #x|ψ" while the momentum space wavefunction is obtained from ψ(p) = #p|ψ". As with a three-dimensional vector space where a · b ≤ |a| |b|, the magnitude of the scalar product is limited by the magnitude of the vectors, " #ψ|φ" ≤ #ψ|ψ"#φ|φ" , a relation known as the Schwartz inequality. Advanced Quantum Physics
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3.1
20
Operators
An operator Aˆ is a “mathematical object” that maps one state vector, |ψ", ˆ into another, |φ", i.e. A|ψ" = |φ". If ˆ A|ψ" = a|ψ" , with a real, then |ψ" is said to be an eigenstate (or eigenfunction) of Aˆ with eigenvalue a. For example, the plane wave state ψp (x) = #x|ψp " = A eipx/! is an eigenstate of the momentum operator, pˆ = −i!∂x , with eigenvalue p. For a free particle, the plane wave is also an eigenstate of the Hamiltonian, ˆ = pˆ2 with eigenvalue p2 . H 2m 2m In quantum mechanics, for any observable A, there is an operator Aˆ which acts on the wavefunction so that, if a system is in a state described by |ψ", the expectation value of A is ˆ #A" = #ψ|A|ψ" =
!
∞
ˆ dx ψ ∗ (x)Aψ(x) .
(3.1)
−∞
Every operator corresponding to an observable is both linear and Hermitian: That is, for any two wavefunctions |ψ" and |φ", and any two complex numbers α and β, linearity implies that ˆ ˆ ˆ A(α|ψ" + β|φ") = α(A|ψ") + β(A|φ") . ˆ the Hermitian conjugate operator Moreover, for any linear operator A, (also known as the adjoint) is defined by the relation ˆ = #φ|Aψ"
!
ˆ = dx φ∗ (Aψ)
!
dx ψ(Aˆ† φ)∗ = #Aˆ† φ|ψ" .
(3.2)
ˆ From the definition, #Aˆ† φ|ψ" = #φ|Aψ", we can prove some useful rela† ˆ ˆ tions: Taking the complex conjugate, #A φ|ψ"∗ = #ψ|Aˆ† φ" = #Aψ|φ", and † ˆ then finding the Hermitian conjugate of A , we have ˆ #ψ|Aˆ† φ" = #(Aˆ† )† ψ|φ" = #Aψ|φ",
i.e. (Aˆ† )† = Aˆ .
Therefore, if we take the Hermitian conjugate twice, we get back to the same ˆ † = Aˆ† + B ˆ † just ˆ † = λ∗ Aˆ† and (Aˆ + B) operator. Its easy to show that (λA) † ˆ ˆ ˆ † Aˆ† from the properties of the dot product. We can also show that (AB) = B † † † ˆ ˆ ˆ ˆ ˆ ˆ from the identity, #φ|ABψ" = #A φ|Bψ" = #B A φ|ψ". Note that operators are associative but not (in general) commutative, ˆ ˆ B|ψ") ˆ ˆ ˆ A|ψ" ˆ AˆB|ψ" = A( = (AˆB)|ψ" = & B . A physical variable must have real expectation values (and eigenvalues). This implies that the operators representing physical variables have some special properties. By computing the complex conjugate of the expectation value of a physical variable, we can easily show that physical operators are their own Hermitian conjugate, #! ∞ $∗ ! ∞ ∗ ∗ ∗ ˆ ˆ ˆ ˆ ψ (x)Hψ(x)dx = ψ(x)(Hψ(x)) dx = #Hψ|ψ" . #ψ|H|ψ" = −∞
−∞
ˆ ˆ ˆ † ψ|ψ", and H ˆ † = H. ˆ Operators that are their i.e. #Hψ|ψ" = #ψ|Hψ" = #H own Hermitian conjugate are called Hermitian (or self-adjoint). Advanced Quantum Physics
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21
ˆ = −i!∇ is Hermitian. Fur' Exercise. Prove that the momentum operator p ther show that the parity operator, defined by Pˆ ψ(x) = ψ(−x) is also Hermitian.
ˆ = Ei |i" form an orthonormal (i.e. Eigenfunctions of Hermitian operators H|i" #i|j" = δij ) complete basis: For % a complete set of states |i", we can expand a state function |ψ" as |ψ" = i |i"#i|ψ". in a coordinate rep% Equivalently, % resentation, we have ψ(x) = #x|ψ" = i #i|ψ"φi (x), where i #x|i"#i|ψ" = φi (x) = #x|i". ' Info. Projection operators and completeness: A ‘ket’ state vector followed by a ‘bra’ state vector is an example of an operator. The operator which projects a vector onto the jth eigenstate is given by |j"#j|. First the bra vector dots into the state, giving the coefficient of |j" in the state, then its multiplied by the unit vector |j", turning it back into a vector, with the right length to be a projection. An operator maps one vector into another vector, so this is an operator. If we sum over a complete set of states, like the eigenstates of a Hermitian operator, we obtain the (useful) resolution of identity & |i"#i| = I . i
% Again, in coordinate form, we can write i φ∗i (x)φi (x" ) = δ(x − x" ). Indeed, we can % form a projection operator into a subspace, Pˆ = subspace |i"#i|.
As in a three-dimensional vector % space, the expansion of the vectors |φ" and % |ψ", as |φ" = i bi |i" and |ψ" = i ci |i",% allows the dot product to be taken by multiplying the components, #φ|ψ" = i b∗i ci . ' Example: The basis states can be formed from any complete set of orthogonal states. In particular, they can' be formed from' the basis states of the position or ∞ ∞ the momentum operator, i.e. −∞ dx|x"#x| = −∞ dp|p"#p| = I. If we apply these definitions, we can then recover the familiar Fourier representation, ! ∞ ! ∞ 1 ψ(x) ≡ #x|ψ" = dp #x|p" #p|ψ" = √ dp eipx/! ψ(p) , ( )* + 2π! −∞ −∞ √ eipx/! / 2π!
where #x|p" denotes the plane wave state |p" expressed in the real space basis.
3.1.1
Time-evolution operator
The ability to develop an eigenfunction expansion provides the means to explore the time evolution of a general wave packet, |ψ" under the action of a Hamiltonian. Formally, we can evolve a wavefunction forward in time by applying the time-evolution operator. For a Hamiltonian which is timeˆ |ψ(0)", where indepenent, we have |ψ(t)" = U ˆ ˆ = e−iHt/! U , 1 denotes % the time-evolution operator. By inserting the resolution of identity, I = i |i"#i|, where the states |i" are eigenstates of the Hamiltonian with eigenvalue Ei , we find that & & ˆ |ψ(t)" = e−iHt/! |i"#i|ψ(0)" = |i"#i|ψ(0)"e−iEi t/! . i
i
ˆ This equation follows from integrating the time-dependent Schr¨ odinger equation, H|ψ! = i!∂t |ψ!. 1
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22
ˆ = ' Example: Consider the harmonic oscillator Hamiltonian H
pˆ2 2m
+ 12 mω 2 x2 . Later in this chapter, we will see that the eigenstates, |n", have equally-spaced eigenvalues, En = !ω(n + 1/2), for n = 0, 1, 2, · · ·. Let us then consider the time-evolution of a general wavepacket, |ψ(0)", under % the action of the Hamiltonian. From the equation above, we find that |ψ(t)" = n |n"#n|ψ(0)"e−iEn t/! . Since the eigenvalues are equally spaced, let us consider what happens when t = tr ≡ 2πr/ω, with r integer. In this case, since e2πinr = 1, we have & |ψ(tr )" = |n"#n|ψ(0)"e−iωtr /2 = (−1)r |ψ(0)" . n
From this result, we can see that, up to an overall phase, the wave packet is perfectly reconstructed at these times. This recurrence or “echo” is not generic, but is a manifestation of the equal separation of eigenvalues in the harmonic oscillator.
' Exercise. Using the symmetry of the harmonic oscillator wavefunctions under parity show that, at times tr = (2r + 1)π/ω, #x|ψ(tr )" = e−iωtr /2 #−x|ψ(0)". Explain the origin of this recurrence. The time-evolution operator is an example of a unitary operator. The latter are defined as transformations which preserve the scalar product, #φ|ψ" = ˆ ψ" =! #φ|ψ", i.e. ˆ φ|U ˆ ψ" = #φ|U ˆ †U #U ˆ †U ˆ = I. U
3.1.2
Uncertainty principle for non-commuting operators
ˆ B] ˆ &= 0, it is straightforward to For non-commuting Hermitian operators, [A, establish a bound on the uncertainty in their expectation values. Given a state |ψ", the mean square uncertainty is defined as ˆ 2 ψ" = #ψ|U ˆ 2 ψ" (∆A)2 = #ψ|(Aˆ − #A") ˆ − #B") ˆ 2 ψ" = #ψ|Vˆ 2 ψ" , (∆B)2 = #ψ|(B ˆ = Aˆ − #ψ|Aψ" ˆ and Vˆ = B ˆ − #ψ|Bψ". ˆ where we have defined the operators U ˆ ˆ ˆ ˆ ˆ ˆ Since #A" and #B" are just constants, [U , V ] = [A, B]. Now let us take the ˆ |ψ" + iλVˆ |ψ" with itself to develop some information about scalar product of U the uncertainties. As a modulus, the scalar product must be greater than or ˆ 2 ψ" + λ2 #ψ|Vˆ 2 ψ" + iλ#U ˆ ψ|Vˆ ψ" − equal to zero, i.e. expanding, we have #ψ|U ˆ ˆ iλ#V ψ|U ψ" ≥ 0. Reorganising this equation in terms of the uncertainties, we thus find ˆ , Vˆ ]|ψ" ≥ 0 . (∆A)2 + λ2 (∆B)2 + iλ#ψ|[U If we minimise this expression with respect to λ, we can determine when the inequality becomes strongest. In doing so, we find ˆ , Vˆ ]|ψ" = 0, 2λ(∆B)2 + i#ψ|[U
λ=−
ˆ , Vˆ ]|ψ" i #ψ|[U . 2 (∆B)2
Substiuting this value of λ back into the inequality, we then find, 1 ˆ , Vˆ ]|ψ"2 . (∆A)2 (∆B)2 ≥ − #ψ|[U 4
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23
We therefore find that, for non-commuting operators, the uncertainties obey the following inequality, i ˆ ˆ ∆A ∆B ≥ #[A, B]" . 2 If the commutator is a constant, as in the case of the conjugate operators [ˆ p, x] = −i!, the expectation values can be dropped, and we obtain the relaˆ B]. ˆ For momentum and position, this result recovers tion, (∆A)(∆B) ≥ 2i [A, Heisenberg’s uncertainty principle, i ! ∆p ∆x ≥ #[ˆ p, x]" = . 2 2 ˆ t] = i!, Similarly, if we use the conjugate coordinates of time and energy, [E, we have ∆E ∆t ≥
3.1.3
! . 2
Time-evolution of expectation values
Finally, to close this section on operators, let us consider how their expectation values evolve. To do so, let us consider a general operator Aˆ which may itself involve time. The time derivative of a general expectation value has three terms. d ˆ ˆ t |ψ") . ˆ ˆ + #ψ|A(∂ #ψ|A|ψ" = ∂t (#ψ|)A|ψ" + #ψ|∂t A|ψ" dt If we then make use of the time-dependent Schr¨odinger equation, i!∂t |ψ" = ˆ H|ψ", and the Hermiticity of the Hamiltonian, we obtain d i, ˆ ˆ A|ψ" ˆ ˆ ˆ #ψ|A|ψ" = #ψ|H − #ψ|AˆH|ψ" +#ψ|∂t A|ψ" . dt ! ( )* + i ˆ A]|ψ" ˆ #ψ|[H, ! This is an important and general result for the time derivative of expectation values which becomes simple if the operator itself does not explicitly depend on time, d i ˆ ˆ A]|ψ" ˆ #ψ|A|ψ" = #ψ|[H, . dt ! From this result, which is known as Ehrenfest’s theorem, we see that expectation values of operators that commute with the Hamiltonian are constants of the motion. ' Exercise. Applied to the non-relativistic Schr¨odinger operator for a single
ˆ = pˆ2 + V (x), show that #x" particle moving in a potential, H ˙ = 2m Show that these equations are consistent with the relations, . / . / d ∂H d ∂H #x" = , #ˆ p" = − , dt ∂p dt ∂x the counterpart of Hamilton’s classical equations of motion.
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%p& ˆ m ,
#pˆ˙ " = −#∂x V ".
Paul Ehrenfest 1880-1933 An Austrian physicist and mathematician, who obtained Dutch citizenship in 1922. He made major contributions to the field of statistical mechanics and its relations with quantum mechanics, including the theory of phase transition and the Ehrenfest theorem.
3.2. SYMMETRY IN QUANTUM MECHANICS
3.2
Symmetry in quantum mechanics
Symmetry considerations are very important in quantum theory. The structure of eigenstates and the spectrum of energy levels of a quantum system reflect the symmetry of its Hamiltonian. As we will see later, the transition probabilities between different states under a perturbation, such as that imposed by an external electromagnetic field, depend in a crucial way on the transformation properties of the perturbation and lead to “selection rules”. Symmetries can be classified into two types, discrete and continuous, according to the transformations that generate them. For example, a mirror symmetry is an example of a discrete symmetry while a rotation in three-dimensional space is continuous. Formally, the symmetries of a quantum system can be represented by a ˆ , that act in the Hilbert group of unitary transformations (or operators), U 2 space. Under the action of such a unitary transformation, operators corresponding to observables Aˆ of the quantum model will then transform as, ˆ † AˆU ˆ. Aˆ → U ˆ †U ˆ = I, i.e. U ˆ† = U ˆ −1 . For unitary transformations, we have seen that U Under what circumstances does such a group of transformations represent a symmetry group? Consider a Schr¨odinger particle in three dimensions:3 ˆ . We The basic observables are the position and momentum vectors, ˆr and p can always define a transformation of the coordinate system, or the observˆ = ˆr or p ˆ 4 If R is an element ˆ is mapped to R[A]. ables, such that a vector A of the group of transformations, then this transformation will be represented ˆ (R), such that by a unitary operator U ˆU ˆ . ˆ †A ˆ = R[A] U Such unitary transformations are said to be symmetries of a general opˆ p, ˆr) if erator O(ˆ ˆ †O ˆU ˆ = O, ˆ U
ˆ U ˆ] = 0 . i.e. [O,
ˆ p, ˆr) ≡ H, ˆ the quantum Hamiltonian, such unitary transformations are If O(ˆ said to be symmetries of the quantum system.
3.2.1
Observables as generators of transformations
ˆ and ˆr for a Schr¨odinger particle are themselves generaThe vector operators p tors of space-time transformations. From the standard commutation relations 2
In quantum mechanics, the possible states of a system can be represented by unit vectors (called “state vectors”) residing in “state space” known as the Hilbert space. The precise nature of the Hilbert space is dependent on the system; for example, the state space for position and momentum states is the space of square-integrable functions. 3 In the following, we will focus our considerations on the realm of “low-energy” physics where the relevant space-time transformations belong to the Galilei group, leaving our discussion of Lorentz invariance to the chapter on relativistic quantum mechanics. 4 e.g., for a clockwise spatial rotation by an angle θ around ez , we have, 0 1 cos θ sin θ 0 R[r] = Rij x ˆj , R = @ − sin θ cos θ 0 A . 0 0 1 Similarly, for a spatial translation by a vector a, R[r] = r + a. (Exercise: construct representations for transformations corresponding to spatial reflections, and inversion.)
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3.2. SYMMETRY IN QUANTUM MECHANICS
25
one can show that, for a constant vector a, the unitary operator # $ i ˆ ˆ , U (a) = exp − a · p ! acting in the Hilbert space of a Schr¨odinger particle performs a spatial transˆ † (a)f (r)U ˆ (a) = f (r + a), where f (r) denotes a general algebraic lation, U function of r. ˆ = −i!∇, ' Info. The proof runs as follows: With p ˆ † (a) = ea·∇ = U
∞ & 1 ai · · · ain ∇i1 · · · ∇in , n! 1 n=0
where summation on the repeated indicies, im is assumed. Then, making use of the Baker-Hausdorff identity (exercise) ˆ ˆ −A ˆ ˆ + [A, ˆ B] ˆ + 1 [A, ˆ [A, ˆ B]] ˆ + ··· , eA Be =B 2!
(3.3)
it follows that ˆ † (a)f (r)U ˆ (a) = f (r) + ai (∇i f (r)) + 1 ai ai (∇i ∇i f (r)) + · · · = f (r + a) , U 1 1 1 2 2! 1 2 where the last identity follows from the Taylor expansion.
' Exercise. Prove the Baker-Hausdorff identity (3.3). Therefore, a quantum system has spatial translation as an invariance group if and only if the following condition holds, ˆ (a)H ˆ =H ˆU ˆ (a), U
ˆ =H ˆp ˆH ˆ. i.e. p
ˆ = H(ˆ ˆ p), This demands that the Hamiltonian is independent of position, H as one might have expected! Similarly, the group of unitary transformaˆ (b) = exp[− i b · ˆr], performs translations in momentum space. tions, U ! ˆ (b) = Moreover, spatial rotations are generated by the transformation U i ˆ ˆ ˆ denotes the angular momentum operator. exp[− ! θen · L], where L = ˆr × p ' Exercise. For an infinitesimal rotation by an angle θ by a fixed axis, eˆn , 2 ˆ = I − i θˆ construct R[r] and show that U ! en · L + O(θ ). Making use of the identity a N −a limN →∞ (1 − N ) = e , show that “large” rotations are indeed generated by the 0 1 ˆ = exp − i θˆ unitary transformations U ! en · L .
As we have seen, time translations are generated by the time evolution opˆ (t) = exp[− i Ht]. ˆ erator, U Therefore, every observable which commutes with ! the Hamiltonian is a constant of the motion (invariant under time translations), ˆ ˆ ˆ Aˆ = AˆH ˆ ⇒ eiHt/! ˆ −iHt/! ˆ H Ae = A,
∀t .
We now turn to consider some examples of discrete symmetries. Amongst these, perhaps the most important in low-energy physics are parity and timereversal. The parity operation, denoted Pˆ , involves a reversal of sign on all coordinates. Pˆ ψ(r) = ψ(−r) . Advanced Quantum Physics
3.2. SYMMETRY IN QUANTUM MECHANICS
26
This is clearly a discrete transformation. Application of parity twice returns the initial state implying that Pˆ 2 = 1. Therefore, the eigenvalues of the parity operation (if such exist) are ±1. A wavefunction will have a defined parity if and only if it is an even or odd function. For example, for ψ(x) = cos(x), Pˆ ψ = cos(−x) = cos(x) = ψ; thus ψ is even and P = 1. Similarly ψ = sin(x) is odd with P = −1. Later, in the next chapter, we will encounter the spherical harmonic functions which have the following important symmetry under parity, Pˆ Y!m = (−1)! Ylm . Parity will be conserved if the Hamiltonian is invariant under the parity operation, i.e. if the Hamiltonian is invariant under a reversal of sign of all the coordinates.5 In classical mechanics, the time-reversal operation involves simply “running the movie backwards”. The time-reversed state of the phase space coordinates (x(t), p(t)) is defined by (xT (t), pT (t)) where xT (t) = x(t) and pT (t) = −p(t). Hence, if the system evolved from (x(0), p(0)) to (x(t), p(t)) in time t and at t we reverse the velocity, p(t) → −p(t) with x(t) → x(t), at time 2t the system would have returned to x(2t) = x(0) while p(2t) = −p(0). If this happens, we say that the system is time-reversal invariant. Of course, this is just the statement that Newton’s laws are the same if t → −t. A notable case where this is not true is that of a charged particle in a magnetic field. As with classical mechanics, time-reversal in quantum mechanics involves the operation t → −t. However, referring to the time-dependent Schr¨odinger ˆ equation, i!∂t ψ(x, t) = Hψ(x, t), we can see that the operation t → −t is ˆ ∗ = H. ˆ equivalent to complex conjugation of the wavefunction, ψ → ψ ∗ if H Let us then consider the time-evolution of ψ(x, t), i
ˆ
c.c.
i
ˆ ∗ (x)t
ψ(x, 0) → e− ! H(x)t ψ(x, 0) → e+ ! H
evolve
i
ˆ
i
ˆ ∗ (x)t
ψ ∗ (x, 0) → e− ! H(x)t e+ ! H
ψ ∗ (x, 0) .
ˆ ∗ (x) = H(x). ˆ If we require that ψ(x, 2t) = ψ ∗ (x, 0), we must have H Therefore, ˆ ˆ H is invariant under time-reversal if and only if H is real. ' Info. Although the group of space-transformations covers the symmetries that pertain to “low-energy” quantum physics, such as atomic physics, quantum optics, and quantum chemistry, in nuclear physics and elementary particle physics new observables come into play (e.g. the isospin quantum numbers and the other quark charges in the standard model). They generate symmetry groups which lack a classical counterpart, and they do not have any obvious relation with space-time transformations. These symmetries are often called internal symmetries in order to underline this fact.
3.2.2
Consequences of symmetries: multiplets
Having established how to identify whether an operator belongs to a group of symmetry transformations, we now consider the consequences. Consider ˆ in the Hilbert space, and an observable Aˆ a single unitary transformation U ˆ ˆ ˆ which commutes with U , [U , A] = 0. If Aˆ has an eigenvector |a", it follows ˆ |a" will be an eigenvector with the same eigenvalue, i.e. that U ˆ A|a" ˆ = AU ˆ |a" = aU |a" . U This means that either: 5
In high energy physics, parity is a symmetry of the strong and electromagnetic forces, but does not hold for the weak force. Therefore, parity is conserved in strong and electromagnetic interactions, but is violated in weak interactions.
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3.3. THE HEISENBERG PICTURE
27
ˆ , or 1. |a" is an eigenvector of both Aˆ and U 2. the eigenvalue a is degenerate: the linear space spanned by the vectors ˆ n |a" (n integer) are eigenvectors with the same eigenvalue. U This mathematical argument leads to the conclusion that, given a group G of ˆ (x), x ∈ G, for any observable which is invariant under unitary operators U these transformations, i.e. ˆ (x), A] ˆ = 0 ∀x ∈ G , [U its discrete eigenvalues and eigenvectors will show a characteristic multiplet structure: there will be a degeneracy due to the symmetry such that the eigenvectors belonging to each eigenvalue form an invariant subspace under the group of transformations. ' Example: For example, if the Hamiltonian commutes with the angular moˆ i , i = x, y, z, i.e. it is invariant under three-dimensional rotamentum operators, L tions, an energy level with a given orbital quantum number , is at least (2, + 1)-fold degenerate. Such a degeneracy can be seen as the result of non-trivial actions of ˆ x and L ˆ y on an energy (and L ˆ z ) eigenstate |E, ,, m" (where m is the the operator L ˆ z ). magnetic quantum number asssociated with L
3.3
The Heisenberg Picture
Until now, the time dependence of an evolving quantum system has been placed within the wavefunction while the operators have remained constant – this is the Schr¨ odinger picture or representation. However, it is sometimes useful to transfer the time-dependence to the operators. To see how, let ˆ us consider the expectation value of some operator B, ˆ ˆ ˆ ˆ ˆ ˆ −iHt/! ˆ −iHt/! #ψ(t)|B|ψ(t)" = #e−iHt/!ψ(0)|B|e ψ(0)" = #ψ(0)|eiHt/!Be |ψ(0)" .
According to rules of associativity, we can multiply operators together beˆ ˆ ˆ ˆ −iHt/! fore using them. If we define the operator B(t) = eiHt/!Be , the timedependence of the expectation values has been transferred from the wavefunction. This is called the Heisenberg picture or representation and in it, the operators evolve with time while the wavefunctions remain constant. In this representation, the time derivative of the operator itself is given by ˆ = ∂t B(t)
ˆ ˆ ˆ iH ˆ ˆ ˆ ˆ −iHt/! ˆ iH e−iHt/! eiHt/!Be − eiHt/!B ! ! i ˆ i ˆ ˆ ˆ ˆ B]e ˆ −iHt/! = eiHt/![H, = [H, B(t)] . ! !
ˆ = ' Exercise. For the general Hamiltonian H
pˆ2 2m
+ V (x), show that the position and momentum operators obey Hamilton’s classical equation of motion.
3.4
Quantum harmonic oscillator
As we will see time and again in this course, the harmonic oscillator assumes a priveledged position in quantum mechanics and quantum field theory finding Advanced Quantum Physics
Werner Heisenberg 1901-76 A German physicist and one of the founders of the quantum theory, he is best known for his uncertainty principle which states that it is impossible to determine with arbitrarily high accuracy both the position and momentum of a particle. In 1926, Heisenberg developed a form of the quantum theory known as matrix mechanics, which was quickly shown to be fully equivalent to Erwin Schr¨ odinger’s wave mechanics. His 1932 Nobel Prize in Physics cited not only his work on quantum theory but also work in nuclear physics in which he predicted the subsequently verified existence of two allotropic forms of molecular hydrogen, differing in their values of nuclear spin.
3.4. QUANTUM HARMONIC OSCILLATOR
28
numerous and somtimes unexpected applications. It is useful to us now in that it provides a platform for us to implement some of the technology that has been developed in this chapter. In the one-dimensional case, the quantum harmonic oscillator Hamiltonian takes the form, 2 ˆ = pˆ + 1 mω 2 x2 , H 2m 2 where pˆ = −i!∂x . To find the eigenstates of the Hamiltonian, we could look for solutions of the linear second order differential equation correspondˆ = Eψ, where H ˆ = ing to the time-independent Schr¨odinger equation, Hψ 1 !2 2 2 2 − 2m ∂x + 2 mω x . The integrability of the Schr¨odinger operator in this case allows the stationary states to be expressed in terms of a set of orthogonal functions known as Hermite polynomials. However, the complexity of the exact eigenstates obscure a number of special and useful features of the harmonic oscillator system. To identify these features, we will instead follow a method based on an operator formalism. The form of the Hamiltonian as the sum of the squares of momenta and position suggests that it can be recast as the “square of an operator”. To this end, let us introduce the operator 2 2 3 4 3 4 mω pˆ mω pˆ † x+i , a = x−i , a= 2! mω 2! mω
where, for notational convenience, we have not drawn hats on the operators a and its Hermitian conjuate a† . Making use of the identity, a† a =
ˆ mω 2 pˆ i H 1 x + + [x, pˆ] = − 2! 2!mω 2! !ω 2
and the parallel relation, aa† = commutation relations
ˆ H !ω
+ 12 , we see that the operators fulfil the
[a, a† ] ≡ aa† − a† a = 1 . Then, setting n ˆ = a† a, the Hamiltonian can be cast in the form ˆ = !ω(ˆ H n + 1/2) . Since the operator n ˆ = a† a must lead to a positive definite result, we see that the eigenstates of the harmonic oscillator must have energies of !ω/2 or higher. Moreover, the ground state |0" can be identified by finding the state for which a|0" = 0. Expressed in the coordinate basis, this translates to the equation,6 2 3 4 ! mω −mωx2 /2! x+ ∂x ψ0 (x) = 0, ψ0 (x) = #x|0" = e . mω π 1/2 ! Since n ˆ |0" = a† a|0" = 0, this state is an eigenstate with energy !ω/2. The higher lying states can be found by acting upon this state with the operator a† . The proof runs as follows: If n ˆ |n" = n|n", we have n ˆ (a† |n") = a† ()*+ aa† |n" = (a† ()*+ a† a +a† )|n" = (n + 1)a† |n" a† a+1
n ˆ
6 " ! Formally, in coordinate basis, we have #x" |a|x! R = δ(x − x)(a + mω ∂x ) and #x|0! = ψ0 (x). Then making use of the resolution of identity dx|x!#x| = I, we have „ « Z ! #x|a|0! = 0 = dx #x|a|x" !#x" |0! = x + ∂x ψ0 (x) . mω
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First few states of the quantum harmonic oscillator. Not that the parity of the state changes from even to odd through consecutive states.
3.4. QUANTUM HARMONIC OSCILLATOR or, equivalently, [ˆ n, a† ] = a† . In other words, if |n" is an eigenstate of n ˆ with eigenvalue n, then a† |n" is an eigenstate with eigenvalue n + 1. From this result, we can deduce that the eigenstates for a “tower” |0", |1" = C1 a† |0", |2" = C2 (a† )2 |0", etc., where Cn denotes the normalization. If #n|n" = 1 we have #n|aa† |n" = #n|(ˆ n + 1)|n" = (n + 1) . 1 Therefore, with |n + 1" = √n+1 a† |n" the state |n + 1" is also normalized, #n + 1|n + 1" = 1. By induction, we can deduce the general normalization,
1 |n" = √ (a† )n |0" , n! ˆ with #n|n% " = δnn" , H|n" = !ω(n + 1/2)|n" and a† |n" =
√
n + 1|n + 1",
a|n" =
√
n|n − 1" .
The operators a and a† represent ladder operators and have the effect of lowering or raising the energy of the state. In fact, the operator representation achieves something quite remarkable and, as we will see, unexpectedly profound. The quantum harmonic oscillator describes the motion of a single particle in a one-dimensional potential well. It’s eigenvalues turn out to be equally spaced – a ladder of eigenvalues, separated by a constant energy !ω. If we are energetic, we can of course translate our results into a coordinate representation ψn (x) = #x|n".7 However, the operator representation affords a second interpretation, one that lends itself to further generalization in quantum field theory. We can instead interpret the quantum harmonic oscillator as a simple system involving many fictitious particles, each of energy !ω. In this representation, known as the Fock space, the vacuum state |0" is one involving no particles, |1" involves a single particle, |2" has two and so on. These fictitious particles are created and annihilated by the action of the raising and lowering operators, a† and a with canonical commutation relations, [a, a† ] = 1. Later in the course, we will find that these commutation relations are the hallmark of bosonic quantum particles and this representation, known as the second quantization underpins the quantum field theory of the electromagnetic field. ' Info. There is evidently a huge difference between a stationary (Fock) state of the harmonic oscillator and its classical counterpart. For the classical system, the equations of motion are described by Hamilton’s equations of motion, P X˙ = ∂P H = , m
P˙ = −∂X H = −∂x U = −mω 2 X ,
where we have used capital letters to distinguish them from the arguments used to describe the quantum system. In the phase space, {X(t), P (t)}, these equations describe a clockwise rotation along an elliptic trajectory specified by the initial conditions {X(0), P (0)}. (Normalization of momentum by mω makes the trajectory circular.) 7 Expressed in real space, the harmonic oscillator wavefunctions are in fact described by the Hermite polynomials, r „r « » – mωx2 1 mω ψn (x) = #x|n! = H x exp − , n 2n n! ! 2! 2
where Hn (x) = (−1)n ex
dn −x2 e . dxn
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3.4. QUANTUM HARMONIC OSCILLATOR
30
On the other hand, the time dependence of the Fock space state, as of any stationary state, is exponential, ψn (x, t) = #x|n"e−iEn t/! , and, as a result, gives time-independent expectation values of x, p, or any function thereof. The best classical image for such a state on the phase plane is a circle of radius r = x0 (2n + 1)1/2 , where x0 = (!/mω)1/2 , along which the wavefunction is uniformly spread as a standing wave. It is natural to ask how to form a wavepacket whose properties would be closer to the classical trajectories. Such states, with the centre in the classical point {X(t), P (t)}, and the smallest possible product of quantum uncertainties of coordinate and momentum, are called Glauber states.8 Conceptually the simplest way to present the Glauber state |α" is as the Fock ground state |0" with the centre shifted from the origin to the classical point {X(t), P (t)}. (After such a shift, the state automatically rotates, following the classical motion.) Let us study how this shift may be implemented in quantum mechanics. The mechanism for such shifts are called the translation operators. Previously, we have seen that the space and momentum translation operators are given by $ # $ # i i ˆ ˆ FP = exp − P x ˆ . FX = exp − pˆX , ! ! A shift by both X and P is then given by # $ † ∗ i ˆ − pˆX) = eαa −α a , Fˆα = exp (P x !
∗ † Fˆα† = eα a−αa ,
where α is the (normalised) complex amplitude of the classical oscillations we are 1 P trying to approximate, i.e. α = √2x (X + i mω ). The Glauber state is then defined 0 by |α" = Fˆα |0". Working directly with the shift operator is not too convenient because of its exponential form. However, it turns out that a much simpler representation for the Glauber state is possible. To see this, let us start with the following general property ˆ B] ˆ = µ (where Aˆ and B ˆ are operators, and µ is a of exponential operators: if [A, c-number), then (exercise – cf. Eq. (3.3)), ˆ −A = B ˆ + µ. eA Be ˆ
ˆ
(3.4)
ˆ ˆ ˆ = I, we If we define Aˆ = α∗ a − αa† , then Fˆα = e−A and Fˆα† = eA . If we then take B † ˆ ˆ have µ = 0, and Fα Fα = I. This merely means that the shift operator is unitary not a big surprise, because if we shift the phase point by (+α) and then by (−α), we certainly come back to the initial position. ˆ = a, using the commutation relations, If we take B
ˆ B] ˆ = [α∗ a − αa† , a] = −α[a† , a] = α , [A,
so that µ = α, and Fˆα† aFˆα = a + α. Now let us consider the operator Fˆα Fˆα† aFˆα . From the unitarity condition, this must equal aFˆα , while application of Eq. (3.4) yields Fˆα a + αFˆα , i.e. aFˆα = Fˆα a + αFˆα .
Applying this equality to the ground state |0" and using the following identities, a|0" = 0 and Fˆα |0" = |α", we finally get a very simple and elegant result: a|α" = α|α" . 8 After R. J. Glauber who studied these states in detail in the mid-1960s, though they were known to E. Schr¨ odinger as early as in 1928. Another popular name, coherent states, does not make much sense, because all the quantum states we have studied so far (including the Fock states) may be presented as coherent superpositions.
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3.5. POSTULATES OF QUANTUM THEORY Thus the Glauber state is an eigenstate of the annihilation operator, corresponding to the eigenvalue α, i.e. to the (normalized) complex amplitude of the classical process approximated by the state. This fact makes the calculations of the Glauber state properties much simpler. Presented as a superposition of Fock states, the Glauber state takes the form (exercise – try making use of the BCH identity (3.3).) |α" =
∞ &
n=0
αn |n",
αn = e−|α|
2
/2
αn . (n!)1/2
This means that the probability of finding the system in level n is given by the Poisson distribution, Pn = |αn |2 = #n"n e−%n& /n! where #n" = |α|2 . More importantly, δn = #n"1/2 / #n" when #n" 0 1 – the Poisson distribution approaches the Gaussian distribution when #n" is large. The time-evolution of Glauber states may be described most easily in the Schr¨odinger representation when the time-dependence is transferred to the wavefunction. In this (t) 1 case, α(t) ≡ √2x (X(t) + i Pmω ), where {X(t), P (t)} is the solution to the classical 0 equations of motion, α(t) ˙ = −iωα(t). From the solution, α(t) = α(0)e−iωt , one may show that the average position and momentum evolve classically while their fluctuations remain stationary, 3 41/2 3 41/2 x0 ! mωx0 !mω ∆x = √ = , ∆p = √ = . 2mω 2m 2 2
In the quantum theory of measurements these expressions are known as the “standard quantum limit”. Notice that their product ∆x ∆p = !/2 corresponds to the lower bound of the Heisenberg’s uncertainty relation.
' Exercise. Show that, in position space, the Glauber state takes the form # $ mω Px #x|α" = ψα (x) = C exp − (x − X)2 + i . 2! !
This completes our abridged survey of operator methods in quantum mechanics. With this background, we are now in a position to summarize the basic postulates of quantum mechanics.
3.5
Postulates of quantum theory
Since there remains no “first principles” derivation of the quantum mechanical equations of motion, the theory is underpinned by a set of “postulates” whose validity rest on experimental verification. Needless to say, quantum mechanics remains perhaps the most successful theory in physics. ' Postulate 1. The state of a quantum mechanical system is completely specified by a function Ψ(r, t) that depends upon the coordinates of the particle(s) and on time. This function, called the wavefunction or state function, has the important property that |Ψ(r, t)|2 dr represents the probability that the particle lies in the volume element dr ≡ dd r located at position r at time t. The wavefunction must satisfy certain mathematical conditions because of this probabilistic interpretation. For the case of a single particle, the net probability of finding it at some ' ∞ point in space must be unity leading to the normalization condition, −∞ |Ψ(r, t)|2 dr = 1. It is customary to also normalize many-particle wavefunctions to unity. The wavefunction must also be single-valued, continuous, and finite. Advanced Quantum Physics
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3.5. POSTULATES OF QUANTUM THEORY ' Postulate 2. To every observable in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics. If we require that the expectation value of an operator Aˆ is real, then it follows that Aˆ must be a Hermitian operator. If the result of a measurement of an operator Aˆ is the number a, then a must be one of the ˆ = aΨ, where Ψ is the corresponding eigenfunction. This eigenvalues, AΨ postulate captures a central point of quantum mechanics – the values of dynamical variables can be quantized (although it is still possible to have a continuum of eigenvalues in the case of unbound states). ' Postulate 3. If a system is in a state described by a normalized wavefunction Ψ, then the average value of the observable corresponding to Aˆ is given by ! ∞ ˆ #A" = Ψ∗ AΨdr . −∞
If the system is in an eigenstate of Aˆ with eigenvalue a, then any measurement of the quantity A will yield a. Although measurements must always yield an eigenvalue, the state does not have to be an eigenstate of Aˆ initially. An arbitrary state can be expanded in the complete set ˆ i = ai Ψi ) as Ψ = %n ci Ψi , where n may go to of eigenvectors of Aˆ ( AΨ i infinity. In this case, the probability of obtaining the result ai from the measurement of Aˆ is given by P (ai ) = |#Ψi |Ψ"|2 = |ci |2 . The expectation value of Aˆ for the state Ψ is the sum over all possible values of the measurement and given by & & #A" = ai |#Ψi |Ψ"|2 = ai |ci |2 . i
i
Finally, a measurement of Ψ which leads to the eigenvalue ai , causes the wavefunction to “collapses” into the corresponding eigenstate Ψi . (In the case that ai is degenerate, then Ψ becomes the projection of Ψ onto the degenerate subspace). Thus, measurement affects the state of the system. ' Postulate 4. The wavefunction or state function of a system evolves in time according to the time-dependent Schr¨odinger equation i!
∂Ψ ˆ = HΨ(r, t) , ∂t
ˆ is the Hamiltonian of the system. If Ψ is an eigenstate of H, ˆ where H it follows that Ψ(r, t) = Ψ(r, 0)e−iEt/!.
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Chapter 4
Quantum mechanics in more than one-dimension Previously, we have explored the manifestations of quantum mechanics in one spatial dimension and discussed the properties of bound and unbound states. The concepts developed there apply equally to higher dimension. However, for a general two or three-dimensional potential, without any symmetry, the solutions of the Schr¨odinger equation are often inaccessible. In such situations, we may develop approximation methods to address the properties of the states (see chapter 7). However, in systems where there is a high degree of symmetry, the quantum mechanics of the system can often be reduced to a tractable lowdimensional theory.
4.1
Rigid diatomic molecule
As a pilot example let us consider the quantum mechanics of a rigid diatomic molecule with nuclear masses m1 and m2 , and (equilibrium) bond length, Re (see figure). Since the molecule is rigid, its coordinates can be specified by 2 r2 its centre of mass, R = m1mr11 +m +m2 , and internal orientation, r = r2 − r1 (with |r| = Re ). Defining the total mass M = m1 + m2 , and moment of inertia, I = µRe2 , where µ = m1 m2 /(m1 + m2 ) denotes the reduced mass, the corresponding Hamiltonian can be then separated into the kinetic energy associated with the centre of mass motion and the rotational kinetic energy, ˆ2 ˆ2 ˆ = P +L , H 2M 2I
(4.1)
ˆ = −i!∇R and L ˆ =r×p ˆ denotes the angular momentum associated where P with the internal degrees of freedom. Since the internal and centre of mass degrees of freedom separate, the wavefunction can be factorized as ψ(r, R) = eiK·R Y (r), where the first factor accounts for the free particle motion of the body, and the second factor relates to the internal angular degrees of freedom. As a result of the coordinate separation, we have reduced the problem of the rigid diatomic molecule to the study of the quantum mechanics of a particle moving on a sphere – the rigid rotor, ˆ2 ˆ rot = L . H 2I The eigenstates of this component of the Hamiltonian are simply the states of the angular momentum operator. Indeed, in any quantum mechanical system involving a radial potential, the angular momentum will be conserved, i.e. Advanced Quantum Physics
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ˆ = 0 meaning that the angular component of the wavefunction can be ˆ L] [H, indexed by the states of the angular momentum operator. We therefore now digress to discuss the quantum mechanics of angular momentum.
4.2
Angular momentum
4.2.1
Commutation relations
Following the usual canonical quantization procedure, the angular momentum ˆ = r׈ ˆ obey the commutation operator is defined by L p where, as usual, r and p relations, [ˆ pi , rj ] = −i!δij .1 Using this relation, one may then show that the angular momentum operators obey the spin commutation relations, (exercise) ˆ i, L ˆ j ] = i!#ijk L ˆk . [L
(4.2)
where, as usual, #ijk denotes the totally antisymmetric tensor — the LeviCivita symbol.2 $ Exercise. Show that the angular momentum operator commutes with the ˆ = pˆ 2 + V (r). Show that Hamiltonian of a particle moving in a central potential, H 2m the Hamiltonian of a free particle of mass m confined to a sphere of radius R is given ˆ = Lˆ 2 2 . by H 2mR
4.2.2
Eigenvalues of angular momentum
In the following, we will construct a basis set of angular momentum states. Since the angular momentum is a vector quantity, it may be characterized by its magnitude and direction. For the former, let us define the operator ˆ2 = L ˆ2 + L ˆ2 + L ˆ 2 . With the latter, since the separate components of the L x y z angular momentum are all mutually non-commuting, we cannot construct a common set of eigenstates for any two of them. They do, however, commute ˆ 2 (exercise). Therefore, in the following, we will look for an eigenbasis with L ˆ 2 and one direction, say L ˆz, of L ˆ 2 |a, b$ = a|a, b$, L
ˆ z |a, b$ = b|a, b$ . L
To find |a, b$, we could simply proceed by looking for a suitable coordinate ˆ 2 and L ˆ z in terms of differential operators. However, basis to represent L although we will undertake such a programme in due course, before getting to this formalism, we can make substantial progress without resorting to an explicit coordinate representation. $ Info. Raising and lowering operators for angular momentum: The set of eigenvalues a and b can be obtained by making use of a trick based on a “ladder operator” formalism which parallels that used in the study of the quantum harmonic oscillator in section 3.4. Specifically, let us define the raising and lowering operators, ˆ± = L ˆ x ± iL ˆy . L 1 In this chapter, we will index the angular momentum operators with a ‘hat’. Later, we will become lazy and the hat may well disappear. 2 Recall that !ijk = 1 if (i, j, k) is an even permutation of (1,2,3), −1 if it is an odd permutation, and 0 if any index is repeated.
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With this definition, one may then show that (exercise) ˆz, L ˆ ± ] = ±!L ˆ± . [L
ˆ 2 , we can deduce Since each component of the angular momentum commutes with L ˆ that the action of L± on |a, b$ cannot affect the value of a relating to the magnitude of the angular momentum. However, they do effect the projection: ˆzL ˆ ± |a, b$ = L ˆ±L ˆ z |a, b$ + [L ˆz, L ˆ ± ]|a, b$ = (b ± !)L ˆ ± |a, b$ . L
ˆ z with eigenvalue b, L ˆ ± |a, b$ is either zero, Therefore, if |a, b$ is an eigenstate of L ˆ ˆ or an eigenstate of Lz with eigenvalue b ± !, i.e. L± |a, b$ = C± (a, b)|a, b ± !$ where C± (a, b) is a normalisation constant. To fix the normalisation, we may note that the norm, !! !!2 !! ˆ !! ˆ† L ˆ ˆ ˆ !!L± |a, b$!! = %a, b|L ± ± |a, b$ = %a, b|L∓ L± |a, b$ ,
ˆ† = L ˆ ∓ . Then, making use of the relation L ˆ∓L ˆ± = where we have used the identity L ± 2 2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ Lx + Ly ± i[Lx Ly ] = L − Lz ± !Lz , and the presumed normalisation, %a, b|a, b$ = 1, one finds !! !!2 " # !! ˆ !! ˆ2 − L ˆ 2z ± !L ˆ z |a, b$ = a − b2 ∓ !b . (4.3) !!L± |a, b$!! = %a, b| L
As a represents the eigenvalue of a sum of squares of Hermitian operators, it is necessarily non-negative. Moreover, b is real. Therefore, for a given a, b must be bounded: there must be a bmax and a (negative or zero) bmin . In particular, !! !!2 !! ˆ !! !!L+ |a, bmax $!! = a − b2max − !bmax !! !!2 !! ˆ !! !!L− |a, bmin $!! = a − b2min + !bmin ,
For a given a, bmax and bmin are determined uniquely — there cannot be two states ˆ + . It also follows immediately that with the same a but different b annihilated by L a = bmax (bmax +!) and bmin = −bmax . Furthermore, we know that if we keep operating ˆ + , we generate a sequence of states with L ˆ z eigenvalues bmin + !, on |a, bmin $ with L bmin + 2!, bmin + 3!, · · ·. This series must terminate, and the only possible way for that to happen is for bmax to be equal to bmin + n! with n integer, from which it follows that bmax is either an integer or half an odd integer times ! At this point, we switch to the standard notation. We have established that the ˆ z form a finite ladder, with spacing !. We write them as m!, and % eigenvalues of L ˆ 2 , a = %(% + 1)!2 . is used to denote the maximum value of m, so the eigenvalue of L Both % and m will be integers or half odd integers, but the spacing of the ladder of m values is always unity. Although we have been writing |a, b$ with a = %(% + 1)!2 , b = m! we shall henceforth follow convention and write |%, m$.
ˆ 2 and L ˆ z have a common set of orthonormal In summary, the operators L eigenstates |%, m$ with ˆ 2 |%, m$ = %(% + 1)!2 |%, m$, L
ˆ z |%, m$ = m!|%, m$ , L
(4.4)
where %, m are integers or half-integers. The allowed quantum numbers m form a ladder with step spacing unity, the maximum value of m is %, and the minimum value is −%. With these results, we may then return to the normalization of the raising and lowering operators. In particular, making use of Eq. (4.3), we have $ ˆ + |%, m$ = %(% + 1) − m(m + 1)!|l, m + 1$ L $ ˆ − |%, m$ = %(% + 1) − m(m − 1)!|l, m − 1$ . L
Advanced Quantum Physics
(4.5)
Figure 4.1: The following is a
schematic showing the angular momentum $ scheme for % = √ 2 with L2 = 2(2 + 1)! = 6! and the five possible values for the Lz projection.
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36
The use of m to denote the component of angular momentum in one direction came about because a Bohr-type electron in orbit is a current loop, with a magnetic moment parallel to its angular momentum. So the m measured the component of magnetic moment in a chosen direction, usually along an external magnetic field. For this reason, m is often termed the magnetic quantum number.
4.2.3
Representation of the angular momentum states
Having established expressions for the eigenvalues of the angular momentum operators, it is now necessary to establish coordinate representations for the corresponding eigenstates, Y!m (θ, φ) = %θ, φ|%, m$. Here the angles θ and φ denote the spherical coordinates parameterising the unit sphere (see figure). Previously, we obtained the eigenvalues of the angular momentum operator by making use of the raising and lowering operators in a manner that parallelled the study of the quantum harmonic oscillator. Similarly, to obtain explicit expressions for the eigenstates, we must make use of the coordinate representation of these operators. With r = rˆ er , the gradient operator can be written in spherical polar coordinates as 1 1 ˆr ∂r + e ˆθ ∂θ + e ˆφ ∇=e ∂φ . r r sin θ From this result, we thus obtain ˆ z = −i!∂φ , L
ˆ ± = !e±iφ (±∂θ + i cot θ∂φ ) , L
(4.6)
and, at least formally, ˆ 2 = −!2 L
%
& 1 1 2 ∂θ (sin θ∂θ ) + ∂ . sin θ sin2 θ φ
ˆ z , the eigenvalue equation (4.4), and Beginning with the eigenstates of L making use of the expression above, we have −i!∂φ Y!m (θ, φ) = m!Y!m (θ, φ) . Since the left hand side depends only on φ, the solution is separable and takes the form Y!m (θ, φ) = F (θ)eimφ . Note that, since m is integer, the continuity of the wavefunction, Y!m (θ, φ + 2π) = Y!m (θ, φ), is ensured. To determine the second component of the eigenstates, F (θ), we could immediately turn to the eigenvalue equation involving the differential operator ˆ 2, for L % & 1 m2 2 ∂θ (sin θ∂θ ) − ∂ F (θ) = %(% + 1)F (θ) . sin θ sin2 θ φ However, to construct the states, it is easier to draw upon the properties of the angular momentum raising and lowering operators (much in the same way that the Hermite polynomials are generated by the action of ladder operators in the harmonic oscillator problem). ˆ + |%, %$ = 0. MakConsider then the state of maximal m, |%, %$, for which L ing use of the coordinate representation of the raising operator above together with the separability of the wavefunction, this relation implies that ˆ + |%, %$ = !eiφ (∂θ + i cot θ∂φ ) Y!! (θ, φ) = !ei(!+1)φ (∂θ − % cot θ) F (θ) . 0 = %θ, φ|L Advanced Quantum Physics
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From this result it follows that ∂θ F (θ) = % cot θF (θ) with the solution F (θ) = C sin! θ, and C a constant of normalization. States with values of m lower than % can then be obtained simply by repeated application of the angular ˆ − to the state |%, %$. This amounts to the momentum lowering operator L relation ' ( ˆ − )!−m sin! θei!φ Y!m (θ, φ) = C(L ' ( = C (−∂θ + i cot θ∂φ )!−m sin! θei!φ . The eigenfunctions produced by this procedure are well known and referred to as the spherical harmonics. In particular, one finds that the normalized eigenstates take the form, Y!m (θ, φ) = (−1)
m+|m|
%
2% + 1 (% − |m|)! 4π (% + |m|)!
&1/2
|m|
P! (cos θ)eimφ ,
(4.7)
where P!m (ξ) =
(1 − ξ 2 )m/2 dm+! 2 (ξ − 1)! , 2! %! dξ m+!
represent the associated Legendre polynomials. In particular, for the first few angular momentum states, we have Y00 = Y10 = Y20 =
√1
)4π
) 3 iφ Y11 = − 8π e sin θ ) ) 5 15 iφ 15 2iφ 2 θ − 1), Y (3 cos = − e sin θ cos θ, Y = sin2 θ 21 22 16π 8π 32π e
3 ) 4π
cos θ,
Figure 4.2 shows a graphical representation of the states for the lowest spherical harmonics. From the colour coding of the states, the symmetry, Y!,−m = ∗ is manifest. (−1)m Y!m As a complete basis set, the spherical harmonics can be used as a resolution of the identity ∞ * ! *
!=0 m=−!
|%, m$%%, m| = I .
Equivalently, expressed in the coordinate basis, we have ∞ * ! *
∗ Y!,m (θ% , φ% )Y!,m (θ, φ) =
!=0 m=−!
1 δ(θ − θ% )δ(φ − φ% ) , sin θ
where the prefactor sin θ derives from the measure. Similarly, we have the orthogonality condition, + π + 2π ∗ Y!,m (θ, φ)Y!! ,m! (θ, φ) = δ!!! δmm! . dθ sin θ 0
0
After this lengthy digression, we may now return to the problem of the quantum mechanical rotor Hamiltonian and the rigid diatomic molecule. From the analysis above, we have found that the eigenstates of the Hamiltonian (4.1) are given by ψ(R, r) = √12π eiK·R Y!,m (θ, φ) with eigenvalues EK,! = Advanced Quantum Physics
!2 !2 K2 + %(% + 1) , 2M 2I
4.3. THE CENTRAL POTENTIAL
Figure 4.2: First four groups of spherical harmonics, Y!m (θ, φ) shown as a function of spherical angular coordinates. Specifically, the plots show the surface generated by |Re Y!m (θ, φ)| to fix the radial coordinate and the colours indicate the relative sign of the real part. where each K, % value has a 2% + 1-fold degeneracy. $ Exercise. Using this result, determine the dependence of the heat capacity of a gas of rigid diatomic molcules on the angular degrees of freedom. How would this result change if the diatomic gas was constrained to just two spatial dimensions, i.e. the axis of rotation was always perpendicular to the plane in which the molecules can move?
4.3
The central potential
In a system where the central force field is entirely radial, the potential energy depends only on r ≡ |r|. In this case, a general non-relativistic Hamiltonian for a single particle is given by ˆ2 ˆ = p H + V (r) . 2m In the classical system, L2 = (r × p)2 = r2 p2 − (r · p)2 . As a result, we can 2 set p2 = Lr2 + p2r , where pr ≡ er · p denotes the radial component of the momentum. In the quantum system, since the space and position coordinates do not commute, we have (exercise) ˆ 2 = r2 p ˆ 2 − (r · p ˆ )2 + i!r · p ˆ. L ˆ = −i!r · ∇ = −i!r∂r , it follows that In spherical coordinates, since r · p 2 ˆ2 ˆ 2 = Lr2 − !r2 [(r∂r )2 + r∂r ]. Equivalently, noting that (r∂r )2 + r∂r = ∂r2 + 2r ∂r , p we can set % & ˆ2 L 2 2 2 2 ˆ = 2 − ! ∂r + ∂r . p r r Advanced Quantum Physics
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Finally, substituted into the Schr¨odinger equation, we obtain the eigenvalue equation , / . ˆ2 2 !2 L (4.8) − ∂r2 + ∂r + + V (r) ψ(r) = Eψ(r) . 2m r 2mr2 ˆ 2 , we can immediately take adSince we already know the eigenstates of L vantage of the separability of the Hamiltonian to find that ψ(r) = R(r)Y!,m (θ, φ), where the radial part of the wavefunction is set by % . & !2 2 !2 2 − ∂r + ∂r + %(% + 1) + V (r) R(r) = ER(r) . 2m r 2mr2 Finally, we can further simplify this expression by setting R(r) = u(r)/r, whereupon we obtain the “one-dimensional” equation % 2 2 & ! ∂r − + Veff (r) u(r) = u(r) , (4.9) 2m ! where the effective potential, Veff (r) = 2mr 2 %(% + 1) + V (r), acquires an additional component due to the centrifugal component of the force. Here the equation must be solved subject to the boundary condition u(0) = 0. From the normalization condition, + + ∞ 1 3 2 d r |ψ(r)| = drr2 2 |u(r)|2 = 1 , r 0 2
for a bound state to exist, limr→∞ |u(r)| ≤ a/r1/2+% with # > 0. From this one-dimensional form of the Hamiltonian, the question of the existence of bound states in higher dimension becomes clear. Since the wavefunction u(r) vanishes at the origin, we may map the Hamiltonian from the half-line to the full line with the condition that we admit only antisymmetric wavefunctions. The question of bound states can then be related back to the one-dimensional case. Previously, we have seen that a symmetric attractive potential always leads to a bound state in one-dimension. However, odd parity states become bound only at a critical strength of the interaction.
4.4
Atomic hydrogen
The Hydrogen atom consists of an electron bound to a proton by the Coulomb potential, V (r) = −
1 e2 . 4π#0 r
We can generalize the potential to a nucleus of charge Ze without complication of the problem. Since we are interested in finding bound states of the proton-electron system, we are looking for solutions with E negative. At large separations, the wave equation (4.9) simplifies to −
!2 ∂r2 u(r) * Eu(r) , 2m
√ having approximate solutions eκr and e−κr , where !κ = −2mE. (Here, strictly speaking, m should denote the reduced mass of electron-proton system.) The bound states we are looking for, of course, have exponentially decreasing wavefunctions at large distances. Advanced Quantum Physics
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To further simplify the wave equation, it is convenient to introduce the dimensionless variable ρ = κr, leading to the equation . 2ν %(% + 1) 2 ∂ρ u(ρ) = 1 − + u(ρ) , ρ ρ2 where (for reasons which will become apparent shortly) we have introduced Ze2 κ . Notice that in transforming from r the dimensionless parameter ν = 4π% 0 2E to the dimensionless variable ρ, the scaling factor depends on energy, so will be different for different energy bound states! Consider now the behaviour of the wavefunction near the origin. The dominant term for sufficiently small ρ is the centrifugal component, so ∂ρ2 u(ρ) *
%(% + 1) u(ρ) , ρ2
for which the solutions are u(ρ) ∼ ρ−! and u(ρ) ∼ ρ!+1 . Since the wavefunction cannot be singular, we must choose the second solution. We have established that the wavefunction decays as e−κr = e−ρ at large distances, and goes as ρ!+1 close to the origin. Factoring out these two asymptotic behaviours, let us then define w(ρ) such that u(ρ) = e−ρ ρ!+1 w(ρ). We leave it as a tedious but straightforward exercise to show that ρ∂ρ2 w(ρ) + 2(% + 1 − ρ)∂ρ w(ρ) + 2(ν − (% + 1))w(ρ) = 0 . Substituting the trial series solution, w(ρ) = rence relation between successive coefficients:
0∞
k k=0 wk ρ ,
we obtain a recur-
2(k + % + 1 − ν) wk+1 = . wk (k + 1)(k + 2(% + 1)) For large values of k, wk+1 /wk → 2/k, so wk ∼ 2k /k! and therefore w(ρ) ∼ e2ρ . This means we have found the diverging radial wavefunction, u(ρ) ∼ eρ , which is in fact the correct behaviour for general values of the energy. To find the bound states, we must choose energies such that the series is not an infinite one. As long as the series stops somewhere, the exponential decrease will eventually take over, and yield a finite (bound state) wavefunction. Just as for the simple harmonic oscillator, this can only happen if for some k, wk+1 = 0. Inspecting the ratio wk+1 /wk , evidently the condition for a bound state is that ν = n,
integer ,
in which case the series for w(ρ) terminates at k = n − % − 1. From now on, since we know that for the functions we’re interested in ν is an integer, we replace ν by n. Finally, making use of the definitions of ν and κ above, we obtain the bound state energies, En = −
-
Ze2 4π#0
.2
m 1 Z2 ≡ − Ry . 2!2 n2 n2
Remarkably, this is the very same series of bound state energies found by Bohr from his model! Of course, this had better be the case, since the series of energies Bohr found correctly accounted for the spectral lines emitted by hot hydrogen atoms. Notice, though, that there are some important differences Advanced Quantum Physics
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with the Bohr model: the energy here is determined entirely by n, called the principal quantum number, but, in contrast to Bohr’s model, n is not the angular momentum. The true ground state of the hydrogen atom, n = 1, has zero angular momentum: since n = k + % + 1, n = 1 means both l = 0 and k = 0. The ground state wavefunction is therefore spherically symmetric, and the function w(ρ) = w0 is just a constant. Hence u(ρ) = ρe−ρ w0 and the actual radial wavefunction is this divided by r, and of course suitably normalized. To write the wavefunction in terms of r, we need to find κ. Putting together $ ρ = κn r, κn = −2mEn /!2 and the expression for En , we find that κn = Ze2 m Z 4π%0 !2 n = na0 , where a0 =
4π#0 !2 = 0.529 × 10−10 m me2
denotes the Bohr radius – the radius of the lowest orbit in Bohr’s model.2 1 (Ze) 1 With this definition, the energy levels can then be expressed as En = − 4π% 2. 0 2a0 n Moving on to the excited states: for n = 2, we have a choice: either the radial function w(ρ) can have one term, as before, but now the angular momentum % = 1 (since n = k + % + 1); or w(ρ) can have two terms (so k = 1), and % = 0. Both options give the same energy, 0.25 Ry, since n is the same, and the energy only depends on n. In fact, there are four states at this energy, since % = 1 has states with m = 1, m = 0 and m = 1, and % = 0 has the one state m = 0. For n = 3, there are 9 states altogether: % = 0 gives one, % = 1 gives 3 and % = 2 gives 5 different m values. In fact, for principal quantum number n there are n2 degenerate states (n2 being the sum of the first n odd integers). From now on, we label the wavefunctions with the quantum numbers, ψn!m (r, θ, φ), so the ground state is the spherically symmetric ψ100 (r). For this state R(r) = u(r)/r, where u(ρ) = e−ρ ρ!+1 w(ρ) = e−ρ ρw0 , with w0 a constant and ρ = κ1 r = Zr/a0 . So, as a function of r, R10 (r) = N e−Zr/a0 with N the normalization constant: - .3 Z 2 −Zr/a0 e . R10 = 2 a0 For n = 2, % = 1 the function w(ρ) is still a single term, a constant, but now u(ρ) = e−ρ ρ!+1 w(ρ) = e−ρ ρ2 w0 , and, for n = 2, ρ = κ2 r = Zr/2a0 , remembering the energy-dependence of κ. After normalization, we find - .3/2 - . 1 Z Zr R21 = √ e−Zr/2a0 . a0 2 6 a0 The other n = 2 state has % = 0. So from n = k + % + 1, we have k = 1 and the series for w has two terms, k = 0 and k = 1, the ratio being wk+1 2(k + % + 1 − n) = = −1 , wk (k + 1)(k + 2(% + 1)) for the relevant values: k = 0, % = 0, n = 2. So w1 = −w0 , w(ρ) = w0 (1 − ρ). For n = 2, ρ = r/2a0 , the normalized wavefunction is given by 1 R20 = √ 2
-
Z a0
.3/2 -
1−
1 Zr 2 a0
.
e−Zr/2a0 .
Note that the zero angular momentum wavefunctions are non-zero and have non-zero slope at the origin. This means that the full three-dimensional wavefunctions have a slope discontinuity there! But this is fine - the potential is Advanced Quantum Physics
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infinite at the origin. (Actually, the proton is not a point charge, so really the kink will be smoothed out over a volume of the size of the proton - a very tiny effect.) In practice, the first few radial functions w(ρ) can be constructed fairly easily using the method presented above, but it should be noted that the differential equation for w(ρ), ρ∂ρ2 w(ρ) + 2(% + 1 − ρ)∂ρ w(ρ) + 2(n − (% + 1))w(ρ) = 0 , is in fact Laplace’s equation, usually written (z∂z2 + (k − 1 − z)∂z + p)Lkp (z) = 0 , where k, p are integers, and Lkp (z) is a Laguerre polynomial. The two equations are the same if z = 2ρ, and the solution to the radial equation is therefore, wn! (ρ) = L2!+1 n−!−1 (2ρ) . The Laguerre polynomials L0p (z), and associated Laguerre polynomials Lkp (z) are given by: L0p (z) = ez
dp −z p (e z ), dz p
Lkp (z) = (−1)k
dp 0 L (z) . dz p p+k
(These representations can be found neatly by solving Laplace’s equation using - surprise - a Laplace transform.) The polynomials satisfy the orthonormality relations (with the mathematicians’ normalization convention) + ∞ [(p + k)!]3 e−z z k Lkp Lkq dz = δpq . p! 0 But what do the polynomials look like? The function e−z z p is zero at the origin (apart from the trivial case p = 0) and zero at infinity, always positive and having non-zero slope except at its maximum value, z = p. The p derivatives bring in p separated zeroes, easily checked by sketching the curves generated by successive differentiation. Therefore, L0p (z), a polynomial of degree p, has p real positive zeroes, and value at the origin L0p (0) = p!, since the only non-zero term at z = 0 is that generated by all p differential operators acting on z p . The associated Laguerre polynomial Lkp (z) is generated by differentiating 0 Lp+k (z) k times. Now L0p+k (z) has p + k real positive zeroes, differentiating it gives a polynomial one degree lower, with zeroes which must be one in each interval between the zeroes of L0p+k (z). This argument remains valid for successive derivatives, so Lkp (z) must have p real separate zeroes. Putting all this together, and translating back from ρ to r, the radial solutions are given by, . Zr ! 2!+1 Rn! (r) = N e−Zr/na0 Ln−!−1 (2Zr/na0 ) , na0 with N the normalization constant. For a given principle quantum number n,the largest % radial wavefunction is given by Rn,n−1 ∝ rn−1 e−Zr/na0 . $ Info. The eigenvalues of the Hamiltonian for the hydrogen exhibit an unexpectedly high degeneracy. The fact that En!m is independent of m is common to all central potentials – it is just a reflection of rotational invariance of the Hamiltonian. Advanced Quantum Physics
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However, the degeneracy of different % values with the same principle quantum number is considered “accidental”, a peculiarity of the 1/r potential. To understand the origin of the degeneracy for atomic hydrogen, it is helpful to reflect first on the classical dynamics. In classical mechanics, central forces also lead to conservation of angular momentum leaving orbits restricted to lie in a plane. However, for 1/r type potentials, these orbits are also closed, i.e. they do not precess. In classical mechanics, this implies that there is an additional conserved vector, since the direction of the major axis of the elliptical orbit is a constant of the motion. This direction is determined by the Runge-Lenz vector, R=
e2 r 1 p×L− . m 4π#0 r
In quantum theory, up to an operator ordering prescription, R becomes an operator, e2 r ˆ = 1 (ˆ ˆ −L ˆ ×p ˆ) − R p×L . 2m 4π#0 r ˆ = 0 (exercise). ˆ R] With this definition, one may confirm that [H, ˆ exhibits the following commutation relations, [R ˆi, L ˆj ] = As a vector operator, R ˆ (−2H) ˆ ˆ ˆ ˆ i!#ijk Rk . Similarly, [Ri , Nj ] = i! m #ijk Lk (exercise). Moreover, one may confirm that - 2 .2 ˆ e 2H ˆ2 = ˆ 2 + !2 ) , R + (L 4π#0 m
ˆ 2 and R ˆ 2. ˆ can be written in terms of the two constants of motion, L showing that H ) −m ˆ ˆ = Focussing on the bound states, if we consider the Hermitian operator, K ˆ R, 2H ˆ i, K ˆ j ] = i!#ijk L ˆ k , [K ˆ i, L ˆj ] = which fulfil the following commutation relations, [K ˆ ˆ ˆ ˆ i!#ijk Kk , and [Li , Lj ] = i!#ijk Lk , we find that ˆ =− H
-
e2 4π#0
.2
m . 2 ˆ ˆ 2 + !2 ) 2(K + L
ˆ = If we now define the “raising and lowering” operators, M following commutation relations emerge (exercise),
ˆ K ˆ L+ 2 ,
ˆ = N
ˆ K ˆ L− 2
the
ˆ i, M ˆ j ] = i!#ijk L ˆk [M ˆ ˆ ˆ [Ni , Nj ] = i!#ijk Kk ˆi , M ˆj] = 0 , [N i.e. we have obtained two commuting angular momentum algebras(!) and - 2 .2 e m ˆ H=− . 2 ˆ ˆ 2 + !2 ) 4π#0 2(2M + 2N ˆ 2, M ˆ 2 and N ˆz, N ˆz , We can simultaneously diagonalize the operators, M ˆ 2 |m, n, µ, ν$ = !2 m(m + 1)|m, n, µ, ν$, M ˆ 2 |m, n, µ, ν$ = !2 m(m + 1)|m, n, µ, ν$, N
ˆ z |m, n, µ, ν$ = !µ|m, n, µ, ν$ M ˆz |m, n, µ, ν$ = !ν|m, n, µ, ν$ . N
where m, n = 0, 1/2, 1, 3/2, · · ·, µ = −m, −m + 1, · · · m and ν = −n, −n + 1, · · · n. ˆ ·L ˆ =L ˆ ·R ˆ = 0, then K ˆ ·L ˆ =L ˆ ·R ˆ = 0 and the only relevant states are those Since R ˆ 2−N ˆ 2 = 0, i.e. m = n. Therefore, for which M - 2 .2 e m ˆ H|m, m, µ, ν$ = − |m, m, µ, ν$ 4π#0 2!2 (4m(m + 1) + 1) - 2 .2 e m =− |m, m, µ, ν$ . 4π#0 2!2 (2m + 1)2
From this result, we can identify 2m + 1 = 1, 2, · · · as the principle quantum number. For a given (2m + 1) value, the degeneracy of the state is (2m + 1)2 as expected.
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Chapter 5
Motion of a charged particle in a magnetic field Hitherto, we have focussed on applications of quantum mechanics to free particles or particles confined by scalar potentials. In the following, we will address the influence of a magnetic field on a charged particle. Classically, the force on a charged particle in an electric and magnetic field is specified by the Lorentz force law: F = q (E + v × B) , where q denotes the charge and v the velocity. (Here we will adopt a convention in which q denotes the charge (which may be positive or negative) and e ≡ |e| denotes the modulus of the electron charge, i.e. for an electron, the charge q = −e = −1.602176487 × 10−19 C.) The velocity-dependent force associated with the magnetic field is quite different from the conservative forces associated with scalar potentials, and the programme for transferring from classical to quantum mechanics - replacing momenta with the appropriate operators - has to be carried out with more care. As preparation, it is helpful to revise how the Lorentz force arises in the Lagrangian formulation of classical mechanics.
5.1
Classical mechanics of a particle in a field
For a system with m degrees of freedom specified by coordinates q1 , · · · qm , the classical action is determined from the Lagrangian L(qi , q˙i ) by S[qi ] =
!
dt L(qi , q˙i ) .
The action is said to be a functional of the coordinates qi (t). According to Hamilton’s extremal principle (also known as the principle of least action), the dynamics of a classical system is described by the equations that minimize the action. These equations of motion can be expressed through the classical Lagrangian in the form of the Euler-Lagrange equations, d (∂q˙ L(qi , q˙i )) − ∂qi L(qi , q˙i ) = 0 . dt i
(5.1)
" Info. Euler-Lagrange equations: According to Hamilton’s extremal principle, for any smooth set of curves wi (t), the variation of the action around the classical solution qi (t) is zero, i.e. lim!→0 1! (S[qi + #wi ] − S[qi ]) = 0. Applied to the action, Advanced Quantum Physics
Hendrik Antoon Lorentz 18531928 A Dutch physicist who shared the 1902 Nobel Prize in Physics with Pieter Zeeman for the discovery and theoretical explanation of the Zeeman effect. He also derived the transformation equations subsequently used by Albert Einstein to describe space and time.
Joseph-Louis Lagrange, born Giuseppe Lodovico Lagrangia 1736-1813 An Italian-born mathematician and astronomer, who lived most of his life in Prussia and France, making significant contributions to all fields of analysis, to number theory, and to classical and celestial mechanics. On the recommendation of Euler and D’Alembert, in 1766 Lagrange succeeded Euler as the director of mathematics at the Prussian Academy of Sciences in Berlin, where he stayed for over twenty years, producing a large body of work and winning several prizes of the French Academy of Sciences. Lagrange’s treatise on analytical mechanics, written in Berlin and first published in 1788, offered the most comprehensive treatment of classical mechanics since Newton and formed a basis for the development of mathematical physics in the nineteenth century.
5.1. CLASSICAL MECHANICS OF A PARTICLE IN A FIELD
45
" the variation implies that, for any i, dt (wi ∂qi L(qi , q˙i ) + w˙ i ∂q˙i L(qi , q˙i )) = 0. Then, integrating the second term by parts, and droping the boundary term, one obtains # $ ! d dt wi ∂qi L(qi , q˙i ) − ∂q˙i L(qi , q˙i ) = 0 . dt Since this equality must follow for any function wi (t), the term in parentheses in the integrand must vanish leading to the Euler-Lagrange equation (5.1).
The canonical momentum is specified by the equation pi = ∂q˙i L, and the classical Hamiltonian is defined by the Legendre transform, H(qi , pi ) =
% i
pi qi − L(qi , q˙i ) .
(5.2)
It is straightforward to check that the equations of motion can be written in the form of Hamilton’s equations of motion, q˙i = ∂pi H,
p˙i = −∂qi H .
From these equations it follows that, if the Hamiltonian is independent of a particular coordinate qi , the corresponding momentum pi remains constant. For conservative forces,1 the classical Lagrangian and Hamiltonian can be written as L = T − V , H = T + V , with T the kinetic energy and V the potential energy. " Info. Poisson brackets: Any dynamical variable f in the system is some function of the phase space coordinates, the qi s and pi s, and (assuming it does not depend explicitly on time) its time-development is given by: d f (qi , pi ) = ∂qi f q˙i + ∂pi f p˙i = ∂qi f ∂pi H − ∂pi f ∂qi H ≡ {f, H}. dt The curly brackets are known as Poisson brackets, and are defined for any dynamical variables as {A, B} = ∂qi A ∂pi B − ∂pi A ∂qi B. From Hamilton’s equations, we have shown that for any variable, f˙ = {f, H}. It is easy to check that, for the coordinates and canonical momenta, {qi , qj } = 0 = {pi , pj }, {qi , pj } = δij . This was the classical mathematical structure that led Dirac to link up classical and quantum mechanics: He realized that the Poisson brackets were the classical version of the commutators, so a classical canonical momentum must correspond to the quantum differential operator in the corresponding coordinate.2
With these foundations revised, we now return to the problem at hand; the infleunce of an electromagnetic field on the dynamics of the charged particle. As the Lorentz force is velocity dependent, it can not be expressed simply as the gradient of some potential. Nevertheless, the classical path traversed by a charged particle is still specifed by the principle of least action. The electric and magnetic fields can be written in terms of a scalar and a vector potential ˙ The corresponding Lagrangian takes the form:3 as B = ∇×A, E = −∇φ− A. 1 L = mv2 − qφ + qv · A. 2
1
i.e. forces that conserve mechanical energy. For a detailed discussion, we refer to Paul A. M. Dirac, Lectures on Quantum Mechanics, Belfer Graduate School of Science Monographs Series Number 2, 1964. 3 In a relativistic formulation, the interaction term R here looks less arbitrary: the relativistic version would have the relativistically invariant q Aµ dxµ added to the action integral, where the four-potential Aµ = (φ, A) and dxµ = (ct, dx1 , dx2 , dx3 ). This is the simplest possible invariant interaction between the R electromagnetic field and R the particle’s four-velocity. Then, in the non-relativistic limit, q Aµ dxµ just becomes q (v · A − φ)dt. 2
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Sim´ eon Denis Poisson 17811840 A French mathematician, geometer, and physicist whose mathematical skills enabled him to compute the distribution of electrical charges on the surface of conductors. He extended the work of his mentors, Pierre Simon Laplace and Joseph Louis Lagrange, in celestial mechanics by taking their results to a higher order of accuracy. He was also known for his work in probability.
5.2. QUANTUM MECHANICS OF A PARTICLE IN A FIELD In this case, the general coordinates qi ≡ xi = (x1 , x2 , x3 ) are just the Cartesian coordinates specifying the position of the particle, and the q˙i are the three components x˙ i = (x˙ 1 , x˙ 2 , x˙ 3 ) of the particle velocities. The important point is that the canonical momentum pi = ∂x˙ i L = mvi + qAi , is no longer simply given by the mass × velocity – there is an extra term! Making use of the definition (5.2), the corresponding Hamiltonian is given by % 1 1 H(qi , pi ) = (mvi + qAi ) vi − mv2 + qφ − qv · A = mv2 + qφ . 2 2 i
Reassuringly, the Hamiltonian just has the familiar form of the sum of the kinetic and potential energy. However, to get Hamilton’s equations of motion, the Hamiltonian has to be expressed solely in terms of the coordinates and canonical momenta, i.e. H=
1 (p − qA(r, t))2 + qφ(r, t) . 2m
Let us now consider Hamilton’s equations of motion, x˙ i = ∂pi H and p˙i = −∂xi H. The first equation recovers the expression for the canonical momentum while second equation yields the Lorentz force law. To understand how, we must first keep in mind that dp/dt is not the acceleration: The A-dependent term also varies in time, and in a quite complicated way, since it is the field at a point moving with the particle. More precisely, & ' p˙i = m¨ xi + q A˙ i = m¨ xi + q ∂t Ai + vj ∂xj Ai ,
where we have assumed a summation over repeated indicies. The right-hand ∂H side of the second of Hamilton’s equation, p˙i = − ∂x , is given by i 1 (p − qA(r, t))q∂xi A − q∂xi φ(r, t) = qvj ∂xi Aj − q∂xi φ . m ' & Together, we obtain the equation of motion, m¨ xi = −q ∂t Ai + vj ∂xj Ai + qvj ∂xi Aj − q∂xi φ. Using the identity, v × (∇ × A) = ∇(v · A) − (v · ∇)A, and the expressions for the electric and magnetic fields in terms of the potentials, one recovers the Lorentz equations −∂xi H =
m¨ x = F = q (E + v × B) . With these preliminary discussions of the classical system in place, we are now in a position to turn to the quantum mechanics.
5.2
Quantum mechanics of a particle in a field
To transfer to the quantum mechanical regime, we must once again implement ˆ = −i!∇, so that [ˆ the canonical quantization procedure setting p xi , pˆj ] = i!δij . However, in this case, pˆi %= mˆ vi . This leads to the novel situation that the velocities in different directions do not commute.4 To explore influence of the magnetic field on the particle dynamics, it is helpful to assess the relative weight of the A-dependent contributions to the quantum Hamiltonian, ˆ = 1 (ˆ H p − qA(r, t))2 + qφ(r, t) . 2m 4
With mˆ vi = −i!∂xi − qAi , it is easy (and instructive) to verify that [ˆ vx , vˆy ] =
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i!q B. m2
46
5.3. ATOMIC HYDROGEN: NORMAL ZEEMAN EFFECT
47
Expanding the square on the right hand side of the Hamiltonian, the cross-term (known as the paramagnetic term) leads to the contribution q! (∇ · A + A · ∇) = iq! − 2im m A · ∇, where the last equality follows from the Coulomb gauge condition, ∇ · A = 0.5 Combined with the diamagnetic (A2 ) contribution, one obtains the Hamiltonian, 2 2 ˆ = − ! ∇2 + iq! A · ∇ + q A2 + qφ . H 2m m 2m For a constant magnetic field, the vector potential can be written as A = −r × B/2. In this case, the paramagnetic component takes the form
iq! iq! q A·∇= (r × ∇) · B = − L · B, m 2m 2m
where L denotes the angular momentum operator (with the hat not shown for brevity!). Similarly, the diamagnetic term leads to ' q2B 2 2 q2 2 q2 & 2 2 A = r B − (r · B)2 = (x + y 2 ) , 2m 8m 8m where, here, we have chosen the magnetic field to lie along the z-axis.
5.3
Atomic hydrogen: Normal Zeeman effect
Before addressing the role of these separate contributions in atomic hydrogen, let us first estimate their relative magnitude. With &x2 + y 2 ' ( a20 , where a0 denotes the Bohr radius, and &Lz ' ( !, the ratio of the paramagnetic and diamagnetic terms is given by (q 2 /8me )&x2 + y 2 'B 2 e a20 B 2 ( 10−6 B/T . = (q/2me )&Lz 'B 4 !B
Therefore, while electrons remain bound to atoms, for fields that can be achieved in the laboratory (B ( 1 T), the diamagnetic term is negligible as compared to the paramagnetic term. Moreover, when compared with the Coulomb energy scale,
2
eB!/2me e! B/T = B( , me c2 α2 /2 (me cα)2 2.3 × 105
e 1 1 where α = 4π# ( 137 denotes the fine structure constant, one may see 0 !c that the paramagnetic term provides only a small perturbation to the typical atomic splittings. 5
The electric field E and magnetic field B of Maxwell’s equations contain only “physical” degrees of freedom, in the sense that every mathematical degree of freedom in an electromagnetic field configuration has a separately measurable effect on the motions of test charges in the vicinity. As we have seen, these “field strength” variables can be expressed in terms of the scalar potential φ and the vector potential A through the relations: E = −∇φ − ∂t A and B = ∇ × A. Notice that if A is transformed to A + ∇Λ, B remains unchanged, since B = ∇ × [A + ∇Λ] = ∇ × A. However, this transformation changes E as E = −∇φ − ∂t A − ∇∂t Λ = −∇[φ + ∂t Λ] − ∂t A . If φ is further changed to φ − ∂t Λ, E remains unchanged. Hence, both the E and B fields are unchanged if we take any function Λ(r, t) and simultaneously transform A → A + ∇Λ φ → φ − ∂t Λ .
A particular choice of the scalar and vector potentials is a gauge, and a scalar function Λ used to change the gauge is called a gauge function. The existence of arbitrary numbers of gauge functions Λ(r, t), corresponds to the U(1) gauge freedom of the theory. Gauge fixing can be done in many ways.
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Splitting of the sodium D lines due to an external magnetic field. The multiplicity of the lines and their “selection rule” will be discussed more fully in chapter 9. The figure is taken from the original paper, P. Zeeman, The effect of magnetization on the nature of light emitted by a substance, Nature 55, 347 (1897).
5.4. GAUGE INVARIANCE AND THE AHARONOV-BOHM EFFECT
48
However, there are instances when the diamagnetic contriubution can play an important role. Leaving aside the situation that may prevail on neutron stars, where magnetic fields as high as 108 T may exist, the diamagnetic contribution can be large when the typical “orbital” scale &x2 + y 2 ' becomes macroscopic in extent. Such a situation arises when the electrons become unbound such as, for example, in a metal or a synchrotron. For a further discussion, see section 5.5 below. Retaining only the paramagnetic contribution, the Hamiltonian for a “spinless” electron moving in a Coulomb potential in the presence of a constant magnetic field then takes the form, ˆ =H ˆ 0 + e BLz , H 2m ˆ 0 , Lz ] = 0, the eigenstates of the unperturbed ˆ 0 = pˆ − e . Since [H where H 2m 4π#0 r ˆ and the corresponding energy Hamiltonian, ψl$m (r) remain eigenstates of H levels are specified by 2
2
En$m = −
Ry + !ωL m n2
eB where ωL = 2m denotes the Larmor frequency. From this result, we expect that a constant magnetic field will lead to a splitting of the (2)+1)-fold degeneracy of the energy levels leading to multiplets separated by a constant energy shift of !ωL . The fact that this behaviour is not recapitulated generically by experiment was one of the key insights that led to the identification of electron spin, a matter to which we will turn in chapter 6.
5.4
Gauge invariance and the Aharonov-Bohm effect
Our derivation above shows that the quantum mechanical Hamiltonian of a charged particle is defined in terms of the vector potential, A. Since the latter is defined only up to some gauge choice, this suggests that the wavefunction is not a gauge invariant object. Indeed, it is only the observables associated with the wavefunction which must be gauge invariant. To explore this gauge freedom, let us consider the influence of the gauge transformation, A )→ A" = A + ∇Λ,
φ )→ φ" − ∂t Λ ,
where Λ(r, t) denotes a scalar function. Under the gauge transformation, one may show that the corresponding wavefunction gets transformed as ( q ) ψ " (r, t) = exp i Λ(r, t) ψ(r, t) . (5.3) ! " Exercise. If wavefunction ψ(r, t) obeys the time-dependent Schr¨odinger equaˆ tion, i!∂t ψ = H[A, φ]ψ, show that ψ " (r, t) as defined by (5.3) obeys the equation ˆ " [A" , φ" ]ψ " . i!∂t ψ " = H The gauge transformation introduces an additional space and time-dependent phase factor into the wavefunction. However, since the observable translates to the probability density, |ψ|2 , this phase dependence seems invisible. " Info. One physical manifestation of the gauge invariance of the wavefunction is found in the Aharonov-Bohm effect. Consider a particle with charge q travelling Advanced Quantum Physics
Sir Joseph Larmor 1857-1942 A physicist and mathematician who made innovations in the understanding of electricity, dynamics, thermodynamics, and the electron theory of matter. His most influential work was Aether and Matter, a theoretical physics book published in 1900. In 1903 he was appointed Lucasian Professor of Mathematics at Cambridge, a post he retained until his retirement in 1932.
5.4. GAUGE INVARIANCE AND THE AHARONOV-BOHM EFFECT
49
Figure 5.1: (Left) Schematic showing the geometry of an experiment to observe the
Aharonov-Bohm effect. Electrons from a coherent source can follow two paths which encircle a region where the magnetic field is non-zero. (Right) Interference fringes for electron beams passing near a toroidal magnet from the experiment by Tonomura and collaborators in 1986. The electron beam passing through the center of the torus acquires an additional phase, resulting in fringes that are shifted with respect to those outside the torus, demonstrating the Aharonov-Bohm effect. For details see the original paper from which this image was borrowed see Tonomura et al., Evidence for Aharonov-Bohm effect with magnetic field completely shielded from electron wave, Phys. Rev. Lett. 56, 792 (1986). along a path, P , in which the magnetic field, B = 0 is identically zero. However, a vanishing of the magnetic field does not imply that the vector potential, A is zero. Indeed, as we have seen, any Λ(r) such that A = ∇Λ will translate to this condition. In traversing the path, the wavefunction of the particle will acquire the phase factor " ϕ = !q P A · dr, where the line integral runs along the path. If we consider now two separate paths P and P " which share the same initial and final points, the relative phase of the wavefunction will be set by ! ! * ! q q q q A · dr = ∆ϕ = A · dr − A · dr = B · d2 r , ! P ! P! ! ! A + " where the line integral runs over the loop involving paths P and P " , and A runs over the area enclosed by the loop. The last relation follows from the application of Stokes’ theorem. This result shows that the" relative phase ∆ϕ is fixed by the factor q/! multiplied by the magnetic flux Φ = A B · d2 r enclosed by the loop.6 In the absence of a magnetic field, the flux vanishes, and there is no additional phase. However, if we allow the paths to enclose a region of non-vanishing magnetic field (see figure 5.1(left)), even if the field is identically zero on the paths P and P " , the wavefunction will acquire a non-vanishing relative phase. This flux-dependent phase difference translates to an observable shift of interference fringes when on an observation plane. Since the original proposal,7 the Aharonov-Bohm effect has been studied in several experimental contexts. Of these, the most rigorous study was undertaken by Tonomura in 1986. Tomomura fabricated a doughnut-shaped (toroidal) ferromagnet six micrometers in diameter (see figure 5.1b), and covered it with a niobium superconductor to completely confine the magnetic field within the doughnut, in accordance with the Meissner effect.8 With the magnet maintained at 5 K, they measured the phase difference from the interference fringes between one electron beam passing though the hole in the doughnut and the other passing on the outside of the doughnut. The results are shown in figure 5.1(right,a). Interference fringes are displaced with just half a fringe of spacing inside and outside of the doughnut, indicating the existence of the Aharonov-Bohm effect. Although electrons pass through regions free of any electromagnetic field, an observable effect was produced due to the existence of vector potentials. 6 Note that the phase difference depends on the magnetic flux, a function of the magnetic field, and is therefore a gauge invariant quantity. 7 Y. Aharonov and D. Bohm, Significance of electromagnetic potentials in quantum theory, Phys. Rev. 115, 485 (1959). 8 Perfect diamagnetism, a hallmark of superconductivity, leads to the complete expulsion of magnetic fields – a phenomenon known as the Meissner effect.
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Sir George Gabriel Stokes, 1st Baronet 1819-1903 A mathematician and physicist, who at Cambridge made important contributions to fluid dynamics (including the NavierStokes equations), optics, and mathematical physics (including Stokes’ theorem). He was secretary, and then president, of the Royal Society.
5.5. FREE ELECTRONS IN A MAGNETIC FIELD: LANDAU LEVELS
50
The observation of the half-fringe spacing reflects the constraints imposed by the superconducting toroidal shield. When a superconductor completely surrounds a magnetic flux, the flux is quantized to an integral multiple of quantized flux h/2e, the factor of two reflecting that fact that the superconductor involves a condensate of electron pairs. When an odd number of vortices are enclosed inside the superconductor, the relative phase shift becomes π (mod. 2π) – half-spacing! For an even number of vortices, the phase shift is zero.9
5.5
Free electrons in a magnetic field: Landau levels
Finally, to complete our survey of the influence of a uniform magnetic field on the dynamics of charged particles, let us consider the problem of a free quantum particle. In this case, the classical electron orbits can be macroscopic and there is no reason to neglect the diamagnetic contribution to the Hamiltonian. Previously, we have worked with a gauge in which A = (−y, x, 0)B/2, giving a constant field B in the z-direction. However, to address the Schr¨odinger equation for a particle in a uniform perpendicular magnetic field, it is convenient to adopt the Landau gauge, A(r) = (−By, 0, 0). " Exercise. Construct the gauge transformation, Λ(r) which connects these two representations of the vector potential.
In this case, the stationary form of the Schr¨odinger equation is given by 1 , ˆ Hψ(r) = (ˆ px + qBy)2 + pˆ2y + pˆ2z ψ(r) = Eψ(r) . 2m
ˆ commutes with both pˆx and pˆz , both operators have a common set of Since H eigenstates reflecting the fact that px and pz are conserved by the dynamics. The wavefunction must therefore take the form, ψ(r) = ei(px x+ipz z)/!χ(y), with χ(y) defined by the equation, . 2 / # $ pˆy p2 1 + mω 2 (y − y0 )2 χ(y) = E − z χ(y) . 2m 2 2m Here y0 = −px /qB and ω = |q|B/m coincides with the cyclotron frequency of the classical charged particle (exercise). We now see that the conserved canonical momentum px in the x-direction is in fact the coordinate of the centre of a simple harmonic oscillator potential in the y-direction with frequency ω. As a result, we can immediately infer that the eigenvalues of the Hamiltonian are comprised of a free particle component associated with motion parallel to the field, and a set of harmonic oscillator states, En,pz = (n + 1/2)!ω +
p2z . 2m
The quantum numbers, n, specify states known as Landau levels. Let us confine our attention to states corresponding to the lowest oscillator (Landau level) state, (and, for simplicity, pz = 0), E0 = !ω/2. What is the degeneracy of this Landau level? Consider a rectangular geometry of area A = Lx × Ly and, for simplicity, take the boundary conditions to be periodic. The centre of the oscillator wavefunction, y0 = −px /qB, must lie 9
The superconducting flux quantum was actually predicted prior to Aharonov and Bohm, by Fritz London in 1948 using a phenomenological theory.
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Lev Davidovich Landau 19081968 A prominent Soviet physicist who made fundamental contributions to many areas of theoretical physics. His accomplishments include the co-discovery of the density matrix method in quantum mechanics, the quantum mechanical theory of diamagnetism, the theory of superfluidity, the theory of second order phase transitions, the GinzburgLandau theory of superconductivity, the explanation of Landau damping in plasma physics, the Landau pole in quantum electrodynamics, and the two-component theory of neutrinos. He received the 1962 Nobel Prize in Physics for his development of a mathematical theory of superfluidity that accounts for the properties of liquid helium II at a temperature below 2.17K.
5.5. FREE ELECTRONS IN A MAGNETIC FIELD: LANDAU LEVELS between 0 and Ly . With periodic boundary conditions eipx Lx /! = 1, so that px = n2π!/Lx . This means that y0 takes a series of evenly-spaced discrete values, separated by ∆y0 = h/qBLx . So, for electron degrees of freedom, q = −e, the total number of states N = Ly /|∆y0 |, i.e. νmax =
Lx Ly B =A , h/eB Φ0
(5.4)
where Φ0 = e/h denotes the “flux quantum”. So the total number of states in the lowest energy level coincides with the total number of flux quanta making up the field B penetrating the area A. The Landau level degeneracy, νmax , depends on field; the larger the field, the more electrons can be fit into each Landau level. In the physical system, each Landau level is spin split by the Zeeman coupling, with (5.4) applying to one spin only. Finally, although we treated x and y in an asymmetric manner, this was merely for convenience of calculation; no physical quantity should dierentiate between the two due to the symmetry of the original problem. " Exercise. Consider the solution of the Schr¨odinger equation when working in the symmetric gauge with A = −r × B/2. Hint: consider the velocity commutation relations, [vx , vy ] and how these might be deployed as conjugate variables.
" Info. It is instructive to infer y0 from purely classical considerations: Writing mv˙ = qv × B in component form, we have m¨ x = qB ˙ m¨ y = − qB ˙ and m¨ z = 0. c y, c x, Focussing on the motion in the xy-plane, these equations integrate straightforwardly qB to give, mx˙ = qB c (y − y0 ), my˙ = − c (x − x0 ). Here (x0 , y0 ) are the coordinates of the centre of the classical circular motion (known as the “guiding centre”) – the velocity vector v = (x, ˙ y) ˙ always lies perpendicular to (r − r0 ), and r0 is given by y0 = y − mvx /qB = −px /qB,
x0 = x + mvy /qB = x + py /qB .
(Recall that we are using the gauge A(x, y, z) = (−By, 0, 0), and px = ∂x˙ L = mvx + qAx , etc.) Just as y0 is a conserved quantity, so is x0 : it commutes with the Hamiltonian since [x + cˆ py /qB, pˆx + qBy] = 0. However, x0 and y0 do not commute with each other: [x0 , y0 ] = −i!/qB. This is why, when we chose a gauge in which y0 was sharply defined, x0 was spread over the sample. If we attempt to localize the point (x0 , y0 ) as much as possible, it is smeared out over an area corresponding to one flux quantum. The0natural length scale of the problem is therefore the magnetic ! length defined by ) = qB .
" Info. Integer quantum Hall effect: Until now, we have considered the impact of just a magnetic field. Consider now the Hall effect geometry in which we apply a crossed electric, E and magnetic field, B. Taking into account both contributions, the total current flow is given by # $ j×B j = σ0 E − , ne where σ0 denotes the conductivity, and n is the electron density. With the electric field oriented along y, and the magnetic field along z, the latter equation may be rewritten as # $# $ # $ σ0 B 1 jx 0 ne = σ0 . σ0 B jy Ey − ne 1 Inverting these equations, one finds that jx =
−σ02 B/ne Ey , 1 + (σ0 B/ne)2 1 23 4 σxy
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jy =
σ0 Ey . 1 + (σ0 B/ne)2 1 23 4 σyy
51
Klaus von Klitzing, 1943German physicist who was awarded the Nobel Prize for Physics in 1985 for his discovery that under appropriate conditions the resistance offered by an electrical conductor is quantized. The work was first reported in the following reference, K. v. Klitzing, G. Dorda, and M. Pepper, New method for high-accuracy determination of the fine-structure constant based on quantized Hall resistance, Phys. Rev. Lett. 45, 494 (1980).
5.5. FREE ELECTRONS IN A MAGNETIC FIELD: LANDAU LEVELS
Figure 5.2: (Left) A voltage V drives a current I in the positive x direction. Normal
Ohmic resistance is V /I. A magnetic field in the positive z direction shifts positive charge carriers in the negative y direction. This generates a Hall potential and a Hall resistance (V H/I) in the y direction. (Right) The Hall resistance varies stepwise with changes in magnetic field B. Step height is given by the physical constant h/e2 (value approximately 25 kΩ) divided by an integer i. The figure shows steps for i = 2, 3, 4, 5, 6, 8 and 10. The effect has given rise to a new international standard for resistance. Since 1990 this has been represented by the unit 1 klitzing, defined as the Hall resistance at the fourth step (h/4e2 ). The lower peaked curve represents the Ohmic resistance, which disappears at each step.
These provide the classical expressions for the longitudinal and Hall conductivities, σyy and σxy in the crossed field. Note that, for these classical expressions, σxy is proportional to B. How does quantum mechanics revised this picture? For the classical model – Drude theory, the random elastic scattering of electrons impurities leads to a con2 τ stant drift velocity in the presence of a constant electric field, σ0 = ne me , where τ denotes the mean time between collisions. Now let us suppose the magnetic field is chosen so that number of electrons exactly fills all the Landau levels up to some N , i.e. nLx Ly = N νmax ⇒ n = N
eB . h
The scattering of electrons must lead to a transfer between quantum states. However, if all states of the same energy are filled,10 elastic (energy conserving) scattering becomes impossible. Moreover, since the next accessible Landau level energy is a distance !ω away, at low enough temperatures, inelastic scattering becomes frozen out. As a result, the scattering time vanishes at special values of the field, i.e. σyy → 0 and σxy →
ne e2 =N . B h
At critical values of the field, the Hall conductivity is quantized in units of e2 /h. Inverting the conductivity tensor, one obtains the resistivity tensor, # $ # $−1 ρxx ρxy σxx σxy = , −ρxy ρxx −σxy σxx where ρxx =
σxx , 2 + σ2 σxx xy
ρxx = −
σxy , 2 + σ2 σxx xy
So, when σxx = 0 and σxy = νe2 /h, ρxx = 0 and ρxy = h/νe2 . The quantum Hall state describes dissipationless current flow in which the Hall conductance σxy is quantized in units of e2 /h. Experimental measurements of these values provides the best determination of fundamental ratio e2 /h, better than 1 part in 107 . 10
Note that electons are subject to Pauli’s exclusion principle restricting the occupancy of each state to unity.
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Chapter 6
Spin Until we have focussed on the quantum mechanics of particles which are “featureless”, carrying no internal degrees of freedom. However, a relativistic formulation of quantum mechanics shows that particles can exhibit an intrinsic angular momentum component known as spin. However, the discovery of the spin degree of freedom marginally predates the development of relativistic quantum mechanics by Dirac and was acheived in a ground-breaking experiement by Stern and Gerlach (1922). In their experiment, they passed a well-collimated beam of silver atoms through a region of inhomogeneous field before allowing the particles to impact on a photographic plate (see figure). The magnetic field was directed perpendicular to the beam, and has a strong gradient, ∂z Bz != 0 so that a beam comprised of atoms with a magnetic moment would be bent towards the z or -z axis. As the magnetic moment will be proportional to the total angular momentum, such an experiment can be thought of as a measurement of its projection along z. At the time of the experiment, there was an expectation that the magnetic moment of the atom was generated in its entirety by the orbital angular momentum. As such, one would expect that there would be a minimum of three possible values of the z-component of angular momentum: the lowest non-zero orbital angular momentum is " = 1, with allowed values of the z-component m!, m = 1, 0, −1. Curiously, Stern and Gerlach’s experiment (right) showed that the beam of silver atoms split into two! This discovery, which caused great discussion and surprise presented a puzzle. However, in our derivation of allowed angular momentum eigenvalues we found that, although for any system the allowed values of m form a ladder with spacing !, we could not rule out half-integral values of m. The lowest such case, " = 1/2, would in fact have just two allowed m values: m = ±1/2. However, such an " value could not translate to an orbital angular momentum because the z-component of the orbital wavefunction, ψ has a factor e±iφ , and therefore acquires a factor −1 on rotating through 2π! This would imply that ψ is not single-valued, which doesn’t make sense for a Schr¨odinger-type wavefunction. Yet the experimental result was irrefutable. Therefore, this must be a new kind of non-orbital angular momentum – spin. Conceptually, just as the Earth has orbital angular momentum in its yearly circle around the sun, and also spin angular momentum from its daily turning, the electron has an analogous spin. But this analogy has obvious limitations: the Earth’s spin is after all made up of material orbiting around the axis through the poles. The electron spin cannot be imagined as arising from a rotating body, since orbital angular momenta always come in integral multiples of !. Fortunately, this lack of a simple quasi-mechanical picture underlying electron spin doesn’t Advanced Quantum Physics
Gerlach’s postcard, dated 8th February 1922, to Niels Bohr. It shows a photograph of the beam splitting, with the message, in translation: “Attached [is] the experimental proof of directional quantization. We congratulate [you] on the confirmation of your theory.” (Physics Today December 2003)
6.1. SPINORS, SPIN PPERATORS, PAULI MATRICES
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prevent us from using the general angular momentum machinery developed ealier, which followed just from analyzing the effect of spatial rotation on a quantum mechanical system.
6.1
Spinors, spin pperators, Pauli matrices
The Hilbert space of angular momentum states for spin 1/2 is two-dimensional. Various notations are used: |", m# becomes |s, m# or, more graphically, |1/2, 1/2# = | ↑#,
|1/2, −1/2# = | ↓# .
A general state of spin can be written as the lienar combination, ! " α α| ↑# + β| ↓# = , β with the normalisation condition, |α|2 + |β|2 = 1, and this two-dimensional ket is called a spinor. Operators acting on spinors are necessarily of the form of 2 × 2 matrices. We shall adopt the usual practice of denoting the angular momentum components Li by Si for spins. (Once again, for clarity, we also drop the hats on the angular momentum operators!) From our definition of the spinor, it is evident that the z-component of the spin can be represented as the matrix, ! " ! 1 0 Sz = σ z , σz = . 0 −1 2 From the general formulae (4.5) for raising and lowering operators S± = Sx ± iSy , with s = 1/2, we have S+ |1/2, −1/2# = !|1/2, 1/2#, S− |1/2, 1/2# = !|1/2, −1/2#, or, in matrix form, ! " ! " 0 1 0 0 Sx + iSy = S+ = ! , Sx − iSy = S− = ! . 0 0 1 0 It therefore follows that an appropriate matrix representation for spin 1/2 is ggiven by the Pauli spin matrices, S = !2 σ where σx =
!
0 1 1 0
"
,
σy =
!
0 −i i 0
"
,
σz =
!
1 0 0 −1
"
.
(6.1)
These matrices are Hermitian, traceless, and obey the relations σi2 = I, σi σj = −σj σi , and σi σj = iσk for (i, j, k) a cyclic permutation of (1, 2, 3). These relations can be summarised by the identity, σi σj = Iδij + i)ijk σk . The total spin S2 =
!2 2 4 σ
= 34 !2 , i.e. s(s + 1)!2 for s = 1/2.
* Exercise. Explain why any 2 × 2 matrix can be written in the form α0 I + ˆ , and (b) α n · σ)2 = I for any unit vector n i i σi . Use your results to show that (a) (ˆ (σ · A)(σ · B) = (A · B)I + σ · (A × B).
#
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Wolfgang Pauli and Niels Bohr demonstrating ‘tippe top’ toy at the inauguration of the new Institute of Physics at Lund, Sweden 1954.
6.2. RELATING THE SPINOR TO THE SPIN DIRECTION
6.2
Relating the spinor to the spin direction
For a general state α| ↑# + β| ↓#, how do α, β relate to which way the spin is pointing? To find out, let us assume that it is pointing up along the unit ˆ = (sin θ cos φ, sin θ sin φ, cos θ), i.e. in the direction (θ, φ). In other vector n ˆ · σ having eigenvalue unity: words, the spin is an eigenstate of the operator n ! "! " ! " nz nx − iny α α = . nx + iny −nz β β From this expression, we find that α/β = (nx − iny )/(1 − nz ) = e−iφ cot(θ/2) (exercise). Then, making use of the normalisation, |α|2 + |β|2 = 1, we obtain (up to an arbitrary phase) !
α β
"
=
!
e−iφ/2 cos(θ/2) eiφ/2 sin(θ/2)
"
.
Since e−iφ cot(θ/2) can be used to specify any complex number with 0 ≤ θ ≤ π, 0 ≤ φ < 2π, so for any possible spinor, there is an associated direction along which the spin points up. * Info. The spin rotation operator: In general, the rotation operator for ˆ is given rotation through an angle θ about an axis in the direction of the unit vector n by eiθnˆ ·J/! where J denotes the angular momentum operator. For spin, J = S = 21 !σ, and the rotation operator takes the form1 eiθnˆ ·J/! = ei(θ/2)(ˆn·σ ) . Expanding the exponential, and making use of the Pauli matrix identities ((n · σ)2 = I), one can show that (exercise) ei(θ/2)(n·σ ) = I cos(θ/2) + in · σ sin(θ/2) . The rotation operator is a 2 × 2 matrix operating on the ket space. The 2 × 2 rotation matrices are unitary and form a group known as SU(2); the 2 refers to the dimensionality, the U to their being unitary, and the S signifying determinant +1. ˆ = (0, 0, 1), it is more natural to replace θ Note that for rotation about the z-axis, n with φ, and the rotation operator takes the form, ! −iφ/2 " e 0 ei(θ/2)(n·σ ) = . 0 eiφ/2 In particular, the wavefunction is multiplied by −1 for a rotation of 2π. Since this is true for any initial wave function, it is clearly also true for rotation through 2π about any axis.
* Exercise. Construct the infinitesimal version of the rotation operator eiδθnˆ ·J/! for spin 1/2, and prove that eiδθnˆ ·J/! σe−iδθnˆ ·J/! = σ + δθˆ n × σ, i.e. σ is rotated in the same way as an ordinary three-vector - note particularly that the change depends on the angle rotated through (as opposed to the half-angle) so, reassuringly, there is no −1 for a complete rotation (as there cannot be - the direction of the spin is a physical observable, and cannot be changed on rotating the measuring frame through 2π). 1 Warning: do not confuse θ – the rotation angle - with the spherical polar angle used to ˆ. parameterise n
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6.3. SPIN PRECESSION IN A MAGNETIC FIELD
6.3
Spin precession in a magnetic field
Consider a magnetized classical object spinning about it’s centre of mass, with angular momentum L and parallel magnetic moment µ, µ = γL. The constant γ is called the gyromagnetic ratio. Now suppose that we impose a magnetic field B along, say, the z-direction. This will exert a torque T = µ × B = γL × B = dL dt . This equation is easily solved and shows that the angular momentum vector L precesses about the magnetic field direction with angular velocity of precession ω 0 = −γB.2 In the following, we will show that precisely the same result appears in the study of the quantum mechanics of an electron spin in a magnetic field. −e The electron has magnetic dipole moment µ = γS, where γ = g 2m and the e 3 gyromagnetic ratio, g, is very close to 2. The Hamiltonian for the interaction ˆ = of the electron’s dipole moment with the magnetic field is given by H −µ · B = −γS · B. Hence the time development is specified by the equation ˆ (t)|ψ(0)#, with the time-evolution operator (or propagator), U ˆ (t) = |ψ(t)# = U ˆ −i Ht/! iγσ·Bt/2 e = e . However, this is nothing but the rotation operator (as shown earlier) through an angle −γBt about the direction of B! For an arbitrary initial spin orientation ! " ! −iφ/2 " α e cos(θ/2) = , β eiφ/2 sin(θ/2) the propagator for a magnetic field in the z-direction is given by ! −iω t/2 " e 0 0 U (t) = eiγσ·Bt/2 = , 0 eiω0 t/2 so the time-dependent spinor is set by ! " ! −i(φ+ω t)/2 " 0 α(t) e cos(θ/2) = . β(t) ei(φ+ω0 t)/2 sin(θ/2) The angle θ between the spin and the field stays constant while the azimuthal angle around the field increases as φ = φ0 +ω0 t, exactly as in the classical case. |e|B The frequency ω0 = gωc , where ωc = 2m denotes the cyclotron frequency. For e 11 a magnetic field of 1 T, ωc ( 10 rads/s. * Exercise. For a spin initially pointing along the x-axis, prove that )Sx (t)# =
(!/2) cos(ω0 t).
6.3.1
Paramagnetic Resonance
The analysis above shows that the spin precession frequency is independent of the angle of the spin with respect to the field direction. Consider then how this looks in a frame of reference which is itself rotating with angular ˆ, since velocity ω about the z-axis. Let us specify the magnetic field B0 = B0 z we’ll soon be adding another component. In the rotating frame, the observed precession frequency is ω r = −γ(B0 + ω/γ), so there is a different effective dL
Proof: From the equation of motion, with L+ = Lx + iLy , dt+ = −iγBL+ , L+ = z Of course, dL = 0, since dL = γL × B is perpendicular to B, which is in the dt dt z-direction. 3 This g-factor terminology is used more widely: the magnetic moment of an atom is e! written µ = gµB , where µB = 2m is the known as the Bohr magneton, and g depends e on the total orbital angular momentum and total spin of the particular atom. 2
L0+ e−iγBt .
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6.3. SPIN PRECESSION IN A MAGNETIC FIELD
57
field (B0 + ω/γ) in the rotating frame. Obviously, if the frame rotates exactly at the precession frequency, ω = ω 0 = −γB0 , spins pointing in any direction will remain at rest in that frame – there is no effective field at all. Suppose we now add a small rotating magnetic field with angular frequency ω in the xy plane, so the total magnetic field, ˆ + B1 (ˆ ˆy sin(ωt)) . ex cos(ωt) − e B = B0 z The effective magnetic field in the frame rotating with the same frequency ω as the small added field is then given by ˆx . Br = (B0 + ω/γ)ˆ z + B1 e Now, if we tune the angular frequency of the small rotating field so that it exactly matches the precession frequency in the original static magnetic field, ω = ω 0 = −γB0 , all the magnetic moment will see in the rotating frame is the small field in the x-direction! It will therefore precess about the x-direction at the slow angular speed γB1 . This matching of the small field rotation frequency with the large field spin precession frequency is the “resonance”. If the spins are lined up preferentially in the z-direction by the static field, and the small resonant oscillating field is switched on for a time such that γB1 t = π/2, the spins will be preferentially in the y-direction in the rotating frame, so in the lab they will be rotating in the xy plane, and a coil will pick up an a.c. signal from the induced e.m.f. * Info. Nuclear magnetic resonance is an important tool in chemical analysis. As the name implies, the method uses the spin magnetic moments of nuclei (particularly hydrogen) and resonant excitation. Magnetic resonance imaging uses the same basic principle to get an image (of the inside of a body for example). In basic NMR, a strong static B field is applied. A spin 1/2 proton in a hydrogen nucleus then has two energy eigenstates. After some time, most of the protons fall into the lower of the two states. We now use an electromagnetic wave (RF pulse) to excite some of the protons back into the higher energy state. The proton’s magnetic moment interacts with the oscillating B field of the EM wave through the Hamiltionian, ˆ = −µ · B = gp e S · B = gp e! σ · B = gp µN σ · B , H 2mp c 4mp c 2 where the gyromagnetic ratio of the proton is about +5.6. The magnetic moment is 2.79µN (nuclear magnetons). Different nuclei will have different gyromagnetic ratios giving more degrees of freedom with which to work. The strong static B field is chosen to lie in the z direction and the polarization of the oscillating EM wave is chosen so that the B field points in the x direction. The EM wave has (angular) frequency ω, ! " Bz Bx cos(ωt) ˆ = gp µN Bz σz + Bx cos(ωt)σx = gp µN H . Bx cos(ωt) −Bz 2 2 ˆ i.e. If we now apply the time-dependent Schr¨odinger equation, i!∂t χ = Hχ, ! " ! "! " a˙ ω0 ωI cos(ωt) a = −i , ωI cos(ωt) −ω0 b b˙ where ω0 = gp µN Bz /2! and ωI = gp µN Bx /2!, we obtain, % ωI $ i(ω−2ω0 )t ∂t (be−iω0 t ) = − e + e−i(ω+2ω0 )t . 2
The second term oscillates rapidly and can be neglected. The first term will only result in significant transitions if ω ≈ 2ω0 . Note that this is exactly the condition that ensures that the energy of the photons in the EM field E = !ω is equal to the energy difference between the two spin states ∆E = 2!ω0 . The conservation of energy
Advanced Quantum Physics
A proton NMR spectrum of a solution containing a simple organic compound, ethyl benzene. Each group of signals corresponds to protons in a different part of the molecule.
6.4. ADDITION OF ANGULAR MOMENTA condition must be satisfied well enough to get a significant transition rate. In NMR, we observe the transitions back to the lower energy state. These emit EM radiation at the same frequency and we can detect it after the stronger input pulse ends (or by more complex methods). NMR is a powerful tool in chemical analysis because the molecular field adds to the external B field so that the resonant frequency depends on the molecule as well as the nucleus. We can learn about molecular fields or just use NMR to see what molecules are present in a sample. In MRI, we typically concentrate on the one nucleus like hydrogen. We can put a gradient in Bz so that only a thin slice of the material has ω tuned to the resonant frequency. Therefore we can excite transitions to the higher energy state in only a slice of the sample. If we vary (in the orthogonal direction!) the B field during the decay, we can recover 3d images.
6.4
Addition of angular momenta
In subsequent chapters, it will be necessary to add angular momentum, be it ˆ = L+S, ˆ the addition of orbital and spin angular momenta, J as with the study of spin-orbit coupling in atoms, or the addition of general angular momenta, ˆ = J ˆ1 + J ˆ 2 as occurs in the consideration of multi-electron atoms. In the J following section, we will explore three problems: The addition of two spin 1/2 degrees of freedom; the addition of a general orbital angular momentum and spin; and the addition of spin J = 1 angular momenta. However, before addressing these examples in turn, let us first make some general remarks. Without specifying any particular application, let us consider the total ˆ = J ˆ1 + J ˆ 2 where J ˆ 1 and J ˆ 2 correspond to distinct angular momentum J ˆ ˆ degrees of freedom, [J1 , J2 ] = 0, and the individual operators obey angular momentum commutation relations. As a result, the total angular momentum also obeys angular momentum commutation relations, [Jˆi , Jˆj ] = i!)ijk Jˆk . For each angular momentum component, the states |j1 , m1 # and |j2 , m2 # where mi = −ji , · · · ji , provide a basis of states of the total angular momentum ˆ 2 and the projection Jˆiz . Together, they form a complete basis operator, J i which can be used to span the states of the coupled spins,4 |j1 , m1 , j2 , m2 # ≡ |j1 , m1 # ⊗ |j2 , m2 # . These product states are also eigenstates of Jˆz with eigenvalue !(m1 + m2 ), ˆ2. but not of J * Exercise. Show that [Jˆ 2 , Jˆiz ] != 0. However, for practical application, we require a basis in which the total angular ˆ 2 is also diagonal. That is, we must find eigenstates momentum operator J ˆ 2 , Jˆz , J ˆ 2 , and J ˆ2. |j, mj , j1 , j2 # of the four mutually commuting operators J 1 2 In general, the relation between the two basis can be expressed as & |j, mj , j1 , j2 # = |j1 , m1 , j2 , m2 #)j1 , m1 , j2 , m2 |j, mj , j1 , j2 # , m1 ,m2
4 Here ⊗ denotes the “direct product” and shows that the two constituent spin states access their own independent Hilbert space.
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High resolution MRI scan of a brain!
6.4. ADDITION OF ANGULAR MOMENTA where the matrix elements are known as Clebsch-Gordon coefficients. In general, the determination of these coefficients from first principles is a somewhat soul destroying exercise and one that we do not intend to pursue in great detail.5 In any case, for practical purposes, such coefficients have been tabulated in the literature and can be readily obtained. However, in some simple cases, these matrix elements can be determined straightforwardly. Moreover, the algorithmic programme by which they are deduced offer some new conceptual insights. Operationally, the mechanism for finding the basis states of the total angular momentum operator follow the strategy: 1. As a unique entry, the basis state with maximal Jmax and mj = Jmax is easy to deduce from the original basis states since it involves the product of states of highest weight, |Jmax , mj = Jmax , j1 , j2 # = |j1 , m1 = j1 # ⊗ |j2 , m2 = j2 # , where Jmax = j1 + j2 . 2. From this state, we can use of the total spin lowering operator Jˆ− to find all states with J = Jmax and mj = −Jmax · · · Jmax . 3. From the state with J = Jmax and mj = Jmax − 1, one can then obtain the state with J = Jmax − 1 and mj = Jmax − 1 by orthogonality.6 Now one can return to the second step of the programme and repeat until J = |j1 − j2 | when all (2j1 + 1)(2j2 + 1) basis states have been obtained.
6.4.1
Addition of two spin 1/2 degrees of freedom
For two spin 1/2 degrees of freedom, we could simply construct and diagonalize the complete 4 × 4 matrix elements of the total spin. However, to gain some intuition for the general case, let us consider the programme above. Firstly, the maximal total spin state is given by |S = 1, mS = 1, s1 = 1/2, s2 = 1/2# = |s1 = 1/2, ms1 = 1/2# ⊗ |s2 = 1/2, ms2 = 1/2# . Now, since s1 = 1/2 and s = 1/2 is implicit, we can rewrite this equation in a more colloquial form as |S = 1, mS = 1# = | ↑1 # ⊗ | ↑2 # . We now follow step 2 of the programme and subject the maximal spin state to the total spin lowering operator, Sˆ− = Sˆ1− + Sˆ1+ . In doing so, making use of Eq. (4.5), we find √ Sˆ− |S = 1, mS = 1# = 2!|S = 1, mS = 0# = ! (| ↓1 # ⊗ | ↑2 # + | ↑1 # ⊗ | ↓2 #) , 5
In fact, one may show that the general matrix element is given by s (j1 + j2 − j)!(j + j1 − j2 )!(j + j2 − j1 )!(2j + 1) $j1 , m1 , j2 , m2 |j, mj , j1 , j2 % = δmj ,m1 +m2 (j + j1 + j2 + 1)! p X (−1)k (j1 + m1 )!(j1 − m1 )!(j2 + m2 )!(j2 − m2 )!(j + m)!(j − m)! × . k!(j1 + j2 − j − k)!(j1 − m1 − k)!(j2 + m2 − k)!(j − j2 + m1 + k)!(j − j1 − m2 + k)! k
6 Alternatively, as a maximal spin state, |J = Jmax − 1, mj = Jmax − 1, j1 , j2 % can be identified by the “killing” action of the raising operator, Jˆ+ .
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6.4. ADDITION OF ANGULAR MOMENTA i.e. |S = 1, mS = 0# =
√1 (| 2
Sˆ− |S = 1, mS = 0# =
↓1 # ⊗ | ↑2 # + | ↑1 # ⊗ | ↓2 #). Similarly,
√
2!|S = 1, mS = −1# =
√
2!| ↓1 # ⊗ | ↓2 # ,
i.e. |S = 1, mS = −1# = | ↓1 # ⊗ | ↓2 #. This completes the construction of the manifold of spin S = 1 states – the spin triplet states. Following the programme, we must now consider the lower spin state. In this case, the next multiplet is the unique total spin singlet state |S = 0, mS = 0#. The latter must be orthogonal to the spin triplet state |S = 1, mS = 0#. As a result, we can deduce that 1 |S = 0, mS = 0# = √ (| ↓1 # ⊗ | ↑2 # − | ↑1 # ⊗ | ↓2 #) . 2
6.4.2
Addition of angular momentum and spin
We now turn to the problem of the addition of angular momentum and spin, ˆ =L ˆ + S. ˆ In the original basis, for a given angular momentum ", one can J identify 2 × (2" + 1) product states |", m% # ⊗ | ↑# and |", m% # ⊗ | ↓#, with ˆ 2, L ˆ 2 and Sˆz , but not J ˆ 2 . From ˆz, S m% = −", · · · ", involving eigenstates of L ˆ 2 , Jˆz , L ˆ 2 and S ˆ 2 . To these basis states, we are looking for eigenstates of J undertake this programme, it is helpful to recall the action of the angular momentum raising and lower operators, ˆ ± |", m% # = ((" ± m% + 1)(" ∓ m% ))1/2 !|", m% ± 1# , L as well as the identity ˆ ·S ˆ 2L ' () * 2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ J = L + S + 2Lz Sz + L+ S− + S+ L− .
ˆ 2 , Jˆz , L ˆ 2 and S ˆ 2 we will adopt the notation |j, mj , "# For the eigenstates of J leaving the spin S = 1/2 implicit. The maximal spin state is given by7 |" + 1/2, " + 1/2, "# = |", "# ⊗ | ↑# . To obtain the remaining states in the multiplet, |j = " + 1/2, mj=%+1/2 , "#, we may simply apply the total spin lowering operator Jˆ− , Jˆ− |", "# ⊗ | ↑# = !(2")1/2 |", " − 1# ⊗ | ↑# + !|", "# ⊗ | ↓# . Normalising the right-hand side of this expression, one obtains the spin state, + + 2" 1 |", " − 1# ⊗ | ↑# + |", "# ⊗ | ↓# . |" + 1/2, " − 1/2, "# = 2" + 1 2" + 1 7
The proof runs as follows: ˆ z + Sˆz )|$, $% ⊗ | ↑% = ($ + 1/2)!|$, $% ⊗ | ↑% , Jˆz |$, $% ⊗ | ↑% = (L
and ˆ 2 |$, $% ⊗ | ↑% = !2 ($($ + 1) + 1/2(1/2 + 1) + 2$ 1 )|$, $% ⊗ | ↑% J 2 = !2 ($ + 1/2)($ + 3/2)|$, $% ⊗ | ↑% .
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6.4. ADDITION OF ANGULAR MOMENTA By repeating this programme, one can develop an expression for the full set of basis states, + " + mj + 1/2 |j = " + 1/2, mj , "# = |", mj − 1/2# ⊗ | ↑# 2" + 1 + " − mj + 1/2 + |", mj + 1/2# ⊗ | ↓# , 2" + 1 with mj = " + 1/2, · · · , −(" + 1/2). In order to obtain the remaining states with j = " − 1/2, we may look for states with mj = " − 1/2, · · · , −(" − 1/2) which are orthogonal to |" + 1/2, mj , "#. Doing so, we obtain + " − mj + 1/2 |", mj − 1/2# ⊗ | ↑# |" − 1/2, mj , "# = − 2" + 1 + " + mj + 1/2 + |", mj + 1/2# ⊗ | ↓# . 2" + 1 Finally, these states can be cast in a compact form by setting |j = " ± 1/2, mj , "# = α± |", mj − 1/2# ⊗ | ↑# + β± |", mj + 1/2# ⊗ | ↓# , (6.2) , %±mj +1/2 where α± = ± = ±β∓ . 2%+1
6.4.3
Addition of two angular momenta J = 1
As mentioned above, for the general case the programme is algebraically technical and unrewarding. However, for completeness, we consider here the explicit example of the addition of two spin 1 degrees of freedom. Once again, the maximal spin state is given by |J = 2, mJ = 2, j1 = 1, j2 = 1# = |j1 = 1, m1 = 1# ⊗ |j2 = 1, m2 = 1# , or, more concisely, |2, 2# = |1# ⊗ |1#, where we leave j1 and j2 implicit. Once again, making use of Eq. (4.5) and an ecomony of notation, we find (exercise) |2, 2# = |1# ⊗ |1# √1 |2, 1# = 2 (|0# ⊗ |1# + |1# ⊗ |0#) |2, 0# = √16 (| − 1# ⊗ |1# + 2|0# ⊗ |0# + |1# ⊗ | − 1#) . |2, −1# = √12 (|0# ⊗ | − 1# + | − 1# ⊗ |0#) |2, 2# = | − 1# ⊗ | − 1#
Then, from the expression for |2, 1#, we can construct the next maximal spin state |1, 1# = √12 (|0# ⊗ |1# − |1# ⊗ |0#), from the orthogonality condition. Once again, acting on this state with the total spin lowering operator, we obtain the remaining members of the multiplet, √1 |1, 1# = 2 (|0# ⊗ |1# − |1# ⊗ |0#) |1, 0# = √12 (| − 1# ⊗ |1# − |1# ⊗ | − 1#) . |1, −1# = √1 (| − 1# ⊗ |0# − |0# ⊗ | − 1#) 2
Finally, finding the state orthogonal to |1, 0# and |2, 0#, we obtain the final state, 1 |0, 0# = √ (| − 1# ⊗ |1# − |0# ⊗ |0# + |1# ⊗ | − 1#) . 3 Advanced Quantum Physics
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Chapter 7
Approximation methods for stationary states 7.1
Time-independent perturbation theory
While we have succeeded in deriving formal analytical solutions for stationary states of the Schr¨odinger operator in a variety of settings, in the majority of practical applications, exact solutions are inaccessible.1 For example, if an atom is placed in an external electric field, the energy levels shift, and the wavefunctions become distorted — the Stark effect. The new energy levels and wavefunctions could in principle be obtained by writing down a complete Hamiltonian, including the external field. Indeed, such a programme may be achieved for the hydrogen atom. But even there, if the external field is small compared with the electric field inside the atom (which is billions of volts per meter) it is easier to compute the changes in the energy levels and wavefunctions within a scheme of successive corrections to the zero-field values. This method, termed perturbation theory, is the single most important method for solving problems in quantum mechanics, and is widely used in atomic physics, condensed matter and particle physics. ! Info. It should be acknowledged that there are – typically very interesting – problems which cannot be solved using perturbation theory, even when the perturbation is very weak; although such problems are the exception rather than the rule. One such case is the one-dimensional problem of free particles perturbed by a localized potential of strength λ. As we found earlier in chapter 2, switching on an arbitrarily weak attractive potential causes the k = 0 free particle wavefunction to drop below the continuum of plane wave energies and become a localized bound state with binding energy of order λ2 . However, on changing the sign of λ to give a repulsive potential, there is no bound state; the lowest energy plane wave state stays at energy zero. Therefore the energy shift on switching on the perturbation cannot be represented as a power series in λ, the strength of the perturbation. This particular difficulty does not typically occur in three dimensions, where arbitrarily weak potentials do not in general lead to bound states.
! Exercise. Focusing on the problem of bound state formation in one-dimension described above, explore the dependence of the ground state energy on λ. Consider why a perturbative expansion in λ is infeasible. 1 Indeed, even if such a solution is formally accessible, its complexity may render it of no practical benefit.
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7.1. TIME-INDEPENDENT PERTURBATION THEORY
7.1.1
63
The Perturbation Series
ˆ (0) , having known eigenLet us then consider an unperturbed Hamiltonian, H (0) states |n(0) ! and eigenvalues En , ˆ (0) |n(0) ! = En(0) |n(0) ! . H
(7.1)
In the following we will address the question of how the eigenstates and ˆ (1) eigenenergies are modified by the imposition of a small perturbation, H (such as that imposed by an external electric or magnetic field on a charged particle, or the deformation of some other external potential). In short, we are interested in the solution of the Schr¨odinger equation, ˆ (0) + H ˆ (1) )|n! = En |n! . (H
(7.2)
ˆ (1) |n(0) ! # En , it seems natural to If the perturbation is small, "n(0) |H (1) ˆ suppose that, on turning on H , the eigenfunctions and eigenvalues will change adiabatically from their unperturbed to their perturbed values, a situation described formally as “adiabatic continuity”, (0)
|n(0) ! $−→ |n!,
En(0) $−→ En .
However, note that this is not always the case. For example, as mentioned above, an infinitesimal perturbation has the capacity to develop a bound state not present in the unperturbed system. For now, let us proceed with the perturbative expansion and return later to discuss its potential range of validity. The basic assumption that underpins the perturbation theory is that, for ˆ (1) small, the leading corrections are of the same order of magnitude as H ˆ (1) itself. The perturbed eigenenergies and eigenvalues can then be obH tained to a greater accuracy by a successive series of corrections, each of order ˆ (1) !/"H ˆ (0) ! compared with the previous. To identify terms of the same "H ˆ (1) !/"H ˆ (0) !, it is convenient to extract from H ˆ (1) a dimensionless order in "H parameter λ, characterising the relative magnitude of the perturbation against ˆ (0) , and then expand |n! and En as a power series in λ, i.e. H |n! = |n(0) ! + λ|n(1) ! + λ2 |n(2) ! + · · · = En = En(0) + λEn(1) + λ2 En(2) + · · · =
∞ !
m=0
∞ !
λm |n(m) !,
λm En(m) .
m=0
One may think of the parameter λ as an artifical book-keeping device to organize the perturbative expansion, and which is eventually set to unity at the end of the calculation. Applied to the stationary form of the Schr¨odinger equation (7.2), an expansion of this sort leads to the relation ˆ (0) + λH ˆ (1) )(|n(0) ! + λ|n(1) ! + λ2 |n(2) ! + · · ·) (H
= (En(0) + λEn(1) + λ2 En(2) + · · ·)(|n(0) ! + λ|n(1) ! + λ2 |n(2) ! + · · ·) .(7.3) From this equation, we must relate terms of equal order in λ. At the lowest order, O(λ0 ), we simply recover the unperturbed equation (7.1). In practical applications, one is usually interested in determining the first non-zero perturbative correction. In the following, we will explore the form of the first and second order perturbative corrections. Advanced Quantum Physics
7.1. TIME-INDEPENDENT PERTURBATION THEORY
7.1.2
64
First order perturbation theory
Isolating terms from (7.3) which are first order in λ, ˆ (0) |n(1) ! + H ˆ (1) |n(0) ! = En(0) |n(1) ! + En(1) |n(0) ! . H
(7.4)
and taking the inner product with the unperturbed states "n(0) |, one obtains ˆ (0) |n(1) ! + "n(0) |H ˆ (1) |n(0) ! = "n(0) |E (0) |n(1) ! + "n(0) |E (1) |n(0) ! . "n(0) |H n n ˆ (0) = "n(0) |En , and exploiting the presumed normalizaNoting that "n(0) |H (0) (0) tion "n |n ! = 1, one finds that the first order shift in energy is given simply by the expectation value of the perturbation taken with respect to the unperturbed eigenfunctions, (0)
ˆ (1) |n(0) ! . En(1) = "n(0) |H
(7.5)
Turning to the wavefuntion, if we instead take the inner product of (7.4) with "m(0) | (with m '= n), we obtain ˆ (0) |n(1) ! + "m(0) |H ˆ (1) |n(0) ! = "m(0) |E (0) |n(1) ! + "m(0) |E (1) |n(0) ! . "m(0) |H n n ˆ (0) = "m(0) |Em and the orthogonality condition on Once again, with "m(0) |H (0) the wavefunctions, "m |n(0) ! = 0, one obtains an expression for the first order shift of the wavefunction expressed in the unperturbed basis, (0)
"m(0) |n(1) ! =
ˆ (1) |n(0) ! "m(0) |H (0)
(0)
En − Em
(7.6)
.
In summary, setting λ = 1, to first order in perturbation theory, we have the eigenvalues and eigenfunctions, ˆ (1) |n(0) ! En ( En(0) + "n(0) |H ! ˆ (1) |n(0) ! "m(0) |H |n! ( |n(0) ! + |m(0) ! . (0) (0) En − Em m"=n Before turning to the second order of perturbation theory, let us first consider a simple application of the method. ! Example: Ground state energy of the Helium atom: For the Helium atom, two electrons are bound to a nucleus of two protons and two neutrons. If one neglects altogether the Coulomb interaction between the electrons, in the ground state, both electrons would occupy the ground state hydrogenic wavefunction (scaled appropriately to accommodate the doubling of the nuclear charge) and have opposite spin. Treating the Coulomb interaction between electrons as a perturbation, one may then use the basis above to estimate the shift in the ground state energy with ˆ (1) = H
1 e2 . 4π$0 |r1 − r2 |
As we have seen, the hydrogenic wave functions are specified by three quantum numbers, n, %, and m. In the ground state, the corresponding wavefunction takes the spatially isotropic form, "r|n = 1, % = 0, m = 0! = ψ100 (r) =
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"
1 πa3
#1/2
e−r/a ,
(7.7)
7.1. TIME-INDEPENDENT PERTURBATION THEORY 2
a0 0 ! where a = 4π" Ze2 me = Z denotes the atomic Bohr radius for a nuclear charge Z. For the Helium atom (Z = 2), the symmetrized ground state of the unperturbed Hamiltonian is then given by the spin singlet (S = 0) electron wavefunction,
1 |g.s.(0) ! = √ (|100, ↑! ⊗ |100, ↓! − |100, ↓! ⊗ |100, ↑!) . 2 Here we have used the direct product ⊗ to discriminate between the two electrons. Then, applying the perturbation theory formula above (7.5), to first order in the Coulomb interaction, the energy shift is given by $ 2 1 e−2(r1 +r2 )/a e2 C0 ˆ (1) |g.s.(0) ! = e En(1) = "g.s.(0) |H dr1 dr2 = , 3 2 4π$0 (πa ) |r1 − r2 | 4π$0 2a % −(z1 +z2 ) 1 where we have defined the dimensionless constant C0 = (4π) dz1 dz1 e|z1 −z2 | . Then, 2 making use of the identity, $ 1 1 1 dΩ1 dΩ2 = , 2 (4π) |z1 − z2 | max(z1 , z2 )
where the integrations runs over the angular coordinates of the vectors z1 and z2 , %∞ %∞ and z1,2 = |z1,2 |, one finds that C0 = 2 0 dz1 z12 e−z1 z1 dz2 z2 e−z2 = 5/4. As a 2
e 1 result, noting that the Rydberg energy, Ry = 4π" , we obtain the first order 0 2a0 5 energy shift ∆E = 4 ZRy ( 34eV for Z = 2. This leads to a total ground state energy of (2Z 2 − 54 Z) Ry = −5.5Ry ( −74.8eV compared to the experimental value of −5.807Ry.
7.1.3
Second order perturbation theory
With the first order of perturbation theory in place, we now turn to consider the influence of the second order terms in the perturbative expansion (7.3). Isolating terms of order λ2 , we have ˆ (0) |n(2) ! + H ˆ (1) |n(1) ! = En(0) |n(2) ! + En(1) |n(1) ! + En(2) |n(0) ! . H
As before, taking the inner product with "n(0) |, one obtains ˆ (0) |n(2) ! + "n(0) |H ˆ (1) |n(1) ! "n(0) |H
= "n(0) |En(0) |n(2) ! + "n(0) |En(1) |n(1) ! + "n(0) |En(2) |n(0) ! .
Noting that the first two terms on the left and right hand sides cancel, we are left with the result ˆ (1) |n(1) ! − En(1) "n(0) |n(1) ! . En(2) = "n(0) |H
Previously, we have made use of the normalization of the basis states, |n(0) !. We have said nothing so far about the normalization of the exact eigenstates, |n!. Of course, eventually, we would like to ensure normalization of these states too. However, to facilitate the perturbative expansion, it is operationally more convenient to impose a normalization on |n! through the condition "n(0) |n! = 1. Substituting the λ expansion for |n!, we thus have "n(0) |n! = 1 = "n(0) |n(0) ! + λ"n(0) |n(1) ! + λ2 "n(0) |n(2) ! + · · · .
From this relation, it follows that "n(0) |n(1) ! = "n(0) |n(2) ! = · · · = 0.2 We can (1) therefore drop the term En "n(0) |n(1) ! from consideration. As a result, we 2 Alternatively, would we suppose that |n! and |n(0) ! were both normalised to unity, to leading order, |n! = |n(0) ! + |n(1) !, and "n(0) |n(1) ! + "n(1) |n(0) ! = 0, i.e. "n(0) |n(1) ! is pure imaginary. This means that if, to first order, |n! has a component parallel to |n(0) !, that component has a small imaginary amplitude allowing us to define |n! = eiφ |n(0) !+orthog. components. However, the corresponding phase factor φ can be eliminated (1) by redefining the phase of |n!. Once again, we can conclude that the term En "n(0) |n(1) ! can be eliminated from consideration.
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7.1. TIME-INDEPENDENT PERTURBATION THEORY
66
obtain ˆ (1) |n(1) ! = "n(0) |H ˆ (1) En(2) = "n(0) |H
!
m"=n
i.e.
|m(0) !
ˆ (1) |n(0) ! "m(0) |H (0)
! |"m(0) |H ˆ (1) |n(0) !|2
En(2) =
(0)
m"=n
(0)
En − Em
(7.8)
.
(0)
En − Em
,
From this result, we can conclude that, ! for the ground state, the second order shift in energy is always negative; ˆ (1) are of comparable magnitude, neighbour! if the matrix elements of H ing levels make a larger contribution than distant levels; ! Levels that lie in close proximity tend to be repelled; ! If a fraction of the states belong to a continuum, the sum in Eq. (7.8) should be replaced by an intergral. Once again, to illustrate the utility of the perturbative expansion, let us consider a concrete physical example. ! Example: The Quadratic Stark Effect: Consider the influence of an external electric field on the ground state of the hydrogen atom. As the composite electron and proton are drawn in different directions by the field, the relative displacement of the electon cloud and nucleus results in the formation of a dipole which serves to lower the overall energy. In this case, the perturbation due to the external field takes the form ˆ (1) = qEz = qEr cos θ , H where q = −|e| denotes the electron charge, and the electric field, E = Eˆez is oriented (0) (0) along the z-axis. With the non-perturbed energy spectrum given by En#m ≡ En = (0) −Ry/n2 , the ground state energy is given by E (0) ≡ E100 = −Ry. At first order in the electric field strength, E, the shift in the ground state energy is given by E (1) = "100|qEz|100! where the ground state wavefunction was defined above (7.7). Since the potential perturbation is antisymmetric in z, it is easy to see that the energy shift vanishes at this order. We are therefore led to consider the contribution second order in the field strength. Making use of Eq. (7.8), and neglecting the contribution to the energy shift from the continuum of unbound positive energy states, we have E (2) =
!
n#=1,#,m
|"n%m|eEz|100!|2 (0)
(0)
E1 − En
,
where |n%m! denote the set of bound state hydrogenic wavefunctions. Although the expression for E (2) can be computed exactly, the programme is somewhat tedious. However, we can place a strong bound on the energy shift through the following (0) (0) (0) (0) argument: Since, for n > 2, |E1 − En | > |E1 − E2 |, we have ! 1 |E (2) | < (0) "100|eEz|n%m!"n%m|eEz|100! . (0) E2 − E1 n#=1,#,m & & Since n,#,m |n%m!"n%m| = I, we have n#=1,#,m |n%m!"n%m| = I−|100!"100|. Finally, since "100|z|100! = 0, we can conclude that |E (2) | < (0) 1 (0) "100|(eEz)2 |100!. With (0)
"100|z 2 |100! = a20 , E1
E2 −E1 (0) (0) = E1 /4,
2
e 1 = − 4π" = −Ry, and E2 0 2a0
|E (2) | <
1 (eE)2 a20 3 2 e /8π$ a 0 0 4
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=
8 4π$0 E 2 a30 . 3
we have
7.2. DEGENERATE PERTURBATION THEORY Furthermore, since all terms in the perturbation series for E (2) are negative, the first 2 term in the series sets a lower bound, |E (2) | > |$210|eEz|100%| . From this result, one (0) (0) E2 −E1
can show that 0.55 × 83 4π$0 E 2 a30 < |E (2) | < 83 4π$0 E 2 a30 (exercise).3
7.2
Degenerate perturbation theory
The perturbative analysis above is reliable providing that the successive terms in the expansion form a convergent series. A necessary condition is that the matrix elements of the perturbing Hamiltonian must be smaller than the corresponding energy level differences of the original Hamiltonian. If it has different states with the same energy (i.e. degeneracies), and the perturbation has non-zero matrix elements between these degenerate levels, then obviously the theory breaks down. However, the problem is easily fixed. To understand how, let us consider a particular example. ˆ = pˆ2 + 1 mω 2 x2 , the Recall that, for the simple harmonic oscillator, H 2m 2 1/4 e−ξ 2 /2 , where ξ = ) ground state wavefunction is given by "x|0! = ( mω π! ' 1/4 ξe−ξ 2 /2 . The wavex mω/! and the first excited state by "x|1! = ( 4mω π! ) functions for the two-dimensional harmonic oscillator, 1 ˆ (0) = 1 (ˆ H p2 + pˆ2y ) + mω 2 (x2 + y 2 ) . 2m x 2 are given simply by the product of two one-dimensional oscillators. So, setting ' ( )1/2 −(ξ2 +η2 )/2 η = y mω/!, the ground state is given by "x, y|0, 0! = mω e , π! and the two degenerate first excited states, an energy !ω above the ground state, are given by, * * + mω ,1/2 ξ "x, y|1, 0! −(ξ 2 +η 2 )/2 = e . "x, y|0, 1! η π! Suppose now we add to the Hamiltonian a perturbation, ˆ (1) = αmω 2 xy = α!ωξη , H controlled by a small parameter α. Notice that, by symmetry, the following ˆ (1) |0, 0! = "1, 0|H ˆ (1) |1, 0! = "0, 1|H ˆ (1) |0, 1! = matrix elements all vanish, "0, 0|H 0. Therefore, according to a na¨ıve perturbation theory, there is no first-order correction to the energies of these states. However, on proceeding to consider the second-order correction to the energy, the theory breaks down. The offˆ (1) |0, 1! = 0 is non-zero, but the two states diagonal matrix element, "1, 0|H |0, 1! and |1, 0! have the same energy! This gives an infinite term in the series (2) for En=1 . Yet we know that a small perturbation of this type will not wreck a twodimensional simple harmonic oscillator – so what is wrong with our approach? To understand the origin of the problem and its fix, it is helpful to plot the original harmonic oscillator potential 12 mω 2 (x2 + y 2 ) together with the perturbing potential αmω 2 xy. The first of course has circular symmetry, the second has two symmetry axes oriented in the directions x = ±y, climbing most steeply from the origin along x = y, falling most rapidly in the directions x = y. If we combine the two potentials into a single quadratic form, " #2 " #2 . 1 1 x + y x − y √ √ mω 2 (x2 + y 2 ) + αmω 2 xy = mω 2 (1 + α) + (1 − α) . 2 2 2 2 3
Energetic readers might like to contemplate how the exact answer of |E (2) | = 94 E 2 a30 can be found exactly. The method can be found in the text by Shankar.
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7.2. DEGENERATE PERTURBATION THEORY the original circles of constant potential become ellipses, with their axes aligned along x = ±y. As soon as the perturbation is introduced, the eigenstates lie in the direction of the new elliptic axes. This is a large change from the original x and y bases, which is not proportional to the small parameter α. But the original unperturbed problem had circular symmetry, and there was no particular reason to choose the x and y axes as we did. If we had instead chosen as our original axes the lines x = ±y, the basis states would not have undergone large changes on switching on the perturbation. The resolution of the problem is now clear: Before switching on the perturbation, one must choose a set of basis states in a degenerate subspace in which the perturbation is diagonal. In fact, for the simple harmonic oscillator example above, the problem can of course be solved exactly √ by rearranging the coordinates to lie along the symmetry axes, (x ± y)/ 2. It is then clear that, despite the results of na¨ıve first order perturbation theory, there is indeed a first order shift in the energy √ levels, !ω → !ω 1 ± α ≈ !ω(1 ± α/2). ! Example: Linear Stark Effect: As with the two-dimensional harmonic oscillator discussed above, the hydrogen atom has a non-degenerate ground state, but degeneracy in its lowest excited states. Specifically, there are four n = 2 states, all having energy − 14 Ry. In spherical coordinates, these wavefunctions are given by , + " #1/2 2 − ar0 ψ200 (r) 1 r ψ210 (r) = e−r/2a0 . θ ar0 cos 32πa30 ±iφ ψ21,±1 (r) sin θ a0 e
ˆ (1) = When perturbing this system with an electric field oriented in the z-direction, H qEr cos θ, a na¨ıve application of perturbation theory predicts no first-order shift in any of these energy levels. However, at second order in E, there is a non-zero matrix ˆ (1) |210!. All the other matrix element between two degenerate levels ∆ = "200|H elements between these basis states in the four-dimensional degenerate subspace are zero. So the only diagonalization necessary is within the two-dimensional degenerate subspace spanned by |200! and |210!, i.e. " # 0 ∆ (1) ˆ H = , ∆ 0 with ∆ = qE
+
1 32πa30
,% +
2−
r a0
,+
r cos θ a0
,2
e−r/a0 r2 dr sin θ dθ dφ = −3qEa0 .
ˆ (1) within this sub-space, the new basis states are given by the Diagonalizing H √ symmetric and antisymmetric combinations, (|200! ±| 210!)/ 2 with energy shifts ±∆, linear in the perturbing electric field. The states |2%, ±1! are not changed by the presence of the field to this level of approximation, so the complete energy map of the n = 2 states in the electric field has two states at the original energy of −Ry/4, one state moved up from that energy by ∆, and one down by ∆. Notice that the new √ eigenstates (|200! ±| 210!)/ 2 are not eigenstates of the parity operator -- a sketch of their wavefunctions reveals that, in fact, they have non-vanishing electric dipole moment µ. Indeed this is the reason for the energy shift, ±∆ = ∓2eEa0 = ∓µ · E.
! Example: As a second and important example of the degenerate perturbation theory, let us consider the problem of a particle moving in one dimension and subject to a weak periodic potential, V (x) = 2V cos(2πx/a) – the nearly free electron model. This problem provides a caricature of a simple crystalline solid in which (free) conduction electrons propagate in the presence of a periodic background lattice potential. Here we suppose that the strength of the potential V is small as compared to the typical energy scale of the particle so that it may be treated as a small perturAdvanced Quantum Physics
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7.3. VARIATIONAL METHOD bation. In the following, we will suppose that the total one-dimensional system is of length L = N a, with periodic boundary conditions. For the unperturbed free particle system, the eigenstates are simply plane waves ψk (x) = "x|k! = √1L eikx indexed by the wavenumber k = 2πn/L, n integer, and (0)
the unperturbed spectrum is given by Ek = !2 k 2 /2m. The matrix elements of the perturbation between states of different wavevector are given by $ " 1 L "k|V |k ! = dxei(k −k)x 2V cos(2πx/a) L 0 $ L+ , " " V = ei(k −k+2π/a)x + ei(k −k−2π/a)x = V δk" −k,±2π/a . L 0 '
Note that all diagonal matrix elements of the perturbation are identically zero. In general, for wavevectors k and k ' separated by G = 2π/a, the unperturbed states are non-degenerate. For these states one can compute the relative energy shift within the framework of second order perturbation theory. However, for states k = −k ' = G/2 ≡ π/a, the unperturbed free particle spectrum is degenerate. Here, and in the neighbourhood of these k values, we must implement a degenerate perturbation theory. For the sinusoidal potential considered here, only states separated by G = 2π/a are coupled by the perturbation. We may therefore consider matrix elements of the full Hamiltonian between pairs of coupled states, |k = G/2 + q! and |k = −G/2 + q! 3 (0) 4 EG/2+q V H= . (0) V E−G/2+q As a result, to leading order in V , we obtain the eigenvalues, Eq =
" #1/2 π 2 !4 q 2 !2 2 (q + (π/a)2 ) ± V 2 + . 2m 4m2 a2
In particular, this result shows that, for k = ±G/2, the degeneracy of the free particle system is lifted by the potential. In the vicinity, |q| # G, the spectrum of eigenvalues is separated by a gap of size 2V . The appearance of the gap mirrors the behaviour found in our study of the Kronig-Penney model of a crystal studied in section 2.2.3. The appearance of the gap has important consequences in theory of solids. Electrons are fermions and have to obey Pauli’s exclusion principle. In a metal, at low temperatures, electrons occupy the free particle-like states up to some (Fermi) energy which lies away from gap. Here, the accessibility of very low-energy excitations due to the continuum of nearby states allows current to flow when a small electric field is applied. However, when the Fermi energy lies in the gap created by the lattice potential, an electric field is unable to create excitations and induce current flow. Such systems are described as (band) insulators.
7.3
Variational method
So far, in devising approximation methods for quantum mechanics, we have focused on the development of a perturbative expansion scheme in which the states of the non-perturbed system provided a suitable platform. Here, by suitable, we refer to situations in which the states of the unperturbed system mirror those of the full system – adiabatic contunity. For example, the states of the harmonic oscillator potential with a small perturbation will mirror those of the unperturbed Hamiltonian: The ground state will be nodeless, the first excited state will be antisymmetric having one node, and so on. However, often we working with systems where the true eigenstates of the problem may not be adiabatically connected to some simple unperturbed reference state. This Advanced Quantum Physics
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7.3. VARIATIONAL METHOD situation is particularly significant in strongly interacting quantum systems where many-particle correlations can effect phase transitions to new states of matter – e.g. the development of superfluid condensates, or the fractional quantum Hall fluid. To address such systems if it is often extremely effective to “guess” and then optimize a trial wavefunction. The method of optimization relies upon a simple theoretical framework known as the variational approach. For reasons that will become clear, the variational method is particularly wellsuited to addressing the ground state. The variational method involves the optimization of some trial wavefunction on the basis of one or more adjustable parameters. The optimization is achieved by minimizing the expectation value of the energy on the trial function, and thereby finding the best approximation to the true ground state wave function. This seemingly crude approach can, in fact, give a surprisingly good approximation to the ground state energy but, it is usually not so good for the wavefunction, as will become clear. However, as mentioned above, the real strength of the variational method arises in the study of many-body quantum systems, where states are more strongly constrained by fundamental symmetries such as “exclusion statistics”. To develop the method, we’ll begin with the problem of a single quantum ˆ = pˆ2 + V (r). If the particle is restricted to particle confined to a potential, H 2m one dimension, and we’re looking for the ground state in any fairly localized potential well, it makes sense to start with a trial wavefunction which belongs 2 to the family of normalized Gaussians, |ψ(α)! = (α/π)1/4 e−αx /2 . Such a trial state fulfils the criterion of being nodeless, and is exponentially localized to the region of the binding potential. It also has the feature that it includes the exact ground states of the harmonic binding potential. The variation approach involves simply minimizing the expectaton value of ˆ the energy, E = "ψ(α)|H|ψ(α)!, with respect to variations of the variational parameter, α. (Of course, as with any minimization, one must check that the variation does not lead to a maximum of the energy!) Not surprisingly, this programme leads to the exact ground state for the simple harmonic oscillator potential, while it serves only as an approximation for other potentials. What is perhaps surprising is that the result is only off by only ca. 30% or so for the attractive δ-function potential, even though the wavefunction looks substantially different. Obviously, the Gaussian family cannot be used if there is an infinite wall anywhere: one must find a family of wavefunctions vanishing at such a boundary. ! Exercise. Using the Gaussian trial state, find the optimal value of the variational state energy, E, for an attractive δ-function potential and compare it with the exact result. To gain some further insight into the approach, suppose the Hamiltonian ˆ has a set of eigenstates, H|n! ˆ H = En |n!. Since the Hamiltonian is Hermitian, these states span the space of possible wave functions, including our & variational family of Gaussians, so we can write, |ψ(α)! = n an (α)|n!. From this expansion, we have ! ˆ "ψ(α)|H|ψ(α)! = |an |2 En ≥ E0 , "ψ(α)|ψ(α)! n for any |ψ(α)!. (We don’t need the denominator if we’ve chosen a family of normalized wavefunctions, as we did with the Gaussians above.) Evidently, Advanced Quantum Physics
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minimizing the left hand side of this equation as function of α provides an upper bound on the ground state energy. We can see immediately that this will probably be better for finding the ground state energy than for the wavefunction: Suppose the optimum state in our family is given by, say, |αmin ! = N (|0!+0.2|1!) with the normalization N ( 0.98, i.e. a 20% admixture of the first excited state. Then the wavefunction is off by ca. 20%, but the energy estimate will be too high by only 0.04(E1 −E0 ), usually a much smaller error. ! Example: To get some idea of of how well the variational approach works, consider its application to the to the ground state of the hydrogen atom. Taking into account the spherical symmetry of the ground state, we may focus on the onedimensional radial component of the wavefunction. Defining the trial radial wavefunction u(ρ) (presumed real), where ρ = r/a0 , the variational energy is given by + 2 , %∞ d 2 + dρ u(ρ) 2 dρ ρ u(ρ) 0 %∞ . E(u) = −Ry 2 dρ u (ρ) 0 For the three families of trial functions, u1 (ρ) = ρe−αρ ,
u2 (ρ) =
ρ , α 2 + ρ2
u3 (ρ) = ρ2 e−αρ ,
and finds αmin = 1, π/4, and 3/2 respectively (exercise). The first family, u1 , includes the exact result, and the minimization procedure correctly finds it. For the three families, the predicted energy of the optimal state is off by 0, 25%, and 21% respectively. The corresponding error in the wavefunction is defined by how far the square of the overlap with the true ground state wavefunction falls short of unity. For the three families, ε = 1 − |"ψ0 |ψvar |2 = 0, 0.21, and 0.05. Notice here that our handwaving argument that the energies would be found much more accurately than the wavefunctions seems to come unstuck. The third family has far better wavefunction overlap than the second, but only a slightly better energy estimate. Why? A key point is that the potential is singular at the origin; there is a big contribution to the potential energy from a rather small region, and the third family of trial states is the least accurate of the three there. The second family of functions are very inaccurate at large distances: the expectation value "r! = 1.5a0 , ∞, 1.66a0 for the three families. But at large distances, both kinetic and potential energies are small, so the result can still look reasonable. These examples reinforce the point that the variational method should be implemented with some caution.
In some cases, one can exploit symmetry to address the properties of higher-lying states. For example, if the one-dimensional attractive potential is symmetric about the origin, and has more than one bound state, the ground state will be even, the first excited state odd. Therefore, we can estimate the energy of the first √ excited state by minimizing a family of odd functions, such π 1/2 −αx2 /2 as ψ(x, α) = ( 2α3/2 ) xe . ! Example: Helium atom addressed by the variational approach: For the hydrogen atom, we know that the ground state energy is 1 Ry, or 13.6 eV. The He+ ion (with just a single electron) has a nuclear charge of Z = 2, so the ground state energy of the electron, being proportional to Z 2 , will now be equal to 4 Ry. Therefore, for the He atom, if we neglect their mutual interaction, the electrons will occupy the ground state wavefunction having opposite spin, leading to a total ground state energy of 8 Ry or 109 eV. In practice, as we have seen earlier, the repulsion between the electrons lowers ground state energy to 79 eV (see page 64). To get a better estimate for the ground state energy, one can retain the form of Z 3 1/2 −Zr/a0 , but rather than setting the nuclear charge e the ionic wavefunction, ( πa 3) 0
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Z = 2, leave it as a variational parameter. In other words, let us accommodate the effects of electron-electron repulsion, which must “push” the wavefunctions to larger radii, by keeping exactly the same wavefunction profile but lessening the effective nuclear charge as reflected in the spread of the wavefunction from Z = 2 to Z < 2. The precise value will be set by varying it to find the minimum total energy, including the term from electron-electron repulsion. To find the potential energy from the nuclear-electron interactions, we of course must use the actual nuclear charge Z = 2, but impose a variable Z for the wavefunction, so the nuclear potential energy for the two electrons is given by, $ ∞ Z 3 e−2Zr/a0 e2 2e2 4πr2 dr 3 = −Z = −8Z Ry. p.e. = −2 × 4π$0 0 πa0 r π$0 a0 This could have been inferred from the formula for the one electron ion, where the potential energy for the one electron is −2Z 2 Ry, one factor of Z being from the nuclear charge, the other from the consequent shrinking of the orbit. The kinetic energy is even easier to determine: it depends entirely on the form of the wavefunction, and not on the actual nuclear charge. So for our trial wavefunction it has to be Z 2 Ry per electron. Finally, making use of our calculation on page 64, we can immediately write down the positive contribution to the energy expectation value from the electronelectron interaction, $ e2 Z3 e−2Z(r1 +r2 )/a0 5 e2 Z 5 dr1 dr2 = = Z Ry . 3 2 4π$0 (πa0 ) |r1 − r2 | 4 4π$0 2a0 4 Collecting all of the terms, the total variational state energy is given by: " # 5 2 E = −2 4Z − Z − Z Ry . 8 5 Minimization of this energy with respect to Z obtains the minimum at Z = 2 − 16 , leading to an energy of 77.5 eV. This result departs from the true value by about 1 eV. So, indeed, the presence of the other electron leads effectively to a shielding of the nuclear charge by an amount of ca. (5/16)e.
This completes our discussion of the principles of the variational approach. However, later in the course, we will find the variational methods appearing in several important applications. Finally, to close this section on approximation methods for stationary states, we turn now to consider a framework which makes explicit the connection between the quantum and classical theory in the limit ! → 0.
7.4
Wentzel, Kramers and Brillouin (WKB) method
The WKB (or Wentzel, Kramers and Brillouin) approximation describes a “quasi-classical” method for solving the one-dimensional time-independent Schr¨odinger equation. Note that the consideration of one-dimensional systems is less restrictive that it may sound as many symmetrical higher-dimensional problems are rendered effectively one-dimensional (e.g. the radial equation for the hydrogen atom). The WKB method is named after physicists Wentzel, Kramers and Brillouin, who all developed the approach independently in 1926.4 Earlier, in 1923, the mathematician Harold Jeffreys had developed a general method of approximating the general class of linear, second-order 4 L. Brillouin, (1926). “La mcanique ondulatoire de Schr¨ odinger: une m´thode g´en´erale de resolution par approximations successives”, Comptes Rendus de l’Academie des Sciences 183: 2426; H. A. Kramers, (1926). “Wellenmechanik und halbz¨ ahlige Quantisierung”, Z. Phys. 39: 828840; G. Wentzel (1926). “Eine Verallgemeinerung der Quantenbedingungen f¨ ur die Zwecke der Wellenmechanik”. Z. Phys. 38: 518529.
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L´ eon Nicolas Brillouin 18891969 A French physicis, his father, Marcel Brillouin, grandfather, ´ Eleuth` ere Mascart, and greatgrandfather, Charles Briot, were physicists as well. He made contributions to quantum mechanics, radio wave propagation in the atmosphere, solid state physics, and information theory.
Hendrik Anthony Kramers 1894-1952
“Hans”
A Dutch physicist who conducted early and important work in quantum theory and electromagnetic dispersion relations, solid-state physics, and statistical mechanics. He was a long-time assistant and friend to Niels Bohr, and collaborated with him on a 1924 paper contending that light consists of probability waves, which became a foundation of quantum mechanics. He introduced the idea of renormalization, a cornerstone of modern field theory, and determined the dispersion formulae that led to Werner Heisenberg’s matrix mechanics. He is not as well known as some of his contemporaries (primarily because his work was not widely translated into English), but his name is still invoked by physicists as they discuss Kramers dispersion theory, Kramers-Heisenberg dispersion formulae, Kramers opacity formula, Kramers degeneracy, or Kramers-Kronig relations.
7.4. WENTZEL, KRAMERS AND BRILLOUIN (WKB) METHOD differential equations, which of course includes the Schr¨odinger equation.5 But since the Schr¨odinger equation was developed two years later, and Wentzel, Kramers, and Brillouin were apparently unaware of this earlier work, the contribution of Jeffreys is often neglected. The WKB method is important both as a practical means of approximating solutions to the Schr¨odinger equation, and also as a conceptual framework for understanding the classical limit of quantum mechanics. The WKB approximation is valid whenever the wavelength, λ, is small in comparison to other relevant length scales in the problem. This condition is not restricted to quantum mechanics, but rather can be applied to any wave-like system (such as fluids, electromagnetic waves, etc.), where it leads to approximation schemes which are mathematically very similar to the WKB method in quantum mechanics. For example, in optics the approach is called the eikonal method, and in general the method is referred to as short wavelength asymptotics. Whatever the name, the method is an old one, which predates quantum mechanics – indeed, it was apparently first used by Liouville and Green in the first half of the nineteenth century. In quantum mechanics, λ is interpreted as the de Broglie wavelength, and L is normally the length scale of the potential. Thus, the WKB method is valid if the wavefunction oscillates many times before the potential energy changes significantly.
7.4.1
Semi-classical approximation to leading order
Consider then the propagation of a quantum particle in a slowly-varying onedimensional potential, V (x). Here, by “slowly-varying” we mean that, in any small region the wavefunction is well-approximated by a plane wave, and that the wavelength only changes over distances that are long compared with the local value of the wavelength. We’re also assuming for the moment that the particle has positive kinetic energy in the region. Under these conditions, we can anticipate that the solution to the time-independent Schr¨odinger equation −
!2 2 ∂ ψ(x) + V (x)ψ(x) = Eψ(x) , 2m x
will take the form A(x)e±ip(x)x/! where p(x) is the “local” value of the momentum set by the classical value, p2 /2m + V (x) = E, and the amplitude, A(x), is slowly-varying compared with the phase factor. Clearly this is a semi-classical limit: ! has to be sufficiently small that there are many oscillations in the typical distance over which the potential varies.6 To develop this idea a little more rigorously, and to emphasize the rapid phase variation in the semi-classical limit, we can parameterize the wavefunction as ψ(x) = eiσ(x)/! . 5
H. Jeffreys, (1924). “On certain approximate solutions of linear differential equations of the second order”, Proc. Lon. Math. Soc. 23: 428436. 6 To avoid any point of confusion, it is of course true that ! is a fundamental constant – not easily adjusted! So what do we mean when we say that the semi-classical limit translates to ! → 0? The validity of the semi-classical approximation relies upon λ/L $ 1. Following the de Broglie relation, we may write this inequality as h/pL $ 1, where p denotes the particle momentum. Now, in this correspondence, both p and L can be considered as “classical” scales. So, formally, we can think of think of accessing the semi-classical limit by adjusting ! so that it is small enough to fulfil this inequality. Alternatively, at fixed !, we can access the semi-classical regime by reaching to higher and higher energy scales (larger and larger p) so that the inequality becomes valid.
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7.4. WENTZEL, KRAMERS AND BRILLOUIN (WKB) METHOD Here the complex function σ(x) encompasses both the amplitude and phase. Then, with −!2 ∂x2 ψ(x) = −i!eiσ(x)/!∂x2 σ(x) + eiσ(x)/!(∂x σ)2 , the Schr¨odinger equation may be rewritten in terms of the phase function as, −i!∂x2 σ(x) + (∂x σ)2 = p2 (x) .
(7.9)
Now, since we’re assuming the system is semi-classical, it makes sense to expand σ(x) as a power series in ! setting, σ = σ0 + (!/i)σ1 + (!/i)2 σ2 + · · · . At the leading (zeroth) order of the expansion, we can drop the first term in'(7.9), leading to (∂x σ0 )2 = p2 (x). Fixing the sign of p(x) by p(x) = + 2m(E − V (x)), we conclude that $ σ0 (x) = ± p(x)dx . For free particle systems – those for which the kinetic energy is proportional to p2 – this expression coincides with the classical action. From the form of the Schr¨odinger equation (7.9), it is evident that this approximate solution is only valid if we can ignore the first term. More precisely, we must have 5 5 5 !∂x2 σ(x) 5 5 5 5 (∂x σ(x))2 5 ≡ |∂x (!/∂x σ)| # 1 . But, in the leading approximation, ∂x σ ( p(x) and p(x) = 2π!/λ(x), so the condition translates to the relation 1 |∂x λ(x)| # 1 . 2π This means that the change in wavelength over a distance of one wavelength must be small. Obviously, this condition can not always be met: In particular, if the particle is confined by an attractive potential, at the edge of the classically allowed region, where E = V (x), p(x) must is zero and the corresponding wavelength infinite. The approximation is only valid well away from these classical turning points, a matter to which we will return shortly.
7.4.2
Next to leading order correction
Let us now turn to the next term in the expansion in !. Retaining terms from Eq. (7.9) which are of order !, we have −i!∂x2 σ0 + 2∂x σ0 (!/i)∂x σ1 = 0 . Rearranging this equation, and integrating, we find ∂x σ1 = −
∂x2 σ0 ∂x p =− , 2∂x σ0 2p
1 σ1 (x) = − ln p(x) . 2
So, to this order of approximation, the wavefunction takes the form, C1 (i/!) R p dx C2 −(i/!) R p dx ψ(x) = ' e +' e , p(x) p(x)
(7.10)
where C1 and C2 denote constants of integration. To interpret the factors of ' p(x), consider the first term: a wave moving to the right. Since p(x) is real Advanced Quantum Physics
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7.4. WENTZEL, KRAMERS AND BRILLOUIN (WKB) METHOD (remember we are currently considering the classically allowed region where E > V (x)), the exponential has modulus unity, and the local probability density is proportional to 1/p(x), i.e. to 1/v(x), where v(x) denotes the velocity of the particle. This dependence has a simple physical interpretation: The probability of finding the particle in any given small interval is proportional to the time it spends there. Hence it is inversely proportional to its speed. We turn now to consider the wavefunction in the classically forbidden region where p2 (x) = E − V (x) < 0 . 2m Here p(x) is of course pure imaginary, but the same formal phase solution of the Schr¨odinger equation applies provided, again, that the particle is remote from the classical turning points where E = V (x). In this region, the wavefunction takes the general form, ψ(x) = '
C1% |p(x)|
e−(1/!)
R
|p| dx
R C% + ' 2 e(1/!) |p| dx . |p(x)|
(7.11)
This completes our study of the wavefunction in the regions in which the semi-classical approach can be formally justified. However, to make use of this approximation, we have to understand how to deal with the regions close to the classical turning points. Remember in our treatment of the Schr¨odinger equation, energy quantization derived from the implementation of boundary conditions.
7.4.3
Connection formulae, boundary conditions and quantization rules
Let us assume that we are dealing with a one-dimensional confining potential where the classically allowed region is unique and spans the interval b ≤ x ≤ a. Clearly, in the classically forbidden region to the right, x > a, only the first term in Eq. (7.11) remains convergent and can contribute while, for x < b it is only the second term that contributes. Moreover, in the classically allowed region, b ≤ x ≤ a, the wavefunction has the oscillating form (7.10). But how do we connect the three regions together? To answer this question, it is necessary to make the assumption that the potential varies sufficiently smoothly that it is a good approximation to take it to be linear in the vicinity of the classical turning points. That is to say, we assume that a linear potential is a sufficiently good approximation out to the point where the short wavelength (or decay length for tunneling regions) description is adequate. Therefore, near the classical turning at x = a, we take the potential to be E − V (x) ( F0 (x − a) , where F0 denotes the (constant) force. For a strictly linear potential, the wavefunction can be determined analytically, and takes the form of an Airy function.7 In particular, it is known that the Airy function to the right of the classical turning point has the asymptotic form Rx C lim ψ(x) = ' e−(1/!) a |p| dx , x&a 2 |p(x)|
7 For a detailed discussion in the present context, we refer to the text by Landau and Lifshitz.
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translating to a decay into the classically forbidden region while, to the left, it has the asymptotic oscillatory solution, 6 $ a 6 7 7 $ C 1 π 1 a π C lim ψ(x) = ' cos p dx − cos p dx . ≡' − b'x
where, for n% even, C % = C and for n odd, C % = −C. Therefore, we have the a condition !1 b p dx = (n + 1/2)π, or when cast in terms of a complete periodic cycle of the classical motion, 8 p dx = 2π!(n + 1/2) . This is just the Bohr-Sommerfeld quantization condition, and n can be interpreted as the number of nodes of the wavefunction. ! Info. Note that the integrated action,
9
p dx, represents the area of the classical path in phase space. This shows that each state is associated with an element of phase space 2π!. From this, we can deduce the approximate energy splitting between levels in the semi-classical limit: The change in the integral with energy ∆E corresponding to one level must be 2π! – one more state and one more node, i.e. 9 9 9 ∆E ∂E p dx = 2π!. Now ∂p E = v, so ∂E p dx = dx/v = T , the period of the orbit. Therefore, ∆E = 2π!/T = !ω: In the semi-classical limit, if a particle emits one photon and drops to the next level, the frequency of the photon emitted is just the orbital frequency of the particle.
! Example: For the quantum harmonic oscillator, H =
the classical momentum is given by : p(x) =
"
mω 2 x2 2m E − 2
#
p2 2m
+ 12 mω 2 x2 = E,
.
The classical turning points are set by E = mω 2 x20 /2, i.e. x0 = ±2E/mω 2 . Over a periodic cycle, the classical action is given by : " # 8 $ x0 mω 2 x2 E p(x)dx = 2 dx 2m E − = 2π . 2 ω −x0 According to the WKB method, the latter must be equated to 2π!(n + 1/2), with the last term reflecting the two turning points. As a result, we find that the energy levels are as expected specified by En = (n + 1/2)!ω. In the WKB approximation, the corresponding wavefunctions are given by " $ x # C 1 π p(x)dx − ψ(x) = ' cos ! −x0 4 p(x) " # $ x C 2π 1 π cos (n + 1/2) + p(x)dx − =' 4 ! 0 4 p(x) : 3 .4 " # C nπ E x x x2 =' cos + arcsin + 1− 2 , 2 !ω x0 x0 x0 p(x)
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The WKB wavefunction (solid) and the exact wavefunction (dashed) for the n = 0 and n = 10 states of the quantum harmonic oscillator.
7.4. WENTZEL, KRAMERS AND BRILLOUIN (WKB) METHOD for 0 < x < x0 and 3 - : " #.4 C E x x2 x ψ(x) = ' exp − . 2 − 1 − arccosh x !ω x x 2 p(x) 0 0 0
for x > a. Note that the failure of the WKB approximation is reflected in the appearance of discontinuities in the wavefunction at the classical turning points (see figures). Nevertheless, the wavefunction at high energies provide a strikingly good approximation to the exact wavefunctions.
! Example: As a second example, let us consider the problem of quantum tunneling. Suppose that a beam of particles is incident upon a localized potential barrier, V (x). Further, let us assume that, over a single continuous region of space, from b to a, the potential rises above the incident energy of the incoming particles so that, classically, all particles would be reflected. In the quantum system, the some particles incident from the left may tunnel through the barrier and continue propagating to the right. We are interested in finding the transmission probability. From the WKB solution, to the left of the barrier (region 1), we expect a wavefunction of the form 6 $ x 6 7 7 $ i i x 1 1 p dx + r(E) √ exp − p dx , ψ1 (x) = √ exp p ! b p ! b ' with p(E) = 2m(E − V (x)), while, to the right of the barrier (region 3), the wavefunction is given by 6 $ x 7 1 i ψ3 (x) = t(E) √ exp p dx . p ! a In the barrier region, the wavefunction is given by 6 7 6 $ x 7 $ C1 1 x C2 1 ψ2 (x) = ' exp − |p| dx + ' exp |p| dx . ! a ! a |p(x)| |p(x)|
Then, applying the continuity condition on the wavefunction and its derivative at the classical turning points, one obtains the transmissivity, -
2 T (E) ( exp − !
$
a
b
.
|p| dx .
! Info. For a particle strictly confined to one dimension, the connection formulae can be understood within a simple picture: The wavefunction “spills over” into the classically forbidden region, and its twisting there collects an π/4 of phase change. So, in the lowest state, the total phase change in the classically allowed region need only be π/2. For the radial equation, assuming that the potential is well behaved at the origin, the wavefunction goes to zero there. A bound state will still spill over beyond the classical turning point at r0 , say, but clearly there must be a total phase change of 3π/4 in the allowed region for the lowest state, since there can be no spill over to negative r. In this case, the general quantization formula will be $ 1 r0 p(r) dr = (n + 3/4)π, n = 0, 1, 2, · · · , ! 0 with the series terminating if and when the turning point reaches infinity. In fact, some potentials, including the Coulomb potential and the centrifugal barrier for % '= 0, are in fact singular at r = 0. These cases require special treatment.
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Chapter 8
Identical Particles Until now, most of our focus has been on the quantum mechanical behaviour of individual particles, or problems which can be “factorized” into independent single-particle systems.1 However, most physical systems of interest involve the interaction of large numbers of particles; electrons in a solid, atoms in a gas, etc. In classical mechanics, particles are always distinguishable in the sense that, at least formally, their “trajectories” through phase space can be traced and their identity disclosed. However, in quantum mechanics, the intrinsic uncertainty in position, embodied in Heisenberg’s principle, demands a careful and separate consideration of distinguishable and indistinguishable particles. In the present section, we will consider how to formulate the wavefunction of many-particle systems, and adress some of the (sometimes striking and often counter-intuitive) implications of particle indistinguishability.
8.1
Quantum statistics
Consider then two identical particles confined to a box in one-dimension. Here, by identical, we mean that the particles can not be discriminated by some internal quantum number. For example, we might have two electrons of the same spin. The normalized two-particle wavefunction ψ(x1 , x2 ), which gives the probability |ψ(x1 , x2 )|2 dx1 dx2 of finding simultaneously one particle in the interval x1 to x1 + dx1 and another between x2 to x2 + dx2 , only makes sense if |ψ(x1 , x2 )|2 = |ψ(x2 , x1 )|2 , since we can’t know which of the two indistinguishable particles we are finding where. It follows from this that the wavefunction can exhibit two (and, generically, only two) possible symmetries under exchange: ψ(x1 , x2 ) = ψ(x2 , x1 ) or ψ(x1 , x2 ) = −ψ(x2 , x1 ).2 If two identical particles have a symmetric wavefunction in some state, particles of that type always have symmetric wavefunctions, and are called bosons. (If in some other state they had an antisymmetric wavefunction, then a linear 1 For example, our treatment of the hydrogen atom involved the separation of the system into centre of mass and relative motion. Each could be referred to an effective single-particle dynamics. 2 We could in principle have ψ(x1 , x2 ) = eiα ψ(x2 , x1 ), with α a constant phase. However, in this case we would not recover the original wavefunction on exchanging the particles twice. Curiously, in some two-dimensional theories used to describe the fractional quantum Hall effect, there exist collective excitations of the electron system — called anyons — that do have this kind of property. For a discussion of this point, one may refer to the seminal paper of J. M. Leinaas and J. Myrheim, On the theory of identical particles. Il Nuovo Cimento B37, 1-23 (1977). Such anyonic systems have been proposed as a strong candidate for the realization of quantum computation. For a pedagogical discussion, we refer to an entertaining discussion by C. Nayak, S. H. Simon, A. Stern, M. Freedman, S. Das Sarma, Non-Abelian Anyons and Topological Quantum Computation, Rev. Mod. Phys. 80, 1083 (2008). However, all ordinary “fundamental” particles are either bosons or fermions.
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8.1. QUANTUM STATISTICS superposition of those states would be neither symmetric nor antisymmetric, and so could not satisfy the relation |ψ(x1 , x2 )|2 = |ψ(x2 , x1 )|2 .) Similarly, particles having antisymmetric wavefunctions are called fermions. To construct wavefunctions for three or more fermions, let first suppose that the particles do not interact with each other, and are confined by a spinindependent potential, such as the Coulomb field of a nucleus. In this case, the Hamiltonian will be symmetric in the fermion degrees of freedom, ˆ2 ˆ2 ˆ 21 p p ˆ = p H + 2 + 3 · · · + V (r1 ) + V (r2 ) + V (r3 ) + · · · , 2m 2m 2m and the solutions of the Schr¨odinger equation will be products of eigenfuncˆs = p ˆ 2 /2m + V (r). However, sintions of the single-particle Hamiltonian H gle products such as ψa (1)ψb (2)ψc (3) do not have the required antisymmetry property under the exchange of any two particles. (Here a, b, c, ... label the ˆ s , and 1, 2, 3,... denote both space and spin single-particle eigenstates of H coordinates of single particles, i.e. 1 stands for (r1 , s1 ), etc.) We could achieve the necessary antisymmetrization for particles 1 and 2 by subtracting the same product wavefunction with the particles 1 and 2 interchanged, i.e. ψa (1)ψb (2)ψc (3) "→ (ψa (1)ψb (2)−ψa (2)ψb (1))ψc (3), ignoring the overall normalization for now. However, the wavefunction needs to be antisymmetrized with respect to all possible particle exchanges. So, for 3 particles, we must add together all 3! permutations of 1, 2, 3 in the state a, b, c, with a factor −1 for each particle exchange necessary to get to a particular ordering from the original ordering of 1 in a, 2 in b, and 3 in c. In fact, such a sum over permutations is precisely the definition of the determinant. So, with the appropriate normalization factor: ! ! ! ψ (1) ψb (1) ψc (1) ! ! 1 !! a ψabc (1, 2, 3) = √ ! ψa (2) ψb (2) ψc (2) !! . 3! ! ψ (3) ψ (3) ψ (3) ! a c b The determinantal form makes clear the antisymmetry of the wavefunction with respect to exchanging any two of the particles, since exchanging two rows of a determinant multiplies it by −1. We also see from the determinantal form that the three states a, b, c must all be different, for otherwise two columns would be identical, and the determinant would be zero. This is just the manifestation of Pauli’s exclusion principle: no two fermions can be in the same state. Although these determinantal wavefunctions (known as Slater determinants), involving superpositions of single-particle states, are only strictly correct for non-interacting fermions, they provide a useful platform to describe electrons in atoms (or in a metal), with the electron-electron repulsion approximated by a single-particle potential. For example, the Coulomb field in an atom, as seen by the outer electrons, is partially shielded by the inner electrons, and a suitable V (r) can be constructed self-consistently, by computing the single-particle eigenstates and finding their associated charge densities (see section 9.2.1). In the bosonic system, the corresponding many-particle wavefunction must be symmetric under particle exchange. We can obtain such a state by expanding all of the contributing terms from the Slater determinant and setting all of the signs to be positive. In other words, the bosonic wave function describes the uniform (equal phase) superposition of all possible permutations of product states.
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8.2. SPACE AND SPIN WAVEFUNCTIONS
8.2
80
Space and spin wavefunctions
Although the metholodology for constructing a basis of many-particle states outlined above is generic, it is not particularly convenient when the Hamiltonian is spin-independent. In this case we can simplify the structure of the wavefunction by factorizing the spin and spatial components. Suppose we have two electrons (i.e. fermions) in some spin-independent potential V (r). We know that the two-electron wavefunction must be antisymmetric under particle exchange. Since the Hamiltonian has no spin-dependence, we must be able to construct a set of common eigenstates of the Hamiltonian, the total spin, and the z-component of the total spin. For two electrons, there are four basis states in the spin space, the S = 0 spin singlet state, |χS=0,Sz =0 % = √12 (| ↑1 ↓2 % − | ↓1 ↑2 %), and the three S = 1 spin triplet states, |χ11 % = | ↑1 ↑2 %,
1 |χ10 % = √ (| ↑1 ↓2 % + | ↓1 ↑2 %) , 2
|χ1,−1 % = | ↓1 ↓2 % .
Here the first arrow in the ket refers to the spin of particle 1, the second to particle 2. # Exercise. By way of revision, it is helpful to recapitulate the discussion of the addition of spin s = 1/2 angular momenta. By setting S = S1 + S2 ,3 where S1 and S2 are two spin 1/2 degrees of freedom, construct the matrix elements of the total spin operator S2 for the four basis states, | ↑↑%, | ↑↓%, | ↓↑%, and | ↓↓%. From the matrix representation of S2 , determine the four eigenstates. Show that one corresponds to a total spin zero state and three correspond to spin 1. It is evident that the spin singlet wavefunction is antisymmetric under the exchange of two particles, while the spin triplet wavefunction is symmetric. For a general state, the total wavefunction for the two electrons in a common ˆ then has the form: eigenstate of S2 , Sz and the Hamiltonian H Ψ(r1 , s1 ; r2 , s2 ) = ψ(r1 , r2 )χ(s1 , s2 ) , where χ(s1 , s2 ) = )s1 , s2 |χ%. For two electron degrees of freedom, the total wavefunction, Ψ, must be antisymmetric under exchange. It follows that a pair of electrons in the spin singlet state must have a symmetric spatial wavefunction, ψ(r1 , r2 ), whereas electrons in the spin triplet states must have an antisymmetric spatial wavefunction. Before discussing the physical consequences of this symmetry, let us mention how this scheme generalizes to more particles. # Info. Symmetry of three-electron wavefunctions: Unfortunately, in seeking a factorization of the Slater determinant into a product of spin and spatial components for three electrons, things become more challenging. There are now 23 = 8 basis states in the spin space. Four of these are accounted for by the spin 3/2 state with Sz = 3/2, 1/2, −1/2, −3/2. Since all spins are aligned, this is evidently a symmetric state, so must be multiplied by an antisymmetric spatial wavefunction, itself a determinant. So far so good. But the other four states involve two pairs of total spin 1/2 states built up of a singlet and an unpaired spin. They are orthogonal to the symmetric spin 3/2 state, so they can’t be symmetric. But they can’t be antisymmetric either, since in each such state, two of the spins must be pointing in the same direction! An example of such a state is presented by |χ% = | ↑1 % ⊗ √1 (| ↑2 ↓3 % − | ↓2 ↑3 %). Evidently, this must be multiplied by a spatial wavefunction 2 3
Here, for simplicity, we have chosen not to include hats on the spin angular momentum operators.
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Hint: begin by proving that, for two spin s = 1/2 degree of freedom, S2 = S21 + S22 + 2S1 · S2 = 2 × s(s + 1)!2 + 2S1z S2z + S1+ S2− + S1− S2+ .
8.3. PHYSICAL CONSEQUENCES OF PARTICLE STATISTICS symmetric in 2 and 3. But to recover a total wave function with overall antisymmetry it is necessary to add more terms: Ψ(1, 2, 3) = χ(s1 , s2 , s3 )ψ(r1 , r2 , r3 ) + χ(s2 , s3 , s1 )ψ(r2 , r3 , r1 ) + χ(s3 , s1 , s2 )ψ(r3 , r1 , r2 ) . Requiring the spatial wavefunction ψ(r1 , r2 , r3 ) to be symmetric in 2, 3 is sufficient to guarantee the overall antisymmetry of the total wavefunction Ψ.4 For more than three electrons, similar considerations hold. The mixed symmetries of the spatial wavefunctions and the spin wavefunctions which together make a totally antisymmetric wavefunction are quite complex, and are described by Young diagrams (or tableaux).5 A discussion of this scheme reaches beyond the scope of these lectures.
# Exercise. A hydrogen atom consists of two fermions, the proton and the electron. By considering the wavefunction of two non-interacting hydrogen atoms under exchange, show that the atom transforms as a boson. In general, if the number of fermions in a composite particle is odd, then it is a fermion, while if even it is a boson. Quarks are fermions: baryons consist of three quarks and so translate to fermions while mesons consist of two quarks and translate to bosons.
8.3
Physical consequences of particle statistics
The overall antisymmetry demanded by the many-fermion wavefunction has important physical implications. In particular, it determines the magnetic properties of atoms. The magnetic moment of the electron is aligned with its spin, and even though the spin variables do not appear in the Hamiltonian, the energy of the eigenstates depends on the relative spin orientation. This arises from the electrostatic repulsion between electrons. In the spatially antisymmetric state, the probability of electrons coinciding at the same position necessarily vanishes. Moreover, the nodal structure demanded by the antisymmetry places the electrons further apart on average than in the spatially symmetric state. Therefore, the electrostatic repulsion raises the energy of the spatially symmetric state above that of the spatially antisymmetric state. It therefore follows that the lower energy state has the electron spins pointing in the same direction. This argument is still valid for more than two electrons, and leads to Hund’s rule for the magnetization of incompletely filled inner shells of electrons in transition metal and rare earths atoms (see chapter 9). This is the first step in understanding ferromagnetism. A gas of hydrogen molecules provides another manifestation of wavefunction antisymmetry. In particular, the specific heat depends sensitively on whether the two protons (spin 1/2) in H2 have their spins parallel or antiparallel, even though that alignment involves only a very tiny interaction energy. If the proton spins occupy a spin singlet configuration, the molecule is called parahydrogen while the triplet states are called orthohydrogen. These two distinct gases are remarkably stable - in the absence of magnetic impurities, paraortho transitions take weeks. The actual energy of interaction of the proton spins is of course completely negligible in the specific heat. The important contributions to the specific heat 4 Particle physics enthusiasts might be interested to note that functions exactly like this arise in constructing the spin/flavour wavefunction for the proton in the quark model (Griffiths, Introduction to Elementary Particles, page 179). 5 For a simple introduction, see Sakurai’s textbook (section 6.5) or chapter 63 of the text on quantum mechanics by Landau and Lifshitz.
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8.3. PHYSICAL CONSEQUENCES OF PARTICLE STATISTICS are the usual kinetic energy term, and the rotational energy of the molecule. This is where the overall (space × spin) antisymmetric wavefunction for the protons plays a role. Recall that the parity of a state with rotational angular momentum $ is (−1)! . Therefore, parahydrogen, with an antisymmetric proton spin wavefunction, must have a symmetric proton spatial wavefunction, and so can only have even values of the rotational angular momentum. Orthohydrogen can only have odd values. The energy of the rotational level with angular momentum $ is E!tot = !2 $($ + 1)/I, where I denotes the moment of inertia of the molecule. So the two kinds of hydrogen gas have different sets of rotational energy levels, and consequently different specific heats. # Exercise. Determine the degeneracy of ortho- and parahydrogen. By expressing the state occupancy of the rotational states through the Boltzmann factor, determine the low temperature variation of the specific heat for the two species.
# Example: As a final example, and one that will feed into our discussion of multielectron atoms in the next chapter, let us consider the implications of particle statistics for the excited state spectrum of Helium. After Hydrogen, Helium is the simplest atom having two protons and two neutrons in the nucleus (Z = 2), and two bound electrons. As a complex many-body system, we have seen already that the Schr¨odinger equation is analytically intractable and must be treated perturbatively. Previously, in chapter 7, we have used the ground state properties of Helium as a vehicle to practice perturbation theory. In the absence of direct electron-electron interaction, the Hamiltonian ˆ0 = H
2 # 2 " ˆ p
n
n=1
2m
$ + V (rn ) ,
V (r) = −
1 Ze2 , 4π&0 r
is separable and the wavefunction can be expressed through the states of the hydrogen atom, ψn!m . In this approximation, the ground state wavefunction involves both electrons occupying the 1s state leading to an antisymmetric spin singlet wavefunction for the spin degrees of freedom, |Ψg.s. % = (|100% ⊗ |100%) ⊗ |χ00 %. In chapter 7, we made use of both the perturbative series expansion and the variational method to determine how the ground state energy is perturbed by the repulsive electron-electron interaction, ˆ1 = H
1 e2 . 4π&0 |r1 − r2 |
Now let us consider the implications of particle statistics on the spectrum of the lowest excited states. From the symmetry perspective, the ground state wavefunction belongs to the class of states with symmetric spatial wavefunctions, and antisymmetric spin (singlet) wavefunctions. These states are known as parahelium. In the absence of electronelectron interaction, the first excited states are degenerate and have the form, 1 |ψp % = √ (|100% ⊗ |2$m% + |2$m% ⊗ |100%) ⊗ |χ00 % . 2 The second class of states involve an antisymmetric spatial wavefunction, and symmetric (triplet) spin wavefunction. These states are known as orthohelium. Once again, in the absence of electron-electron interaction, the first excited states are degenerate and have the form, 1 |ψo % = √ (|100% ⊗ |2$m% − |2$m% ⊗ |100%) ⊗ |χ1Sz % . 2 The perturbative shift in the ground state energy has already been calculated within the framework of first order perturbation theory. Let us now consider the shift
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8.4. IDEAL QUANTUM GASES
83
in the excited states. Despite the degeneracy, since the off-diagonal matrix elements vanish, we can make use of the first order of perturbation theory to compute the shift. In doing so, we obtain % 1 e2 1 p,o ∆En! = d3 r1 d3 r2 |ψ100 (r1 )ψn!0 (r2 ) ± ψn!0 (r1 )ψ100 (r2 )|2 , 2 4π&0 |r1 − r2 |
with the plus sign refers to parahelium and the minus to orthohelium. Since the matrix element is independent of m, the m = 0 value considered here applies to all p,o values of m. Rearranging this equation, we thus obtain ∆En! = Jn! ± Kn! where % 2 2 2 |ψ100 (r1 )| |ψn!0 (r2 )| e Jn! = d 3 r1 d 3 r 2 4π&0 |r1 − r2 | 2 % ∗ ∗ e ψ (r1 )ψn!0 (r2 )ψ100 (r2 )ψn!0 (r1 ) Kn! = d3 r1 d3 r2 100 . 4π&0 |r1 − r2 |
Physically, the term Jn! represents the electrostatic interaction energy associated with the two charge distributions |ψ100 (r1 )|2 and |ψn!0 (r2 )|2 , and it is clearly positive. By contrast, the exchange term, which derives from the antisymmetry of the wavefunction, leads to a shift with opposite signs for ortho and para states. In fact, one may show that, in the present case, Kn! is positive leading to a positive energy shift for parahelium and a negative shift for orthohelium. Moreover, noting that ' ( & 3 1/2 triplet 2 2 2 2 2 =! 2S1 · S2 = (S1 + S2 ) − S1 − S2 = ! S(S + 1) − 2 × −3/2 singlet 4 the energy shift can be written as p,o ∆En!
1 = Jn! − 2
&
' 4 1 + 2 S1 · S2 Kn! . !
This result shows that the electron-electron interaction leads to an effective ferromagnetic interaction between spins – i.e. the spins want to be aligned.6 In addition to the large energy shift between the singlet and triplet states, electric dipole decay selection rules ∆$ = ±1, ∆s = 0 (whose origin is discussed later in the course) cause decays from triplet to singlet states (or vice-versa) to be suppressed by a large factor (compared to decays from singlet to singlet or from triplet to triplet). This caused early researchers to think that there were two separate kinds of Helium. The diagrams (right) shows the levels for parahelium (singlet) and for orthohelium (triplet) along with the dominant decay modes.
8.4
Ideal quantum gases
An important and recurring example of a many-body system is provided by the problem of free (i.e. non-interacting) non-relativistic quantum particles in a closed system – a box. The many-body Hamiltonian is then given simply by ˆ0 = H
N " ˆ 2i p , 2m i=1
ˆ i = −i!∇i and m denotes the particle mass. If we take the dimensions where p of the box to be Ld , and the boundary conditions to be periodic7 (i.e. the box 6
A similar phenomenology extends to the interacting metallic system where the exchange interaction leads to the phenomenon of itinerant (i.e. mobile) ferromagnetism – Stoner ferromagnetism. 7 It may seem odd to consider such an unphysical geometry – in reality, we are invariably dealing with a closed system in which the boundary conditions translate to “hard walls” – think of electrons in a metallic sample. Here, we have taken the boundary conditions to be periodic since it leads to a slightly more simple mathematical formulation. We could equally well consider closed boundary conditions, but we would have to separately discriminate between “even and odd” states and sum them accordingly. Ultimately, we would arrive to the same qualitative conclusions!
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Energy level diagram for orthoand parahelium showing the first order shift in energies due to the Coulomb repulsion of electrons. Here we assume that one of the electrons stays close to the ground state of the unperturbed Hamiltonian.
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Figure 8.1: (Left) Schematic showing the phase space volume associated with each
plane wave state in a Fermi gas. (Right) Schematic showing the state occupancy of a filled Fermi sea.
has the geometry of a d-torus) the normalised eigenstates of the single-particle 1 Hamiltonian are simply given by plane waves, φk (r) = )r|k% = Ld/2 eik·r , with 8 wavevectors taking discrete values, k=
2π (n1 , n2 , · · · nd ), L
ni integer .
To address the quantum mechanics of the system, we start with fermions.
8.4.1
Non-interacting Fermi gas
In the (spinless) Fermi system, Pauli exclusion inhibits the multiple occupancy of single-particle states. In this case, the many-body ground state wavefunction is obtained by filling states sequentially up to the Fermi energy, EF = !2 kF2 /2m, where the Fermi wavevector, kF , is fixed by the number of particles. All the plane wave states φk with energies lower than EF are filled, while all states with energies larger than EF remain empty. Since each state is associated with a k-space volume (2π/L)d (see Fig. 8.1), in the three-dimensional system, the total number of occupied states is given by L 34 N = ( 2π ) 3 πkF3 , i.e. defining the particle density n = N/L3 = kF3 /6π 2 , EF =
2 !2 (6π 2 n) 3 . 2m
The density of states per unit volume, 1 dN 1 d g(E) = 3 = 2 L dE 6π dE
&
2mE !2
'3/2
1 = 2 4π
&
2m !2
'3/2
E 1/2 .
# Exercise. Obtain an expression for the density of states, g(E) in dimension d. In particular, show that the density of states varies as g(E) ∼ E (d−2)/2 . 8
ˆ+ The quantization condition follows form the periodic boundary condition, φ[r+L(mx x ˆ + mz z ˆ)] = φ(r), where m = (mx , my , mz ) denotes an arbitrary vector of integers. my y
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Note that the volume of a ddimensional sphere is given by 2π d/2 Sd = Γ(d/2) .
8.4. IDEAL QUANTUM GASES
85
We can also integrate to obtain the total energy density of all the fermions, Etot 1 = 3 3 L L
%
kF
0
!2 3 4πk 2 dk !2 k 2 = (6π 2 n)5/3 = nEF . 3 2 (2π/L) 2m 20π m 5
# Info. Degeneracy pressure: The pressure exerted by fermions squeezed into a small box is what keeps cold stars from collapsing. A White dwarf is a remnant of a normal star which has exhausted its fuel fusing light elements into heavier ones (mostly 56 Fe). As the star cools, it shrinks in size until it is arrested by the degeneracy pressure of the electrons. If the white dwarf acquires more mass, the Fermi energy EF rises until electrons and protons abruptly combine to form neutrons and neutrinos, an event known as a supernova. The neutron star left behind is supported by degeneracy pressure of the neutrons. We can compute the degeneracy pressure from analyzing the dependence of the energy on volume for a fixed number of particles (fermions). From thermodynamics, we have dE = F · ds = P dV , i.e. the pressure P = −∂V Etot . The expression for Etot given above shows that the pressure depends on the volume and particle number N only through the density, n. To determine the point of collapse of stars, we must compare this to the pressure exerted by gravity. We can compute this approximately, ignoring general relativity and, more significantly, the variation of gravitational pressure with radius. The mass contained within a shell of width dr at radius r is given by dm = 4πρr2 dr, where ρ denotes the density. This mass experiences a gravitational force from the mass contained within the shell, M = 4 3 3 πr ρ. The resulting potential energy is given by EG = −
%
GM dm =− r
%
R
d3 r
0
G( 43 πr3 ρ)4πr2 ρ (4π)2 2 5 3GM 2 =− Gρ R = − ,. r 15 5R
The mass of the star is dominated by nucleons, M = N MN , where MN denotes the nucleon mass and N their number. Substituting this expression into the formula for 4π 13 the energy, we find EG = − 35 G(N MN )2 ( 3V ) , from which we obtain the pressure, 1 PG = −∂V EG = − G(N MN )2 5
&
4π 3
'1/3
V −4/3 .
For the point of instability, this pressure must perfectly balance with the degeneracy pressure. For a white dwarf, the degeneracy pressure is associated with electrons and given by, Pe = −∂V Etot =
!2 (6π 2 Ne )5/3 V −5/3 . 60π 2 me
Comparing PG and Pe , we can infer the critical radius, 5/3
R≈
!2 N e 2 N2 . Gme MN
Since there are about two nucleons per electron, NN ≈ 2Ne , R .
!2 2 Gme MN 57
1
N−3 ,
showing that radius decreases as we add mass. For one solar mass, N = 10 , we get a radius of ca. 7200 km, the size of the Earth while EF . 0.2 MeV. During the cooling period, a white dwarf star will shrink in size approaching the critical radius. Since the pressure from electron degeneracy grows faster than the pressure of gravity, the star will stay at about Earth size even when it cools. If the star is more massive, the Fermi energy goes up and it becomes possible to absorb the electrons into the nucleons, converting protons into neutrons. If the electrons disappear this way, the star collapses suddenly down to a size for which the Fermi pressure of the neutrons stops the collapse. Some white dwarfs stay at Earth size for a long time as they suck in mass from their surroundings. When they have just enough mass, they collapse forming a neutron star and making a supernova. The supernovae are all nearly identical since the dwarfs are gaining mass very slowly. The brightness
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The image shows the crab pulsar, a magnetized neutron star spinning at 30 times per second, that resides at the centre of the crab nebula. The pulsar powers the X-ray and optical emission from the nebula, accelerating charged particles and producing the glowing X-ray jets. Ring-like structures are X-ray emitting regions where the high energy particles slam into the nebular material. The innermost ring is about a light-year across. With more mass than the Sun and the density of an atomic nucleus; the spinning pulsar is the collapsed core of a massive star that exploded, while the nebula is the expanding remnant of the star’s outer layers. The supernova explosion was witnessed in the year 1054. The lower image shows the typical size of a neutron star against Manhattan!
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86
of this type of supernova has been used to measure the accelerating expansion of the universe. An estimate the neutron star radius using the formulae above leads to Rneutron me . . 10−3 , Rwhite dwarf MN i.e. ca. 10 km. If the pressure at the center of a neutron star becomes too great, it collapses to become a black hole.
8.4.2
Ideal Bose gas
In a system of N spinless non-interacting bosons, the ground state of the many-body system is described simply by a wavefunction in which all N particles occupy the lowest energy single-particle state, i.e. in this case, the fully symmetrized ) wavefunction can be expressed as the product state, ψB (r1 , r2 , · · · rN ) = N i=1 φk=0 (ri ). (More generally, for a confining potential V (r), φk (r) denote the corresponding single-particle bound states with k the associated quantum numbers.) However, in contrast to the Fermi system, the transit to the ground state from non-zero temperatures has an interesting feature with intriguing experimental ramifications. To understand why, let us address the thermodynamics of the system. For independent bosons, the number of particles in plane wave state k with energy &k is given by the Bose-Einstein distribution, nk =
1 , e(#k −µ)/kB T − 1
* where the chemical potential, µ, is fixed by the condition N = k nk with N the total number of particles. In a three-dimensional system, for N large, we the sum by an integral over momentum space setting + * may Lapproximate 3 d3 k. As a result, we find that → " ( ) k 2π % N 1 1 = n = d3 k (# −µ)/k T . B L3 (2π)3 e k −1 For a free particle system, where &k = n=
!2 k2 2m ,
this means that
1 Li3/2 (µ/kB T ) , λ3T
(8.1)
* zk h2 1/2 where Lin (z) = ∞ k=1 kn denotes the polylogarithm, and λT = ( 2πmkB T ) denotes the thermal wavelength, i.e. the length scale at which the corresponding energy scale becomes comparable to temperature. As the density of particles increase, or the temperature falls, µ increases from negative values until, at some critical value of density, nc = λ−3 T ζ(3/2), µ becomes zero (note that Lin (0) = ζ(n)). Equivalently, inverting, this occurs at a temperature, kB Tc = α
!2 2/3 n , m
α=
2π . ζ 2/3 (3/2)
Clearly, since nk ≥ 0, the Bose distribution only makes sense for µ negative. So what happens at this point? Consider first what happens at zero temperature. Since the particles are bosons, the ground state consists of every particle sitting in the lowest energy
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Satyendra Nath Bose 1894-1974 An Indian physicist who is best known for his work on quantum mechanics in the early 1920s, providing the foundation for Bose-Einstein statistics and the development of the theory of the Bose-Einstein condensate. He is honored as the namesake of the boson.
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87
state (k = 0). But such a singular distribution was excluded by the replacement of the sum by the integral. Suppose that, at T < Tc , we have a thermodynamic fraction f (T ) of particles sitting in this state.9 Then the chemical potential may stay equal to zero and Eq. (8.1) becomes n = λ13 ζ(3/2)+f (T )n, T
where f (T ) denotes the fraction of particles in the ground state. But, since n = λ13 ζ(3/2), we have Tc
f (T ) = 1 −
&
λTc λT
'3
=1−
&
T Tc
'3/2
.
The unusual, highly quantum degenerate state emerging below Tc is known as a Bose-Einstein condensate (BEC).10 # Exercise. Ideal Bose gas in a harmonic trap: Show that, in an harmonic T )3 ], trap, V (r) = 21 mω 2 r2 , the corresponding relation is given by f (T ) = N [1 − ( TBEC 1/3 with kB Tc = !ω(N/ζ(3)) . # Info. Although solid state systems continue to provide the most “accessible” arena in which to study the properties of quantum liquids and gases, in recent years, a new platform has been realized through developments in atomic physics – in the field of ultracold atom physics, dilute atomic vapours are maintained at temperatures close to absolute zero, typically below some tenths of microkelvins, where their behaviour are influenced by the effects of quantum degneracy. The method of cooling the gas has a long history which it would be unwise to detail here. But in short, alkali atoms can be cooled by a technique known as laser cooling. Laser beams, in addition to carrying heat, also carry momentum. As a result, photons impart a pressure when they collide with atoms. The acceleration of an atom due to photons can be some four orders of magnitude larger than gravity. Consider then a geometry in which atoms are placed inside two counterprogating laser beams. To slow down, an atom has to absorb a photon coming towards it, and not from behind. This can be arranged by use of the Doppler shift. By tuning the laser frequency a little bit towards the lowfrequency (“red”) side of a resonance, the laser beam opposing the atom is Doppler shifted to a higher (more “blue”) frequency. Thus the atom is more likely to absorb that photon. A photon coming from behind the atom is now a little bit redder, which means the atom is less likely to absorb that photon. So in whichever direction the atom is moving, the laser beam opposing the motion seems stronger to the atom, and it slows the atom down. If you multiply this by three and have laser beams coming from the north, south, east, west, up, and down, you get an “optical molasse”. If you walk around in a pot full of molasses, whichever direction you go, the molasses somehow knows that is the direction to push against. It’s the same idea. In the study of ultracold atomic gases, experimentalists are usually concerned with addressing the properties of neutral alkali atoms. The number of atoms in a typical experiment ranges from 104 to 107 . The atoms are conned in a trapping potential of magnetic or optical origin, with peak densities at the centre of the trap ranging from 1013 cm3 to 1015 cm3 . The development of quantum phenomena such as Bose-Einstein condensation requires a phase-space density of order unity, or nλ3T ∼ 1 where λT denotes the thermal de Broglie wavelength. These densities correspond to temperatures, T ∼
!2 n2/3 ∼ 100nK to a few µK . mkB
9
Here, by thermodynamic, we mean that the fraction scales in proportion to the density. The Bose-Einstein condensate was first predicted by Satyendra Nath Bose and Albert Einstein in 1924-25. Bose first sent a paper to Einstein on the quantum statistics of light quanta (now called photons). Einstein was impressed, translated the paper himself from English to German and submitted it for Bose to the Zeitschrift fr Physik which published it. Einstein then extended Bose’s ideas to material particles (or matter) in two other papers. 10
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Note that, % ∞ x2 dx x = 2ζ(3) ∼ 2.404 . e −1 0
Schematic of a Magneto-Optical Trap (MOT). The invention of the MOT in 1987 at Bell Labs and optical molasses was the basis for the 1997 Nobel Prize in Physics.
8.4. IDEAL QUANTUM GASES
88
Figure 8.2: (Left) Observation of BEC by absorption imaging. The top row shows
shadow pictures, which are rendered in a 3d plot below. The “sharp peak” is the BEC, characterized by its slow expansion observed after 6 ms time of flight. The total number of atoms at the phase transition is about 7 × 105 , and the temperature at the transition point is 2 µK. (Figure from Ketterle.) (right) Figure shows the shrinking of the atom cloud in a magnetic as the temperature is reduced by evaporative cooling. Comparison between bosonic 7 Li (left) and fermionic 6 Li (right) shows the distinctive signature of quantum statistics. The fermionic cloud cannot shrink below a certain size determined by the Pauli exclusion principle. This is the same phenomenon that prevents white dwarf and neutron stars from shrinking into black holes. At the highest temperature, the length of the clouds was about 0.5mm. (Figure from J. R. Anglin and W. Ketterle, Bose-Einstein condensation of atomic gases, Nature 416, 211 (2002).) , At these temperatures the atoms move at speeds of ∼ kB T /m ∼ 1 cm s−1 , which should be compared with around 500 ms1 for molecules at room temperature, and ∼ 106 ms−1 for electrons in a metal at zero temperature. Achieving the regime n3T ∼ 1, through sufficient cooling, is the principle experimental advance that gave birth to this new field of physics. It should be noted that such low densities of atoms are in fact a necessity. We are dealing with systems whose equilibrium state is a solid (that is, a lump of Sodium, Rubidium, etc.). The rst stage in the formation of a solid would be the combination of pairs of atoms into diatomic molecules, but this process is hardly possible without the involvement of a third atom to carry away the excess energy. The rate per atom of such three-body processes is 1029 − 10−30 cm6 s−1 , leading to a lifetime of several seconds to several minutes. These relatively long timescales suggest that working with equilibrium concepts may be a useful first approximation. Since the alkali elements have odd atomic number Z, we readily see that alkali atoms with odd mass number are bosons, and those with even mass number are fermions. Thus bosonic and fermionic alkalis have half-integer and integer nuclear spin respectively. Alkali atoms have a single valence electron in an ns state, so have electronic spin J = S = 1/2. The experimental star players: are shown in the table (right). The hyperfine coupling between electronic and nuclear spin splits the ground state manifold into two multiplets with total spin F = I ± 1/2. The Zeeman splitting of these multiplets by a magnetic field forms the basis of magnetic trapping. So, based on our discussion above, what happens when a Bose or Fermi gas is confined by an harmonic trapping potential, V (r) = 12 mω 2 r2 . At high temperatures, Bose and Fermi gases behave classically and form a thermal (Gaussian) distribution, 2 2 P (r) . e−mω r /2kB T . As the system is cooled towards the point of quantum degeneracy, i.e. when the typical separation between particles, n−1/3 , becomes comparable to the thermal wavelength, λT , quantum statistics begin to impact. In the Fermi system, Pauli exclusion leads to the development of a Fermi surface and the cloud size becomes arrested. By contrast, the Bose system can form a BEC, with atoms condensing into the ground state of the harmonic potential. Both features are shown in Figure 8.2.
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Bosons Li I=3/2 23 Na I=3/2 87 Rb I=3/2 7
Fermions Li I=1 23 K I=4 6
Chapter 9
Atomic structure Previously, we have seen that the quantum mechanics of atomic hydrogen, and hydrogen-like atoms is characterized by a large degeneracy with eigenvalues separating into multiplets of n2 -fold degeneracy, where n denotes the principle quantum number. However, although the idealized Schr¨odinger Hamiltonian, ˆ2 ˆ0 = p H + V (r), 2m
V (r) = −
1 Ze2 , 4π"0 r
(9.1)
provides a useful platform from which develop our intuition, there are several important effects which mean that the formulation is a little too na¨ıve. These “corrections”, which derive from several sources, are important as they lead to physical ramifications which extend beyond the realm of atomic physics. Here we outline some of the effects which need to be taken into account even for atomic hydrogen, before moving on to discuss the quantum physics of multielectron atoms. In broad terms, the effects to be considered can be grouped into those caused by the internal properties of the nucleus, and those which derive from relativistic corrections. To orient our discussion, it will be helpful to summarize some of key aspects ˆ 0 ψ = Eψ on of the solutions of the non-relativisitic Schr¨odinger equation, H which we will draw: Hydrogen atom revisited: $ As with any centrally symmetric potential, the solutions of the Schr¨odinger equation take the form ψ!m (r) = R(r)Y!m (θ, φ), where the spherical harmonic functions Y!m (θ, φ) depend only on spherical polar coordinates, and R(r) represents the radial component of the wavefunction. Solving the radial wave equation introduces a radial quantum number, nr ≥ 0. In the case of a Coulomb potential, the energy depends on the principal quantum number n = nr + ' + 1 ≥ 1, and not on nr and ' separately. $ For atomic hydrogen (Z = 1), the energy levels of the Hamiltonian (9.1) are given by ! 2 " Ry e m e2 1 1 En = − 2 , Ry = = = mc2 α2 , 2 n 4π"0 2! 4π"0 2a0 2 4π#0 !2 e2 m
e2 1 4π#0 !c
where a0 = is the Bohr radius, α = denotes the fine structure constant, and strictly speaking m represents the reduced mass of the electron and proton. Applied to single electron ions with higher atomic weight, such as He+ , Li2+ , etc., the Bohr radius is reduced by a factor 1/Z, where Z denotes the nuclear charge, and the energy is 2 given by En = − Zn2 Ry = − 2n1 2 mc2 (Zα)2 . Advanced Quantum Physics
The fine structure constant is known to great accuracy and is given by, α = 7.297352570(5) × 10−3 1 = . 137.035999070(9)
9.1. THE “REAL” HYDROGEN ATOM $ Since n ≥ 1 and nr ≥ 0, the allowed combinations of quantum numbers are shown on the right, where we have introduced the conventional notation whereby values of ' = 0, 1, 2, 3, 4 · · · are represented by letters s, p, d, f, g · · · respectively. $ Since En depends only on n, this implies, for example, an exact degeneracy of the 2s and 2p, and of the 3s, 3p and 3d levels. These results emerge from a treatment of the hydrogen atom which is inherently non-relativistic. In fact, as we will see later in our discussion of the Dirac equation in chapter 15, the Hamiltonian (9.1) represents only the leading term in an expansion in v 2 /c2 $ (Zα)2 of the full relativistic Hamiltonian (see below). Higher order terms provide relativistic corrections, which impact significantly in atomic and condensed matter physics, and lead to a lifting of the degeneracy. In the following we will discuss and obtain the heirarchy of leading relativistic corrections.1 This discussion will provide a platform to describe multi-electron atoms.
9.1
The “real” hydrogen atom
The relativistic corrections (sometimes known as the fine-structure corrections) to the spectrum of hydrogen-like atoms derive from three different sources: $ relativistic corrections to the kinetic energy; $ coupling between spin and orbital degrees of freedom; $ and a contribution known as the Darwin term. In the following, we will discuss each of these corrections in turn.
9.1.1
Relativistic correction to the kinetic energy
Previously, we have taken the kinetic energy to have the familiar non-relativistic ˆ2 p form, 2m . However, from the expression for the relativistic energy-momentum invariant, we can already anticipate that the leading correction to the nonrelativistic Hamiltonian appears at order p4 , E = (pµ pµ )1/2 =
# p2 1 (p2 )2 p2 c2 + m2 c4 = mc2 + − + ··· 2m 8 m3 c2
As a result, we can infer the following perturbation to the kinetic energy of the electron, p ) ˆ 1 = − 1 (ˆ H . 8 m3 c2 2 2
When compared with the non-relativistic kinetic energy, p2 /2m, one can see that the perturbation is smaller by a factor of p2 /m2 c2 = v 2 /c2 $ (Zα)2 , i.e. ˆ 1 is only a small perturbation for small atomic number, Z ( 1/α $ 137. H 1 It may seem odd to discuss relativistic corrections in advance of the Dirac equation and the relativistic formulation of quantum mechanics. However, such a discussion would present a lengthy and unnecessarily complex digression which would not lead to further illumination. We will therefore follow the normal practice of discussing relativistic corrections as perturbations to the familiar non-relativistic theory.
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90
n 1 2 3 4 n
' 0 0, 1 0, 1, 2 0, 1, 2, 3 0 · · · (n − 1)
Subshell(s) 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f ns · · ·
To see that v 2 /c2 $ (Zα)2 , we may invoke the virial theorem. The latter shows that the average kinetic energy is related to the potential energy as %T & = − 12 %V &. Therefore, the average energy is given by %E& = %T & + %V & = −%T & ≡ − 21 mv 2 . We therefore have that 12 mv 2 = Ry ≡ 21 mc2 (Zα)2 from which follows the relation v 2 /c2 $ (Zα)2 .
9.1. THE “REAL” HYDROGEN ATOM
91
We can therefore make use of a perturbative analysis to estimate the scale of the correction. In principle, the large-scale degeneracy of the hydrogen atom would demand an analysis based on the degenerate perturbation theory. However, fortunately, since the off-diagonal matrix elements vanish,2 ˆ 1 |n'! m! & = 0 %n'm|H
for ' )= '!
or m )= m! ,
degenerate states are uncoupled and such an approach is unnecessary. Then ˆ 1 = − 1 2 [H ˆ 0 − V (r)]2 , the scale of the resulting making use of the identity, H 2mc energy shift can be obtained from first order perturbation theory, '2 & 2 ˆ 1 &n!m ≡ %n'm|H ˆ 1 |n'm& = − 1 %H En − 2En %V (r)&n! + %V 2 (r)&n! . 2 2mc
Since the calculation of the resulting expectation values is not particularly illuminating, we refer to the literature for a detailed exposition3 and present here only the required identities (right). From these considerations, we obtain the following expression for the first order energy shift, ! " ! " mc2 Zα 4 n 3 ˆ %H1 &n!m = − − . (9.2) 2 n ' + 1/2 4 From this term alone, we expect the degerenacy between states with different values of total angular momentum ' to be lifted. However, as well will see, this conclusion is a little hasty. We need to gather all terms of the same order of perturbation theory before we can reach a definite conclusion. We can, however, confirm that (as expected) the scale of the correction is of order ˆ 1 #n!m "H En
9.1.2
2 ∼ ( Zα n ) . We now turn to the second important class of corrections.
Spin-orbit coupling
As well as revealing the existence of an internal spin degree of freedom, the relativistic formulation of quantum mechanics shows that there is a further relativistic correction to the Schr¨odinger operator which involves a coupling between the spin and orbital degrees of freedom. For a general potential V (r), this spin-orbit coupling takes the form, ˆ2 = H
1 1 ˆ ·L ˆ. (∂r V (r)) S 2m2 c2 r 2
1 Ze For a hydrogen-like atom, V (r) = − 4π# , and 0 r
ˆ2 = H
1 Ze2 1 ˆ ˆ S · L. 2m2 c2 4π"0 r3
$ Info. Physically, the origin of the spin-orbit interaction can be understoon from the following considerations. As the electron is moving through the electric field of the nucleus then, in its rest frame, it will experience this as a magnetic field. There will be an additional energy term in the Hamiltonian associated with the orientation of the spin magnetic moment with respect to this field. We can make an estimate of the spin-orbit interaction energy as follows: If we have a central field determined by an electrostatic potential V (r), the corresponding electric field is given 2 ˆ 2 ] = 0, !2 [!! (!! + 1) − !(! + 1)] "n!m|H ˆ1, L ˆ 1 |n!! m! # = The proof runs as follows: Since [H ˆ1, L ˆ z ] = 0, !(m! − m)"n!m|H ˆ 1 |n!! m! # = 0. 0. Similarly, since [H 3 see, e.g., Ref [1].
Advanced Quantum Physics
By making use of the form of the radial wavefunction for the hydrogen atom, one may obtain the identities, $ % 1 Z = r n! a0 n2 $ % 1 Z2 . = 2 3 2 r n! a0 n (' + 1/2)
9.1. THE “REAL” HYDROGEN ATOM
92
by E = −∇V (r) = −er (∂r V ). For an electron moving at velocity v, this translates to an effective magnetic field B = c12 v × E. The magnetic moment of the electron q e associated with its spin is equal to µs = gs 2m S ≡ −m S, and thus the interaction energy is given by −µs · B =
e e 1 e S · (v × E) = − S · (p × er (∂r V )) = (∂r V ) L · S , 2 2 mc (mc) (mc)2 r
where we have used the relation p × er = p × rr = − Lr . In fact this isn’t quite correct; there is a relativistic effect connected with the precession of axes under rotation, called Thomas precession which multiplies the formula by a further factor of 12 .
ˆ2 Once again, we can estimate the effect of spin-orbit coupling by treating H as a perturbation. In the absence of spin-orbit interaction, one may express the eigenstates of hydrogen-like atoms in the basis states of the mutually ˆ 2, L ˆ 2 , and Sˆz . However, in the presence ˆ 0, L ˆz, S commuting operators, H ˆz of spin-orbit coupling, the total Hamiltonian no longer commutes with L or Sˆz (exercise). It is therefore helpful to make use of the degeneracy of the unperturbed Hamiltonian to switch to a new basis in which the angular momentum components of the perturbed system are diagonal. This can be ˆ 2 , Jˆz , ˆ 0, J achieved by turning to the basis of eigenstates of the operators, H 2 2 ˆ , and S ˆ , where J ˆ =L ˆ +S ˆ denotes the total angular momentum. (For a L discussion of the form of these basis states, we refer back to section 6.4.2.) ˆ2 = L ˆ2 + S ˆ 2 + 2L ˆ · S, ˆ in this basis, it follows Making use of the relation, J that, ˆ ·S ˆ = 1 (J ˆ2 − L ˆ2 − S ˆ 2) . L 2 Combining the spin and angular momentum, the total angular momentum takes values j = ' ± 1/2. The corresponding basis states |j = ' ± 1/2, mj , '& (with s = 1/2 implicit) therefore diagonalize the operator, ! " !2 ' ˆ ˆ S · L|j = ' ± 1/2, mj , '& = |' ± 1/2, mj , '& , −' − 1 2
where the brackets index j = ' + 1/2 (top) and j = ' − 1/2 (bottom). As for the radial dependence of the perturbation, once again, the off-diagonal matrix elements vanish circumventing the need to invoke degenerate perturbation theory. As a result, at first order in perturbation theory, one obtains ! " $ % 1 !2 Ze2 1 ' . %H2 &n,j=!±1/2,mj ,! = 2 2 −' − 1 2m c 2 4π"0 r3 n! Then making use of the identity (right),4 one obtains ( ) ! " 1 1 2 Zα 4 n j ˆ %H2 &n,j=!±1/2,mj ,! = mc . 1 − j+1 4 n j + 1/2
Note that, for ' = 0, there is no orbital angular momentum with which to ˆ 1 & (9.2) in the new basis, couple! Then, if we rewrite the expression for %H ! "4 ( 1 ) 1 Zα j ˆ 1 &n,j=!±1/2,m ,! = − mc2 %H n , 1 j 2 n j+1 and combining both of these expressions, for ' > 0, we obtain ! " ! " 1 2 Zα 4 3 n ˆ ˆ %H1 + H2 &n,j=!±1/2,mj ,! = mc − , 2 n 4 j + 1/2
while for ' = 0, we retain just the kinetic energy term (9.2). 4
For details see, e.g., Ref [1].
Advanced Quantum Physics
For ' > 0, $
1 r3
%
= n!
!
mcαZ !n
"3
1 . '(' + 12 )(' + 1)
9.1. THE “REAL” HYDROGEN ATOM
9.1.3
93
Darwin term
The final contribution to the Hamiltonian from relativistic effects is known as the Darwin term and arises from the “Zitterbewegung” of the electron – trembling motion – which smears the effective potential felt by the electron. Such effects lead to a perturbation of the form, ! " 2 2 e Ze2 !2 ˆ 3 = ! ∇2 V = ! H Q (r) = 4πδ (3) (r) , nuclear 2 2 2 2 8m c 8m c "0 4π"0 8(mc)2 where Qnuclear (r) = Zeδ (3) (r) denotes the nuclear charge density. Since the perturbation acts only at the origin, it effects only states with ' = 0. As a result, one finds that 2 !2 1 ˆ 3 &njm ! = Ze %H 4π|ψ!n (0)|2 = mc2 j 2 4π"0 8(mc) 2
!
Zα n
"4
nδ!,0 .
Intriguingly, this term is formally identical to that which would be obtained ˆ 2 & at ' = 0. As a result, combining all three contributions, the total from %H energy shift is given simply by 1 ∆En,j=!±1/2,mj ,! = mc2 2
!
Zα n
"4 !
3 n − 4 j + 1/2
"
,
(9.3)
a result that is independent of ' and mj . To discuss the predicted energy shifts for particular states, it is helpful to introduce some nomenclature from atomic physics. For a state with principle quantum number n, total spin s, orbital angular momentum ', and total angular momentum j, one may use spectroscopic notation n2s+1 Lj to define the state. For a hydrogen-like atom, with just a single electron, 2s + 1 = 2. In this case, the factor 2s + 1 is often just dropped for brevity. If we apply our perturbative expression for the relativistic corrections (9.3), how do we expect the levels to shift for hydrogen-like atoms? As we have seen, for the non-relativistic Hamiltonian, each state of given n exhibits a 2n2 -fold degeneracy. For a given multiplet specified by n, the relativistic corrections depend only on j and n. For n = 1, we have ' = 0 and j = 1/2: Both 1S1/2 states, with mj = 1/2 and −1/2, experience a negative energy shift by an amount ∆E1,1/2,mj ,0 = − 41 Z 4 α2 Ry. For n = 2, ' can take the values of 0 or 1. With j = 1/2, both the former 2S1/2 state, and the latter 2P1/2 states share 5 4 2 the same negative shift in energy, ∆E2,1/2,mj ,0 = ∆E2,1/2,mj ,1 = − 64 Z α Ry, 1 4 2 while the 2P3/2 experiences a shift of ∆E2,3/2,mj ,1 = − 64 Z α Ry. Finally, for n = 3, ' can take values of 0, 1 or 2. Here, the pairs of states 3S1/2 and 3P1/2 , and 3P3/2 and 2D3/2 each remain degenerate while the state 3D5/2 is unique. These predicted shifts are summarized in Figure 9.1. This completes our discussion of the relativistic corrections which develop from the treatment of the Dirac theory for the hydrogen atom. However, this does not complete our discription of the “real” hydrogen atom. Indeed, there are further corrections which derive from quantum electrodynamics and nuclear effects which we now turn to address.
9.1.4
Lamb shift
According to the perturbation theory above, the relativistic corrections which follow from the Dirac theory for hydrogen leave the 2S1/2 and 2P1/2 states degenerate. However, in 1947, a careful experimental study by Willis Lamb Advanced Quantum Physics
Willis Eugene Lamb, 1913-2008 A physicist who won the Nobel Prize in Physics in 1955 “for his discoveries concerning the fine structure of the hydrogen spectrum”. Lamb and Polykarp Kusch were able to precisely determine certain electromagnetic properties of the electron.
9.1. THE “REAL” HYDROGEN ATOM
94
Figure 9.1:
Figure showing the heirarchy of energy shifts of the spectra of hydrogen-like atoms as a result of relativistic corrections. The first column shows the energy spectrum predicted by the (non-relativistic) Bohr theory. The second column shows the predicted energy shifts from relativistic corrections arising from the Dirac theory. The third column includes corrections due quantum electrodynamics and the fourth column includes terms for coupling to the nuclear spin degrees of freedom. The H-α line, particularly important in the astronomy, corresponds to the transition between the levels with n = 2 and n = 3.
and Robert Retherford discovered that this was not in fact the case:5 2P1/2 state is slightly lower in energy than the 2S1/2 state resulting in a small shift of the corresponding spectral line – the Lamb shift. It might seem that such a tiny effect would be deemed insignificant, but in this case, the observed shift (which was explained by Hans Bethe in the same year) provided considerable insight into quantum electrodynamics. In quantum electrodynamics, a quantized radiation field has a zero-point energy equivalent to the mean-square electric field so that even in a vacuum there are fluctuations. These fluctuations cause an electron to execute an oscillatory motion and its charge is therefore smeared. If the electron is bound by a non-uniform electric field (as in hydrogen), it experiences a different potential from that appropriate to its mean position. Hence the atomic levels are shifted. In hydrogen-like atoms, the smearing occurs over a length scale, %(δr)2 & $
2α π
!
! mc
"2
ln
1 , αZ
some five orders of magnitude smaller than the Bohr radius. This causes the electron spin g-factor to be slightly different from 2, ! " α α2 gs = 2 1 + − 0.328 2 + · · · . 2π π
Hans Albrecht Bethe 1906-2005 A GermanAmerican physicist, and Nobel laureate in physics “for his work on the theory of stellar nucleosynthesis.” A versatile theoretical physicist, Bethe also made important contributions to quantum electrodynamics, nuclear physics, solid-state physics and particle astrophysics. During World War II, he was head of the Theoretical Division at the secret Los Alamos laboratory developing the first atomic bombs. There he played a key role in calculating the critical mass of the weapons, and did theoretical work on the implosion method used in both the Trinity test and the “Fat Man” weapon dropped on Nagasaki.
There is also a slight weakening of the force on the electron when it is very close to the nucleus, causing the 2S1/2 electron (which has penetrated all the way to the nucleus) to be slightly higher in energy than the 2P1/2 electron. Taking into account these corrections, one obtains a positive energy shift ∆ELamb
! "4 ! " Z 8 1 2 $ nα Ry × α ln δ!,0 , n 3π αZ
for states with ' = 0. 5
W. E. Lamb and R. C. Retherfod, Fine Structure of the Hydrogen Atom by a Microwave Method, Phys. Rev. 72, 241 (1947).
Advanced Quantum Physics
Hydrogen fine structure and hyperfine structure for the n = 3 to n = 2 transition (see Fig. 9.1).
9.1. THE “REAL” HYDROGEN ATOM
9.1.5
Hyperfine structure
So far, we have considered the nucleus as simply a massive point charge responsible for the large electrostatic interaction with the charged electrons which surround it. However, the nucleus has a spin angular momentum which is associated with a further set of hyperfine corrections to the atomic spectra of atoms. As with electrons, the protons and neutrons that make up a nucleus are fermions, each with intrinsic spin 1/2. This means that a nucleus will have some total nuclear spin which is labelled by the quantum number, I. The latter leads to a nuclear magnetic moment, µN = gN
Ze I. 2MN
where MN denotes the mass of the nucleus, and gN denotes the gyromagnetic ratio. Since the nucleus has internal structure, the nuclear gyromagnetic ratio is not simply 2 as it (nearly) is for the electron. For the proton, the sole nuclear constituent of atomic hydrogen, gP ≈ 5.56. Even though the neutron is charge neutral, its gyromagnetic ratio is about −3.83. (The consitituent quarks have gyromagnetic ratios of 2 (plus corrections) like the electron but the problem is complicated by the strong interactions which make it hard to define a quark’s mass.) We can compute (to some accuracy) the gyromagnetic ratio of nuclei from that of protons and neutrons as we can compute the proton’s gyromagnetic ratio from its quark constituents. Since the nuclear mass is several orders of magnitude higher than that of the electron, the nuclear magnetic moment provides only a small perturbation. According to classical electromagnetism, the magnetic moment generates a magnetic field B=
µ0 2µ0 (3(µN · er )er − µN ) + µ δ (3) (r) . 4πr3 3 N
To explore the effect of this field, let us consider just the s-electrons, i.e. ' = 0, for simplicity.6 In this case, the interaction of the magnetic moment of the electrons with the field generated by the nucleus, gives rise to the hyperfine interaction, ˆ · B. ˆ hyp = −µe · B = e S H mc For the ' = 0 state, the first contribution to the first order correction, ! "4 Z %Hhyp &n,1/2,0 = nα2 Ry × n
B vanishes while second leads to 8 m 1 gN S · I. 3 MN ! 2
Once again, to evaluate the expectation values on the spin degrees of freedom, it is convenient to define the total spin F = I + S. We then have 1 1 1 S · I = 2 (F2 − S2 − I2 ) = (F (F + 1) − 3/4 − I(I + 1)) 2 ! 2 *2! 1 I F = I + 1/2 = 2 −I − 1 F = I − 1/2 Therefore, the 1s state of Hydrogen is split into two, corresponding to the two possible values F = 0 and 1. The transition between these two levels has frequency 1420 Hz, or wavelength 21 cm, so lies in the radio waveband. It 6
For a full discussion of the influence of the orbital angular momentum, we refer to Ref. [6].
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9.2. MULTI-ELECTRON ATOMS
96
is an important transition for radio astronomy. A further contribution to the hyperfine structure arises if the nuclear shape is not spherical thus distorting the Coulomb potential; this occurs for deuterium and for many other nuclei. Finally, before leaving this section, we should note that the nucleus is not point-like but has a small size. The effect of finite nuclear size can be estimated perturbatively. In doing so, one finds that the s (' = 0) levels are those most effected, because these have the largest probability of finding the electron close to the nucleus; but the effect is still very small in hydrogen. It can be significant, however, in atoms of high nuclear charge Z, or for muonic atoms. This completes our discussion of the “one-electron” theory. We now turn to consider the properties of multi-electron atoms.
9.2
Multi-electron atoms
To address the electronic structure of a multi-electron atom, we might begin with the hydrogenic energy levels for an atom of nuclear charge Z, and start filling the lowest levels with electrons, accounting for the exclusion principle. The degeneracy for quantum numbers (n, ') is 2 × (2' + 1), where (2' + 1) is the number of available m! values, and the factor of 2 accounts for the spin degeneracy. Hence, the number of electrons accommodated in shell, n, would be 2 × n2 , n 1 2 3 4
' 0 0, 1 0, 1, 2 0, 1, 2, 3
Degeneracy in shell 2 (1 + 3) × 2 = 8 (1 + 3 + 5) × 2 = 18 (1 + 3 + 5 + 7) × 2 = 32
Cumulative total 2 10 28 60
We would therefore expect that atoms containing 2, 10, 28 or 60 electrons would be especially stable, and that in atoms containing one more electron than this, the outermost electron would be less tightly bound. In fact, if we look at data (Fig. 9.2) recording the first ionization energy of atoms, i.e. the minimum energy needed to remove one electron, we find that the noble gases, having Z = 2, 10, 18, 36 · · · are especially tightly bound, and the elements containing one more electron, the alkali metals, are significantly less tightly bound. The reason for the failure of this simple-minded approach is fairly obvious – we have neglected the repulsion between electrons. In fact, the first ionization energies of atoms show a relatively weak dependence on Z; this tells us that the outermost electrons are almost completely shielded from the nuclear charge.7 Indeed, when we treated the Helium atom as an example of the variational method in chapter 7, we found that the effect of electron-electron repulsion was sizeable, and really too large to be treated accurately by perturbation theory. 7
In fact, the shielding is not completely perfect. For a given energy shell, the effective nuclear charge varies for an atomic number Z as Zeff ∼ (1 + α)Z where α > 0 characterizes 2 the ineffectiveness of screening. This implies that the ionization energy IZ = −EZ ∼ Zeff ∼ (1 + 2αZ). The near-linear dependence of IZ on Z is reflected in Fig. 9.2.
Advanced Quantum Physics
A muon is a particle somewhat like an electron, but about 200 times heavier. If a muon is captured by an atom, the corresponding Bohr radius is 200 times smaller, thus enhancing the nuclear size effect.
9.2. MULTI-ELECTRON ATOMS
97
Figure 9.2: Ionization energies of the elements.
9.2.1
Central field approximation
Leaving aside for now the influence of spin or relativistic effects, the Hamiltonian for a multi-electron atom can be written as - + + , !2 1 e2 1 Ze2 2 ˆ + H= − ∇i − , 2m 4π"0 ri 4π"0 rij i<j
i
where rij ≡ |ri − rj |. The first term represents the “single-particle” contribution to the Hamiltonian arising from interaction of each electron with the nucleus, while the last term represents the mutual Coulomb interaction between the constituent electrons. It is this latter term that makes the generic problem “many-body” in character and therefore very complicated. Yet, as we have already seen in the perturbative analysis of the excited states of atomic Helium, this term can have important physical consequences both on the overall energy of the problem and on the associated spin structure of the states. The central field approximation is based upon the observation that the electron interaction term contains a large central (spherically symmetric) component arising from the “core electrons”. From the following relation, ! +
m=−!
|Ylm (θ, φ)|2 = const.
it is apparent that a closed shell has an electron density distribution which is isotropic (independent of θ and φ). We can therefore develop a perturbative ˆ =H ˆ0 + H ˆ 1 , where scheme by setting H , + 1 e2 + + 1 Ze2 !2 2 ˆ1 = ˆ ∇i − + Ui (ri ) , H − Ui (ri ) . H0 = − 2m 4π"0 ri 4π"0 rij i
i<j
i
Here the one-electron potential, Ui (r), which is assumed central (see below), incorporates the “average” effect of the other electrons. Before discussing how ˆ 0 is separable into a sum of to choose the potentials Ui (r), let us note that H terms for each electron, so that the total wavefunction can be factorized into components for each electron. The basic idea is first to solve the Schr¨odinger ˆ 0 , and then to treat H ˆ 1 as a small perturbation. equation using H ˆ 0 continues to commute with On general grounds, since the Hamiltonian H ˆ = 0, we can see that the eigenfuncˆ 0 , L] the angular momentum operator, [H ˆ 0 will be characterized by quantum numbers (n, ', m! , ms ). However, tions of H Advanced Quantum Physics
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since the effective potential is no longer Coulomb-like, the ' values for a given n need not be degenerate. Of course, the difficult part of this procedure is to estimate Ui (r); the potential energy experienced by each electron depends on the wavefunction of all the other electrons, which is only known after the Schr¨odinger equation has been solved. This suggests that an iterative approach to solving the problem will be required. To understand how the potentials Ui (r) can be estimated – the selfconsistent field method – it is instructive to consider a variational approach due originally to Hartree. If electrons are considered independent, the wavefunction can be factorized into the product state, Ψ({ri }) = ψi1 (r1 )ψi2 (r2 ) · · · ψiN (rN ) , where the quantum numbers, ik ≡ (n'm! ms )k , indicate the individual state occupancies. Note that this product state is not a properly antisymmetrized Slater determinant – the exclusion principle is taken into account only in so far as the energy of the ground state is taken to be the lowest that is consistent with the assignment of different quantum numbers, n'm! ms to each electron. Nevertheless, using this wavefunction as a trial state, the variational energy is then given by ! 2 2 " +. ! ∇ 1 Ze2 3 ∗ ˆ E = %Ψ|H|Ψ& = d r ψi − − ψi 2m 4π"0 r i . . e2 1 + 3 + ψj (r! )ψi (r) . d r d3 r! ψi∗ (r)ψi∗ (r! ) 4π"0 |r − r! | i<j
Now, according to the variational principle, we must minimize the energy functional by varying E[{ψi }] with respect to the complex wavefunction, ψi , subject to the normalization condition, %ψi |ψi & = 1. The latter can be imposed using a set of Lagrange multipliers, εi , i.e. , !. "δ 3 2 E − εi d r|ψi (r)| − 1 = 0. δψi∗
Following the variation,8 one obtains the Hartree equations, ! 2 2 " . ! ∇ 1 Ze2 1 + e2 − − ψi + d3 r! |ψj (r! )|2 ψi (r) 2m 4π"0 r 4π"0 |r − r! | j&=i
= εi ψi (r) .
(9.4)
Then according to the variational principle, amongst all possible trial functions ψi , the set that minimizes the energy are determined by the effective potential, . 1 + e2 Ui (r) = d3 r! |ψj (r! )|2 . 4π"0 |r − r! | j&=i
Equation (9.4) has a simple interpretation: The first two terms relate to the nuclear potential experienced by the individual electrons, while the third term represents the electrostatic potential due to the other electrons. However, to simplify the procedure, it is useful to engineer the radial symmetry of the potential by replacing Ui (r) by its spherical average, . dΩ Ui (r) -→ Ui (r) = Ui (r) . 4π Note that, in applying the variation, the wavefunction ψi∗ can be considered independent of ψi – you might like to think why. 8
Advanced Quantum Physics
Douglas Rayner Hartree FRS 1897-1958 An English mathematician and physicist most famous for the development of numerical analysis and its application to atomic physics. He entered St John’s College Cambridge in 1915 but World War I interrupted his studies and he joined a team studying anti-aircraft gunnery. He returned to Cambridge after the war and graduated in 1921 but, perhaps because of his interrupted studies, he only obtained a second class degree in Natural Sciences. In 1921, a visit by Niels Bohr to Cambridge inspired him to apply his knowledge of numerical analysis to the solution of differential equations for the calculation of atomic wavefunctions.
9.2. MULTI-ELECTRON ATOMS Finally, to relate the Lagrange multipliers, εi (which have the appearance of one-electron energies), to the total energy, we can multiply Eq. (9.4) by ψi∗ (r) and integrate, ! 2 2 " . ! ∇ 1 Ze2 3 ∗ "i = d r ψi − − ψi 2m 4π"0 r . 1 + e2 + d3 r! d3 r |ψj (r! )|2 |ψi (r)|2 . 4π"0 |r − r! | j&=i
If we compare this expression with the variational state energy, we find that . + 1 + e2 E= "i − d3 r! d3 r |ψj (r! )|2 |ψi (r)|2 . (9.5) 4π"0 |r − r! | i
i<j
To summarize, if we wish to implement the central field approximation to determine the states of a multi-electron atom, we must follow the algorithm: 1. Firstly, one makes an initial “guess” for a (common) central potential, U (r). As r → 0, screening becomes increasingly ineffective and we ex1 (Z−1)e2 pect U (r) → 0. As r → ∞, we anticipate that U (r) → 4π# , r 0 corresponding to perfect screening. So, as a starting point, we make take some smooth function U (r) interpolating between these limits. For this trial potential, we can solve (numerically) for the eigenstates of the single-particle Hamiltonian. We can then use these states as a platform to build the product wavefunction and in turn determine the selfconsistent potentials, Ui (r). 2. With these potentials, Ui (r), we can determine a new set of eigenstates for the set of Schr¨odinger equations, , !2 2 1 Ze2 − ∇ − + Ui (r) ψi = εi ψi . 2m 4π"0 r 3. An estimate for the ground state energy of an atom can be found by filling up the energy levels, starting from the lowest, and taking account of the exclusion principle. 4. Using these wavefunctions, one can then make an improved estimate of the potentials Ui (ri ) and return to step 2 iterating until convergence. Since the practical implemention of such an algorithm demands a large degree of computational flair, if you remain curious, you may find it useful to refer to the Mathematica code prepared by Ref. [4] where both the Hartree and the Hartree-Fock procedures (described below) are illustrated. $ Info. An improvement to this procedure, known the Hartree-Fock method, takes account of exchange interactions. In order to do this, it is necessary to ensure that the wavefunction, including spin, is antisymmetric under interchange of any pair of electrons. This is achieved by introducing the Slater determinant. Writing the individual electron wavefunction for the ith electron as ψk (ri ), where i = 1, 2 · · · N and k is shorthand for the set of quantum numbers (n'm! ms ), the overall wavefunction is given by / / / ψ1 (r1 ) ψ1 (r2 ) ψ1 (r3 ) · · · / / / 1 // ψ2 (r1 ) ψ2 (r2 ) ψ2 (r3 ) · · · // Ψ = √ / ψ3 (r1 ) ψ3 (r2 ) ψ3 (r3 ) · · · / . / N ! // .. .. .. . . // / . . . .
Advanced Quantum Physics
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9.2. MULTI-ELECTRON ATOMS
100
Note that each of√the N ! terms in Ψ is a product of wavefunctions for each individual electron. The 1/ N ! factor ensures the wavefunction is normalized. A determinant changes sign if any two columns are exchanged, corresponding to ri ↔ rj (say); this ensures that the wavefunction is antisymmetric under exchange of electrons i and j. Likewise, a determinant is zero if any two rows are identical; hence all the ψk s must be different and the Pauli exclusion principle is satisfied.9 In this approximation, a variational analysis leads to the Hartree-Fock equations (exercise), , !2 2 Ze2 εi ψi (r) = − ∇i − ψi (r) 2m 4π"ri 0 1 +. 1 e2 ∗ # # # d 3 rj + ψ (r ) ψ (r )ψ (r) − ψ (r)ψ (r )δ . j i j i m ,m j s s i j 4π"0 |r − r# | j"=i
The first term in the last set of brackets translates to the ordinary Hartree contribution above and describes the influence of the charge density of the other electrons, while the second term describes the non-local exchange contribution, a manifestation of particle statistics.
The outcome of such calculations is that the eigenfunctions are, as for hydrogen, characterized by quantum numbers n, ', m! , with ' < n, but that the states with different ' for a given n are not degenerate, with the lower values of ' lying lower. This is because, for the higher ' values, the electrons tend to lie further from the nucleus on average, and are therefore more effectively screened. The states corresponding to a particular value of n are generally referred to as a shell, and those belonging to a particular pair of values of n, ' are usually referred to as a subshell. The energy levels are ordered as below (with the lowest lying on the left): Subshell name n= '= Degeneracy Cumulative
1s 1 0 2 2
2s 2 0 2 4
2p 2 1 6 10
3s 3 0 2 12
3p 3 1 6 18
4s 4 0 2 20
3d 3 2 10 30
4p 4 1 6 36
5s 5 0 2 38
4d 4 2 10 48
··· ··· ··· ··· ···
Note that the values of Z corresponding to the noble gases, 2, 10, 18, 36, at which the ionization energy is unusually high, now emerge naturally from this filling order, corresponding to the numbers of electrons just before a new shell (n) is entered. There is a handy mnemonic to remember this filling order. By writing the subshells down as shown right, the order of states can be read off along diagonals from lower right to upper left, starting at the bottom. We can use this sequence of energy levels to predict the ground state electron configuration of atoms. We simply fill up the levels starting from the lowest, accounting for the exclusion principle, until the electrons are all accommodated (the aufbau principle). Here are a few examples: Z 1 2 3 4 5 6 7 8 9 10 11
Element H He Li Be B C N O F Ne Na
Configuration (1s) (1s)2 He (2s) He (2s)2 He (2s)2 (2p) He (2s)2 (2p)2 He (2s)2 (2p)3 He (2s)2 (2p)4 He (2s)2 (2p)5 He (2s)2 (2p)6 Ne (3s)
9
2S+1 2
LJ
S1/2 S0 2 S1/2 1 S0 2 P1/2 3 P0 4 S3/2 3 P2 2 P3/2 1 S0 2 S1/2 1
Ioniz. Pot. (eV) 13.6 24.6 5.4 9.3 8.3 11.3 14.5 13.6 17.4 21.6 5.1
Note that for N = 2, the determinant reduces to the familiar antisymmetric wavefunction, √12 [ψ1 (r1 )ψ2 (r2 ) − ψ2 (r1 )ψ1 (r2 )].
Advanced Quantum Physics
Vladimir Aleksandrovich Fock 1898-1974 A Soviet physicist, who did foundational work on quantum mechanics and quantum electrodynamics. His primary scientific contribution lies in the development of quantum physics, although he also contributed significantly to the fields of mechanics, theoretical optics, theory of gravitation, physics of continuous medium. In 1926 he derived the Klein-Gordon equation. He gave his name to Fock space, the Fock representation and Fock state, and developed the HartreeFock method in 1930. Fock made significant contributions to general relativity theory, specifically for the many body problems.
! " ! ! " ! ! ··· 7p 7s ! ! 7d ! " ! 6p ! 6s ! 6d 6f · · · ! ! ! ! " ! 5p 5d 5s ! ! 5f ! 5g ! " ! ! ! ! ! 4s 4p ! 4d ! 4f ! ! " ! ! ! !! 3p 3s ! 3d ! ! ! " ! ! !! 2p 2s ! ! ! " ! !! 1s ! ! !
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Figure 9.3: Periodic table of elements. Since it is generally the outermost electrons which are of most interest, contributing to chemical activity or optical spectra, one often omits the inner closed shells, and just writes O as (2p)4 , for example. However, the configuration is not always correctly predicted, especially in the heavier elements, where levels may be close together. It may be favourable to promote one or even two electrons one level above that expected in this simple picture, in order to achieve a filled shell. For example, Cu (Z = 29) would be expected to have configuration · · · (4s)2 (3d)9 , and actually has configuration · · · (4s)1 (3d)10 . There are several similar examples in the transition elements where the d subshells are being filled, and many among the lanthanides (rare earths) and actinides where f subshells are being filled. $ Info. Since the assignment of an electron configuration requires only the enumeration of the values of n and ' for all electrons, but not those of m! and ms , each configuration will be accompanied by a degeneracy g. If νn! denotes the number of electrons occupying a given level En,! , and δ! = 2 × (2' + 1) is the degeneracy of that level, there are dn! =
δ! ! νn! !(δ! − νn! )!
(9.6)
ways of distributing the νn! electrons among the δ! individual states. The total degeneracy, g, is then obtained from the product.
This scheme provides a basis to understand the periodic table of elements (see Fig. 9.3). We would expect that elements which have similar electron configurations in their outermost shells (such as Li, Na, K, Rb, Cs, Fr which all have (ns)1 or F, Cl, Br, I, which all have (np)5 ) would have similar chemical properties, such as valency, since it is the unpaired outer electrons which especially participate in chemical bonding. Therefore, if one arranges the atoms in order of increasing atomic number Z (which equals the number of electrons in the atom), periodic behaviour is seen whenever a new subshell of a given ' is filled. Advanced Quantum Physics
9.3. COUPLING SCHEMES
9.3
Coupling schemes
The procedure outlined above allows us to predict the occupation of subshells in an atomic ground state. This is not in general sufficient to specify the ground state fully. If there are several electrons in a partially filled subshell, then their spins and orbital angular momenta can combine in several different ways, to give different values of total angular momentum, with different energies. In order to deal with this problem, it is necessary to consider the spin-orbit interaction as well as the residual Coulomb interaction between the outer electrons. Schematically we can write the Hamiltonian for this system as follows: ˆ ≈H ˆ0 + H
+ + 1 e2 + ˆi · S ˆi , − Ui (r) + ξi (ri )L 4π"0 rij i i i<j 34 5 2 34 5 2 ˆ ˆ H H1 2
ˆ 0 includes the kinetic energy and central field terms, H ˆ 1 is the residual where H 1 1 ˆ Coulomb interaction, and (with ξi (ri ) = 2m2 c2 r (∂r V (r))) H2 is the spin-orbit interaction. We can then consider two possible scenarios: ˆ1 2 H ˆ 2 : This tends to apply in the case of light atoms. In this situation, H ˆ0 + H ˆ 1 , and then treats H ˆ 2 as a one considers first the eigenstates of H perturbation. This leads to a scheme called LS (or Russell-Saunders) coupling. ˆ2 2 H ˆ 1 : This can apply in very heavy atoms, or in heavily ionized light H atoms, in which the electrons are moving at higher velocities and relativistic effects such as the spin-orbit interaction are more important. In this case, a scheme called jj coupling applies. It is important to emphasise that both of these scenarios represent approximations; real atoms do not always conform to the comparatively simple picture which emerges from these schemes, which we now discuss in detail.
9.3.1
LS coupling scheme
ˆ0 + H ˆ 1 . We In this approximation, we start by considering the eigenstates of H note that this Hamiltonian must commute with the total angular momentum ˆ 2 (because of invariance under rotations in space), and also clearly commutes J ˆ 2 . It also commutes with the total orbital angular momenwith the total spin S 2 ˆ ˆ tum L , since H1 only involves internal interactions, and must therefore be invariant under global rotation of all the electrons. Therefore the energy levels can be characterised by the corresponding total angular momentum quantum numbers L, S, J. Their ordering in energy is given by Hund’s rules: 1. Combine the spins of the electrons to obtain possible values of total spin S. The largest permitted value of S lies lowest in energy. 2. For each value of S, find the possible values of total angular momentum L. The largest value of L lies lowest in energy. 3. Couple the values of L and S to obtain the values of J (hence the name of the scheme). If the subshell is less than half full, the smallest value of J lies lowest; otherwise, the largest value of J lies lowest.
Advanced Quantum Physics
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In deciding on the permitted values of L and S, in addition to applying the usual rules for adding angular momenta, one also has to ensure that the exclusion principle is respected, as we will see later when considering some examples. These rules are empirical; there are exceptions, especially to the L and J rules (2 and 3). Nevertheless, Hund’s rules are a useful guide, and we should try to understand their physical origin. 1. Maximising S makes the spin wavefunction as symmetric as possible. This tends to make the spatial wavefunction antisymmetric, and hence reduces the Coulomb repulsion, as we saw when discussing the exchange interactions in Helium. 2. Maximising L also tends to keep the electrons apart. This is less obvious, though a simple classical picture of electrons rotating round the nucleus in the same or different senses makes it at least plausible. 3. The separation of energies for states of different J arises from treating ˆ 2 as a perturbation (fine structure). It can be the spin-orbit term H shown (using the Wigner-Eckart theorem – beyond the scope of these lectures) that + ˆi · S ˆ i |J, mJ , L, S& ξi (ri )L %|J, mJ , L, S| i
ˆ · S|J, ˆ mJ , L, S& = ζ(L, S)%J, mJ , L, S|L ζ(L, S) = [J(J + 1) − L(L + 1) − S(S + 1)] , 2
(9.7)
where the matrix element ζ(L, S) depends on the total L and S values. Since one may show that the sign of ζ(L, S) changes according to the whether the subshell is more or less than half-filled, the third Hund’s rule is established. To understand the application of LS coupling, it is best to work through some examples. Starting with the simplest multi-electron atom, helium, the ground state has an electron configuration (1s)2 , and must therefore have L = S = J = 0. In fact, for any completely filled subshell, we have L = S = 0 and hence J = 0, since the total mL and mS must equal zero if all substates are occupied. Consider now an excited state of helium, e.g. (1s)1 (2p)1 , in which one electron has been excited to the 2p level. We can now have S = 1 or S = 0, with the S = 1 state lying lower in energy according to Hund’s rules. Combining the orbital angular momenta of the electrons yields L = 1 and thus, with S = 0, J = 1, while with S = 1, J = 0, 1, 2 with J = 0 lying lowest in energy. Once again, as with the hydrogen-like states, we may index the states of multi-electron atoms by spectroscopic term notation, 2S+1 LJ . The superscript 2S + 1 gives the multiplicity of J values into which the level is split by the spin-orbit interaction; the L value is represented by a capital letter, S, P , D, etc., and J is represented by its numerical value. Thus, for the (1s)1 (2p)1 state of helium, there are four possible states, with terms: 3
P0 3 P1 3 P2
1
P1 ,
where the three 3 P states are separated by the spin-orbit interaction, and the singlet 1 P state lies much higher in energy owing to the Coulomb interaction. The separations between the 3 P2 and 3 P1 and the 3 P1 and 3 P0 should be in the Advanced Quantum Physics
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ratio 2:1. This is an example of the Land´ e interval rule, which states that the separation between a pair of adjacent levels in a fine structure multiplet is proportional to the larger of the two J values involved. This is easily shown using Eq. (9.7) – the separation in energy between states J and J − 1 is ∝ J(J + 1) − (J − 1)J = 2J . Actually in the case of helium the situation is a bit more complicated, because it turns out that the spin-orbit interaction between different electrons makes a non-negligible additional contribution to the fine structure. Other excited states of helium, of the form (1s)1 (n')1 , can be handled similarly, and again separate into singlet and triplet states. $ Exercise. For the case of boron, with the electron configuration (1s)2 (2s)2 (2p), use Hund’s rules to show that the ground state is 2 P1/2 .
We next consider the case of carbon, which has ground state electron configuration (1s)2 (2s)2 (2p)2 . This introduces a further complication; we now have two identical electrons in the same unfilled subshell, and we need to ensure that their wavefunction is antisymmetric with respect to electron exchange. The total spin can either be the singlet S = 0 state, which has an antisymmetric wavefunction √12 [| ↑1 & ⊗ | ↓2 & − | ↓1 & ⊗ | ↑2 &], or one of the triplet
S = 1 states, which are symmetric, √12 [| ↑1 & ⊗ | ↓2 & + | ↓1 & ⊗ | ↑2 &], | ↑1 & ⊗ | ↑2 & or | ↓1 & ⊗ | ↓2 &. We must therefore choose values of L with the appropriate symmetry to partner each value of S. To form an antisymmetric state, the two electrons must have different values of m! , so the possibilities are as shown right. Inspecting the values of mL we can deduce that L = 1.10 By contrast, to form a symmetric total angular momentum state, the two electrons may have any values of m! , leading to the possibilities shown right. Inspecting the values of mL we infer that L = 2 or 0. We must therefore take S = 1 with L = 1 and S = 0 with L = 2 or 0. Finally, to account for the fine structure, we note that the states with S = 1 and L = 1 can be combined into a single J = 0 state, three J = 1 states, and five J = 2 states leading to the terms 3 P0 , 3 P1 , and 3 P2 respectively. Similarly the S = 0, L = 2 state can be combined to give five J = 2 states, 1 D2 , while the S = 0, L = 0 state gives the single J = 0 state, 1 S0 . Altogther we recover the 1 + 3 + 5 + 5 + 1 = 15 possible states (cf. Eq. (9.6) with the ordering in energy given by Hund’s rules (shown to the right). The experimental energy values are given using the conventional spectroscopic units of inverse wavelength. Note that the Land´e interval rule is approximately obeyed by the fine structure triplet, and that the separation between L and S values caused by the electronelectron repulsion is much greater than the spin-orbit effect. In an excited state of carbon, e.g. (2p)1 (3p)1 , the electrons are no longer equivalent, because they have different radial wavefunctions. So now one can combine any of S = 0, 1 with any of L = 0, 1, 2, yielding the following terms (in order of increasing energy, according to Hund’s rules): 3
D1,2,3
3
P0,1,2
3
S1
1
D2
1
P1
1
S0 .
For nitrogen, the electron configuration is given by (1s)2 (2s)2 (2p)3 . The maximal value of spin is S = 3/2 while L can take values 3, 2, 1 and 0. Since 10 This result would also be apparent if we recall the that angular momentum states are eigenstates of the parity operator with eigenvalue (−1)L . Since there are just two electrons, this result shows that both the L = 0 and L = 2 wavefunction must be symmetric under exchange.
Advanced Quantum Physics
(1)
m!
1 1 0
(1)
m!
1 1 1 0 0 −1
S0 D2 3 P2 3 P1 3 P0 1
1
(2)
m!
0 −1 −1 (2)
m!
1 0 −1 0 −1 −1
mL 1 0 −1 mL 2 1 0 0 −1 −2
E /cm−1 20649 10195 43 16 0
9.3. COUPLING SCHEMES
105
the spin wavefunction (being maximal) is symmetric, the spatial wavefunction must be completely antisymmetric. This demands that all three states with m! = 1, 0, −1 must be involved. We must therefore have L = 0, leading to J = 3/2 and the term, 4 S3/2 . $ Exercise. Construct the L = 0 state involving the addition of three ' = 1 angular momentum states. Hint: make use of the total antisymmetry condition. As a final example, let us consider the ground state of oxygen, which has electron configuration (2p)4 . Although there are four electrons in the (2p) subshell, the maximum value of S = 1. This is because there are only three available values of m! = ±1, 0, and therefore one of these must contain two electrons with opposite spins. Therefore, the maximum value of mS = 1, achieved by having electrons with ms = + 12 in both the other m! states. By pursuing this argument, it is quite easy to see that the allowed values of L, S and J are the same as for carbon (2p)2 . This is in fact a general result – the allowed quantum numbers for a subshell with n electrons are the same as for that of a subshell with n “holes”. Therefore, the energy levels for the oxygen ground state configuration are the same as for carbon, except that the fine structure multiplet is inverted, in accordance with Hund’s third rule.
9.3.2
jj coupling scheme
When relativitic effects take precedence over electron interaction effects, we ˆi · S ˆ i. ˆ0 + H ˆ2 = H ˆ 0 + 6 ξi (ri )L must start by considering the eigenstates of H i 2 ˆ These must be eigenstates of J as before, because of the overall rotational ˆ 2 for each electron. Therefore, in this case, the couinvariance, and also of J i pling procedure is to find the allowed j values of individual electrons, whose energies will be separated by the spin-orbit interaction. Then these individual j values are combined to find the allowed values of total J. The effect of the residual Coulomb interaction will be to split the J values for a given set of js. Sadly, in this case, there are no simple rules to parallel those of Hund. As an example, consider a configuration (np)2 in the jj coupling scheme, to be compared with the example of carbon which we studied in the LS scheme. Combining s = 1/2 with ' = 1, each electron can have j = 1/2 or 3/2. If the electrons have the same j value, they are equivalent, so we have to take care of the symmetry of the wavefunction. We therefore have the following possibilities: $ j1 = j2 = 3/2 ⇒ J = 3, 2, 1, 0, of which J = 2, 0 are antisymmetric. $ j1 = j2 = 1/2 ⇒ J = 1, 0, of which J = 0 is antisymmetric. $ j1 = 1/2, j2 = 3/2 ⇒ J = 2, 1. In jj coupling, the term is written (j1 , j2 )J , so we have the following terms in our example: (1/2, 1/2)0
(3/2, 1/2)1 (3/2, 1/2)2
(3/2, 3/2)2 (3/2, 3/2)0
in order of increasing energy. Note that both LS and jj coupling give the same values of J (in this case, two states with J = 0, two with J = 2 and one with J = 1) and in the same order. However, the pattern of levels is different; in LS coupling we found a triplet and two singlets, while in this ideal jj scenario, we have two doublets and a singlet. The sets of states in the two coupling Advanced Quantum Physics
Level scheme of the carbon atom (1s)2 (2s)2 (2p)2 . Drawing is not to scale. On the left the energy is shown without any two-particle interaction. The electron-electron interaction leads to a three-fold energy splitting with L and S remaining good quantum numbers. Spinorbit coupling leads to a further splitting of the states with J remaining a good quantum number. Finally on the right, the levels show Zeeman splittings in an external magnetic field. In this case, the full set of 15 levels become non-degenerate.
9.4. ATOMIC SPECTRA
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schemes must be expressible as linear combinations of one another, and the physical states for a real atom are likely to differ from either approximation. In fact, this idealized form of jj coupling is not seen in the heaviest such atom in the periodic table, lead (6p)2 . However, it is seen in some highly ionized states, for example in Cr18+ , which has the same electron configuration (2p)2 as carbon, but where, because of the larger unscreened charge on the nucleus, the electrons are moving more relativistically, enhancing the spin-orbit effect. However, a classic example of the transition from LS to jj coupling is seen in the series C-Si-Ge-Sn-Pb in the excited states (2p)(3s), (3p)(4s), · · · (6p)(7s) (see figure right). Here, the electrons are not in the same subshell, so their wavefunctions overlap less, and the Coulomb repulsion is reduced compared to the spin-orbit interaction. Analysing this situation in the LS coupling approximation, one expects a triplet and a singlet: 3
P0,1,2
1
P1 ,
while in the jj scheme one expects two doublets: (1/2, 1/2)0,1
(1/2, 3/2)2,1 .
Experimentally, C and Si conform to the LS expectation and Pb to the jj scheme, while Ge and Sn are intermediate.
9.4
Atomic spectra
Atomic spectra result from transitions between different electronic states of an atom via emission or absorption of photons. In emission spectra, an atom is excited by some means (e.g. thermally through collisions, or by an electric discharge), and one observes discrete spectral lines in the light emitted as the atoms relax. In absorption spectra, one illuminates atoms using a broad waveband source, and observes dark absorption lines in the spectrum of transmitted light. Of course the atoms excited in this process subsequently decay by emitting photons in random directions; by observing in directions away from the incident light this fluorescence radiation may be studied. The observation of these spectral lines is an important way of probing the atomic energy levels experimentally. In the case of optical spectra and the nearby wavebands, the excited states responsible generally involve the excitation of a single electron from the ground state to some higher level. In some cases, it may simply involve a different coupling of the angular momenta within the same electron configuration. These are the kinds of excitations which we are about to discuss. However, other types of excitations are also possible. For example, X-ray emission occurs when an electron has been removed from one of the innermost shells of a heavy atom; as electrons cascade down to fill the hole, high energy photons may be emitted. The basic theory governing stimulated emission and absorption, and spontaneous emission of photons will be outlined in detail when we study radiative transitions in chapter 13. Here we must anticipate some of the basic conclusions of that study. In the electric dipole approximation, the rate of transitions is proportional to the square of the matrix element of the electric ˆ i &|2 . In addition, dipole operator between the initial and final states, |%ψf |d|ψ 3 the rate of spontaneous transitions is proportional to ω , where ω = |Ef − Ei | denotes the energy separation between the states. ˆ means that the matrix elements may The form of the dipole operator, d vanish identically. This leads to a set of selection rules defining which transitions are allowed. Here we consider the simplest case of a single electron, but Advanced Quantum Physics
9.4. ATOMIC SPECTRA the principles can be generalized. Referring to chapter 13 for a more detailed discussion, one finds that, for a transition to take place: $ Parity must change; $ ∆J = ±1, 0 (but 0 → 0 is not allowed) and ∆MJ = ±1, 0 . Atomic states are always eigenstates of parity and of total angular momentum, J, so these selection rules can be regarded as absolutely valid in electric dipole transitions. It should be emphasized again, though, that the electric dipole approximation is an approximation, and higher order processes may occur, albeit at a slower rate, and have their own selection rules. In specific coupling schemes, further selection rules may apply. In the case of ideal LS coupling, we also require: $ ∆S = 0 and ∆MS = 0; $ ∆L = ±1, 0 (but 0 → 0 is not allowed) and ∆ML = ±1, 0; $ and ∆'i = ±1 if only electron i is involved in the transition. In LS coupling, the states are eigenstates of total spin; since the dipole operator does not operate on the spin part of the wavefunction, the rules on ∆S and ∆MS follow straightforwardly. This, and the absolute rules relating to J, imply the rules for L and ML . The rule for ∆'i follows from the parity change rule, since the parity of the atom is the product of the parities of the separate electron wavefunctions, given by (−1)!i . However, since LS coupling is only an approximation, these rules should themselves be regarded as approximate. With this preparation, we now turn to the consequences of the selection rules on the atomic spectra of atoms.
9.4.1
Single electron atoms
In this context, “single electron atoms” refer to atoms whose ground state consists of a single electron in an s level, outside closed shells; it is this electron which is active in optical spectroscopy. Our discussion therefore encompasses the alkali metals, such as sodium, and also hydrogen. We take sodium, whose ground state configuration is (3s)1 , as our example: $ The ground state has term 2 S1/2 . The excited states are all doublets with J = L ± 1/2, except for the s states, which are obviously restricted to J = 1/2. $ The parity is given by (−1)!i , so the allowed transitions involve changes in ' by ±1 unit, i.e. s ↔ p, p ↔ d, d ↔ f , etc. Changes of more than one unit in ' would fall foul of the ∆J rule. $ The s ↔ p transitions are all doublets. All the doublets starting or ending on a given p state have the same spacing in energy. The transition 3s ↔ 3p gives rise to the familiar yellow sodium “D-lines” at 589 nm (see right). $ The p ↔ d transitions involve two doublets, 2 P1/2,3/2 and 2 D3/2,5/2 . However, the 2 P1/2 ↔2 D5/2 transition is forbidden by the ∆J rule, so the line is actually a triplet. In practice, the spin-orbit interaction falls quite rapidly with increasing ' (and with increasing n) as the effect of screening increases, so that the effect of the 2 D3/2,5/2 splitting may not be resolved experimentally. Advanced Quantum Physics
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9.4. ATOMIC SPECTRA $ As n increases, the energy levels approach (from below) those for hydrogen, because the nuclear charge is increasingly effectively screened by the inner electrons. This happens sooner for the higher ' values, for which the electron tends to lie further out from the nucleus. $ In an absorption spectrum, the atoms will start from the ground state, so only the 3s → np lines will be seen. In emission, the atoms are excited into essentially all their excited levels, so many more lines will be seen in the spectrum. The comments above for sodium also apply for hydrogen, except that, in this case, (2s, 2p), (3s, 3p, 3d), etc. are degenerate. One consequence is that the 2s state in hydrogen is metastable – it cannot decay to the only lower lying level (1s) by an electric dipole transition. In fact its favoured spontaneous decay is by emission of two photons; a process which is described by second-order perturbation theory. In practice, hydrogen atoms in a 2s state are more likely to deexcite through collision processes. During an atomic collision, the atoms are subject to strong electric fields, and we know from our discussion of the Stark effect that this will mix the 2s and 2p states, and decay from 2p to 1s is readily possible.
9.4.2
Helium and alkali earths
We next discuss atoms whose ground state consists of two electrons in an s level. Our discussion therefore covers helium (1s)2 , and the alkali earths: beryllium (2s)2 , magnesium (3s)2 , calcium (4s)2 , etc. We start with helium. $ The ground state has term 1 S0 . The excited states are of the form (1s)(n') (the energy required to excite both of the 1s electrons to higher states is greater than the first ionization energy, and therefore these form discrete states within a continuum of ionized He+ +e− states). The excited states can have S = 0 or S = 1, with S = 1 lying lower in energy (Hund). $ The LS coupling approximation is pretty good for helium, so the ∆S = 0 selection rule implies that the S = 0 and S = 1 states form completely independent systems as far as spectroscopy is concerned. $ The lines in the S = 0 system are all singlets. They can be observed in emission, and those starting from the ground state can be seen in absorption. $ The lines in the S = 1 system are all multiplets. They can be observed in emission only. Transitions of the form 3 S1 ↔3 P2,1,0 are observed as triplets, spaced according to the Land´e interval rule. Transitions of the form 3 P2,1,0 ↔3 D3,2,1 are observed as sextuplets, as is easily seen by application of the ∆J = ±1, 0 rule. Actually, as mentioned above, the fine structure is a little more subtle in the case of helium. The alkali earths follow the same principles. In the case of calcium, the triplet 4p state is the lowest lying triplet state, and therefore metastable. In fact a faint emission line corresponding to the 3 P1 →1 S0 decay to the ground state may be observed; this violates the ∆S = 0 rule, and indicates that the LS coupling approximation is not so good in this case. A more extreme example is seen in Mercury, ground state (6s)2 (5d)10 . Excited states involving promotion of one of the 6s electrons to a higher level can be treated just like the alkali Advanced Quantum Physics
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9.5. ZEEMAN EFFECT earths. In this case the “forbidden” 3 P1 →1 S0 is actually a prominent feature of the emission spectrum in the visible, implying a significant breakdown of the LS approximation.
9.4.3
Multi-electron atoms
Similar principles can be used to make sense of the spectra of more complicated atoms, though unsurprisingly their structure is more complex. For example, carbon, with ground state (2s)2 (2p)2 , corresponds to terms 3 P0,1,2 , 1 D2 and 1 S as discussed above. The excited states are of the form (2s)2 (2p)1 (n')1 , 0 and can be separated into singlets and triplets, and in addition excitations of the form (2s)1 (2p)3 can arise. Nitrogen, with ground state (2s)2 (2p)3 , has three unpaired electrons, so the ground state and excited states form doublets (S = 1/2) and quartets (S = 3/2) with correspondingly complex fine structure to the spectral lines.
9.5 9.5.1
Zeeman effect Single-electron atoms
Before leaving this chapter on atomic structure, we are now in a position to revisit the question of how atomic spectra are influenced by a magnetic field. To orient our discussion, let us begin with the study of hydrogen-like atoms involving just a single electron. In a magnetic field, the Hamiltonian of such ˆ = H ˆ0 + H ˆ rel. + H ˆ Zeeman , where H ˆ 0 denotes the a system is described by H ˆ non-relativistic Hamiltonian for the atom, Hrel. incorporates the relativistic corrections considered earlier in the chapter, and ˆ Zeeman = − eB (L ˆ z + 2Sˆz ) , H 2mc denotes the Zeeman energy associated with the coupling of the spin and orbital angular momentum degrees of freedom to the magnetic field. Here, since we are dealing with confined electrons, we have neglected the diamagnetic contribution to the Hamiltonian. Depending on the scale of the magnetic field, ˆ rel. or the Zeeman term may dominate the spectrum the spin-orbit term in H of the atom. Previously, we have seen that, to leading order, the relativistic corrections lead to a fine-structure energy shift of ! " ! " 1 2 Zα 4 3 n rel. ∆En,j = mc − , 2 n 4 j + 1/2
for states |n, j = ' ± 1/2, mj , '&. For weak magnetic fields, we can also treat the Zeeman energy in the framework of perturbation theory. Here, although states with common j values (such as 2S1/2 and 2P1/2 ) are degenerate, the two spatial wavefunctions have different parity (' = 0 and 1 in this case), ˆ Zeeman coupling these states vanishes. and the off-diagonal matrix element of H We may therefore avoid using degenerate perturbation theory. Making use of the relation (exercise – refer back to the discussion of the addition of angular momenta and spin in section 6.4.2),
!mj , 2' + 1 we obtain the following expression for the first order energy shift, ! " 1 Zeeman ∆Ej=!±1,mj ,! = ' ± 1/2, mj , '& = µB Bmj 1 ± , 2' + 1 %n, j = ' ± 1/2, mj , '|Sz |n, j = ' ± 1/2, mj , '& = ±
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9.5. ZEEMAN EFFECT
Figure 9.4: The well known doublet which is responsible for the bright yellow light
from a sodium lamp may be used to demonstrate several of the influences which cause splitting of the emission lines of atomic spectra. The transition which gives rise to the doublet is from the 3p to the 3s level. The fact that the 3s state is lower than the 3p state is a good example of the dependence of atomic energy levels on orbital angular momentum. The 3s electron penetrates the 1s shell more and is less effectively shielded than the 3p electron, so the 3s level is lower. The fact that there is a doublet shows the smaller dependence of the atomic energy levels on the total angular momentum. The 3p level is split into states with total angular momentum J = 3/2 and J = 1/2 by the spin-orbit interaction. In the presence of an external magnetic field, these levels are further split by the magnetic dipole energy, showing dependence of the energies on the z-component of the total angular momentum.
where µB denotes the Bohr magneton. Therefore, we see that all degenerate levels are split due to the magnetic field. In contrast to the “normal” Zeeman effect, the magnitude of the splitting depends on '. $ Info. If the field is strong, the Zeeman energy becomes large in comparison with the spin-orbit contribution. In this case, we must work with the basis states ˆ 0 and H ˆ Zeeman are diagonal. Within |n, ', m! , ms & = |n, ', m! & ⊗ |ms & in which both H first order of perturbation theory, one then finds that (exercise) ! "4 ! " 1 Zα 3 n nm! ms ∆En,!,m! ,ms = µB (m! + ms ) + mc2 − − , 2 n 4 ' + 1/2 '(' + 1/2)(' + 1) ˆ rel. . At the first term arising from the Zeeman energy and the remaining terms from H intermediate values of the field, we have to apply degenerate perturbation theory to the states involving the linear combination of |n, j = '±1/2, mj , '&. Such a calculation reaches beyond the scope of these lectures and, for details, we refer to the literature (see, e.g., Ref. [6]. Let us instead consider what happens in multi-electron atoms.
9.5.2
Multi-electron atoms
For a multi-electron atom in a weak magnetic field, the appropriate unperturbed states are given by |J, MJ , L, S&, where J, L, S refer to the total angular momenta. To determine the Zeeman energy shift, we need to determine the matrix element of Sˆz . To do so, we can make use of the following Advanced Quantum Physics
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ˆ ·S ˆ=J ˆ2 − L ˆ2 − S ˆ 2 , this operator is diagonal in the basis of argument. Since 2L states, |J, MJ , L, S&. Therefore, the matrix element of the operator (exercise, ˆ i , Sˆk ] = 0), hint: recall that [Sˆi , Sˆj ] = i!"ijk Sˆk and [L ˆ ×L ˆ ≡ S( ˆ L ˆ · S) ˆ − (L ˆ · S) ˆ S ˆ −i!S ˆ · S, ˆ J] ˆ = 0, it follows that the must vanish. Moreover, from the identity [L matrix element of the vector product, ˆ × L) ˆ ×J ˆ=S ˆ × J( ˆ L ˆ · S) ˆ − (L ˆ · S) ˆ S ˆ ×J ˆ, −i!(S must also vanish. If we expand the left hand side, we thus find that the matrix element of ˆ × L) ˆ ×J ˆ = L( ˆ S ˆ · J) ˆ − S( ˆ L ˆ · J) ˆ (S
ˆ J− ˆ S ˆ L=
=
ˆ S ˆ · J) ˆ −S ˆJ ˆ2 , J(
ˆJ ˆ 2 & = %J( ˆ S ˆ · J)&, ˆ where the expecalso vanishes. Therefore, it follows that %S ˆ ˆ = 1 (J ˆ2 + S ˆ2 − L ˆ 2 ), tation value is taken over the basis states. Then, with S · J 2 we have that %Sˆz & = %Jˆz &
J(J + 1) + S(S + 1) − L(L + 1) . 2J(J + 1)
As a result, we can deduce that, at first order in perturbation theory, the energy shift arising from the Zeeman term is given by ∆EJ,MJ ,L,S = µB gJ MJ B , where gJ = 1 +
J(J + 1) + S(S + 1) − L(L + 1) , 2J(J + 1)
denotes the effective Land´ e g-factor, which lies between 1 and 2. Note that, in the special case of hydrogen, where S = 1/2 and J = L ± 1/2, we recover our previous result. The predicted Zeeman splitting for sodium is shown in figure 9.4. $ Info. In the strong field limit, where the influence of Zeeman term dominates, the appropriate basis states are set by |L, ML , S, MS &, in which the operators ˆ2, L ˆ z , Sˆ2 , Sˆz , and H ˆ Zeeman are diagonal. In this case, the energy splitting takes the L form ! "4 1 Zα nML MS ∆EL,ML ,S,MS = µB B(ML + 2MS ) + mc2 , 2 n '(' + 1/2)(' + 1) where the second term arises from the spin-orbit interaction.
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Figure 9.5: In the weak field
case, the vector model (top) implies that the coupling of the orbital angular momentum L to the spin angular momentum S is stronger than their coupling to the external field. In this case where spin-orbit coupling is dominant, they can be visualized as combining to form a total angular momentum J which then precesses about the magnetic field direction. In the strong field case, S and L couple more strongly to the external magnetic field than to each other, and can be visualized as independently precessing about the external field direction.
Chapter 10
From molecules to solids In the previous section, we studied the quantum mechanics of multi-electron – the subject of atomic physics. In this section, we will begin to explore how these concepts get translated into systems with many atoms, from simple molecular structures to the solid state. As with atomic physics, the subjects of molecular and solid state physics represent fields in their own right and it would fanciful to imagine that we could do more than touch on the basic principles. Nevertheless, in establishing the foundations of these subjects, we will see two things: firstly, by organising the heirarchy of approximation schemes, much be can be understood about the seemingly complex quantum dynamics of many-particle systems. Secondly, we will find that constraints imposed by symmetries (such as translation) allow a simple phenomenology to emerge from complex solid state systems. We begin our discussion with the molecular system. A molecule consists of electrons moving in the complicated potential set up by all the constituent electric charges. Even in classical mechanics, it would be extremely difficult to solve the equations of motion of the internal molecular degrees of freedom. Fortunately, for most purposes, we can treat the motion of the electrons and nuclei separately, due to their very different masses. As the forces on a nucleus are similar in magnitude to those that act on an electron, so the electrons and nuclei must have comparable momenta. Therefore the nuclei are typically moving much more slowly than the electrons. In studying the motion of the electrons, we can therefore treat the nuclei as being “nailed down” in fixed positions. Conversely, in studying the nuclear motion (vibrations and rotations of the molecule) we can assume, as a first approximation, that the electrons adjust instantly to changes in the molecular conformation defined by the positions of the nuclei. This picture forms the basis of the Born-Oppenheimer approximation. In quantum mechanics, the wavefunction of a molecule, Ψ({rn }, {RN }), is a function of the positions of all the electrons and nuclei, and the Hamiltonian has the form ˆ = H
! p ! p ˆ 2N ˆ 2n + + V ({rn }, {RN }) , 2me 2mN n N
ˆ n and p ˆ N , act only on the corwhere, as usual, the momentum operators, p responding coordinates. Here we have labelled electrons by lower case letters n = 1, 2, · · · to distinguish them from nuclei which are denoted by capitals, N = a, b, · · ·. The statement that the electrons and nuclei have comparable momenta translates into the fact that ∇2n Ψ and ∇2N Ψ will be comparable. Therefore the second term in the Hamiltonian above, the sum over nuclear Advanced Quantum Physics
J. Robert Oppenheimer 19041967 An American theoretical physicist best known for his role as the scientific director of the Manhattan Project: the World War II effort to develop the first nuclear weapons at Los Alamos National Laboratory. In reference to the Trinity test in New Mexico, where his Los Alamos team first tested the bomb, Oppenheimer famously recalled the Bhagavad Gita: “If the radiance of a thousand suns were to burst at once into the sky, that would be like the splendor of the mighty one.” and “Now I am become Death, the destroyer of worlds.” As a scientist, Oppenheimer is remembered most for being the chief founder of the American school of theoretical physics while at the University of California, Berkeley. As director of the Institute for Advanced Study he held Einstein’s old position of Senior Professor of Theoretical Physics. Oppenheimer’s notable achievements in physics include the Born-Oppenheimer approximation, work on electron-positron theory, the Oppenheimer-Phillips process, quantum tunneling, special relativity, quantum mechanics, quantum field theory, black holes, and cosmic rays.
10.1. THE H+ 2 ION kinetic energies, can be neglected as a first approximation when we solve for the dependence of Ψ on the electron position vectors {rn }. In the Born-Oppenheimer approximation, the time-independent Schr¨odinger equation for the electronic motion is therefore given by " # ! !2 ∇2 n − + V ({rn }, {RN }) ψk ({rn }, {RN }) = Ek ({RN }) ψk ({rn }, {RN }) , 2m e n where the eigenfunctions ψk , with k = 0, 1, 2, · · ·, describe the electronic ground state and excited states with the nuclei nailed down at positions Ra , Rb , . . .; Ek are the corresponding energy levels. Notice that the nuclear positions appear as parameters in ψk and Ek . As the molecular conformation is varied by changing Ra , Rb , . . ., the ground state energy E0 follows a curve called the molecular potential energy curve and the minimum of this curve defines the equilibrium conformation of the molecule. We shall discuss later how the molecular potential energy curve can be used to predict the vibrational and rotational energy levels of the molecule when we go beyond the approximation that the nuclei are nailed down.
10.1
The H+ 2 ion
The simplest system that exhibits molecular properties is the hydrogen ion H+ 2 , which consists of two protons with positions Ra , Rb and one electron at r. With the potential energy $ % e2 1 1 1 V (r, Ra , Rb ) = − − , 4π#0 |Ra − Rb | |r − Ra | |r − Rb | the Schr¨odinger equation takes the form, & 2 2 ' ! ∇r − + V (r, Ra , Rb ) ψ(r; Ra , Rb ) = Eψ(r; Ra , Rb ) . 2me Although this equation can actually be solved exactly using elliptical polar coordinates, it will be more instructive for our purposes to seek an approximate method of solution. Since there is no obvious parameter in which to develop a perturbative expansion, we will instead follow a variational route to explore the low energy states of system. If the electron is close to one of the protons, one would expect the other proton to have a small influence on its dynamics, and that the wavefunction in this region would be close to that of a hydrogen atomic orbital. Therefore, in seeking the ground state of the H+ 2 ion, we may take a trial wavefunction that is a linear combination of the ground state (1s) wavefunctions centred on the two protons, ψ(r; Ra , Rb ) = αψa (r; Ra ) + βψb (r; Rb ) , where ψa,b = (πa30 )−1/2 exp[−|r − Ra,b |/a0 ], represents the corresponding hydrogenic wavefunction with a0 the atomic Bohr radius. In this case, the coefficients α and β can be taken as real. The variational expression to be minimized in order to estimate the ground state energy is given by #E$ =
ˆ |ψ$ #ψ| H α2 Haa + β 2 Hbb + 2αβHab = #ψ|ψ$ α2 + β 2 + 2αβS
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ˆ a $ = Hbb = #ψb |H|ψ ˆ b $, and Hab = #ψa |H|ψ ˆ b $ = #ψb |H|ψ ˆ a $. where Haa = #ψa |H|ψ ˆ is symmetric with reNote that the matrix elements Haa = Hbb because H spect to Ra and Rb . Moreover, since ψa and ψb are not orthogonal we have to introduce the overlap integral, S = #ψa |ψb $, which measures the overlap between the two atomic wavefunctions. In fact we can simplify this expression further, because the potential is symmetric about the mid-point between the two protons. The wavefunction must therefore be either symmetric or antisymmetric, α = ±β, and hence, E0 ≤ #E$ =
Haa ± Hab . 1±S
The matrix elements in this expression can be evaluated in closed form, though the calculation is rather tedious.1 We have, therefore, found two possible wavefunctions for the H+ 2 ion, ψa + ψb ψg = ( , 2(1 + S)
ψa − ψb ψu = ( , 2(1 − S)
−Hab +Hab with energies Eg = Haa1+S , Eu = Haa1−S . The subscript g refers to the term gerade (German for even) used in molecular physics to denote a state that is even under the operation of inverting the electronic wavefunction through the centre of symmetry of the molecule, without changing the positions of the nuclei. Such an inversion changes r → Ra + Rb − r, which interchanges ψa and ψb . Note that this is not the same as parity inversion, which would also affect the nuclear coordinates. The ungerade (odd) state is denoted by subscript u. Note that ψg and ψu are orthogonal, even though ψa and ψb are not. In fact, ψg and ψu are just the orthonormal states that diagonalize the Hamiltonian, if we limit ourselves to linear combinations of ψa and ψb . In chemistry, ψg and ψu are called molecular orbitals and the assumption that they are linear combinations of atomic stationary states is called the linear combination of atomic orbitals (LCAO) approximation. The state ψg has the lower energy, while ψu represents an excited state of the molecular ion. Physically, the reason for this is that, in the ψg state, the two atomic wavefunctions interfere constructively in the region between the protons, giving an enhanced electron density in the region where the electron is attracted strongly by both protons, which serves to screen the two protons from each other. Conversely, in ψu we have destructive interference in the region between the protons. If we plot Eg and Eu as functions of the nuclear separation R = |Ra − Rb |, the results are as shown in the figure right. For both curves, we have plotted E + Ry since −Ry is the ground state energy of the hydrogen atom. The curve of Eg + Ry in the LCAO approximation has a minimum value of −1.8eV at R = R0 = 130pm, which is the predicted equilibrium nuclear separation. The predicted energy required to dissociate the molecular ion into a hydrogen atom and a free proton is thus 1.8eV. The curve of Eu + Ry does not have a minimum, suggesting that the odd wavefunction ψu does not correspond to a bound molecular state. 1
We simply quote the results here: „ « „ « R R2 e2 R S = 1+ + 2 e−R/a0 , Haa = −Ry + 1+ e−2R/a0 a0 3a0 4π"0 R a0 „ « „ « e2 e2 R Hab = 1+ e−R/a0 + S −Ry + , 4π"0 a0 a0 4π"0 R
where R = |Ra − Rb | and Ry is the Rydberg constant, the binding energy of the Hydrogen atom in its ground state.
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E/eV 4! 3 2 1 0 100 –1 –2 –3
# # $
Eu + Ry
200
" R/pm
300
% & % % & %% LCAO Eg + Ry % Exact E + Ry g
Molecular potential + curves for H2 ion.
energy
10.1. THE H+ 2 ION As the variational method provides an upper limit on the ground state energy, it is no surprise that the true molecular potential energy curve (shown dotted) lies below the LCAO one. The true values of the equilibrium separation and dissociation energy are 106pm and 2.8eV respectively. Clearly the LCAO wavefunction ψg is not a very accurate approximation to the true ground state. We could improve it by introducing further variational parameters or additional atomic orbitals. For example, when R becomes very small the true wavefunction should approach that of a He+ ion; we could include such a term in the trial function. We could also include an effective charge parameter in ψa and ψb , which is equivalent to replacing the Bohr radius a0 by a free parameter. Although not very reliable quantitatively, the LCAO wavefunction ψg does however exhibit a number of important features of the true ground state: (i) it is even (g) with respect to inversion of the electron wavefunction; (ii) there is constructive interference which leads to an enhanced probability of finding the electron in the region between the two nuclei. These features will be important when we come to discuss bonding. Since the odd u-states are orthogonal to the even g-states, and the true ground state is a g-state, the curve of Eu actually represents an upper limit on the energy of the lowest u-state. Thus the fact that the curve has no minimum does not really prove that there are no bound u-states; but this does turn out to be the case. The LCAO wavefunction ψu shows the characteristic feature of an anti-bonding state: there is destructive interference in the region between the two nuclei, so that the electron is actually forced out of the region of overlap of the two atomic wavefunctions. & Info. At this stage, it is helpful to introduce some notation to label the molecular orbitals. Although the wavefunctions ψg,u that we have been discussing are formed in the LCAO approximation from linear combinations of atomic 1s (n = 1, ' = 0) states, with no orbital angular momentum, they do not themselves necessarily have zero orbital angular momentum. An ' = 0 state must be proportional to Y00 , i.e., it must have no dependence on θ and φ, giving an isotropic probability distribution. But these states are certainly not isotropic: they have a ‘dumbbell’ shape, concentrated around the two protons. They do not have unique electronic orbital ˆ 2 for the electron does not commute with angular momentum because the operator L the Hamiltonian on account of the non-central terms 1/|r − Ra | and 1/|r − Rb | in the potential. ˆ that does commute with these terms is L ˆ z , provided The only component of L we choose the z-axis parallel to the internuclear axis Ra − Rb . Therefore instead of classifying the states as s, p, d, . . . orbitals according to whether ' = 0, 1, 2, . . ., we call them σ, π, δ, . . . orbitals according to whether Λ = 0, 1, 2, . . ., where Λ ≡ |m! |. A subscript u or g denotes whether the state is even or odd under inversion; this notation can be applied to all homonuclear diatomic molecules, in which the potential is symmetric about the median plane of the molecule. Thus the ground state of the hydrogen molecular ion is σg and the corresponding odd state is σu∗ , where the star signifies an antibonding orbital. In the LCAO approximation used above, these molecular orbitals are linear combinations of 1s atomic orbitals and so they can be written as 1sσg and 1sσu∗ . To get some insight into how this denotation applies, it is helpful to refer to the figures on the right.
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Approximation Experiment MO fixed charge MO variable charge VB variable charge Variable λ and charge 13 variable parameters
B.E./eV 4.75 2.70 3.49 3.78 4.03 4.72
R0 /pm 74.1 85.2 73.0 74.6 75.7 74.1
Table 10.1: Binding energy and equilibrium nuclear separation of the hydrogen molecule in various approximations.
10.2
The H2 Molecule
We now turn to consider the hydrogen molecule. One might imagine that this would be a trivial extension of the H+ 2 ion, but in fact several new features arise when we consider this simple molecule. For two electrons with positions r1,2 and two protons at Ra,b , in the Born-Oppenheimer approximation, the Hamiltonian is given by & ' 2 2 ) * e 1 1 1 1 1 1 ! 2 2 ˆ =− H ∇r 1 + ∇ r 2 + + − − − − , 2me 4π#0 rab r12 r1a r1b r2a r2b
where r1b = |r1 − Rb |, etc. This is just the sum of two Hamiltonians for e2 1 1 H+ 2 ions, plus the additional term 4π"0 ( r12 − rab ). It is plausible that the expectation values of 1/r12 and 1/rab will be comparable and therefore the extra term can be treated as a perturbation. Thus, as a first approximation we neglect it and assign each electron to one of the H+ 2 molecular orbitals defined above. There are four ways of filling the two orbitals σg and σu∗ , which we can represent by ψg (r1 )ψg (r2 ),
ψg (r1 )ψu∗ (r2 ),
ψu∗ (r1 )ψg (r2 ),
ψu∗ (r1 )ψu∗ (r2 ) .
Of these, we would expect the first to be the ground state. However, at this stage, we have given no consideration to the constraints imposed by particle statistics. In fact, since the electrons are identical fermions, the wavefunction must be antisymmetric with respect to their interchange. Taking into account the spin degree of freedom, for both electrons to occupy the bonding σg molecular orbital, they must must have opposite spins and occupy the singlet spin state, X0,0 = √12 (χ+ (1)χ− (2) − χ− (1)χ+ (2)). If we calculate the energy of the state ψg (r1 )ψg (r2 )X0,0 as a function of the nuclear separation R, the minimum ground state energy occurs at R0 = 85pm and corresponds to a binding energy of 2.7eV. The true molecule is smaller and more tightly bound. Allowing for more variation in the atomic orbitals, in the form of a variable effective charge, gives an equilibrium value of R0 much closer to experiment, but a binding energy that is still not high enough (see Table 10.2). The reason is that the σg2 configuration alone is not a very good representation of the ground state. To understand why, consider the following. If we multiply out the spatial part of the σg2 wavefunction in the LCAO approximation, we see that it has a rather strange form, ψg (r1 )ψg (r2 ) ∝ [ψa (r1 )ψb (r2 ) + ψb (r1 )ψa (r2 )] + [ψa (r1 )ψa (r2 ) + ψb (r1 )ψb (r2 )] . The terms in the first square bracket correspond to the two electrons being shared between the two hydrogen atoms. This is the covalent bonding picture of the bound state. In the other square bracket, however, both electrons Advanced Quantum Physics
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are assigned to the same atom, corresponding to ionic bonding. Since all the terms have equal coefficients, the ionic and covalent contributions are equal, which seems rather constraining, if not implausible. For example, it means that, when the two protons are pulled apart, the system is just as likely to be found to consist of an H+ and an H− ion as two neutral atoms. If we go to the pure valence bonding (VB) approximation and drop the ionic part of the wavefunction altogether, we find that the predicted binding energy and nuclear separation are both improved (see Table 10.2). Including a small parameter λ for the amplitude of the ionic component, i.e., taking ψ V B ∝ [ψa (r1 )ψb (r2 ) + ψb (r1 )ψa (r2 )] + λ[ψa (r1 )ψa (r2 ) + ψb (r1 )ψb (r2 )] , we find that the variational method gives an optimal value of λ of about 1/6, meaning only about a 3% probability of finding the ionic configuration. Still, even with this refinement, the agreement with experiment is far from perfect. However, this doesn’t that mean quantum mechanics is failing; by taking enough free parameters in the trial function an excellent result can be obtained, as shown in the Table. & Info. As mentioned above, there are four possible ways of putting two electrons into the 1sσg and 1sσu∗ molecular orbitals. From these four two-electron states we can make three states that are symmetric under interchange of the positions of the electrons, all of which need to be combined with the antisymmetric spin state X0,0 , Σg
:
Σu
:
Σg
:
1 1
1
X0,0 σg (r1 )σg (r2 )
√ X0,0 [σg (r1 )σu∗ (r2 ) + σu∗ (r1 )σg (r2 )]/ 2 X0,0 σu∗ (r1 )σu∗ (r2 ) .
In addition we can make a triplet of states from an antisymmetric spatial wavefunction and the symmetric triplet spin states X1,mS (mS = 0, ±1), √ 3 Σu : X1,mS [σg (r1 )σu∗ (r2 ) − σu∗ (r1 )σg (r2 )]/ 2 . We have introduced to the left of these equations a new notation, called the molecular term, to describe the overall quantum numbers of the molecule, all of which must be good quantum numbers because they correspond to operators which commute with the molecular Hamiltonian. It derives from a historic spectroscopic notation used in atomic physics. The term is written 2S+1 Λu/g . The prefix 2S + 1 denotes the multiplicity of the total spin state of the electrons, hence 1 for a singlet and 3 for a triplet state. The central greek capital letter represents the magnitude of the ˆ z quantum number, Λ = 0, 1, 2 . . . being represented by Σ, Π, ∆ . . .. In the case total L of the molecular orbitals above based on the 1s atomic orbitals, all the wavefunctions clearly have zero orbital angular momentum about the internuclear axis and hence they are all Σ states. The g or u suffix means even or odd under inversion, and is only meaningful for homonuclear molecules. Notice that since (±1)2 = +1 we get a g-state by combining two g or two u-states, and a u-state by combining a g and a u-state. In terms of the molecular orbital approach, the VB approximation implies that the ground state is not simply the σg2 configuration but rather a mixture of the two 1 Σg states, ψ V B ∝ (1 + λ)(1 + S)σg (r1 )σg (r2 ) − (1 − λ)(1 − S)σu∗ (r1 )σu∗ (r2 ) . Thus there is configuration mixing in the hydrogen molecule. The two states can mix because they have the same overall quantum numbers. At large nuclear separations the energy eigenstates are clearly separated into those that are almost a pair of neutral atoms and those consisting of an H+ and an H− ion. In the case of the u-states, we can see by expanding the spatial wave functions that 3 Σu is of the former type and 1 Σu of the latter. Since these configurations have different resultant
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E !
H+ +H− σ ∗2 u H+H
σg2
" !" R
Molecular potential energy and configuration mixing in H2 .
10.3. FROM MOLECULES TO SOLIDS electron spins they do not mix significantly (just like singlet and triplet states of atomic helium). Also the u- and g-states are prevented from mixing by their different inversion symmetry. However, the two 1 Σg configurations have the same electron spin and symmetry and can mix to give the above covalent-bonded ground state and an orthogonal excited state that is more ionic.
10.3
From molecules to solids
With these basic principles in hand, we could go on to discuss the orbital and electronic structure of more complicated molecules. In doing so, we would sink deeper into the realm of quantum chemistry. Instead, we will use these simple ideas of molecular bonding to develop a surprisingly versatile and faithful description of the ordered solid state. With ca. 1023 nuclei and electrons now involved such an enterprise seems foolhardy. However, we will see that, by exploiting symmetries, we can capture many of the basic principles of the solid state. Let us then consider the electronic structure of an ordered crystalline array of equivalent atoms. (The consideration of more complicated periodic structures would not present conceptual challenges, but would bring unnecessary complications to the discussion.) Once again, we can draw upon the Born-Oppenheimer approximation and focus solely on the motion of electrons around an otherwise ordered array of nuclei. Even then, we are confronted with a many-particle Hamiltonian of apparantly great complexity, ˆ =− H
! !2 ∇2
n
n
2me
−
! e2 Ze2 ! e2 e2 + . 4π#0 rnN m,n 4π#0 rmn n,N
Here the second term represents the interaction of the electrons with the consituent nuclei, and the third term involves the electron-electron interaction. In a physical system, we would have to take into account the influence of further relativistic corrections which would introducte additional spin-orbit couplings. To address the properties of such a complex interacting system, we will have to draw upon many of the insights developed previously. To begin, it is helpful to partition the electrons into those which are bound to the core and those which are able to escape the potential of the individual atomic nuclei and propagate “freely” through the lattice. The electrons which are tightly bound to the nuclei screen the nuclear charge leading to a modified nuclear potential, Veff (r). Focussing on those electrons which are free, the effective + e2 e2 ˆ *+ H ˆ Hamiltonian can be written as, H n n+ m,n 4π"0 rmn , where ˆ n = − ! ∇n + Veff (rn ) H 2me 2
2
represents the single-particle Hamiltonian experienced by each of the electrons ˆ n describes the motion of an electron moving in a periodic lattice – i.e. H potential, Veff (r) = Veff (r + R) with R belonging to the set of periodic lattice vectors. Despite engineering this approximation, we are still confronted by a challenging many-particle problem. Firstly, the problem remains coupled through the electron-electron Coulomb interaction. Secondly, the electrons move in a periodic potential. However, if we assume that the electrons remain mobile – the jargon is itinerant, free to propagate through the lattice, they screen each other and diminish the effect of the Coulomb interaction. Therefore, we can Advanced Quantum Physics
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10.3. FROM MOLECULES TO SOLIDS proceed by neglecting the Coulomb interaction altogether, when the Hamiltonian is said to be free. Of course, we still have to contend with the constraints placed by Pauli exclusion and wavefunction antisymmetry. But we are now in the realm of the molecular orbital theory, and can proceed analogously using the variational LCAO approach. In particular, we can building a trial wavefunction by combining orbital + states of the single ion, Vion (r), where Veff (r) = R Vion (r − R). As with the ˆ 0 = !2 ∇2 + hydrogen molecule, the Hamiltonian for the individual nuclei, H 2me Vion (r) are associated with a set of atomic orbitals, ψq , characterized by a set of quantum numbers, q. In the atomic limit, when the atoms are far-separated, these will mirror the simple hydrogenic states. To find the variational ground state of the system, we can then build a trial state from a linear combination of these atomic orbitals. Taking only the lowest orbital, q = 0, into account we have, ! ψ(r) = αR ψ(r − R) , R
where, as before, αR represent the set of variational coefficients, one for each site (and, in principle, each atomic orbital if we had taken more than one!). Once again, we can construct the variational state energy, + ∗ ˆ 0 |ψ$ #ψ|H R,R" αR HRR" αR" E= = + , ∗ #ψ|ψ$ R,R" αR SRR" αR" , ˆ 0 ψ(r − R% ) denote the matrix elewhere, as before, HRR" = dd rψ ∗ (r − R)H , ments of the orbital wavefunction on the Hamiltonian and SRR" = dd rψ ∗ (r− R)ψ(r − R% ) represent the overlap integrals. Then, varying the energy with ∗ , we find that the coefficients obey the secular equation (exrespect to αR ercise) ! [HRR" − ESRR" ] αR" = 0 . R"
Note that, if the basis functions were orthogonal, this would just be an eigenvalue equation. & Exercise. To develop this idea, let us first see how the method relates to back to the problem of H+ 2 : In this case, the secular equation translates to the 2 × 2 matrix equation, $ % Haa − E Hab − ES α = 0, Hab − ES Haa − ES where the notation and symmetries of matrix elements follow from section 10.1. As a result, we find that the states α divide into even and odd states as expected.
Now let us consider a one-dimensional periodic lattice system. If we assume that the atoms are well-separated, it is evident that both the overlap integrals and matrix elements will decay exponentially fast with orbital separation. The dominant contribution to the energy will then derive from matrix elements coupling only neighbouring states. In this case, the secular equation separates into the coupled sequence of equations, (ε − E)αn − (t + ES)(αn+1 + αn−1 ) = 0 , for each n, where Hnn = ε denotes the atomic orbital energy, Hn,n+1 = Hn+1,n = −t is the matrix element between neighbouring states, Sn,n = 1 Advanced Quantum Physics
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and Sn+1,n = Sn+1,n = S. All other matrix elements are exponentially small. Here we consider a system with N lattice sites and impose the periodic boundary condition, αn+N = αn . The periodicity of the secular equation suggests a solution of the from αn = √1N eikna , where k = 2πm/N a denote a discrete set of N reciprocal lattice vectors with m integer lying in the range −N/2 < m ≤ N/2. Substitution confirms that this is a solution with energies (exercise), Ek =
ε − 2t cos(ka) . 1 + S cos(ka)
The reciprocal lattice vector k parameterizes a band of states. Then, according to the LCAO approximation, for a single electron, distributed throughout the system, the lowest energy state is predicted to be the uniform state αn = √1N with the energy E0 = ε−2t 1+S . (Here we have assumed that the matrix element, t is positive – in the atomic limit, consider why this is a sound assumption.) However, if we suppose that each atom contributes a non-zero fraction of electrons to the system, we must consider the influence of Pauli exclusion and particle statistics. Since the electrons are fermions, each state k can host two electrons in a total spin singlet configuration. The lowest energy state is then obtained by adding electron pairs sequentially into states of increasing k until all electrons are accounted for. If the maximum k value, known as the Fermi wavevector, kF , lies within the band of states, elementary excitations of the electrons cost vanishingly small energy leading to metallic behaviour. On the other hand, if the system is stoichiometric, with each atom contributing an integer number of electrons, the Fermi wavevector may lie at a band gap between two different band of states. In this case, the system has a gap to excitations and the material forms a band insulator.
One dimensional band structure for Ek for S = 0.3, t = 1 and ε = 0. Here we have shown schematically a band-filling with kF a = π/2.
& Exercise. Consider how the calculation above can be extended to a twodimensional square lattice system. What would be the corresponding band dispersion?
& Exercise. In the one-dimensional system, extend the calculation to compute the electron band structure for a periodic lattice with two atomic orbitals on each site, one with s-wave symmetry, and another with p-wave. In particular, consider how the dispersion would change with the orientation of the p state relative to the chain direction.
& Info. To add flesh to these ideas, let us then consider a simple, but prominent problem from the realm of quantum condensed matter physics. In recent years, there has been great interest in the properties of graphene, a single layer of graphite. Remarkably, high quality single crystals of graphene can be obtained by running graphite – a pencil! – over an adhesive layer. The resulting electron states of the single layer compound have been of enormous interest to physicists. To understand why, let us implement the LCAO technology to explore the valence electron structure of graphene. Graphene forms a periodic two-dimensional honeycomb lattice structure with two atoms per each unit cell. With an electron configuration 1s2 2s2 2p2 , the two 1s electrons are bound tightly to the nucleus. The remaining 2s electrons hybridize with one of the p orbitals to form three sp2 hydridized orbitals. These three orbitals form the basis of a strong covalent bond of σ orbitals that constitute the honeycomb lattice. The remaining electron, which occupies the out-of-plane pz orbital, is then Advanced Quantum Physics
Atomically resolved STM image of graphene sheet at 77K.
10.3. FROM MOLECULES TO SOLIDS
Figure 10.1: Left: dispersion relation of graphene Ek obtained with the LCAO approximation. Notice that near the centre of the band, the dispersion becomes point-like. Around these points, the dispersion takes form of a linear (Dirac) cone. Right: band structure of a sample of multilayer epitaxial graphene. Linear bands emerge from the K-points in the band structure. Three “cones” can be seen from three layers of graphene in the MEG sample.
capable of forming an itinerant band of electron states. It is this band which we now address. Once again, let us suppose that the wavefunction of this band involves the basis of single pz orbital states localized to each lattice site, ! ψ(r) = [αR ψ1 (r − R) + βR ψ2 (r − R)] . R
Here the pz orbital wavefunction ψ1 is centred on one of the atoms in the unit cell, and ψ2 is centred on the other. Once again, taking into account matrix elements involving only nearest neighbours, the trial wavefunction translates to the secular equation, (ε − E)αR − (t + ES)(βR + βR−a1 + βR−a2 ) = 0
(ε − E)βR − (t + ES)(αR + αR+a1 + αR+a2 ) = 0 , √ √ where the lattice vectors a1 = ( 3/2, 1/2)a and a2 = ( 3/2, −1/2)a, with a the lattice spacing, are shown in the figure to the right. Note that the off-diagonal matrix elements involve only couplings between atoms on different sublattices. Once again, we can make the ansatz that the solutions are of the form of plane waves with αR = αk ik·R k √ e and βR = √βN eik·R . Notice that, in this case, we must allow for different N relative weights, αk and βk . Substituting, we find that this ansatz is consistent if (ε − E)αk − (t + ES)fk βk = 0 √
(ε − E)βk − (t + ES)fk∗ αk = 0
where fk = 1+2e−i 3kx a/2 cos(ky a/2). Although this equation can be solved straightforwardly, it takes a particularly simple form when the overlap integral between neighbouring sites, S, is neglected. In this case, one obtains, Ek = ε ± |fk |t . The corresponding band structure is shown right as a function of k. In particular, one may note that, at the band centre, the dispersion relation for the electrons becomes point-like (see Fig. 10.3). In the half-filled system, where each carbon atom contributes a single electron to the band, the Fermi level lies precisely at the centre of the band where the dispersion, Ek is point like. Doping electrons into (or removing electrons from) the system results in (two copies) of a linear dispersion, Ek * c|k|, where is c is a constant (velocity). Such a linear dispersion relation is the hallmark of a relativistic particle (cf. a photon). Of course, in this case, the electrons are not moving at relativistic velocities. But their properties will mirror the behaviour of relativistic particles. And herein lies the interest that these materials have drawn.
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10.4. MOLECULAR SPECTRA Finally, let us comment on the influence of the electron-electron interaction effects that were neglected in our treatment above. In principle, we could adopt a Hartree or Hartree-Fock scheme to address the effects of electron interaction in a perturbative manner. Indeed, such a programme would lead to interaction corrections which would modify the electronic band structure derived above. It is a surprising yet robust feature of Fermi systems that the properties of the non-interacting ground state wavefunction remain qualitatively correct over an unreasonably wide range of interaction strengths.2 This rigidity can be ascribed to the contraints implied on the nodal structure of the wavefunction imposed by particle statistics. However, in some cases, the manifestations of electron interactions translate to striking modifications in the experimental behaviour leading to effects such as interaction-driven electron localization – the Mott transition, local moment and itinerant magnetism, quantum Hall fluids, and superconductivity. Such phases, which by their nature, lie outside any perturbative scheme built around the non-interacting ground state, form the field of quantum condensed matter and solid state physics. With this detour, we now return to the simple molecular system, and consider the question of excitations and transitions.
10.4
Molecular spectra
In the atomic system, our consideration of radiative transitions was limited to the problem of electronic transitions between states. In the molecular system, the internal structure allows for transitions involving rotational and vibrational excitations of the consituent nuclei. As with atoms, electronic transitions are typically of order eV, and so corresponds to wavelengths in or near the optical region. However, it is unlikely that an electronic transition will occur without inducing motion of the nuclei as well, because the equilibrium distances between the nuclei will be different in the initial and final electron states. The typical energies of rotational states of molecules are much smaller than those of electronic excited states, of order !2 /2I, where I is the molecular moment of inertia. Substituting typical values for the interatomic spacing and atomic masses, one finds that rotational energies are O(10−4 eV), corresponding to the far infra-red or microwave regions. Typical energies for vibrational excitations of molecules are O(10−1 eV) (see section 10.4.2 below), corresponding to the infra-red waveband. All of these types of transitions can occur radiatively, i.e. through the emission or absorption of a photon of the appropriate frequency ν = ∆E/h.3 As in the case of atoms, the most probable radiative transitions are usually electric dipole transitions. In an electric dipole transition, the photon carries one unit of angular momentum and negative parity, so there will be the usual selection rule for the change in the total angular momentum quantum number of the molecule: ∆J = 0, ±1,
but not 0 → 0 ,
accompanied by a change in the parity of the molecular state (which may impose further restrictions on ∆J). & Info. In a gas or a liquid, transitions can also be produced by collisions 2 The feature is embodied in the theory of the electron liquid known as Landau’s Fermi liquid theory. 3 Spectroscopists always quote wavenumbers ν˜ = ν/c instead of frequencies, and measure them in cm−1 .
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between molecules. The appropriate energy change can be provided by excitation or de-excitation of the other molecule, or as part of the kinetic energy of the collision. Such non-radiative transitions do not have to obey the selection rules above. Thus, for example, a molecule in a metastable state, i.e. one that cannot return to the ground state via an allowed radiative transition, can be de-excited by a collision. In the absence of any incident radiation or other non-thermal sources of excitation, collisions will bring about a thermal distribution of molecular energy levels, with the number of molecules in state i being given by ni ∝ gi exp[− kEBiT ], where gi is the degeneracy and Ei the energy. At room temperature, kB T ∼ 2 × 10−2 eV, so typically many rotational states of molecules are excited, but not electronic or vibrational.
To study the motion of nuclei in the Born-Oppenheimer Approximation and the associated types of transitions more formally, we must go back to the molecular Schr¨odinger equation considered above. Having found the electronic eigenfunctions, ψk , in the Born-Oppenheimer approximation, we can use the completeness property of these functions to express the full stationary state wavefunction of the molecule as ! Ψ({rn }, {RN }) = φk ({RN })ψk ({rn }, {RN }) , k
where φk represents the nuclear part of the wavefunction. Substituting this expression into the full time-independent Schr¨odinger equation, we obtain ! !2 ∇2 ! !2 ∇2 n N − − + V ({rn }, {RN }) Ψ({rn }, {RN }) 2me 2mN n=1,2,...
N =a,b,...
= EΨ({rn }, {RN }) .
Using the electronic energy levels Ek , we have 2 ! ! ! ! − ∇2N + Ek ({RN }) φk ψk = E φk ψk . 2mN k
N =a,b,...
k
Now another important aspect of the Born-Oppenheimer approximation comes into play: the dependence of ψk ({rn }, {RN }), the electronic part of the wavefunction, on the nuclear coordinates {RN } is weak compared with that of the nuclear part φk ({RN }). We shall verify this explicitly in section 10.4.2 below. Therefore we can write ∇2N φk ψk * ψk ∇2N φk , to a good approximation. Using the orthogonality of the ψk s, we can now pick out the k = 0 term, for example, by multiplying by ψ0∗ and integrating over all electron positions, to obtain 2 ! ! − ∇2 + E0 ({RN }) φ0 = Eφ0 . 2mN N N =a,b,...
Thus the nuclear part of the wavefunction, φ0 , satisfies a Schr¨odinger equation in which the full potential V ({rn }, {RN }) is replaced by the molecular potential energy curve E0 ({RN }), which we derived earlier in the approximation that the nuclei were nailed down at positions {RN }. The replacement of the full potential by E0 ({RN }) in the nuclear Schr¨odinger equation corresponds physically to assuming that the electrons react instantly to changes in the molecular conformation.
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10.4. MOLECULAR SPECTRA
10.4.1
Molecular rotation
In general, the Schr¨odinger equation for the nuclear motion has many solutions, which give the various molecular energy levels for a given electronic configuration. Here, for simplicity, we focus on diatomic molecules which capture the main features of the general phenomenology. In this case, E0 (R1 , R2 ) = E0 (R) where R = |R1 − R2 | and the usual separation of variables can be performed. One equation describes the translational motion of the overall centre of mass, and leads to a continuum of possible energies for a free molecule, or the usual energies of a particle in a box if it is confined. The equation for the relative coordinate, R = R1 − R2 , is given by ' & 2 ! − ∇2R + E0 (R) φ0 = Eφ0 , 2µ
where µ is the reduced mass of the two atoms. In this case, E0 (R) acts as a central potential, and the usual separation into angular and radial equations can be carried out. The simplest solutions are the purely rotational states, in which the whole molecule rotates around its centre of mass. The solutions will be the spherical harmonic functions YJ,mJ . Conventionally the quantum numbers are labelled J and mJ (mJ taking values J, J − 1, · · · , −J), and the corresponding energy is given by EJ =
!2 J(J + 1) , 2I
where I = µR02 is the moment of inertia of the molecule about an axis through the centre of mass orthogonal to the bond, and R0 is the equilibrium bond length. As mentioned earlier, the typical energies of rotational states of molecules are much smaller than those of electronic excited states. Since molecular dimensions are determined by the electronic wavefunction, their scale is set by the Bohr radius a0 . Thus moments of inertia are of order mN a20 and the scale of rotational energies is !2 /mN a20 . For the electronic states, the uncertainty relation implies momenta of order !/a0 and hence electron energies around !2 /me a20 , a factor of mN /me ∼ 104 greater. To bring about a radiative rotational transition, an emitted or absorbed photon must interact with the electric dipole moment of the molecule. Since the initial and final electronic states are the same, this state needs to have a permanent electric dipole moment. Thus we can have purely rotational radiative transitions in heteronuclear diatomic molecules like HCl and CO, which have permanent dipole moments, but not in homonuclear ones like H2 and O2 .4 The usual electric dipole selection rules apply; ∆J = ±1, 0 with a parity change. In a rotational state with angular momentum quantum numbers J and mJ , the nuclear wavefunction φ({rN }) is proportional to the spherical harmonic YJmJ , which has parity (−1)J . (For simplicity, we consider only molecular states in which the electronic wavefunction has zero angular momentum and even parity, such as the diatomic 1 Σg states.) Then the fact 4 However, we can produce purely rotational transitions in all types of molecules by the process of Raman scattering, in which a photon is effectively absorbed and then reemitted by a molecule, since the virtual intermediate state can have a dipole moment.
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that the parity must change in a radiative transition excludes the possibility ∆J = 0. Therefore the possible energy changes in emission (J + 1 → J; J = 0, 1, 2 · · ·) are given by: ∆E =
!2 !2 [(J + 1)(J + 2) − J(J + 1)] = (J + 1) . 2I I
In fact, the rate for spontaneous emission between rotational states is very small, because of the small energy release (varying as ω 3 ), and so rotational transitions are more conveniently studied by absorption spectroscopy. The same formula for the energies of transition clearly applies to the J → (J + 1) absorption case. Therefore the spectrum is expected to consist of equally spaced lines, separated by energy !2 /I. Observation of this spacing can be used to determine the moment of inertia and hence the bond length of the molecule. Strictly speaking the spacing isn’t quite uniform, because the radial Schr¨odinger equation acquires a centrifugal potential term !2 J(J + 1)/2µR2 which means that the equilibrium separation slightly increases with J and consequently the moment of inertia increases and the line spacing decreases. & Exercise. Use perturbation theory to estimate the strength of the effect of the centrifugal term.
& Info. The intensities of rotational spectral lines show some interesting features. Although the transition matrix element depends on the quantum numbers, the dominant factor is usually the population of the initial state. As mentioned earlier, non-radiative transitions due to molecular collisions bring about a thermal distribution, nJ ∝ gJ exp[−EJ /kB T ] ∝ (2J + 1) exp[−!2 J(J + 1)/2IkB T ] . This increases with J up to some value, which depends on the temperature, and then decreases. Thus successive spectral lines increase and then decrease in intensity.
10.4.2
Vibrational transitions
Another important type of molecular motion is vibration, in which the nuclei oscillate around their equilibrium positions. For a diatomic molecule, we can Taylor expand the molecular potential E0 (R) around the equilibrium nuclear separation R0 to obtain 1 2 E0 |R0 + . . . . E0 (R) = E0 (R0 ) + (R − R0 )2 ∂R 2 Keeping only the two terms shown on the right-hand side, we have the poten2 E | )1/2 tial for a simple harmonic oscillator with classical frequency ω = ((1/µ)∂R 0 R0 where µ is the reduced mass. Thus we expect the energy levels including nuclear vibration to be given by E = E0 (R0 ) + (n + 1/2)!ω,
n = 0, 1, 2, . . . .
The excitation energies of molecular vibrational (states are typically larger than those of rotational states by a factor of about ( mN /me and smaller than electronic excitation energies by a factor of about me /mN . As we discussed earlier, E0 (R0 ) will be of the same order of magnitude as atomic energies, i.e. of the order of !2 /me a20 where a0 is the Bohr radius. Thus, on dimensional Advanced Quantum Physics
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2E | 2 2 4 2 √ grounds, ∂R 0 R0 will be of order ! /me a0 , and !ω ∼ ! / me mN a0 . Therefore ( the vibrational energy is smaller than the electronic by a factor of order me /mN . This puts vibrational spectra in the wavelength region around 10µm, which is in the infra-red. We can now check explicitly that the Born-Oppenheimer approximation is valid for nuclear vibrational states, as follows. The mean square nuclear ( vibrational momentum is of order mN !ω ∼ mN /me (!/a0 )2 , which means ( 2 that ∇N φk ∼ mN /me φk /a20 , where φk is the nuclear part of the wavefunction. On the other hand ∇2N ψk ∼ ψk /a20 , where ψ (k is the electronic part. Thus φk ∇2N ψk is smaller than ψk ∇2N φk by a factor of me /mN , and it is legitimate to neglect the former. For vibrational transitions we have the selection rule (exercise)
∆n = ±1 . This implies only a single energy in the spectrum, ∆E = (En+1 − En ) = !ω , corresponding to the classical frequency of oscillation. In practice the Taylor expansion around R = R0 has non-negligible terms of higher than second order and the harmonic oscillator approximation is not very reliable: there is anharmonicity. The flattening of the molecular potential energy curve at larger separations has the effect of bringing the energy levels closer together. Thus transitions at larger n have lower energies than that given above. Also, since the true stationary state wavefunctions are not precisely harmonic oscillator eigenfunctions, our selection rule is not exactly valid, and transitions with |∆n| > 1 become possible. & Info. Vibrational modes for polyatomic molecules can be quite complicated. If there are N atoms, in general there are 3N − 6 normal modes (3N coordinates, less 3 to define the overall position of the centre of mass, and less 3 to define the overall orientation of the molecule) or 3N − 5 in the case of a diatomic molecule. Thus, in the simple case of the linear CO2 molecule, there are four modes, two with the atoms remaining collinear (one with the two O atoms moving in antiphase with the C stationary, and one with the O atoms moving in phase in the opposite direction to the C atom) and two degenerate orthogonal bending modes. In the following chapter, we will turn to consider the vibrational motion of periodic lattice systems where, once again, a degree of simplicity is restored!
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Figure 10.2: Generally vibra-
tional transitions occur in conjunction with rotational transitions. Consequently, it is possible to observe both rotational and vibrational transitions in the vibrational spectrum. The top figure shows an energy level diagram demonstrating some of the transitions involved in the IR vibration-rotation spectrum of a linear molecule: P branch (where ∆J = 1), Q branch (not always allowed, ∆J = 0) and R branch (∆J = +1). The lower figure shows the vibration-rotation spectrum of HCl. The left hand branch of the spectrum represents the P branch and the right the R. The Q branch is not allowed. The splitting of the lines is associated with the two isotopes 35 Cl and 37 Cl.
Chapter 11
Field theory: from phonons to photons In our survey of single- and “few”-particle quantum mechanics, it has been possible to work with a discrete representation in which we index individual constituent particles. However, when the “elementary excitations” of the system involve the coherent collective motion of many individual discrete particle degrees of freedom – such as the wave-like atomic vibrations of an ordered elastic solid, or where discrete underlying classical particles can not even be identified – such as the electromagnetic field, such a representation is inconvenient or even inaccessible. In such cases, it is useful to turn to a continuum formulation of quantum mechanics. In the following, we will develop these fundamental ideas on the background of the simplest continuum theory: lattice vibrations of the atomic chain. As we will see, this study will provide a platform to investigate the quantum mechanics of the electromagnetic field – the subject of quantum electrodynamics – and will pave the way to the development of quantum field theory of relativistic particles.
11.1
Quantization of the classical atomic chain
As a simplified model of an ordered (one-dimensional) crystal, let us consider a chain of point particles each of mass m (atoms) which are elastically connected by springs with spring constant ks (chemical bonds) (see Fig. 11.1). Although our target will be to construct a quantum theory of the elementary vibrational excitations, it is helpful to begin our analysis by reviewing the classical properties of the system.
11.1.1
Classical chain
For reasons that will become clear, it is instructive to consider the Lagrangian formulation of the problem. For the N -atom chain, the classical Lagrangian is given by, L=T −V =
N " ! m
n=1
2
x˙ 2n
# ks 2 − (xn+1 − xn − a) , 2
(11.1)
where the first term accounts for the kinetic energy of the particles whilst the second describes their coupling.1 For convenience, we adopt periodic boundary 1 In real solids, the inter-atomic potential is, of course, more complex than our quadratic approximation. Yet, for “weak coupling”, the harmonic contribution dominantes (cf. our discussion of molecular vibrations). For the sake of simplicity we, therefore, neglect the effects caused by higher order contributions.
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11.1. QUANTIZATION OF THE CLASSICAL ATOMIC CHAIN
128
Figure 11.1: Toy model of a one-dimensional solid: a chain of point-like particles each of mass m coupled elastically by springs with spring constant ks .
conditions such that xN +1 ≡ N a + x1 where a denotes the “natural” equilibrium lattice spacing. Anticipating that the effect of lattice vibrations on the solid is weak (i.e. long-range atomic order is maintained) we will assume that (a) the n-th atom has its equilibrium position at x ¯n ≡ na, and (b) that the deviation from the equilibrium position remains small (|xn (t) − x ¯n | # a), i.e. the integrity of the solid is maintained. With xn (t) = x ¯n + φn (t) (φN +1 = φ1 ) the Lagrangian (11.1) then takes the form L=
N " ! m
n=1
# ks 2 2 ˙ φ − (φn+1 − φn ) . 2 n 2
Now, typically, we are not concerned with the behaviour of a given system on ‘atomic’ length scales. (For such purposes, our model is in any case much to primitive!) Rather, we are interested in universal features, i.e. experimentally observable behaviour, common to a wide range of physical systems, that manifests itself on macroscopic length scales where the detailed form of the model is inessential. For example, we might wish to study the specific heat of the solid in the limit of infinitely many atoms (or at least a macroscopically large number, O(1023 )). Under these conditions, microscopic models can usually be substantially simplified. In particular it is often permissible to subject a discrete lattice model to a continuum approximation, i.e. to neglect the discreteness of the microscopic entities of the system and to describe it in terms of effective continuum degrees of freedom. In the present case, taking a continuum limit amounts to describing the lattice displacements φn in terms of smooth functions, φ(x) of a continuous variable x (see figure). Clearly such a description makes sense only if relative fluctuations on atomic scales are weak (for otherwise the smoothness condition would be violated).
Joseph-Louis Lagrange 17361813: A mathematician who excelled in all fields of analysis, number theory, and celestial mechanics. In 1788 he published M´ecanique Analytique, which summarised all of the work done in the field of mechanics since the time of Newton, and is notable for its use of the theory of differential equations. In it he transformed mechanics into a branch of mathematical analysis.
" Exercise. Starting with the discrete form of the Lagrangian, or otherwise, show the classical equations of motion take the form, mφ¨n = ks a2 (φn+1 − 2φn + φn−1 ) . Remembering that the boundary conditions are periodic, obtain the normal modes. From this result, determine the condition under which the continuum approximation can be justified.
Introducing continuum degrees of freedom, φ(x), and applying a first order Taylor expansion,2 we can define $ $ φn → φ(x)$$
, x=na
$ $ φn+1 − φn → a∂x φ(x)$$
, x=na
N !
n=1
1 → a
%
L
dx,
0
2 Indeed, for reasons that will become clear, higher order contributions to the Taylor expansion do not contribute to the low-energy properties of the system where the continuum approximation is valid.
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Hint: consider the φn (t) = ei(kna−ωt) .
ansatz,
11.1. QUANTIZATION OF THE CLASSICAL ATOMIC CHAIN
129
where L = N a (not to be confused with the Lagrangian itself!) denotes the total length of the chain. Expressed in terms of the new°rees of freedom, L ˙ φ), where the continuum limit of the Lagrangian then reads L[φ] = 0 dx L(φ, ˙ φ) = L(φ,
ρ ˙ 2 κs a2 φ − (∂x φ)2 , 2 2
(11.2)
denotes the Lagrangian density, ρ = m/a denotes the mass per unit length and κs = ks /a. The corresponding classical action is given by % S[φ] = dt L[φ] . (11.3) Thus, we have succeeded in trading the N -point particle description in for one involving continuous degrees of freedom, φ(x), a (classical) field. The dynamics of the latter are specified by “functionals” L[φ] and S[φ] which represent the continuum generalizations of the discrete classical Lagrangian and action, respectively.3 However, although we have achieved a continuum formulation, we have yet to extract concrete physical information from the action. To do so, we need to derive equations of motion. At first sight, it may not be entirely clear what is meant by ‘equations of motion’ in the context of an infinite dimensional model. The answer to this question lies in Hamilton’s extremal principle of classical mechanics: " Info. Hamilton’s extremal principle: Suppose that the dynamics of a classical point particle with& coordinate x(t) is described by the classical Lagrangian L(x, x), ˙ and action S[x] = dtL(x, x). ˙ Hamilton’s extremal principle states that the configurations x(t) that are actually realized are those that extremize the action. This means that, for any smooth curve y(t), 1 lim (S[x + &y] − S[x]) = 0 , &
"→0
(11.4)
i.e. to first order in &, the action has to remain invariant. Applying this condition, one finds that it is fulfilled if and only if x(t) obeys the Euler-Lagrange equation of motion (exercise), d (∂x˙ L) − ∂x L = 0 . dt
(11.5)
Now, in Eq. (11.3), we are dealing with a system of infinitely many degrees of freedom, φ(x, t). Yet Hamilton’s principle is general, and we may see what happens if (11.3) is subjected to an extremal principle analogous to Eq. (11.4). To do so, we must effect the substitution φ(x, t) → φ(x, t) + &η(x, t) into Eq. (11.3) and demand that the contribution first order in & vanishes. When applied to the specific Lagrangian (11.2), a substitution of the of the ‘varied’ field leads to % % L ' ( S[φ + &η] = S[φ] + & dt dx ρ φ˙ η˙ − κs a2 ∂x φ∂x η + O(&2 ). 0
Integrating by parts (with respect to time for the first term under the integral, and space in the second) and demanding that the contribution linear
3 In the mathematics and physics literature, mappings of functions into the real or complex numbers are generally called functionals. The argument of a functional is commonly indicated in rectangular brackets [ · ]. For example, in this case, S maps the ‘functions’ ˙ ∂x φ(x, t) and φ(x, t) to the real number S[φ].
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Sir William Rowan Hamilton 1805-1865: A mathematician credited with the discovery of quaternions, the first noncommutative algebra to be studied. He also invented important new methods in Mechanics.
11.1. QUANTIZATION OF THE CLASSICAL ATOMIC CHAIN & &L in & vanishes, one obtains dt 0 dx(ρφ¨ − κs a2 ∂x2 φ)η = 0. (Notice that the boundary terms associated with both t and x vanish identically.4 Now, since η was defined to be any arbitrary smooth function, the integral above can only vanish if the term in parentheses is globally vanishing. Thus the equation of motion takes the form of a wave equation, ρφ¨ = κs a2 ∂x2 φ .
(11.6)
The solutions of Eq. (11.6) have the general form φ+ (x + vt) + φ− (x − vt) ) where v = a κs /ρ, and φ± are arbitrary smooth functions of their argument. From this we can deduce that the basic low energy elementary excitations of our model are lattice vibrations propagating as sound waves to the left or right at a constant velocity v (see figure). The trivial behaviour of our model is of course a direct consequence of its simplistic definition — no dissipation, dispersion or other non-trivial ingredients. Adding these refinements leads to the general classical theory of lattice vibrations. With this background, let us now turn to the consider the quantization of the quantum mechanical chain.
11.1.2
Quantum chain
In addressing the quantum description, the first question to ask is a conceptual one: is there a general methodology to quantize models of the form described by the atomic chain (11.2)? Indeed, there is a standard procedure to quantize continuum theories which closely resembles the quantization of point mechanics. The first step is to introduce canonical momenta conjugate to the continuum degrees of freedom (coordinates), φ, which will later be used to introduce canonical commutation relations. The natural generalization of the definition pn ≡ ∂x˙ n L of point mechanics to a continuum suggests setting ˙ φ) . π = ∂φ˙ L(φ,
(11.7)
In common with φ(x, t), the canonical momentum, π(x, t), is a continuum degree of freedom. At each space point it may take an independent value. & From the Lagrangian, we can define the Hamiltonian, H[φ, π] ≡ dx H(φ, π), ˙ ˙ φ). represents the Hamiltonian density. Applied where H(φ, π) ≡ π φ−L( φ, to the atomic chain (11.2), the canonical momentum π = ρφ˙ and H(φ, π) = κs a2 π2 2 2ρ + 2 (∂x φ) . In this form, the Hamiltonian can be quantized according to the following ˆ π %→ π rules: (a) promote the fields φ(x) and π(x) to operators: φ %→ φ, ˆ , and (b) generalise the canonical commutation relations of one-particle quantum mechanics, [ˆ pm , xn ] = −i!δmn , according to the relation5 ˆ " )] = −i!δ(x − x" ) . [ˆ π (x), φ(x
(11.8)
Operator-valued functions like φˆ and π ˆ are generally referred to as quantum fields. Employing these definitions, we obtain the quantum Hamiltonian density 1 2 κs a2 ˆ π ˆ 2. ˆ φ, H( ˆ) = π ˆ + (∂x φ) 2ρ 2 4 If we assume that the function φ already obeys the boundary conditions, we must have η(0, t) = η(L, t) = η(x, 0) = η(x, T ) = 0). 5 Note that the dimensionality of both the quantum and classical continuum fields is compatible with the dimensionality of the Dirac δ-function, [δ(x − x! )] = [Length]−1 .
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11.1. QUANTIZATION OF THE CLASSICAL ATOMIC CHAIN " Exercise. To develop this field theoretical formulation of the Hamiltonian, we have pursued a Lagrangian formulation. If you feel uncertain about this methodology, ˆ directly from the discrete atomic formulation. you should explore the derivation of H First, show that, for the discrete harmonic chain, the classical Hamiltonian is given by # ! " p2 ks n ˆ = H + (φn+1 − φn )2 . 2m 2 n Promoting the displacements and momenta to operators, and applying the canonical quanitization conditions, [ˆ pn , φn! ] = −i!δnn! , obtain the discrete form of the Hamiltonian. Taking the continuum limit, show that the Hamiltonian recovers the continuum form derived through the Lagrangian formulation.
The Hamiltonian represents a quantum field theoretical formulation of the problem but not yet a solution. To address the quantum properties of the system, it is helpful now to switch to a Fourier representation. As with any function, operator-valued functions can be represented in a variety of different ways. In particular they can be subjected to Fourier expansion, * * * * % L ˆ ˆ 1 ! {±ikx φˆk φ(x) φ(x) φˆk ≡ 1 {∓ikx dx e , = e ,(11.9) π ˆ (x) π ˆ (x) L1/2 0 L1/2 k π ˆk π ˆk + where k represents the sum over all Fourier coefficients indexed by quantized wavevectors k = 2πm/L, m integer. Note that, since the classical field φ(x) ˆ is real, the quantum field φ(x) is Hermitian, i.e. φˆk = φˆ†−k (and similarly for π ˆk ). In the Fourier representation, the transformed field operators obey the canonical commutation relations (exercise), [ˆ πk , φˆk! ] = −i!δkk! . " Exercise. Making use of Eqs. (11.8) and (11.9) derive the canonical commutation relation above. When expressed in the Fourier representation, making use of the identity δk+k! ,0 , % -. / L L ! ! 1 ! ˆ2= dx (∂ φ) (ik φˆk )(ik " φˆk! ) dx ei(k+k )x = k 2 φˆk φˆ−k , L 0 0 !
%
k
k,k
&L together with the parallel relation for 0 dx π ˆ 2 , the Hamiltonian assumes the “near diagonal” form, # !" 1 1 2ˆ ˆ ˆ H= π ˆk π ˆ−k + ρωk φk φ−k , (11.10) 2ρ 2 k
where ωk = v|k|, and v = a(κs /ρ)1/2 denotes the classical sound wave velocity. In this form, the Hamiltonian can be identified as nothing more than a superposition of independent quantum harmonic oscillators. The only difference between (11.10) and the canonical form of an oscillator Hamiltonian p2 H = 2m + 12 mω 2 x2 is the presence of sub-indices k and −k (a consequence † of φˆk = φˆ−k ). As we will show shortly, this difference is inessential. This Advanced Quantum Physics
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11.1. QUANTIZATION OF THE CLASSICAL ATOMIC CHAIN result is actually not difficult to understand (see figure): Classically, the system supports a discrete set of wave-like excitations, each indexed by a wave number k = 2πm/L. Within the quantum picture, each of these excitations is described by an oscillator Hamiltonian with a k-dependent frequency. However, it is important not to confuse the atomic constituents, also oscillators (albethey coupled), with the independent collective oscillator modes described ˆ by H. The description above, albeit perfectly valid, still suffers from a deficiency: Our analysis amounts to explicitly describing the effective low energy excitations of the system (the waves) in terms of their microscopic constituents (the ˆ keeps track of details of the atoms). Indeed the different contributions to H microscopic oscillator dynamics of individual k-modes. However, it would be much more desirable to develop a picture where the relevant excitations of the system, the waves, appear as fundamental units, without explicit account of underlying microscopic details. (As with hydrodynamics, information is encoded in terms of collective density variables rather than through individual molecules.) To understand how this programme can be achieved let us recall the properties of the quantum harmonic oscillator. " Info. In quantum mechanics, the harmonic oscillator has the status of a single-particle problem. However, the fact that the energy levels, &n = !ω(n + 1/2), are equidistant suggests an alternative interpretation: One can think of a given energy state &n as an accumulation of n elementary entities, or quasi-particles, each having energy !ω. What can be said about the features of these new objects? First, they are structureless, i.e. the only ‘quantum number’ identifying the quasi-particles is their energy !ω (otherwise n-particle states formed of the quasi-particles would not be equidistant). This implies that the quasi-particles must be bosons. (The same state !ω can be occupied by more than one particle — see figure.) This idea can be formulated in quantitative terms by employing the formalism of ladder operators in which pˆ and ) x ˆ are traded for the pair of Hermitian adjoint operators )the operators i mω i † a ≡ mω (ˆ x + p ˆ ), a ≡ x − mω pˆ). Up to a factor of i, the transformation 2! mω 2! (ˆ (ˆ x, pˆ) → (a, a† ) is canonical, i.e. the new operators obey the canonical commutation relation, [a, a† ] = 1. More importantly, in the a-representation, the Hamiltonian ˆ = !ω(a† a + 1/2), as can be checked by direct substitution. takes the simple form, H The complete hierarchy of higher energy states can be generated by setting |n& ≡ √1 (a† )n |0&. n! While the a-representation provides another way of constructing eigenstates of the quantum harmonic oscillator, its real advantage is that it naturally affords a many-particle interpretation. Temporarily forgetting about the original definition of the oscillator, we can declare |0& to be a ‘vacuum’ state, i.e. a state with no particles present. a† |0& then represents a state with a single featureless particle (the operator a† does not carry any quantum number labels) of energy !ω. Similarly, (a† )n |0& is considered as a many-body state with n particles, i.e. within the new picture, a† is an operator that creates particles. The total energy of these states is given by !ω × (occupation number). Indeed, it is straightforward to verify that a† a|n& = n|n&, i.e. the Hamiltonian basically counts the number of particles. While, at first sight, this may look unfamiliar, the new interpretation is internally consistent. Moreover, it fulfils our objective: it allows an interpretation of the excited states of the harmonic oscillator as a superposition of independent structureless entities. With this background, we may return to the harmonic atomic chain (11.10)
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11.1. QUANTIZATION OF THE CLASSICAL ATOMIC CHAIN
133
and, inspired by the ladder operator formalism, define6 0 0 2 2 1 1 mωk ˆ i mωk ˆ i † π ˆ−k , ak ≡ π ˆk . ak ≡ φk + φ−k − 2! mωk 2! mωk With this definition, one finds that the ladder operators obey the commutation relations (exercise) [ak , a†k! ] = δkk! ,
[ak , ak! ] = [a†k , a†k! ] = 0 ,
(11.11)
and the Hamiltonian assumes the diagonal form 1 2 ! 1 † ˆ !ωk ak ak + H= . 2
(11.12)
k
Eqs. (11.11) and (11.12) represent the final result of our analysis: The low-lying elementary excitations of the discrete atomic chain are described by oscillator wave-like modes – known as phonons – each characterised by a wavevector k and a linear dispersion, ωk = v|k|. A generic state of the system is given by 1 |{nk } = (n1 , n2 , · · ·)& = )3
i ni !
(a†k1 )n1 (a†k2 )n2 · · · |0& .
The representation derived above illustrates the capacity to think about quantum problems in different complementary “pictures”, a principle that finds innumerable applications. The existence of different interpretations of a given system is by no means heretic but, rather, is consistent with the spirit of quantum mechanics. Indeed, it is one of the prime principles of quantum theories that there is no such thing as ‘the real system’ which underpins the phenomenology. The only thing that matters is observable phenomena. For example, the ‘fictitious’ quasi-particle states of the harmonic chain, the phonons, behave as ‘real’ particles, i.e. they have dynamics, can interact, be detected experimentally, etc. From a quantum point of view there is actually no fundamental difference between these objects and ‘real’ particles. " Example: Debye theory of solids: Our analysis above focussed on the longitudinal vibrations of the one-dimensional atomic chain. In three-dimensions, each mode is associated with three possible polarizations, λ: two transverse and one longitudinal. Taking into account all degrees of freedom, it is straightforward to show that the generalized Hamiltonian takes the form, 1 2 ! 1 † ˆ H= !ωk ak,λ ak,λ + , 2 kλ
where, for simplicity, we assume that the dispersion, ωk = v|k| is independent of polarization. Let us use this result to obtain the internal energy and specific heat due to phonons. Now, for an equilibrium thermal distribution, the average phonon occupancy of state (k, λ) is given simply by the Bose-Einstein distribution, nB (!ωk ) ≡ 1 . The internal energy is therefore given by e!ωk /kB T −1 E=
!
!ωk (nB (!ωk ) + 1/2) =
kλ
! kλ
!ωk
"
1 e!ωk /kB T − 1
+
# 1 . 2
In the thermodynamic limit, where N ≡ V /a3 → ∞, we may replace the sum over & kD d3 k & kD 2 + V modes by an integral, k → V 0 (2π)3 = 2π 2 0 k dk, where kD , denotes the As for the consistency of these definitions, recall that φˆ†k = φˆ−k and π ˆk† = π ˆ−k . Under these conditions the second of the definitions below indeed follows from the first upon taking the Hermitian conjuate. 6
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Figure shows a typical measured phonon dispersion of an ordered crystalline solid obtained by neutron scattering. The x-axis indexes wavenumbers along a lattice direction (specified in units of π/a). Three generic aspects are visible: (1) near k = 0, the dispersion is, as expected, linear. (2) The several branches are associated with different “polarizations” of the lattice fluctuations. (3) For wavelengths comparable to the lattice spacing, k ∼ π/a, non-universal features specific to the particular material become visible.
Peter Josephus Wilhelmus Debye 1884-1966: Dutch-American physicist renowned for his work on molecular structure, especially dipole moments and the diffraction of X-rays and electrons in gases. Debye was awarded a Nobel Prize in Chemistry, 1936, “for his contributions to our knowledge of molecular structure through his investigations on dipole moments and on the diffraction of X-rays and electrons in gases”.
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134
largest wave vector accessible in the crystal. We can fix kD by ensuring that the total number of modes (for each polarization) matches the total number of degrees 2 1/3
3 = N , i.e. kD = (6π a) . The corresponding frequency of freedom, i.e. (2π)13 /V 43 πkD scale, ωD = vkD , is known as the Debye frequency. In this limit, dropping the temperature independent contribution from zero point fluctuations, the internal energy per particle is given by % !vk E 3a3 kD 2 k dk !vk/k T ε≡ = . B N 2π 2 0 e −1
& T /T Then, defining the Debye temperature, TD = !vkD /kB , we have ε = 9kB T ( TTD )3 0 D The corresponding specific heat per particle can be obtained from the temperature derivative and leads to 1 23 % TD /T T z 4 dz cV = 9kB . TD (ez − 1)2 0 In particular, at high temperatures, we recover the Dulong-Petit law, cV = 3kB following from the equipartition theorem – each degree of freedom is associated with an energy kB T /2. At low temperatures, T # TD , we may replace the upper limit on the integral by ∞ from which we find that cV ∼ T 3 . Both limits compare well with experiment (see figure).
This completes our discussion of the classical and quantum field theory of the harmonic atomic chain. In this example, we have seen how we can effect a quantum formulation of a continuum system. Using the insights obtained in this example, we now turn to consider the quantization of the electromagnetic field.
11.2
Quantum electrodynamics
In common with the continuous formulation of the atomic chain, in vacua, the electromagnetic (EM) field satisfies a wave equation. The generality of the procedure outlined above suggests that the quantization of the EM field might therefore proceed in an entirely analogous manner. However, there are a number of practical differences that make quantization a slightly more difficult enterprise: Firstly, the vector character of the vector potential A, alongside relativistic covariance, gives the problem a non-trivial internal geometry. Moreover, the gauge freedom of the vector potential introduces redundant degrees of freedom whose removal on the quantum level is not straightforward. To circumvent a lengthy discussion of these issues, we will not address the problem of EM field quantization in all its detail.7 On the other hand, the photon field plays a much too important role in all branches of physics for us to drop the problem altogether. We will therefore aim at an intermediate exposition, largely insensitive to the problems outlined above but sufficiently general to illustrate the main principles. As with the harmonic chain, to prepare the way, we begin by developing the classical field theory of the EM field.
11.2.1
Classical theory of the electromagnetic field
In vacuum, the Lagrangian density of the EM field is given by L = − 4µ1 0 Fµν F µν (summation convention implied) where µ0 = 4π ×10−7 Hm−1 denotes the vac7 Readers interested in a more thorough and illuminating exposition are referred to the literature, e.g., L. H. Ryder, Quantum Field Theory (Cambridge University Press, 1996), or the excellent lecture notes of Eduardo Fradkin that have been made available online at http://webusers.physics.illinois.edu/ efradkin/phys582/physics582.html.
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z 3 dz ez −1 .
11.2. QUANTUM ELECTRODYNAMICS
135
uum permeability,
Fµν
0 −Ex /c −Ey /c −Ez /c Ex /c 0 Bz −By = ∂µ Aν − ∂ν Aµ = Ey /c −Bz 0 Bx Ez /c By −Bx 0 µν
˙ is the electric field, and B = ∇ × A is the denotes the EM field tensor, E = A magnetic field. As a first step towards quantization, we must specify a gauge. In the absence of charge, a particularly convenient choice is the Coulomb gauge, ∇ · A = 0, with the scalar component φ = 0.8 Using these gauge conditions, one may verify that the classical Lagrangian assumes the form, L[A(x, t)] =
1 2µ0
%
d3 x
"
# 1 ˙2 2 A − (∇ × A) . c2
(11.13)
The corresponding classical Euler-Lagrange equations of motion, ∂µ F µν = 0, translate to the wave equation (exercise) 1 ¨ A = ∇2 A . c2 The structural similarity between the EM field and the continuous formulation of the harmonic chain is clear. By analogy with our discussion above, we should now switch to the Fourier representation and quantize the classical field. However, in contrast to our analysis of the chain, we are now dealing (i) with the full three-dimensional Laplacian acting upon (ii) the vector field A that is (iii) subject to the constraint ∇ · A = 0. It is these aspects which lead to the complications outlined above. We can circumvent these difficulties by considering cases where the geometry of the system reduces the complexity of the eigenvalue problem while still retaining the key conceptual aspects of the problem. This restriction is less artificial than it might appear. For example, just as the field φ in the classical atomic chain can be expanded in Fourier harmonics, in long waveguides, the EM vector potential can be expanded in solutions of the eigenvalue equation9 −∇2 uk (x) = λk uk (x),
(11.14)
where k denotes a discrete one-dimensional & index, and the vector-valued functions uk are real and orthonormalized, d3 x uk · uk! = δkk! . The dependence of the eigenvalues λk on k depends on details of the geometry and need not be specified for the moment. " Info. An electrodynamic waveguide is a quasi one-dimensional cavity with metallic boundaries (see Fig. 11.2). The practical advantage of waveguides is that they are good at confining EM waves. At large frequencies, where the wavelengths are of order meters or less, radiation loss in conventional conductors is high. In these frequency domains, hollow conductors provide the only practical way of transmitting radiation. EM field propagation inside a waveguide is constrained by boundary conditions. Assuming the walls of the system to be perfectly conducting, E$ (xb ) = 0, 8
B⊥ (xb ) = 0 ,
(11.15)
Keep in mind that, once a gauge has been set, we cannot expect further results to display “gauge invariance.” 9 More precisely, one should say that Eq. (11.14) defines the set of eigenfunctions relevant for the low-energy dynamics of the waveguide. More complex eigenfunctions of the Laplace operator exist but they involve much higher energy.
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136
Figure 11.2: EM waveguide with rectangular cross-section. The structure of the
eigenmodes of the EM field is determined by boundary conditions at the walls of the cavity. where xb parameterize points on the boundary of the system, and E$ (B⊥ ) is the parallel (perpendicular) component of the electric (magnetic) field. Applied to the problem at hand, let us consider a long cavity with uniform rectangular cross-section Ly × Lz . To conveniently represent the Lagrangian of the system, we wish to express the vector potential in terms of eigenfunctions uk that are consistent with the boundary conditions (11.15). A complete set of functions fulfilling this condition is given by c1 cos(kx x) sin(ky y) sin(kz z) uk = Nk c2 sin(kx x) cos(ky y) sin(kz z) . c3 sin(kx x) sin(ky y) cos(kz z) Here ki = ni π/Li , with i = x, y, z and ni is integer, Nk is a factor normalizing uk to unit modulus, and the coefficients ci are subject to the condition c1 kx + c2 ky + c3 kz = 0 (reflecting the gauge choice ∇ · A = 0). Indeed, + it is straightforward to verify that a general superposition of the type A(x, t) ≡ k αk (t)uk (x), αk (t) ∈ R, is divergenceless, and generates an EM field compatible with (11.15). Substitution of uk into (11.14) identifies the eigenvalues as λk = kx2 + ky2 + kz2 . In the physics and electronic engineering literature, eigenfunctions of the Laplace operator in a quasi-one-dimensional geometry are commonly described as modes. As we will see shortly, the energy of a mode (i.e. the Hamiltonian evaluated on a specific mode configuration) grows with λk . In cases where one is interested in the low-energy dynamics of the EM field, only configurations with small λk are relevant. For example, let us consider a massively anisotropic waveguide with Lz < Ly # Lx . In this case the modes with smallest λk are those with kz = 0, ky = π/Ly , and kx ≡ k # L−1 z,y . With this choice, 2 ˆz , uk = √ sin(πy/Ly ) sin(kx) e V
λk = k + 2
1
π Ly
22
,
(11.16)
and a scalar index k suffices to label both eigenvalues and eigenfunctions uk . A caricature of the spatial structure of the functions uk is shown in Fig. 11.2.
Returning to the problem posed by (11.13) and (11.14), one + can expand the vector potential in terms of eigenfunctions uk as A(x, t) = k αk (t)uk (x), where the sum runs over all allowed values of the index parameter k. (In a waveguide of length L, k = πn/L with n integer.) Substituting this expansion into (11.13), and using the normalization properties of uk , we obtain the Lagrangian, " # 1 ! 1 2 2 L[α, ˙ α] = α ˙ − λ α k k , 2µ0 c2 k k
i.e. a decoupled representation where the system is described in terms of independent dynamical systems with coordinates αk . From this point on, the quantization procedure mirrors that of the atomic chain.
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11.2. QUANTUM ELECTRODYNAMICS
11.2.2
137
Quantum field theory of the electromagnetic field
To achieve the electromagnetic field quantization, we first define the canonical momenta through the relation, πk = ∂α˙ k L = &0 α˙ k , where &0 = 1/µ0 c2 denotes+the vacuum permittivity, which leads to the classical Hamiltonian H = k ( 2'10 πk2 + 12 &0 c2 λk αk2 ). Next we quantize the theory by promoting fields to operators αk → α ˆ k and πk → π ˆk , and declare the canonical commutation relations [ˆ πk , α ˆ k! ] = −i!δkk! . The quantum Hamiltonian operator, again of harmonic oscillator type, then reads # !" π ˆk2 1 2 2 ˆ H= + &0 ωk α ˆk , 2&0 2 k
where ωk2 = c2 λk . Then, guided by the analysis of the atomic chain, we now introduce the ladder operators, 0 0 2 2 1 1 &0 ωk i &0 ωk i † π ˆk , ak = π ˆk , α ˆk + α ˆk − ak = 2! &0 ωk 2! &0 ωk whereupon the Hamiltonian assumes the now familiar form 1 2 ! 1 † ˆ H= !ωk ak ak + . 2
(11.17)
k
For the specific problem of the first excited mode in a waveguide of width Ly , !ωk = c[k 2 + (π/Ly )2 ]1/2 . Eq. (11.17) represents our final result for the quantum Hamiltonian of the EM waveguide. Before concluding this section let us make a few comments on the structure of the result. " Firstly, notice that the construction above almost completely paralleled our previous discussion of the atomic chain.10 The structural similarity between the two systems finds its origin in the fact that the free field Lagrangian (11.13) is quadratic in the fields and, therefore, bound to map onto an oscillator-type Hamiltonian. That we obtained a simple one-dimensional superposition of oscillators is due to the boundary conditions specific to a narrow waveguide. For less restrictive geometries, e.g. free space, a more complex superposition of vectorial degrees of freedom in three-dimensional space would have been obtained (see below). However, the principal mapping of the free EM field onto a superposition of oscillators is independent of geometry. " Physically, the quantum excitations described by (11.17) are, of course, the photons of the EM field. The unfamiliar appearance of the dispersion ωk is again a peculiarity of the waveguide. However, in the limit of large longitudinal wave numbers k - L−1 y , the dispersion approaches ωk . c|k|, i.e. the familiar linear (relativistic) dispersion of the photon field. Also notice that, due to the equality of the Hamiltonians (11.12) and (11.17), all that has been said about the behavior of the phonon modes of the atomic chain carries over to the photon modes of the waveguide. 10 Technically, the only difference is that, instead of index pairs (k, −k), all indices (k, k) are equal and positive. This can be traced back to the fact that we have expanded in terms of the real eigenfunctions of the closed waveguide instead of the complex eigenfunctions of the periodic oscillator chain.
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11.2. QUANTUM ELECTRODYNAMICS
138
" As with their phonon analogue, the oscillators described by (11.17) exhibit zero-point fluctuations. It is a fascinating aspect of quantum electrodynamics that these oscillations, caused by quantization of the most relativistic field, surface at various points of non-relativistic physics, e.g. the attraction of two conducting plates in vacuum – the Casimir effect. With the analysis of the waveguide complete, let us go back and consider the quantitization of the full three-dimensional system. For the waveguide, we have found + that the vector potential can be expanded in modes of the cavity as ˆ A(x) = ˆ k uk where, rearranging the expressions for the ladder operators, : kα α ˆk = may
† ! 2'0 ωk (ak + ak ). show that11
ˆ A(x) =
!
More generally, in a fully three-dimensional cavity, one
kλ=1,2
;
< = ! ˆkλ akλ eik·x + e ˆ∗kλ a†kλ e−ik·x , e 2&0 ωk V
where V denotes the volume of the system, ωk = c|k|, and the two sets of ˆkλ , are in general complex and normalized to unity, polarization vectors, e ˆ∗kλ · e ˆkλ = 1. To ensure that the vector potential satifies the Coulomb gauge e ˆkλ · k = e ˆ∗kλ · k = 0, i.e. the two polarization condition, we require that e ˆkλ correspond vectors are orthogonal to the wave vector. Two real vectors, e to two linear polarizations while, for circular polarization, the vectors are complex. It is also convenient to assume that the two polarization vectors are ˆkλ · e ˆkµ = δµν . The corresponding operators obey the mutually orthogonal, e commutation relations, [akλ , a†k! λ! ] = δk,k! δλλ! ,
[akλ , ak! λ! ] = 0 = [a†kλ , a†k! λ! ] .
With these definitions, the Hamiltonian then takes the familiar form ˆ = H
! kλ
< = !ωk a†kλ akλ + 1/2 ,
(11.18)
while, defining the vacuum, |Ω&, the eigenstates involve photon number states, |{nkλ }& ≡ | · · · , nkλ , · · ·& =
> (a† )nkλ √kλ |Ω& . nkλ ! kλ
Finally, in practical applications (including our forthcoming study of radiative transitions in atoms), it is convenient to transfer the time-dependence to the operators by turning to the Heisenberg representation. In this representation, the field operators obey the Heisenberg equations of motion (exercise), a˙ kλ =
i ˆ [H, akλ ] = −iωk akλ . !
Integrating, we have akλ (t) = akλ (0)e−iωk t , which translates to the relation ˆ A(x, t) =
!
kλ=1,2
11
;
< = ! ˆkλ akλ ei(k·x−ωk t) + e ˆ∗kλ a†kλ e−i(k·x−ωk t) . e 2&0 ωk V
In the infinite system, the mode sum becomes replaced by an integral,
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P
k
→
V (2π)3
R
d3 k.
Chapter 12
Time-dependent perturbation theory So far, we have focused largely on the quantum mechanics of systems in which the Hamiltonian is time-independent. In such cases, the time dependence of a wavepacket can be developed through the time-evolution operator, ˆ ˆ = e−iHt/! U or, when cast in terms of the eigenstates of the Hamiltonian, ! ˆ ˆ H|n! = En |n!, as |ψ(t)! = e−iHt/!|ψ(0)! = n e−iEn t/!cn (0)|n!. Although this framework provides access to any closed quantum mechanical system, it does not describe interaction with an external environment such as that imposed by an external electromagnetic field. In such cases, it is more convenient ˆ 0 , through a to describe the induced interactions of a small isolated system, H time-dependent interaction V (t). Examples include the problem of magnetic resonance describing the interaction of a quantum mechanical spin with an external time-dependent magnetic field, or the response of an atom to an external electromagnetic field. In the following, we will develop a formalism to treat time-dependent perturbations.
12.1
Time-dependent potentials: general formalism
ˆ = H ˆ 0 + V (t), where all time-dependence Consider then the Hamiltonian H enters through the potential V (t). In the Schr¨ odinger representation, the dynamics of the system are specified by the time-dependent wavefunction, ˆ |ψ(t)!S through the Schr¨odinger equation i!∂t |ψ(t)!S = H|ψ(t)! S . However, in many cases, and in particular with the current application, it is convenient to work in the Interaction representation,1 defined by ˆ
|ψ(t)!I = eiH0 t/!|ψ(t)!S where |ψ(0)!I = |ψ(0)!S . With this definition, one may show that the wavefunction obeys the equation of motion (exercise) i!∂t |ψ(t)!I = VI (t)|ψ(t)!I ˆ
(12.1)
ˆ
where VI (t)!= eiH0 t/!V e−iH0 t/!. Then, if we form the eigenfunction expansion, |ψ(t)!I = n cn (t)|n!, and contract the equation of motion with a general state, "n|, we obtain " i!c˙m (t) = Vmn (t)eiωmn t cn (t) , (12.2) n
1
Note how this definition differs from that of the Heisenberg representation, |ψ!H = |ψ(t)!S in which all time-dependence is transferred into the operators.
ˆ iHt/!
e
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12.1. TIME-DEPENDENT POTENTIALS: GENERAL FORMALISM where the matrix elements Vmn (t) = "m|V (t)|m!, and ωmn = (Em − En )/! = −ωnm . To develop some intuition for the action of a time-dependent potential, it is useful to consider first a periodically-driven two-level system where the dynamical equations can be solved exactly. $ Info. The two-level system plays a special place in the modern development of quantum theory. In particular, it provides a platform to encode the simplest quantum logic gate, the qubit. A classical computer has a memory made up of bits, where each bit holds either a one or a zero. A quantum computer maintains a sequence of qubits. A single qubit can hold a one, a zero, or, crucially, any quantum superposition of these. Moreover, a pair of qubits can be in any quantum superposition of four states, and three qubits in any superposition of eight. In general a quantum computer with n qubits can be in an arbitrary superposition of up to 2n different states simultaneously (this compares to a normal computer that can only be in one of these 2n states at any one time). A quantum computer operates by manipulating those qubits with a fixed sequence of quantum logic gates. The sequence of gates to be applied is called a quantum algorithm. An example of an implementation of qubits for a quantum computer could start with the use of particles with two spin states: |↓! and |↑!, or |0! and |1!). In fact any system possessing an observable quantity A which is conserved under time evolution and such that A has at least two discrete and sufficiently spaced consecutive eigenvalues, is a suitable candidate for implementing a qubit. This is true because any such system can be mapped onto an effective spin-1/2 system.
$ Example: Dynamics of a driven two-level system: Let us consider a two-state system with $ # # $ 0 δeiωt E1 0 ˆ0 = , V (t) = H . 0 E2 δe−iωt 0 Specifying the wavefunction by the two-component vector, c(t) = (c1 (t) c2 (t)), Eq. (12.2) translates to the equation of motion (exercise) # $ 0 ei(ω−ω21 )t i!∂t c = δ c(t) , e−i(ω−ω21 )t 0 where ω21 = (E2 − E1 )/!. With the initial condition c1 (0) = 1, and c2 (0) = 0, this equation has the solution, |c2 (t)|2 =
δ2 sin2 Ωt, δ 2 + !2 (ω − ω21 )2 /4
|c1 (t)|2 = 1 − |c2 (t)|2 ,
where Ω = ((δ/!)2 +(ω −ω21 )2 /4)1/2 is known as the Rabi frequency. The solution, which varies periodically in time, describes the transfer of probability from state 1 to state 2 and back. The maximum probability of occupying state 2 is a Lorentzian with |c2 (t)|2max =
γ2 , γ 2 + !2 (ω − ω21 )2 /4
taking the value of unity at resonance, ω = ω21 . $ Exercise. Derive the solution from the equations of motion for c(t). Hint: eliminate c1 from the equations to obtain a second order differential equation for c2 .
$ Info. The dynamics of the driven two-level system finds practical application in the Ammonia maser: The ammonia molecule NH3 has a pryramidal structure with an orientation characterised by the position of the “lone-pair” of electrons sited Advanced Quantum Physics
140
12.2. TIME-DEPENDENT PERTURBATION THEORY
141
on the nitrogen atom. At low temperature, the molecule can occupy two possible states, |A! and |S!, involving symmetric (S) or an antisymmetric (A) atomic configurations, separated by a small energy splitting, ∆E. (More precisely, along the axis of three-fold rotational symmetry, the effective potential energy of the nitrogen atom takes the form of a double-well. The tunneling of the nitrogen atom through the double well leads to the symmetric and asymmetric combination of states.) In a timedependent uniform electric field the molecules experience a potential V = −µd · E, where E = Eˆ ez cos ωt, and µd denotes the electric dipole moment. Since µd is odd under parity transformation, P µd P = −µd , and P |A! = −|A! and P |S! = |S!, the matrix elements of the electric dipole moment are off-diagonal: "S|µd |S! = "A|µd |A! = 0 and "S|µd |A! = "S|µd |A! = & 0. If we start with all of the molecules in the symmetric ground state, we have shown above that the action of an oscillating field for a particular time can can drive a collection of molecules from their ground state into the antisymmetric first excited state. The ammonia maser works by sending a stream of ammonia molecules, traveling at known velocity, down a tube having an oscillating field for a definite length, so the molecules emerging at the other end are all (or almost all, depending on the precision of ingoing velocity, etc.) in the first excited state. Application of a small amount of electromagnetic radiation of the same frequency to the outgoing molecules will cause some to decay, generating intense radiation and therefore a much shorter period for all to decay, emitting coherent radiation.
12.2
Time-dependent perturbation theory
We now turn to consider a generic time-dependent Hamiltonian for which an analytical solution is unavailable – sadly the typical situation! In this case, we must turn to a perturbative analysis, looking for an expansion of the basis coefficients cn (t) in powers of the interaction, (1) (2) cn (t) = c(0) n + cn (t) + cn (t) + · · · , (m)
(0)
where cn ∼ O(V m ) and cn is some (time-independent) initial state. The programme to complete this series expansion is straightforward but technical. $ Info. In the interaction representation, the state |ψ(t)!I can be related to an inital state |ψ(t0 )!I through the time-evolution operator, UI (t, t0 ), i.e. |ψ(t)!I = UI (t, t0 )|ψ(t0 )!I . Since this is true for any initial state |ψ(t0 )!I , from Eq. (12.1), we must have i!∂t UI (t, t0 ) = VI (t)UI (t, t0 ) , with the boundary condition UI (t0 , t0 ) = I. Integrating this equation from t0 to t, formally we obtain, % i t " UI (t, t0 ) = I − dt VI (t" )UI (t" , t0 ) . ! t0 This result provides a self-consistent equation for UI (t, t0 ), i.e. if we take this expression and substitute UI (t" , t0 ) under the integrand, we obtain UI (t, t0 ) = I −
i !
%
t
t0
# $2 % t % t! i dt" VI (t" ) + − dt" VI (t" ) dt"" VI (t"" )UI (t"" , t0 ) . ! t0 t0
Iterating this procedure, we thus obtain UI (t, t0 ) =
$n % t % tn−1 ∞ # " i dt1 · · · dtn VI (t1 )VI (t2 ) · · · VI (tn ) , − ! t0 t0 n=0
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(12.3)
Charles (left)
Hard
Townes
1915-
is an American Nobel prizewinning physicist and educator. Townes is known for his work on the theory and application of the maser – microwave amplification by stimulated emission of radiation, on which he got the fundamental patent, and other work in quantum electronics connected with both maser and laser devices. He received the Nobel Prize in Physics in 1964.
12.2. TIME-DEPENDENT PERTURBATION THEORY
142
where the term n = 0 translates to I. Note that the operators VI (t) are organised in a time-ordered sequence, with t0 ≤ tn ≤ tn−1 ≤ · · · t1 ≤ t. With this understanding, we can write this expression more compactly as & i Rt ! ! ' − dt VI (t ) UI (t, t0 ) = T e ! t0 ,
where “T” denotes the time-ordering operator and its action is understood by Eq. (12.3).
If a system is prepared in an initial state, |i! at time t = t0 , at a subsequent time, t, the system will be in a final state,
|i, t0 , t! = UI (t, t0 )|i! =
" n
cn (t) ( )* + |n! "n|UI (t, t0 )|i! .
Making use of Eq. (12.3), and the resolution of identity, obtain c(0) n ()*+ (
cn (t) = δni
!
m |m!"m|
= I, we
c(2) n )* + % t % t % t! " i 1 − dt" "n|VI (t" )|i! − 2 dt" dt"" "n|VI (t" )|m!"m|VI (t"" )|i! + · · · . ! t0 ! t0 t0 m c(1) n )*
ˆ
+(
ˆ
Recalling that VI = eiH0 t/!V e−iH0 t/!, we thus find that % i t " iωni t! dt e Vni (t" ) ! t0 % % ! 1 " t " t "" iωnm t! +iωmi t!! (2) cn (t) = − 2 dt dt e Vnm (t" )Vmi (t"" ) , ! m t0 t0
c(1) n (t) = −
(12.4)
where Vnm (t) = "n|V (t)|m! and ωnm = (En − Em )/!, etc. In particular, the probability of effecting a transition from state |i! to state |n! for n &= i is given (1) (2) by Pi→n = |cn (t)|2 = |cn (t) + cn (t) + · · · |2 . $ Example: The kicked oscillator: Suppose a simple harmonic oscillator is prepared in its ground state |0! at time t = −∞. If it is perturbed by a small time2 2 dependent potential V (t) = −eEx e−t /τ , what is the probability of finding it in the first excited state, |1!, at t = +∞? Working to the first order of perturbation theory, the probability is given by ,t !2 2 ! (1) (1) P0→1 * |c1 |2 where c1 (t) = − !i t0 dt" eiω10 t V10 (t" ), V10 (t" ) = −eE"1|x|0!e−t /τ and ω10 = ω. Using the ladder operator formalism, with |1! = a† |0! and x = -
! ! (a + a† ), we have "1|x|0! = 2mω . Therefore, making use of the identity , ∞2mω " √ 2 dt exp[iωt" − t" /τ 2 ] = πτ exp[−ω 2 τ 2 /4], we obtain the transition amplitude, −∞ . π −ω2 τ 2 /4 (1) c1 (t → ∞) = ieEτ 2m!ω e . As a result, we obtain the transition probabil2 −ω 2 τ 2 /2 ity, P0→1 * (eEτ ) (π/2m!ω)e . Note that the probability is maximized for τ ∼ 1/ω.
$ Exercise. Considering the same perturbation, calculate the corresponding transition probability from the ground state to the second excited state. Hint: note that this calculation demands consideration of the second order of perturbation theory.
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12.3. “SUDDEN” PERTURBATION
12.3
143
“Sudden” perturbation
To further explore the time-dependent perturbation theory, we turn now to consider the action of fast or “sudden” perturbations. Here we define sudden as a perturbation in which the switch from one time-independent Hamiltonian ˆ 0 to another H ˆ " takes place over a time much shorter than any natural H 0 period of the system. In this case, perturbation theory is irrelevant: if the ˆ 0 , its time evolution following the system is initially in an eigenstate |n! of H " ˆ switch will follow that of H0 , i.e. one simply has to expand the initial state as ˆ " , |n! = ! ! |n" !"n" |n!. The non-trivial part a sum over the eigenstates of H 0 n of the problem lies in establishing that the change is sudden enough. This is achieved by estimating the actual time taken for the Hamiltonian to change, and the periods of motion associated with the state |n! and with its transitions to neighboring states.
12.3.1
Harmonic perturbations: Fermi’s Golden Rule
Let us then consider a system prepared in an initial state |i! and perturbed by a periodic harmonic potential V (t) = V e−iωt which is abruptly switched on at time t = 0. This could represent an atom perturbed by an external oscillating electric field, such as an incident light wave. What is the probability that, at some later time t, the system lies in state |f!? From Eq. (12.4), to first order in perturbation theory, we have (1)
cf (t) = −
i !
%
0
t
i ei(ωfi −ω)t − 1 ! dt" "f|V |i!ei(ωfi −ω)t = − "f|V |i! . ! i(ωfi − ω)
The probability of effecting the transition after a time t is therefore given by # $2 1 (1) 2 2 sin((ωfl − ω)t/2) Pi→f (t) * |cf (t)| = 2 |"f|V |i!| . ! (ωfl − ω)/2 Setting α = (ωfl − ω)/2, the probability takes the form sin2 (αt)/α2 with a peak at α = 0, with maximum value t2 and width of order 1/t giving a total weight of order t. The function has more peaks positioned at αt = (n + 1/2)π. These are bounded by the denominator at 1/α2 . For large t their contribution comes from a range of order 1/t also, and as t → ∞ the function tends towards a δ-function centred at the origin, but multiplied by t, i.e. the likelihood of transition is proportional to time elapsed. We should therefore divide by t to get the transition rate. ,∞ 2 Finally, with the normalisation, −∞ dα( sin(αt) α ) = πt, we may effect the 2 replacement, limt→∞ 1t ( sin(αt) α ) = πδ(α) = 2πδ(2α) leading to the following expression for the transition rate,
Ri→f (t) = lim
t→∞
Pi→f (t) 2π = 2 |"f|V |i!|2 δ(ωfl − ω) . t !
(12.5)
This expression is known as Fermi’s Golden Rule.2 One might worry that, in the long time limit, we found that the probability of transition is in fact 2
Curiously, although named after Fermi, most of the work leading to the Golden Rule was undertaken in an earlier work by Dirac, (P. A. M. Dirac, The quantum theory of emission and absorption of radiation. Proc. Roy. Soc. (London) A 114, 243265 (1927)) who formulated an almost identical equation, including the three components of a constant, the matrix element of the perturbation and an energy difference. It is given its name due to the fact that, being such a useful relation, Fermi himself called it “Golden Rule No. 2” (E. Fermi, Nuclear Physics, University of Chicago Press, 1950).
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Plot of sin2 (αt)/α2 for t = 1. Note that, as t → ∞, this function asymptotes to a δ-function, πtδ(α).
Enrico Fermi 1901-1954: An Italian physicist most noted for his work on the development of the first nuclear reactor, and for his contributions to the development of quantum theory, nuclear and particle physics, and statistical mechanics. Fermi was awarded the Nobel Prize in Physics in 1938 for his work on induced radioactivity and is today regarded as one of the most influential scientists of the 20th century. He is acknowledged as a unique physicist who was highly accomplished in both theory and experiment. Fermium, a synthetic element created in 1952 is named after him.
12.3. “SUDDEN” PERTURBATION
144
diverging — so how can we justify the use of perturbation theory? For a transition with ωfl &= ω, the “long time” limit is reached when t - 1/(ωfl − ω), a value that can still be very short compared with the mean transition time, which depends on the matrix element. In fact, Fermi’s Rule agrees extremely well with experiment when applied to atomic systems. $ Info. Alternative derivation of the Golden Rule: When light falls on an atom, the full periodic potential is not suddenly imposed on an atomic time scale, but builds up over many cycles (of the atom and of the light). If we assume that V (t) = eεt V e−iωt , with ε very small, V is switched on very gradually in the past, and we are looking at times much smaller than 1/ε. We can then take the initial time to be −∞, that is, % ! 1 ei(ωfl −ω−iε)t i t (1) "f|V |i!ei(ωfl −ω−i%)t dt" = − "f|V |i! , cf (t) = − ! −∞ ! ωfl − ω − iε 2εt
e 2 i.e. |cf (t)|2 = !12 (ωfl −ω) 2 +ε2 |"f|V |i!| . Applied to the transition rate 2ε identity limε→0 (ωfl −ω)2 +ε2 → 2πδ(ωfl − ω) leads to the Golden Rule.
d (1) 2 dt |cf (t)| ,
the
From the expression for the Golden rule (12.5) we see that, for transitions to occur, and to satisfy energy conservation: (a) the final states must exist over a continuous energy range to match ∆E = !ω for fixed perturbation frequency ω, or (b) the perturbation must cover a sufficiently wide spectrum of frequency so that a discrete transition with a fixed ∆E = !ω is possible. For two discrete states, since |Vfi |2 = |Vif |2 , we have the semiclassical result Pi→f = Pf→i – a statement of detailed balance.
12.3.2
Info: Harmonic perturbations: second-order transitions
Although the first order perturbation theory is often sufficient to describe transition probabilities, sometimes first order matrix element, "f|V |i! is identically zero due to symmetry (e.g. under parity, or through some selection rule, etc.), but other matrix elements are non-zero. In such cases, the transition may be accomplished by an indirect route. We can estimate the transition probabilities by turning to the second order of perturbation theory (12.4), % ! % 1 " t " t "" iωfm t! +iωmi t!! (2) dt e Vfm (t" )Vmi (t"" ) . dt cf (t) = − 2 ! m t0 t0
If, as above, we suppose that a harmonic potential perturbation is gradually switched on, V (t) = eεt V e−iωt , with the initial time t0 → −∞, we have % t % t! 1 " ! !! (2) cf (t) = − 2 "f|V |m!"m|V |i! dt" dt"" ei(ωfm −ω−iε)t ei(ωmi −ω−iε)t . ! m −∞ −∞ The integrals are straightforward, and yield c(2) n =−
" "f|V |m!"m|V |i! 1 i(ωfi −2ω)t e2εt e . 2 ! ωfi − 2ω − 2iε m ωm − ωi − ω − iε
Then, following our discussion above, we obtain the transition rate: / /2 d (2) 2 2π //" "f|V |m!"m|V |i! // |c | = 4 / / δ(ωfi − 2ω). dt n ! / m ωm − ωi − ω − iε / Advanced Quantum Physics
12.3. “SUDDEN” PERTURBATION This is a transition in which the system gains energy 2!ω from the harmonic perturbation, i.e. two “photons” are absorbed in the transition, the first taking the system to the intermediate energy ωm , which is short-lived and therefore not well defined in energy – indeed there is no energy conservation requirement for the virtual transition into this state, only between initial and final states. Of course, if an atom in an arbitrary state is exposed to monochromatic light, other second order processes in which two photons are emitted, or one is absorbed and one emitted (in either order) are also possible.
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145
Chapter 13
Radiative transitions Previously, we have addressed the quantum theory of atoms coupled to a classical time-independent electromagneic field, cf. our discussion of the Zeeman and Stark effects. However, to develop a complete quantum mechanical description of light-matter interaction, we have to address both the quantum theory of the electromagnetic field and the coupling of light to matter. In the following section, we will address both of these issues in turn. Our motivation for developing such a consistent theory is that it will (a) provide us with a platform to address the problem of radiative transitions in atoms and (b) it forms the basis of the field of quantum optics.
13.1
Coupling of matter to the electromagnetic field
Let us then consider the Hamiltonian of a single-electron atom subject to a time-dependent external electromagnetic field, ˆ atom = 1 (ˆ H p + eA(r, t))2 − eφ(r, t) + V (r) . 2m Here V (r) denotes the binding potential associated with the atomic nucleus. To keep our discussion of a complex problem as simple as possible, we have focussed on the single electron system. However, a generalization of the methodology to multi-electron atoms would not present significant challenges. Exˆ atom = panding the kinetic energy, the atomic Hamiltonian can be recast as H ˆ ˆ ˆ H0 + Hpara + Hdia. , where ˆ2 ˆ0 = p H + V (r) , 2m denotes the usual non-interacting Hamiltonian of the isolated atom, ˆ para (t) = e A(t) · p ˆ, H m represents the time-dependent paramagnetic term arising from the coupling of ˆ dia = (eA)2 /2m represents the the electron to the electromagnetic field, and H diamagnetic term. Since we will be interested in the absorption and emission of single photons, we can neglect the influence of diamagnetic term which presents only a tiny perturbation in the atomic system. Previously, in chapter 11.2, we have see that the quantum Hamiltonian for the electromagnetic field can be expressed as, " # ! 1 ˆ rad = H , !ωk a†kλ akλ + 2 k,λ=1,2
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147
where the operators a†kλ and akλ create and annihilate photons with wavevector k and polarization λ, and ωk = c|k|. These ladder operators obey the (bosonic) commutation relations, [akλ , a†k! λ! ] = δk,k! δλ,λ! , with [akλ , ak! λ! ] = [a†kλ , a†k! λ! ] = 0, and act on photon number states as akλ |nkλ " =
a†kλ |nkλ "
=
√
√
nkλ |nkλ − 1"
nkλ + 1|Nkλ + 1" .
Here |nkλ " represents a photon number state with nk,λ photons in the mode (kλ). Finally, in the Heisenberg representation, we have seen that the vector potential can be expanded in field operators as A(r, t) =
! k,λ
$
% & ! ˆkλ akλ ei(k·r−ωk t) + e ˆ∗kλ a†kλ e−i(k·r−ωk t) . e 2%0 ωk V
ˆ =H ˆ0 +H ˆ rad + H ˆ para (t) specify the full quantum mechanical Taken together, H Hamiltonian of the atom light system.
13.1.1
Spontaneous emission
With this background, let us now consider the probability for an atom, initially in a state |i" to make a transition to a state |f" leading to the emission of a photon of wavevector k and polarization λ – a process of spontaneous emission. If we suppose that the radiation field is initially prepared in the vacuum state, |0", then the final state involves one photon, |kλ" = a†kλ |0". Therefore, making ˆ ˆ para = e A(t) ˆ, use of Fermi’s Golden rule (12.5), with the perturbation H ·p m we have the transition probability Γi→f (t) =
2π ˆ para |i" ⊗ |Ω"|2 δ(ωif − ωk ) |$f| ⊗$ kλ|H !2
ˆ where ωif = (Ei − Ef )/!. Then substituting the field operator expansion of A, we have $ ' '2 ' 2π '' e ! ' ˆ∗kλ a†kλ e−ik·r · p ˆ |i" ⊗ |Ω"' δ(ωif − ωk ) Γi→f,kλ = e '$f| ⊗$ Ω|akλ ' ! ' m 2%0 ωk V
As a result, we finally obtain Γi→f,kλ
2π = !
$ ' '2 ' e ' ! ' ' ∗ −ik·r ˆkλ e ˆ |i"' δ(Ei − Ef − !ωk ) e ·p '$f| ' m 2%0 ωk V '
To determine the transition rate, we have to analyse matrix elements of the ˆ∗kλ · p ˆ |i". Let us begin by estimating its magnitude. For a typical form $f|eik·r e ˆ " & p & Zmcα, where we have included a general nuclear atomic state, $ˆ e∗kλ · p charge, Z. But what about the exponential factor? With r ∼ !/p & !/mZcα, p2 and ωk = c|k| & 2m (for electronic transitions), we have k·r&
ωk ! !p & & Zα . c p mc
This means that, for Zα ( 1, we can expand the exponential as a power series in k · r with the lowest terms being dominant. Taking the zeroth order ˆ 0 , r] which follows from ˆ = !i [H term, and making use of the operator identity, p Advanced Quantum Physics
Schematic showing spontaneous emission from an initial state at energy Ei = E2 to a final state at energy Ef = E1 .
13.1. COUPLING OF MATTER TO THE ELECTROMAGNETIC FIELD
148
the Heisenberg equations of motion for operators, the matrix element may be written as i ˆ Ef − Ei ∗ ˆ |i" = mˆ ˆkλ · $f|r|i" = −imωk $f|ˆ $f|ˆ e∗kλ · p e∗kλ · $f| [H e e∗kλ · r|i" . 0 , r]|i" = im ! ! This result, which emerges from the leading approximation in Zα, is known as the electric dipole approximation. Effectively, we have set (exercise) e ˆ ˆ t) · r , ˆ & eE(r, A(r, t) · p m translating to the potential energy of a dipole, with moment d = −er, in an oscillating electric field.
13.1.2
Absorption and stimulated emission
Let us now consider the absorption of a photon with wave number, k, and polarization, λ. If we assume that, in the initial state, there are nkλ photons in state (kλ) then, after the transition, there will be nkλ − 1. Then, if the initial state of the atom is i" and the final state is |f", the transition amplitude involves the matrix element, ˆ para |i" ⊗ |nkλ " $f| ⊗$ (nk,λ − 1)|H $ e ! ˆkλ akλ eik·r · p ˆ |i" ⊗ |nkλ " = $f| ⊗$ (nk,λ − 1)| e m 2%0 ωk V Then, using the relation akλ |nkλ " =
√
nkλ |(nkλ − 1)", $ e !nkλ ik·r ˆ para |i" ⊗ |nkλ " = $f| ˆkλ · p ˆ |i" $f| ⊗$ (nk,λ − 1)|H e e m 2%0 ωk V
Schematic showing absorption from an initial state at energy Ei = E1 to a final state at energy Ef = E2 .
As a result, using Fermi’s Golden rule, we obtain the transition amplitude, Γi→f,kλ
2π = !
$ ' '2 ' e ' !nkλ ik·r ' ' ˆkλ · p ˆ |i"' δ(Ef − Ei − !ωk ) e e '$f| ' m 2%0 ωk V '
In particular, we find that the absorption rate increases linearly with photon number, nkλ . Similarly, if we now consider the emission process in which there is are already nk,λ photons in the initial state, we have the revised transition rate, Γi→f,kλ
' '2 $ ' ' ' 2π '' e !(nk,λ + 1) −ik·r ∗ ˆkλ · p ˆ |i"'' δ(Ef − Ei − !ωk ) . = $f| e e ' ! ' m 2%0 ωk V '
This enhancement of the transition rate by the photon occupancy is known as stimulated emission. Altogether, in the dipole approximation, we have the transition rates, Γi→f,kλ =
πωk |$f|ˆ ekλ · d|i"|2 %0 V
(
nkλ δ(Ef − Ei − !ωk ) absorption (nkλ + 1) δ(Ei − Ef − !ωk ) emission
If there are no photons present initially, this expression reduces to that obtained from spontaneous emission. The nkλ -independent component of the Advanced Quantum Physics
Schematic showing the stimulated emission from an initial state at energy Ei = E2 to a final state at energy Ef = E1 .
13.1. COUPLING OF MATTER TO THE ELECTROMAGNETIC FIELD expression for absorption and emission coincide, an equality known as detailed balance. If we are interested in the total rate, dΓλ at which photons of)polarization λ are scattered into the solid angle dΩ, we must compute dRλ = k∈dΩ Γi→f,kλ . Since, in) the elemental d3 k = k 2 dk dΩ, there are d3 kV /(2π)3 states, we * volume V may set k = (2π)3 k 2 dk dΩ. Finally, if we assume that the photon occupation of state (kλ) is isotropic, dependent only on |k|, we find that the integrated * k2 dk λ transition rate per unit solid angle is given by dR Γ from dΩ = V (2π)3 i→f,kλ which we obtain dRλ 1 ω3 = |$f|ˆ ekλ · d|i"|2 dΩ 4π%0 2π!c3
(
nλ (ω) absorption nλ (ω) + 1 emission
Here, in carrying out the integral, we have used the relation ωk = c|k| and !ω = |Ef − Ei |. For a thermal distribution of photons, with the energy density specified by the Planck formula, u(ω) =
!ω 3 n ¯ λ (ω), πc3
n ¯ λ (ω) =
1 e!ωk /kB T
−1
,
this equates a stimulated absorption/emission rate, dRλ 1 1 = |$f|ˆ ekλ · d|i"|2 u(ω) dΩ 4π%0 2!2 From these expressions, we can obtain the power loss as Pλ = !ωRλ . Before discussing the selection rules implied by the form of the dipolar coupling, it is first helpful to digress and discuss connections of this result to a famous result due to Einstein. ( Info. Einstein’s A and B coefficients: In fact, the frequency dependence of the spontaneous emission rate can be inferred without invoking quantum field theoretic methods by means of an ingenious argument due to Einstein which showed that the stimulated and spontaneous transitions must be related. Consider an ensemble of atoms exposed to a black-body radiation field at temperature T . Let us consider transitions between two states |ψj " and |ψk ", with Ek − Ej = !ω. Suppose the numbers of atoms in the two states are nj and nk . The possible transitions and their rates per atom are given by: absorption stimulated emission spontaneous emission
j→k k→j k→j
Bj→k u(ω) Bk→j u(ω) Ak→j (ω)
where u(ω), the energy density of radiation per unit ω. A and B are known as Einstein’s A and B coefficients, and, as we have seen, are properties of the atomic states concerned. Now, in thermodynamic equilibrium, the rates must balance, so that nk [Ak→j (ω) + Bk→j u(ω)] = nj Bj→k u(ω) . At the same time, the relative populations of the two states (assumed non-degenerate for simplicity), are given by a Boltzmann factor nj e−Ej /kB T = −E /k T = e!ω/kB T . nk e k B Thus we have:
+ , Ak→j (ω) = Bj→k e!ω/kB T − Bk→j u(ω) .
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(13.1)
149
13.2. SELECTION RULES
150
For a black-body, the energy density u(ω) is just given by Planck’s formula, u(ω) = !ω 3 1 π 2 c3 e!ω/kB T −1 . The Ak→j coefficient in Eq. (13.1) certainly cannot depend on temperature, so T must cancel on the right hand side. Hence, Bk→j = Bj→k
and
Ak→j (ω) = Bk→j
!ω 3 . π 2 c3
So, the A and B coefficients are related, and if we can calculate the B coefficient for stimulated emission from Fermi’s golden rule, we can infer A, and vice versa.
13.2
Selection Rules
It is clear from the formulae for the transition rates that no transition, either spontaneous or stimulated, will occur between the states |i" and |f" unless at ˆ is nonleast one component of the dipole transition matrix element $f|d|i" zero. It is often possible to show that the matrix elements are zero for certain pairs of states. If so, the transition is not allowed (at least in the electric dipole approximation), and the results can often be summarised in terms of simple selection rules governing the allowed changes in quantum numbers in transitions. ˆ = qr changes sign under parity (r → −r), the Since the dipole operator d ˆ matrix element $f|d|i" will trivially vanish if the states |f" and |i" have the same parity. Therefore, the parity of the wavefunction must change in an electric dipole transition. Moreover, in the absence of spin-orbit interaction, since the wavefunction can be separated into spatial and spin components, |f" = |φf " ⊗ |χf ", with χf being the spin wavefunction, and the dipole operator only acts on the spatial part of the wavefunction, so the matrix element becomes ˆ = $χf |χi " d3 r φ∗ (r) qrφi (r) . $f|d|i" f The spin term $χf |χi " (and therefore the matrix element) vanishes unless |χi " and |χf " are identical. This can be expressed by the selection rule ∆s = 0,
∆ms = 0 .
The spin state is not altered in an electric dipole transition. Let us now consider the selection rules for the orbital angular momenta. ˆ i , rj ] = i!%ijk rk (exercise), it follows that From the operator identity, [L ˆ z , z] = 0, [L
ˆ z , x ± iy] = ±(x ± iy)! , [L
We therefore obtain the relation, ˆ z , z]|+, m" = (m% − m)!$+% , m% |z|+, m" = 0 . $+% , m% |[L ˆ z , x ± iy]|+, m" = ±$+% , m% |x ± iy|+, m", it follows that Similarly, since $+% , m% |[L (m% − m ∓ 1)$+% , m% |x ± iy|+, m" = 0 . Therefore, to get non-zero component of the dipole matrix element, we require. ∆m$ = 0, ±1 . Advanced Quantum Physics
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151
ˆ 2 , [L ˆ 2 , r]] = 2!2 (rL ˆ2 + L ˆ 2 r) (exercise), Similarly, using operator identity [L we have ˆ 2 , [L ˆ 2 , r]]|+, m" = [+% (+% + 1) − +(+ + 1)]2 $+% , m% |r|+, m" $+% , m% |[L = 2[+% (+% + 1) + +(+ + 1)]$+% , m% |r|+, m"
i.e. (+ + +% )(+ + +% + 2)[(+% − +)2 − 1]$+% , m% |r|+, m" = 0. Since +, +% ≥ 0, we can conclude that, to effect an electric dipole transition, we must have ∆+ = ±1 . One may summarize the selection rules for + and m$ is by saying that the photon carries off (or brings in, in an absorption transition) one unit of angular momentum. It should be noted, however, that these rules were derived for the specific case of an electric dipole transition of the system. It is possible, though much less likely in the case of an atom, for the electromagnetic field to interact with some other observable such as the magnetic dipole moment or the electric quadrupole moment. In such transitions the selection rules are different. For ˆ (or −2µB S/! ˆ for the ˆ = −µB L/! example, the magnetic dipole operator is µ spin) and since the angular momentum does not change sign under the parity transformation, there is no change of parity in a magnetic dipole transition. To avoid confusion, we shall continue to confine the discussion to electric dipole transitions, which are responsible for the prominent lines in atomic spectra. ˆz and For transitions with ∆m$ = 0, the dipole matrix element $f|d|i" ∼ e there is no component of polarization along z-direction. Similarly, for electric dipole transitions with m% = m ± 1, $+% , m% |x ∓ iy|+, m" = 0 = $+% , m% |z|+, m", and $f|d|i" ∼ (1, ∓i, 0). In this case, if the wavevector of photon lies along z, the emitted light is circularly polarized with a polarization which depends on helicity. Conversely, if the wavevector lies in xy place, the emitted light is linearly polarized, while in general the polarization is elliptical. Finally, in the presence of spin-orbit coupling, stationary states are labelled ˆ=L ˆ + S. ˆ In this case, the selection by quantum numbers J, mJ , +, s where J rules can be inferred by looking for the conditions for non-zero matrix elements $J % , mJ ! , +% , s% |r|J, mJ , +, s". By expanding states |J, mJ , +, s" in the basis states |+, m$ "⊗|s, ms ", one may uncover the following set of selection rules: For dipole transitions to take place, we require that ∆mj = 0, ±1
∆j = 0, ±1
not
0→0
( Info. As another example of selection rules, consider a charged particle moving in a one-dimensional harmonic potential. The wavefunctions are characterised by the quantum number n, so the state |n" corresponds to energy (n+1/2)!ω. An oscillating electric field in the x-direction can induce transitions between states |n" and |n# ", governed by matrix elements of the form . $n# | x |n". These can be evaluated by use of the ladder operators a ˆ and a ˆ† . Since x = $n | x |n" = #
/
! † 2mω (a + a ),
the matrix element becomes
1 √ ! 0√ n + 1$n# |n + 1" + n$n# |n − 1" , 2mω
and therefore vanishes unless n = n# ± 1. Hence the selection rule, in the electric dipole approximation, is ∆n = ±1.
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152
Lasers
Finally, to close this section, we will consider a principle application of light matter interaction – the laser. The laser provides a light source which enables modern spectroscopy. The term “laser” is an acronym for “light amplification by stimulated emission of radiation”. However, a laser not only amplifies light, but it acts as a special kind of light source which is characterised by a number of properties: Monochromaticity: The emission of the laser generally corresponds to just one of the atomic transitions of the gain medium (in contrast to discharge lamps, which emit on all transitions). The spectral line width can be much smaller than that of the atomic transition. This is because the emission is affected by the optical cavity. In certain cases, the laser can be made to operate on just one of the modes of the cavity. Since the Q of the cavity1 is generally rather large, the mode is usually much narrower than the atomic transition, and the spectral line width is orders of magnitude smaller than the atomic transition. Coherence: In discussing the coherence of an optical beam, we must distinguish between spatial and temporal coherence – laser beams have a high degree of both. Spatial coherence refers to whether there are irregularities in the optical phase in a cross-sectional slice of the beam. Temporal coherence refers to the time duration over which the phase of the beam is well defined. In general, the temporal coherence time, tcoh is given as the reciprocal of the spectral line width, ∆ν. Thus the coherence length +coh is given by, +coh = ctcoh =
c . ∆ν
Typical values of the coherence length for a number of light sources are given in the table below: Source Na discharge lamp (D-lines at 589nm) Multi-mode HeNe laser (632.8nm line) Single-mode HeNe laser (632.8nm line)
∆ν (Hz) 5 × 1011
tcoh (s) 2 × 10−12
+coh (m) 6 × 10−4
1.5 × 109
6 × 10−10
0.2
1 × 106
1 × 10−6
300
These figures explain why it is much easier to conduct interference experiments with a laser than with a discharge lamp. If the path difference exceeds +coh you will not get interference fringes, because the light is incoherent. Brightness: The brightness of lasers arises from two factors. First of all, the fact that the light is emitted in a well-defined beam means that the power per unit area is very high, even though the total amount of power can be rather low. Then we must consider that all the energy is concentrated within the narrow spectrum of the active atomic transition. This means that the spectral brightness (i.e. the intensity in the beam divided by the width of the emission line) is even higher in comparison with a white light source like a light bulb. For example, the spectral brightness of a 1 mW laser beam could easily be millions of time greater than that of a 100 W light bulb. Ultra-short pulse generation: In some cases, lasers can be made to operate in pulses. The time duration of the pulses tp is linked to the spectral 1 Recall that the Q-factor is approximately the number of oscillations required for a freely oscillating system’s energy to fall by a factor of 1/e2π of its original energy.
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Arthur Leonard Schawlow 19211999 American physicist and corecipient, with Nicolaas Bloembergen of the US and Kai Manne Borje Siegbahn of Sweden, of the 1981 Nobel Prize for Physics for his work in developing the laser and in laser spectroscopy. In 1949 he went to Columbia University, where he began collaborating with Charles Townes on the development of masers, lasers, and laser spectroscopy. Schawlow worked on the project that led to the construction of the first working maser in 1953 (for which Townes received a share of the 1964 Nobel Prize for Physics). Schawlow was a research physicist at Bell Telephone Laboratories from 1951 to 1961. In 1958 he and Townes published a paper in which they outlined the working principles of the laser, though the first such working device was built by another American physicist, Theodore Maiman, in 1960. In 1961 Schawlow became a professor at Stanford University. He became a world authority on laser spectroscopy, and he and Bloembergen earned their share of the 1981 Nobel Prize by using lasers to study the interactions of electromagnetic radiation with matter.
13.3. LASERS band width of the laser light ∆ν by the “uncertainty” product ∆t∆ν ∼ 1. This follows from taking the Fourier transform of a pulse of duration tp . As an example, the bandwidth of the 632.8nm line in the HeNe laser is 1.5 GHz (see above), so that the shortest pulses that a HeNe laser can produce would be 0.67 ns long. This is not particularly short by modern standards. Dye lasers typically have gain bandwidths greater than 1013 Hz, and can be used to generate pulses shorter than 100fs. This is achieved by a technique called “mode-locking”. These short pulsed lasers are very useful for studying fast processes in physics, chemistry and biology.
13.3.1
Operating principles of a laser
Light amplification is achieved by stimulated emission. Ordinary optical materials do not amplify light. Instead, they tend to absorb or scatter the light, so that the light intensity out of the medium is less than the intensity that went in. To get amplification you have to drive the material into a non-equilibrium state by pumping energy into it. Positive optical feedback is achieved by inserting the amplifying medium inside a resonant cavity. Light in the cavity passes through the gain medium and is amplified. It then bounces off the end mirrors and passes through the gain medium again, getting amplified further. This process repeats itself until a stable equilibrium condition is achieved when the total round trip gain balances all the losses in the cavity. The losses in the cavity fall into two categories: useful, and useless. The useful loss comes from the output coupling. One of the mirrors (called the “output coupler”) has reflectivity less than unity, and allows some of the light oscillating around the cavity to be transmitted as the output of the laser. The value of the transmission is chosen to maximise the output power. If the transmission is too low, very little of the light inside the cavity can escape, and thus we get very little output power. On the other hand, if the transmission is too high, there may not be enough gain to sustain oscillation, and there would be no output power. The optimum value is somewhere between these two extremes. Useless losses arise from absorption in the optical components (including the laser medium), scattering, and the imperfect reflectivity of the other mirror (the “high reflector”). In general we expect the gain to increase as we pump more energy into the laser medium. At low pump powers, the gain will be small, and there will be insufficient gain to reach the oscillation condition. The laser will not start to oscillate until there is enough gain to overcome all the losses. This implies that the laser will have a threshold in terms of the pump power.
13.3.2
Gain mechanism
Laser operation relies upon the phenomenon of stimulated emission. In a gas of atoms in thermal equilibrium, the population of lower levels will always be greater than the population of upper levels. Therefore, if a light beam is incident on the medium, there will always be more upward transitions due to absorption than downward transitions due to stimulated emission. Hence there will be net absorption, and the intensity of the beam will diminish on progressing through the medium. To amplify the beam, we require that the rate of stimulated emission exceeds the rate of absorption. If the light beam is sufficiently intense that we can ignore spontaneous emission, and the levels are non-degenerate, this implies that the number of atoms in some upper level, N2 , must exceed that of the lower level N1 . This is a highly non-equilibrium situation, and is called popuAdvanced Quantum Physics
153
13.3. LASERS
154
lation inversion. Once we have population inversion, we have a mechanism for generating gain in the laser medium. The art of making a laser operate is to work out how to get population inversion for the relevant transition. To develop a theory of the laser threshold, we can consider separately the rate equations for the photon and atomic excitation. Starting with photons, let us consider excitations created by the transitions between just two levels of the atom – a lower level 1, and an excited state 2. If the dipole matrix elements, W , between the two levels are independent of position and frequency, the net downwards transition rate is given by W (N2 (n + 1) − N1 n) where n denotes the total number of photons in the cavity, and N1,2 is the number of atoms in states 1, 2. The first term represents the contribution from stimulated and spontaneous emission, while the latter is associated with absorption. Taking into account photon loss from the leaky cavity, the rate of change of photon number is therefore given by n˙ = DW n + N2 W −
n , τph
(13.2)
where D = N2 − N1 represents the population imbalance and 1/τph is the photon loss rate. This equation shows that the gain in a laser medium is directly proportional to the degree of population inversion. Laser operation will occur when there is enough gain to overcome the losses in the cavity. This implies that a minimum amount of population inversion must be obtained before the laser will oscillate. To achieve population inversion atoms must be “pumped” into the upper level 2. This can be achieved by a variety of techniques: Lasers are classified as being either three-level of four-level systems. In the following, we will consider the case of a three-level laser, although four-level lasers are more common. Examples of four-level lasers include Helium Neon or Nd:YAG. In a four-level laser, the levels comprise the ground state (0), the two lasing levels (1 and 2), and a fourth level (3) which is used as part of the pumping mechanism. In the three-level system, such as the first laser, ruby, level 1 is the ground state, and pumping is achieved by exiting atoms to level 3 with a bright flash lamp or by an electrical discharge, and then allowing them to decay rapidly to level 2. In this case, the corresponding rate equations for the populations of levels 1 and 2 can be written as N˙ 2 & −w21 N2 + w12 N1 − (N2 − N1 )W n & −N˙ 1 , where w12 , w21 denote the “effective” transition rates between states 1 and 2 due to the pumping via the third state, and we have dropped the small contribution from spontaneous emission. From this equation, we can deduce that N1 + N2 = N , a constant, i.e. the decay from state 3 is so rapid that its population is always negligible. In this case, we obtain D0 − D D˙ = − 2DW n . T
(13.3)
where D0 = N (w12 − w21 )/(w12 + w21 ) denotes the unsaturated inversion (i.e. the degree of population inversion that would exist if there were no photons in the cavity, n = 0) and 1/T = w12 + w21 .
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In steady-state, n˙ = D˙ = 0, and Eq. (13.3) translates to a population imbalance, D ≡ N2 − N1 =
D0 . 1 + 2T W n
From this result, we find the steady state photon number is given by n=
D0 W − 1/τph . 2T W/τph
When D0 > 1/W τph , the laser threshold condition, there is a rapid increase in the number of photons in the cavity and the system starts lasing. Although this analysis addresses the threshold conditions, it does not provide any insight into the coherence properties of the radiation field. In fact, one may show that the radiation field generated by the laser cavity forms a coherent or Glauber state. The proof of this statement and the coherence properties that follow would take us too far into the realm of laser physics. However, we can gain some insight into this statement by studying a toy example.
13.4
Driven harmonic oscillator
Consider a quantum harmonic oscillator Hamiltonian driven by some external classical field, " # % & 1 † ˆ H = !ω a a + + ! f ∗ (t)a + f (t)a† . 2
Here f (t), which represents an (as yet) arbitrary function of time, t, characterises the coupling between the harmonic oscillator and the classical pump. For example, if f (t) is real, the function couples directly to the displacement, a + a† ∼ x. If the system is prepared in the ground state of the harmonic oscillator, the perturbation drives the system into a coherent state. To understand how, let us consider the time-evolution in the interaction ˆ representation, i!∂t |ψ(t)"I = VI |ψ(t)"I where |ψ(t)"I = eiH0 t/!|ψ(t)"S and, iωt 2 ˜ defining f = f e , % & % & ˆ ˆ VI (t) = eiH0 t/!! f ∗ (t)a + f (t)a† e−iH0 t/! = ! f˜∗ (t)a + f˜(t)a† .
To solve for the time-evolution operator, i!∂t UI (t) = VI UI (t), let us consider ˆ (α)|0" where U ˆ (α) = exp[αa† − α∗ a], where α = the coherent state, |α" = U α(t). Equivalently, making use of the BCH identity, the unitary operator may be written as ˆ (α) = e−α∗ α/2 eαa† e−α∗ a . U †
†
Then, taking the time derivative, and making use of the identity, [eαa , a]e−αa = −α, one obtains 2 3 i † ∗ ∗ ˆ ˆ (α) . ∂t U (α) = αa ˙ − α˙ a + Im(α˙ α) U 2 *t Therefore, setting α(t) = −i 0 dt% f˜(t% ), we obtain the solution + , UI (t) = exp α(t)a† − α∗ (t)a + iϕ(t) , 2
†
†
Here we have made use of the identity (exercise), eiωta a ae−iωta
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a
= e−iωt a.
13.4. DRIVEN HARMONIC OSCILLATOR
156
*t where ϕ(t) = 0 dt% 12 Im(α˙ ∗ α). *t If the driving force f (t) = f0 e−iωt (with f0 real), we have α(t) = −i 0 dt% f0 = −if0 t and ϕ(t) = 0 leading to the solution, UI (t) = exp[−if0 (a† + a)t]. Therefore, if the system was prepared in the harmonic oscillator ground state |0" at time t = 0, the solution at time t is given by |ψ(t)"I = exp[−if0 (a† + a)t]|0" = 2 † e−(f0 t) /2 e−if0 a t |0". Then, reexpressed in the Schr¨odinger representation, ˆ
|ψ(t)"S = e−iH0 t/!|ψ(t)"I = e−(f0 t)
2 /2
−iωt a† t
e−if0 e
|0" .
As a result, we can conclude that a classical oscillatory force drives a system prepared in the vacuum into a coherent state. Applied to an optical cavity, an oscillating classical dipole generates a coherent state of light – the principle that underlies the operation of a laser.
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Chapter 14
Scattering theory Almost everything we know about nuclei and elementary particles has been discovered in scattering experiments, from Rutherford’s surprise at finding that atoms have their mass and positive charge concentrated in almost pointlike nuclei, to the more recent discoveries, on a far smaller length scale, that protons and neutrons are themselves made up of apparently point-like quarks. More generally, the methods that we have to probe the properties of condensed matter systems rely fundamentally on the notion of scattering. In this section, we will provide a brief introduction to the concepts and methodology of scattering theory. As preparation for the quantum mechanical scattering problem, let us first consider the classical problem. This will allow us to develop (hopefully a revision!) some elementary concepts of scattering theory, and to introduce some notation. In a classical scattering experiment, one considers particles of energy E = 12 mv02 (mass m and asymptotic speed v0 ), incident upon a target with a central potential V (r). For a repulsive potential, particles are scattered through an angle θ (see figure). The scattering cross-section, σ, can be inferred from the number of particles dn scattered into some element of solid angle, dΩ, at angle (θ, φ), i.e. for an incident flux ji (number of particles per unit time per ! unit area), dn ! π = ji σ dΩ. ! 2πThe total cross-section is then obtained as σT = dΩ σ(θ, φ) = 0 sin θdθ 0 dφ σ(θ, φ). The angle of deflection of the beam depends on the impact parameter, b (see figure right). We therefore have that dn = ji bdb dφ = ji σ sin θdθdφ and σ(θ, φ) =
b db . sin θ dθ
$ Example: Let us consider then the case of classical Coulomb scattering from a repulsive potential V (r) = κr where κ > 0. From classical physics, we know that the particle will follow a hyperbolic trajectory with r=
L2 , mκ(e cos ϕ − 1)
where r = (r, ϕ) parameterises the relative coordinates of the particle and target,1 2 1/2 and e = (1 + 2EL > 1 denotes the eccentricity. Since the potential is central, the κ2 m ) angular momentum L is conserved and can be fixed asymptotically by the condition L = mv0 b. To obtain the scattering angle, θ, we can use the relation above to find the limiting angle, cos ϕ0 = 1/e, where ϕ0 = (π − θ)/2. We therefore have tan(θ/2) = cot ϕ0 = 1
Note that the angle ϕ is distinct form the azimuthal angle φ associated with the axis of scattering.
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14.1. BASICS
158
√ mκ2 1/2 1/ e2 − 1 = ( 2EL = 2)
κ 2Eb .
σ=
Then, from this relation, we obtain the cross-section
b db κ2 1 = , sin θ dθ 16E 2 sin4 θ/2
known as the Rutherford formula.
14.1
Basics
Let us now turn to the quantum mechanical problem of a beam of particles incident upon a target. The potential of the target, V (r), might represent that experienced by a fast electron striking an atom, or an α particle colliding with a nucleus. As in the classical problem, the basic scenario involves directing a stream or flux of particles, all at the same energy, at a target and detect how many particles are deflected into a battery of detectors which measure angles of deflection. In principle, if we assume that all the in-going particles are represented by wavepackets of the same shape and size, our challenge is to solve the full time-dependent Schr¨odinger equation for such a wavepacket, " # !2 2 i!∂t Ψ(r, t) = − ∇ + V (r) Ψ(r, t) , 2m
and find the probability amplitudes for out-going waves in different directions at some later time after scattering has taken place. However, if the incident beam of particles is switched on for times very long as compared with the time a particle would take to cross the interaction region, steady-state conditions apply. Moreover, if we assume that the wavepacket has a well-defined energy (and hence momentum), so it is many wavelengths long, and we may consider it a plane wave. Setting Ψ(r, t) = ψ(r)e−iEt/!, we may therefore look for solutions ψ(r) of the time-independent Schr¨odinger equation, # " !2 2 Eψ(r) = − ∇ + V (r) ψ(r) , 2m
subject to the boundary condition that the incoming component of the wavefunction is a plane wave, eik·x . Here E = p2 /2m = !2 k2 /2m denotes the energy of the incoming particles while their flux is given by ! !k j = −i (ψ ∗ ∇ψ − ψ∇ψ ∗ ) = . 2m m In the one-dimensional geometry, the impact of a plane wave with the localized target resulted in a portion of the wave being reflected and a portion transmitted through the potential region. From energy conservation, we may deduce that both components of the outgoing scattered wave are plane waves with wavevector ±k, while the influence of the potential are encoded in the amplitude of the reflected and transmitted beams, and a potential phase shift. Both amplitudes and phase shifts are then determined by solving the timeindependent Schr¨odinger equation subject to the boundary conditions which ensure energy and flux conservation. In the three-dimensional system, the phenomenology is similar: In this case, the wavefunction well outside the localized target region will involve a superposition of the incident plane wave and the scattered (spherical wave),2 ψ(r) $ eik·r + f (θ, φ)
eikr , r
2 Here, by localized, we mean a potential which is sufficiently short-ranged. At this stage, it is not altogether clear what constraint this implies. But it will turn out that it excludes the Coulomb potential!
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14.1. BASICS
159
where the function f (θ, φ) records the relative amplitude and phase of the scattered components along the direction (θ, φ) relative to the incident beam. To place these ideas on a more formal footing, consider the following: If we define the direction of the incoming wave k to lie along the z-axis, a plane wave can be recast in the form of an incoming and an outgoing spherical wave, % & ∞ e−i(kr−"π/2) ei(kr−"π/2) i $ " ik·r i (2+ + 1) e = − P" (cos θ) , 2k r r "=0
4π 1/2 where P" (cos θ) = ( 2"+1 ) Y"0 (θ) denote the Legendre polynomials. If we assume that the potential perturbation, V (r) depends only on the radial coordinate (i.e. that it is spherically symmetric) and that the number of particles are conserved by the potential (the flux of incoming particles is matched by the flux of outgoing),3 when the potential is sufficient short-ranged (decreasing faster than 1/r), the scattering wavefunction takes the asymptotic form % & ∞ e−i(kr−"π/2) i $ " ei(kr−"π/2) i (2+ + 1) ψ(r) $ − S" (k) P" (cos θ) , 2k r r "=0
subject to the constraint |S" (k)| = 1 following from the conservation of particle flux (i.e. S" (k) = e2iδ! (k) ). Physically, the incoming component of the spherical wave is undisturbed by the potential while the separate components of the outgoing spherical wave are subject to a set of phase shifts, δ" (k). Recast in the form of a perturbation, the asymptotic form of the wavefunction can be straightforwardly rewritten as ψ(r) $ eik·r + f (θ)
eikr , r
where the second component of the wavefunction denotes the change in the outgoing spherical wave due to the potential, and f (θ) =
∞ $
(2+ + 1)f" (k)P" (cos θ) ,
(14.1)
"=0
1 with the coefficents f" (k) = 2ik (S" (k) − 1) defining the partial wave scattering amplitudes. The corresponding asymptotic flux is then given by '" #∗ " #( ! eikr eikr ik·r ik·r j = −i Re e + f (θ) ∇ e + f (θ) . m r r
In general, an expansion then leads to a formidible collection of contributing terms. However, for most of these contributions, there remains an exponential factor, e±ikr(1−cos θ) where θ denotes the angle between k and r. For r → ∞, the small angular integration implied by any physical measurement leads to a fast oscillation of this factor. As a result, such terms are strongly suppressed and can be neglected. Retaining only those terms where the phase cancellation is complete, one obtains, j=
!k !k |f (θ)|2 ˆr + e + O(1/r3 ) . m m r2
3
Note that this assumption is not innocent. In a typical high energy physics experiment, the collision energies are high enough to lead to particle production.
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14.2. METHOD OF PARTIAL WAVES The first term represents the incident flux, while the remainder describes the radial flux of scattered particles. In particular, the number of particles crossing the area that subtends a solid angle dΩ at the origin (the target) is given by ˆr dA = j·e
!k |f (θ)|2 2 r dΩ + O(1/r) . m r2
Dropping terms of order 1/r, negligible in the asymptotic limit, one thus obtains the differential cross-section, the ratio of the scattered flux to the m ˆr dA = |f (θ)|2 dΩ, i.e. incident flux, dσ = !k j·e dσ = |f (θ)|2 . dΩ ! ! The total cross-section is then given by σtot = dσ = |f (θ)|2 dΩ. Then, ! 4π δ""! , and Eq. (14.1) making use of the identity dΩP" (cos θ)P"! (cos θ) = 2"+1 one obtains (exercise) σtot
∞ 4π $ (2+ + 1) sin2 δ" (k) . = 2 k "=0
In particular, noting that P" (1) = 1, from Eq. (14.1) it follows that Im f (0) = k 4π σtot , a relation known as the optical theorem.
14.2
Method of partial waves
Having established the basic concepts for the scattering problem, we turn now to consider operationally how the scattering characteristics can be computed. Here, for simplicity, we will focus on the properties of a centrally symmetric potential, V (r), where the scattering wavefunction, ψ(r) (and indeed that scattering amplitudes, f (θ)) must be symmetrical about the axis of incidence, and hence independent of the azimuthal angle, φ. In this case, the wavefunction can be expanded in a series of Legendre polynomials, ψ(r, θ) =
∞ $
R" (r)P" (cos θ) .
"=0
Each term in the series is known as a partial wave, and is a simultaneous ˆ 2 and L ˆ z having eigenvalue eigenfunction of the angular momentum operators L 2 ! +(+ + 1) and 0 respectively. Following standard spectroscopic notation, + = 0, 1, 2, · · · are referred to as s, p, d, · · · waves. The partial wave amplitudes, f" are determined by the radial functions, R" (r), defined by " # 2 +(+ + 1) 2 ∂r2 + ∂r − − U (r) + k R" (r) = 0 , r r2 where U (r) = 2mV (r)/!2 represents the effective potential. $ Example: To develop the partial wave scattering method, we will consider the problem of quantum scattering from an attractive square well potential, U (r) = −U0 θ(R − r). In this case, the radial wave equation takes the form " # +(+ + 1) 2 2 2 ∂r + ∂r − + U0 θ(R − r) + k R" (r) = 0 . r r2 Advanced Quantum Physics
160
14.2. METHOD OF PARTIAL WAVES
161
At high energies, many channels contribute to the total scattering amplitude. However, at low energies, the scattering is dominated by the s-wave (+ = 0) channel. In this case, setting u(r) = rR0 (r), the radial equation takes the simple form, (∂r2 + U0 θ(R − r) + k 2 )u(r) = 0, with the boundary condition that u(0) = 0. We therefore obtain the solution ) C sin Kr r R where K 2 = k 2 + U0 > k 2 . The continuity condition of the wavefunction and its derivative at r = R translates to the relation K cot(KR) = k cot(kR + δ0 ). From this k tan(KR)) − kR,4 i.e. expression, we obtain the + = 0 phase shift, δ0 = tan−1 ( K tan δ0 (k) =
k tan(KR) − K tan(kR) , K + k tan(kR) tan(KR)
Then, unless tan(KR) = ∞ (see below), an expansion at low energy (small k) obtains δ0 $ kR( tan(KR) − 1), and the + = 0 partial cross-section, KR * +2 4π 4π 1 4π 2 tan(KR) 2 2 σ0 = 2 sin δ0 (k) = 2 $ 2 δ0 = 4πR −1 . k k 1 + cot2 δ0 (k) k KR From this result, we find that when tan(KR) = 1, the scattering cross-section vanishes. KR An expansion in small k obtains k cot δ0 = −
1 1 + r0 k 2 + · · · , a0 2 1/2
γ where a0 = (1 − tan γ )R, with γ = U0 R, defines the scattering length, and r0 is the effective range of the interaction. At low energies, k → 0, the scattering crosssection, σ0 = 4πa20 is fixed by the scattering length alone. If γ ' 1, a0 is negative. As γ is increased, when γ = π/2, both a0 and σ0 diverge – there is said to be a zero energy resonance. This condition corresponds to a potential well that is just able to support an s-wave bound state. If γ is further increased, a0 turns positive – as it would be for an effective repulsive interaction until γ = π when σ0 = 0 and the process repeats with the appearance of a second bound state at γ = 3π/2, and so on. More generally, the +-th partial cross-section
σ" =
4π 1 (2+ + 1) , 2 k 1 + cot2 δ" (k)
Scattering wavefunction, u(r), for three-dimensional square well potential for kR = 0.1 and γ = 1 (top), π/2 (middle) and 2 (bottom). Note that the scattering length, a0 changes from negative to positive as system passes through bound state.
takes its maximum value is there is an energy at which cot δ" vanishes. If this occurs as a result of δ" (k) increasing rapidly through an odd multiple of π/2, the cross-section exhibits a narrow peak as a function of energy and there is said to be a resonance. Near the resonance, cot δ" (k) =
ER − E , Γ(E)/2
where ER is the resonance energy. If Γ(E) varies slowly in energy, the partial crosssection in the vicinity of the resonance is given by the Breit-Wigner formula, σ" (E) = 4
4π Γ (ER )/4 (2+ + 1) . k2 (E − ER )2 + Γ2 (ER )/4 2
(14.2)
More generally, choosing the solution to be finite at the origin, we find that R! (r) = N! (K)j! (Kr),
r < R,
where N! (K) is a normalization constant. In the exterior region, the general solution can be written as R! (r) = B! (k)[j! (kr) − tan δ! (k)η! (kr)]. Continuity of R! and the derivative ∂r R! at the boundary, r = R, lead to the following expression for the phase shifts tan δ! (k) =
kj!! (kR)j! (KR) − Kj! (kR)j!! (KR) . kη!! (kR)j! (KR) − Kη! (kR)j!! (KR)
Here j!! (x) = ∂x j! (x) and similarly η!! .
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Scattering phase shift for kR = 0.1 as a function of γ.
14.3. THE BORN APPROXIMATION
162
$ Exercise. For a hard-core interaction, U (r) = U0 θ(R − r), with U0 → ∞, show that at low energy δ0 = −kR. As k → 0, show that the differential cross-section k→0 dσ is isotropic and given by dΩ = R2 and σtot → 4πR2 . $ Info. Ultracold atomic gases provide a topical arena in which resonant scattering phenomena are exploited. In particular, experimentalists make use of Feshbach resonance phenomena to tune the effective interaction between atoms. This tunability arises from the coupling of free unbound atoms to a molecular state in which the atoms are tightly bound. The closer this molecular level lies with respect to the energy of two free atoms, the stronger the interaction between them. In the example on the left, the two free atoms are both “spin up”, whereas the molecular state is a “singlet”, in which the atoms have opposite spin, adding up to zero total magnetic moment. Thus, a magnetic field shifts the energies of two free atoms relative to the molecular state and thereby controls the interatomic interaction strength. The interaction between two atoms can be described by the scattering length, shown right as a function of magnetic field close to a Feshbach resonance. On the side where the scattering length is positive, the molecular energy level is lower in energy than the energy of two unbound atoms. The molecular state is thus “real” and stable, and atoms tend to form molecules. If those atoms are fermions, the resulting molecule is a boson. A gas of these molecules can thus undergo Bose-Einstein condensation (BEC). This side of the resonance is therefore called “BEC-side”. On the side of the resonance where the scattering length is negative, isolated molecules are unstable. Nevertheless, when surrounded by the medium of others, two fermions can still form a loosely bound pair, whose size can become comparable to or even larger than the average distance between particles. A Bose-Einstein condensate of these fragile pairs is called a “BCS-state”, after Bardeen, Cooper and Schrieffer. This is what occurs in superconductors, in which current flows without resistance thanks to a condensate of electron pairs (“Cooper pairs”).
14.3
The Born approximation
The partial wave expansion is tailored to the consideration of low-energy scattering processes. At higher energies, when many partial waves contribute, the expansion is not very convenient and it is helpful to develop a different methodology. By developing a general expansion of the scattering wavefunction, ψk (r), in terms of the Green function of the scattering potential one may show that, ψk (r) = eik·r −
1 4π
,
d3 r $
!
eik|r−r | U (r$ )ψk (r$ ) . |r − r$ |
(14.3)
Here the subscript k reminds us that the solution is for a particular incoming plane wave. This integral representation of the scattering wavefunction, known as the Lippmann-Schwinger equation, provides a more useful basis to address situations where the energy of the incoming particles is large and the scattering potential is weak. The elements of the derivation of this equation are summarised in the info box below: $ Info. Lippmann-Schwinger equation: For the time-independent Schr¨odinger Advanced Quantum Physics
14.3. THE BORN APPROXIMATION
163
equation (∇2 + k 2 )ψ(r) = U (r)ψ(r), the general solution can be written formally as , ψ(r) = φ(r) + d3 r# G0 (r, r# )U (r# )ψ(r# ) , where φ(r) is a solution of the homogeneous (free particle) Schr¨odinger equation, (∇2 + k 2 )φ(r) = 0, and G0 (r, r# ) is a Green function of the Laplace operator, (∇2 + k 2 )G0 (r, r# ) = δ 3 (r − r# ). From the asymptotic behaviour of the boundary condition, it is evident that φ(r) = eik·r . In the Fourier basis, the Green function is diagonal and given by G0 (k, k# ) = (2π)3 δ 3 (k − k# ) k12 . Transformed back into real space, we have G0 (r, r# ) = −
!
1 eik|r−r | . 4π |r − r# |
Substituted back into the expression for the scattering wavefunction, we obtain the Lippmann-Schwinger equation (14.3).
In the far-field region, |r − r$ | $ r − ˆr · r$ + · · ·, i.e. !
eikr −ik! ·r! eik|r−r | $ e , |r − r$ | r where the vector k$ = kˆ er is oriented along the direction of the scattered particle. We therefore find that the scattering wavefunction ψk (r) = eik·r + ikr f (θ, φ) e r can be expressed in integral form, with the scattering amplitude given by f (θ, φ) = −
1 1 (φk! |U |ψk ) ≡ − 4π 4π
,
!
d3 r$ e−ik ·r U (r$ )ψk (r$ ) .
(14.4)
The corresponding differential cross-section can then be expressed as dσ m2 = |f |2 = |Tk,k! |2 , dΩ (2π)2 !4 where, cast in terms of the original scattering potential, V (r) = !2 U (r)/2m, Tk,k! = (φk! |V |ψk ) denotes the transition matrix element. Eq. (14.3) provides a natural means to expand the scattering wavefunction in powers of the interaction potential. At zeroth order in V , the scattering (0) wavefunction is specified by the unperturbed incident plane wave, φk (r) = φk (r). Using this approximation, Eq. (14.3) leads to the first order correction, , (1) (0) ψk (r) = φk (r) + d3 r$ G0 (r, r$ )U (r$ )ψk (r$ ) . From this equation, we can use (14.3) to obtain the next term in the series, , (2) (1) ψk (r) = φk (r) + d3 r$ G0 (r, r$ )U (r$ )ψk (r$ ) , and so on, i.e. f =−
1 (φk! |U + U G0 U + U G0 U G0 U + · · · |φk ) . 4π
Physically, an incoming particle undergoes a sequence of multiple scattering events from the potential (see schematic on the right). This series expansion is
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14.4. INFO: SCATTERING OF IDENTICAL PARTICLES
164
known as the Born series, and the leading term in known as the first Born approximation to the scattering amplitude, fBorn = −
1 (φk! |U |φk ) . 4π
(14.5)
Setting ∆ = k − k$ , where !∆ denotes the momentum transfer, the Born scattering amplitude for a central potential is given by (exercise) , , ∞ 1 sin(∆r) 3 i∆·r fBorn (∆) = − d re U (r) = − rdr U (r) , 4π ∆ 0 where, noting that |k$ | = |k|, ∆ = 2k sin(θ/2). Coulomb scattering: Due to the long range nature of the Coulomb scattering potential, the boundary condition on the scattering wavefunction does not apply. We can, however, address the problem by working with the −r/α screened (Yukawa) potential, U (r) = U0 e r , and taking α → ∞. For this potential, one may show that (exercise) fBorn = −U0 /(α−2 + ∆2 ). Therefore, for α → ∞, we obtain σ(θ) = |f (θ)|2 =
U02 , 16k 4 sin4 θ/2
which is just the Rutherford formula. $ Info. Previously, we have used time-dependent perturbation theory to develop an expression for the transition rate between states. In the leading order of perturbation theory, we found that the transition rate between states and i and f induced by a potential V is given by Fermi’s Golden rule, Γi→f =
2π |(f|V |i)|2 δ(E − (Ef − Ei )) . !
In a three-dimensional scattering problem, we should consider the initial state as a plane wave state of wavevector k and the final state as the continuum of states with wavevectors k# . In this case, the total transition (or scattering) rate into a fixed solid angle, dΩ, is given by Γk→k! =
$ 2π 2π # |(k# |V |k)|2 δ(E − (Ek! − Ek )) = |(k |V |k)|2 g(E) , ! ! !
k ∈dΩ
dn where g(E) = dE denotes the density of states and both states |k) and |k# ) have 2 energy E = !2 k 2 /2m = !2 k # /2m – they are said to be “on-shell”. As a result, we dk k2 dΩ m obtain the density of states g(E) = dn dk dE = (2π/L)3 !2 k while the incident flux per unit volume is given by !k/mL3 . As a result, we obtain the scattering cross-section, Γk→k! dσ dΩ = !k/mL3
dσ 1 2mV = |(k# | 2 |k)|2 . dΩ (4π)2 ! We can therefore recognize that Fermi’s Golden rule is equivalent to the first order Born approximation.
14.4
Info: Scattering of identical particles
Until now, we have assumed that the particles involved in the scattering process, the incoming particle and the target, are distinguishable. However, very often we are interested in the scattering of identical quantum particles. In such cases, we
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14.5. SCATTERING BY AN ATOMIC LATTICE have to consider the influence of quantum statistics on the scattering process. As a preliminary exercise, consider the classical picture of scattering between two identical positively charged particles, e.g. α-particles viewed in the center of mass frame. If an outgoing α particle is detected at an angle θ to the path of the ingoing α-particle, it could be (a) deflected through an angle θ, or (b) deflected through π − θ. Classically, we could tell which one it was by watching the collision as it happened, and keeping track. However, in a quantum mechanical scattering process, we cannot keep track of the particles unless we bombard them with photons having wavelength substantially less than the distance of closest approach. This is just like detecting an electron at a particular place when there are two electrons in a one dimensional box: the probability amplitude for finding an α particle coming out at angle θ to the ingoing direction of one of them is the sum of the amplitudes (not the sum of the probabilities!) for scattering through θ and π − θ. Writing the asymptotic scattering wavefunction in the standard form for scatterikr ing from a fixed target, ψ(r) ≈ eikz + f (θ) e r , the two-particle wavefunction in the center of mass frame, in terms of the relative coordinate, is given by symmetrizing: ψ(r) ≈ eikz + e−ikz + (f (θ) + f (π − θ))
eikr . r
How does the particle symmetry affect the actual scattering rate at an angle θ? If dσ the particles were distinguishable, the differential cross section would be ( dΩ )dist. = 2 2 |f (θ)| + |f (π − θ)| , but quantum mechanically we must compute, * + dσ = |f (θ) + f (π − θ)|2 . dΩ indist. This makes a big difference! For example, for scattering through 90o , where f (θ) = f (π−θ), the quantum mechanical scattering rate is twice the classical (distinguishable) prediction. Furthermore, if we make the standard expansion of the scattering amplitude f (θ) -∞ in terms of partial waves, f (θ) = "=0 (2+ + 1)a" P" (cos θ), then f (θ) + f (π − θ) =
∞ $ l=0
(2+ + 1)a" (P" (cos θ) + P" (cos(π − θ))) .
Since P" (−x) = (−1)" P" (x), the scattering only takes place in even partial wave states. This is the same thing as saying that the overall wavefunction of two identical bosons is symmetric. So, if they are in an eigenstate of total angular momentum, from P" (−x) = (−1)" P" (x) it has to be a state of even +. For fermions in an antisymmetric spin state, such as proton-proton scattering with the two proton spins forming a singlet, the spatial wavefunction is symmetric, and the argument is the same as for the boson case above. For parallel spin protons, however, the spatial wavefunction has to be antisymmetric, and the scattering amplitude will then be f (θ) − f (π − θ). In this case there is zero scattering at 90o ! Note that for (non-relativistic) equal mass particles, the scattering angle in the center of mass frame is twice the scattering angle in the fixed target (lab) frame.
14.5
Scattering by an atomic lattice
Finally, to close this section, let us say a few words about scattering phenomena in solid state systems. If we ignore spin degrees of freedom, so that we do not have to worry whether an electron does or does not flip its spin during the scattering process, then at low energies, the scattering amplitude of particles from a crystal f (θ) becomes independent of angle (s-wave). In this case, the solution of the Schr¨odinger equation by a single atom i located at a point Ri has the asymptotic form, ψ(r) = eik·(r−Ri ) + f Advanced Quantum Physics
eik|r−Ri | . |r − Ri |
165
14.5. SCATTERING BY AN ATOMIC LATTICE
166
Now, since .
k|r − Ri | = k r − 2r · Ri + 2
and kˆ er = k$ , we have
/1/2 R2i
−ikRi
ψ(r) = e
*
2r · Ri $ kr 1 − r2
+1/2
$ kr − kˆ er · Ri ,
# " ikr ik·r −i(k! −k)·Ri e e + fe . r
As a result, we can deduce the effective scattering amplitude, f (θ) = f e−i∆·Ri ,
∆ = k$ − k .
If we consider scattering from a crystal lattice, we must sum over all atoms. In this case, the total differential scattering cross-section is given by 0 02 0 0 $ 0 0 dσ −i∆·Ri 0 0 = 0f e 0 . dΩ 0 0 Ri
In the case of a periodic crystal, the sum over atoms translates to the Bragg condition, (2π)3 dσ = |f |2 3 δ (3) (k$ − k − 2πn/L) , dΩ L
where L represents the size of the (cubic) lattice, and n denote a vector of integers – the Miller indices of the Bragg planes. We therefore expect that the differential cross-section is very small expect when k$ − k = 2πn/L. These relations can be generalised straightforwardly to address more complicated crystal structures.
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X-ray diffraction pattern of a quasi-crystal.
Chapter 15
Relativistic Quantum Mechanics The aim of this chapter is to introduce and explore some of the simplest aspects of relativistic quantum mechanics. Out of this analysis will emerge the KleinGordon and Dirac equations, and the concept of quantum mechanical spin. This introduction prepares the way for the construction of relativistic quantum field theories, aspects touched upon in our study of the quantum mechanics of the EM field. To prepare our discussion, we begin first with a survey of the motivations to seek a relativistic formulation of quantum mechanics, and some revision of the special theory of relativity. Why study relativistic quantum mechanics? Firstly, there are many experimental phenomena which cannot be explained or understood within the purely non-relativistic domain. Secondly, aesthetically and intellectually it would be profoundly unsatisfactory if relativity and quantum mechanics could not be united. Finally there are theoretical reasons why one would expect new phenomena to appear at relativistic velocities. When is a particle relativistic? Relativity impacts when the velocity approaches the speed of light, c or, more intrinsically, when its energy is large compared to its rest mass energy, mc2 . For instance, protons in the accelerator at CERN are accelerated to energies of 300GeV (1GeV= 109 eV) which is considerably larger than their rest mass energy, 0.94 GeV. Electrons at LEP are accelerated to even larger multiples of their energy (30GeV compared to 5 × 10−4 GeV for their rest mass energy). In fact we do not have to appeal to such exotic machines to see relativistic effects – high resolution electron microscopes use relativistic electrons. More mundanely, photons have zero rest mass and always travel at the speed of light – they are never non-relativistic. What new phenomena occur? To mention a few: ! Particle production: One of the most striking new phenomena to emerge is that of particle production – for example, the production of electron-positron pairs by energetic γ-rays in matter. Obviously one needs collisions involving energies of order twice the rest mass energy of the electron to observe production. Astrophysics presents us with several examples of pair production. Neutrinos have provided some of the most interesting data on the 1987 supernova. They are believed to be massless, and hence inherently relativistic; moreover the method of their production is the annihilation of electron-positron pairs in the hot plasma at the core of the supernova. High temperatures, of the order of 1012 K are also inferred to exist in the nuclei of some galaxies (i.e. kB T " 2mc2 ). Thus electrons and positrons Advanced Quantum Physics
168
Figure 15.1: Anderson’s cloud chamber picture of cosmic radiation from 1932 show-
ing for the first time the existence of the positron. A cloud chamber contains a gas supersaturated with water vapour (left). In the presence of a charged particle (such as the positron), the water vapour condenses into droplets – these droplets mark out the path of the particle. In the picture a charged particle is seen entering from the bottom at high energy. It then looses some of the energy in passing through the 6 mm thick lead plate in the middle. The cloud chamber is placed in a magnetic field and from the curvature of the track one can deduce that it is a positively charged particle. From the energy loss in the lead and the length of the tracks after passing though the lead, an upper limit of the mass of the particle can be made. In this case Anderson deduces that the mass is less that two times the mass of the electron. Carl Anderson (right) won the 1936 Nobel Prize for Physics for this discovery. (The cloud chamber track is taken from C. D. Anderson, The positive electron, Phys. Rev. 43, 491 (1933).
are produced in thermal equilibrium like photons in a black-body cavity. Again a relativistic analysis is required. ! Vacuum instability: Neglecting relativistic effects, we have shown that the binding energy of the innermost electronic state of a nucleus of charge Z is given by, ! 2 "2 m Ze . E=− 4π$0 2!2 If such a nucleus is created without electrons around it, a peculiar phenomenon occurs if |E| > 2mc2 . In that case, the total change in energy of producing an electron-positron pair, subsequently binding the electron in the lowest state and letting the positron escape to infinity (it is repelled by the nucleus), is negative. There is an instability! The attractive electrostatic energy of binding the electron pays the price of producing the pair. Nuclei with very high atomic mass spontaneously “screen” themselves by polarising the vacuum via electron-positron production until the they lower their charge below a critical value Zc . This implies that objects with a charge greater than Zc are unobservable due to screening. ! Info. An estimate based on the non-relativistic formula above gives Zc $ 270. Taking into account relativistic effects, the result is renormalised downwards to 137, while taking into account the finite size of the nucleus one finally obtains Zc ∼ 165. Of course, no such nuclei exist in nature, but they can be manufactured, fleetingly, in uranium ion collisions where Z = 2 × 92 = 184. Indeed, the production rate of positrons escaping from the nucleus is seen to increase dramatically as the total Z of the pair of ions passes 160. ! Spin: Finally, while the phenomenon of electron spin has to be grafted Advanced Quantum Physics
169 artificially onto the non-relativistic Schr¨odinger equation, it emerges naturally from a relativistic treatment of quantum mechanics. When do we expect relativity to intrude into quantum mechanics? According to the uncertainty relation, ∆x∆p ≥ !/2, the length scale at which the kinetic energy is comparable to the rest mass energy is set by the Compton wavelength h ≡ λc . mc We may expect relativistic effects to be important if we examine the motion of particles on length scales which are less than λc . Note that for particles of zero mass, λc = ∞! Thus for photons, and neutrinos, relativity intrudes at any length scale. What is the relativistic analogue of the Schr¨odinger equation? Non-relativistic quantum mechanics is based on the time-dependent Schr¨odinger equation ˆ = i!∂t ψ, where the wavefunction ψ contains all information about a given Hψ system. In particular, |ψ(x, t)|2 represents the probability density to observe a particle at position x and time t. Our aim will be to seek a relativistic version of this equation which has an analogous form. The first goal, therefore, is to find the relativistic Hamiltonian. To do so, we first need to revise results from Einstein’s theory of special relativity: ∆x ≥
! Info. Lorentz Transformations and the Lorentz Group: In the special theory of relativity, a coordinate in space-time is specified by a 4-vector. A contravariant 4-vector x = (xµ ) ≡ (x0 , x1 , x2 , x3 ) ≡ (ct, x) is transformed into the covariant 4-vector xµ = gµν xν by the Minkowskii metric 1 −1 (gµν ) = gµν gνλ = δµλ , , −1 −1
Here, by convention, summation is assumed over repeated indicies. Indeed, summation covention will be assumed throughout this chapter. The scalar product of 4-vectors is defined by x · y = xµ y µ = xµ y ν gµν = xµ yµ .
The Lorentz group consists of linear Lorentz transformations, Λ, preserving x·y, µ i.e. for xµ )→ x! = Λµν xν , we have the condition gµν Λµα Λνβ = gαβ .
(15.1)
Specifically, a Lorentz transformation along the x1 direction can be expressed in the form γ −γv/c γ −γv/c Λµν = 1 0 0 1
where γ = (1 − v 2 /c2 )−1/2 .1 With this definition, the Lorentz group splits up into four components. Every Lorentz transformation maps time-like vectors (x2 > 0) into 1
Equivalently the Lorentz transformation can be represented in the form 0 1 0 −1 B C −1 0 C, Λ = exp[ωK1 ], [K1 ]µν = B @ 0 0A 0 0
where ω = tanh−1 (v/c) is known as the rapidity, and K1 is the generator of velocity transformations along the x1 -axis.
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170
time-like vectors. Time-like vectors can be divided into those pointing forwards in time (x0 > 0) and those pointing backwards (x0 < 0). Lorentz transformations do not always map forward time-like vectors into forward time-like vectors; indeed Λ does so if and only if Λ00 > 0. Such transformations are called orthochronous. (Since Λµ0 Λµ0 = 1, (Λ00 )2 − (Λj0 )2 = 1, and so Λ00 += 0.) Thus the group splits into two according to whether Λ00 > 0 or Λ00 < 0. Each of these two components may be subdivided into two by considering those Λ for which det Λ = ±1. Those transformations Λ for which det Λ = 1 are called proper. Thus the subgroup of the Lorentz group for which det Λ = 1 and Λ00 > 0 is called the proper orthochronous Lorentz group, sometimes denoted by L↑+ . It contains neither the time-reversal nor parity transformation, −1 1 1 −1 T = P = (15.2) , . 1 −1 1 −1
We shall call it the Lorentz group for short and specify when we are including T or P . In particular, L↑+ , L↑ = L↑+ ∪ L↑− (the orthochronous Lorentz group), L+ = L↑+ ∪ L↓+ (the proper Lorentz group), and L0 = L↑+ ∪ L↓− are subgroups, while L↓− = P L↑+ , L↑− = T L↑+ and L↓+ = T P L↑+ are not. Special relativity requires that theories should be invariant under Lorentz transformations xµ )−→ Λµν xν , and, more generally, Poincar´ e transformations xµ → Λµν xν + aµ . The proper orthochronous Lorentz transformations can be reached continuously from identity.2 Loosely speaking, we can form them by putting together infinitesimal Lorentz transformations Λµν = δ µν + ω µν , where the elements of ω µν - 1. Applying the identity gαβ = Λµα Λµβ = gαβ + ωαβ + ωβα + O(ω 2 ), we obtain the relation ωαβ = −ωβα . ωαβ has six independent components: L↑+ is a six-dimensional (Lie) group, i.e. it has six independent generators: three rotations and three boosts. Finally, according to the definition of the 4-vectors, the covariant and contravari∂ , ∇), ∂ µ = ∂x∂ µ = ant derivative are respectively defined by ∂µ = ∂x∂ µ = ( 1c ∂t ∂ ( 1c ∂t , −∇). Applying the scalar product to the derivative we obtain the d’Alembertian ∂2 2 operator (sometimes denoted as !), ∂ 2 = ∂µ ∂ µ = c12 ∂t 2 − ∇ .
15.1
Klein-Gordon equation
Historically, the first attempt to construct a relativistic version of the Schr¨odinger equation began by applying the familiar quantization rules to the relativistic Oskar Benjamin Klein 1894energy-momentum invariant. In non-relativistic quantum mechanics the cor1977 A Swedish theorespondence principle dictates that the momentum operator is associated with retical physicist, ˆ = −i!∇, and the energy operator with the time derivathe spatial gradient, p Klein is credited ˆ = i!∂t . Since (pµ ≡ (E/c, p) transforms like a 4-vector under Lorentz for inventing tive, E the idea, part transformations, the operator pˆµ = i!∂ µ is relativistically covariant. of Kaluza-Klein Non-relativistically, the Schr¨odinger equation is obtained by quantizing theory, that extra dimensions may the classical Hamiltonian. To obtain a relativistic version of this equation, be physically real one might apply the quantization relation to the dispersion relation obtained but curled up and very small, an idea essential to from the energy-momentum invariant p2 = (E/c)2 − p2 = (mc)2 , i.e. string theory/M-theory. ) 2 4 * + 2 4 , 2 2 1/2 2 2 2 1/2 E(p) = + m c + p c ⇒ i!∂t ψ = m c − ! c ∇ ψ where m denotes the rest mass of the particle. However, this proposal poses a dilemma: how can one make sense of the square root of an operator? Interpreting the square root as the Taylor expansion, i!∂t = mc2 ψ − 2
!2 ∇2 !4 (∇2 )2 ψ− ψ + ··· 2m 8m3 c2
They are said to form the path component of the identity.
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15.1. KLEIN-GORDON EQUATION
171
we find that an infinite number of boundary conditions are required to specify the time evolution of ψ.3 It is this effective “non-locality” together with the asymmetry (with respect to space and time) that suggests this equation may be a poor starting point. A second approach, and one which circumvents these difficulties, is to apply the quantization procedure directly to the energy-momentum invariant: ) * E 2 = p2 c2 + m2 c4 , −!2 ∂t2 ψ = −!2 c2 ∇2 + m2 c4 ψ.
Recast in the Lorentz invariant form of the d’Alembertian operator, we obtain the Klein-Gordon equation )
* ∂ 2 + kc2 ψ = 0 ,
(15.3)
where kc = 2π/λc = mc/!. Thus, at the expense of keeping terms of second order in the time derivative, we have obtained a local and manifestly covariant equation. However, invariance of ψ under global spatial rotations implies that, if applicable at all, the Klein-Gordon equation is limited to the consideration of spin-zero particles. Moreover, if ψ is the wavefunction, can |ψ|2 be interpreted as a probability density? To associate |ψ|2 with the probability density, we can draw intuition from the consideration of the non-relativistic Schr¨odinger equation. Applying the 2 ∇2 ψ) = 0, together with the complex conjugate of this identity ψ ∗ (i!∂t ψ + !2m equation, we obtain ∂t |ψ|2 − i
! ∇ · (ψ ∗ ∇ψ − ψ∇ψ ∗ ) = 0 . 2m
Conservation of probability means that density ρ and current j must satisfy the continuity relation, ∂t ρ + ∇ · j = 0, which states simply that the rate of decrease of density in any volume element is equal to the net current flowing out of that element. Thus, for the Schr¨odinger equation, we can consistently ! define ρ = |ψ|2 , and j = −i 2m (ψ ∗ ∇ψ − ψ∇ψ ∗ ). Applied to the Klein-Gordon equation (15.3), the same consideration implies !2 ∂t (ψ ∗ ∂t ψ − ψ∂t ψ ∗ ) − !2 c2 ∇ · (ψ ∗ ∇ψ − ψ∇ψ ∗ ) = 0 , from which we deduce the correspondence, ρ=i
! (ψ ∗ ∂t ψ − ψ∂t ψ ∗ ) , 2mc2
j = −i
! (ψ ∗ ∇ψ − ψ∇ψ ∗ ) . 2m
The continuity equation associated with the conservation of probability can be expressed covariantly in the form ∂µ j µ = 0 ,
(15.4)
where j µ = (ρc, j) is the 4-current. Thus, the Klein-Gordon density is the time-like component of a 4-vector. From this association it is possible to identify three aspects which (at least initially) eliminate the Klein-Gordon equation as a wholey suitable candidate for the relativistic version of the wave equation: 3
You may recognize that the leading correction to the free particle Schr¨ odinger equation is precisely the relativistic correction to the kinetic energy that we considered in chapter 9.
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172
! The first disturbing feature of the Klein-Gordon equation is that the density ρ is not a positive definite quantity, so it can not represent a probability. Indeed, this led to the rejection of the equation in the early years of relativistic quantum mechanics, 1926 to 1934. ! Secondly, the Klein-Gordon equation is not first order in time; it is necessary to specify ψ and ∂t ψ everywhere at t = 0 to solve for later times. Thus, there is an extra constraint absent in the Schr¨odinger formulation. ! Finally, the equation on which the Klein-Gordon equation is based, E 2 = m2 c4 + p2 c2 , has both positive and negative solutions. In fact the apparently unphysical negative energy solutions are the origin of the preceding two problems. To circumvent these difficulties one might consider dropping the negative energy solutions altogether. For a free particle, whose energy is thereby constant, we can simply supplement the Klein-Gordon equation with the condition p0 > 0. However, such a definition becomes inconsistent in the presence of local interactions, e.g. ) 2 * ∂ + kc2 ψ = F (ψ) self − interaction . 2 2 (∂ + iqA/!c) + kc ψ = 0 interaction with EM field.
The latter generate transitions between positive and negative energy states. Thus, merely excluding the negative energy states does not solve the problem. Later we will see that the interpretation of ψ as a quantum field leads to a resolution of the problems raised above. Historically, the intrinsic problems confronting the Klein-Gordon equation led Dirac to introduce another equation.4 However, as we will see, although the new formulation implied a positive norm, it did not circumvent the need to interpret negative energy solutions.
15.2
Dirac Equation
Dirac attached great significance to the fact that Schr¨odinger’s equation of motion was first order in the time derivative. If this holds true in relativistic quantum mechanics, it must also be linear in ∂. On the other hand, for free particles, the equation must imply pˆ2 = (mc)2 , i.e. the wave equation must be consistent with the Klein-Gordon equation (15.3). At the expense of introducing vector wavefunctions, Dirac’s approach was to try to factorise this equation: (γ µ pˆµ − m) ψ = 0 .
(15.5)
(Following the usual convention we have, and will henceforth, adopt the shorthand convention and set ! = c = 1.) For this equation to be admissible, the following conditions must be enforced: ! The components of ψ must satisfy the Klein-Gordon equation. 4
The original references are P. A. M. Dirac, The Quantum theory of the electron, Proc. R. Soc. A117, 610 (1928); Quantum theory of the electron, Part II, Proc. R. Soc. A118, 351 (1928). Further historical insights can be obtained from Dirac’s book on Principles of Quantum mechanics, 4th edition, Oxford University Press, 1982.
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Paul A. M. Dirac 1902-1984 Dirac was born on 8th August, 1902, at Bristol, England, his father being Swiss and his mother English. He was educated at the Merchant Venturer’s Secondary School, Bristol, then went on to Bristol University. Here, he studied electrical engineering, obtaining the B.Sc. (Engineering) degree in 1921. He then studied mathematics for two years at Bristol University, later going on to St. John’s College, Cambridge, as a research student in mathematics. He received his Ph.D. degree in 1926. The following year he became a Fellow of St.John’s College and, in 1932, Lucasian Professor of Mathematics at Cambridge. Dirac’s work was concerned with the mathematical and theoretical aspects of quantum mechanics. He began work on the new quantum mechanics as soon as it was introduced by Heisenberg in 1928 – independently producing a mathematical equivalent which consisted essentially of a noncommutative algebra for calculating atomic properties – and wrote a series of papers on the subject, leading up to his relativistic theory of the electron (1928) and the theory of holes (1930). This latter theory required the existence of a positive particle having the same mass and charge as the known (negative) electron. This, the positron was discovered experimentally at a later date (1932) by C. D. Anderson, while its existence was likewise proved by Blackett and Occhialini (1933) in the phenomena of “pair production” and “annihilation”. Dirac was made the 1933 Nobel Laureate in Physics (with Erwin Schr¨ odinger) for the discovery of new productive forms of atomic theory.
15.2. DIRAC EQUATION
173
! There must exist a 4-vector current density which is conserved and whose time-like component is a positive density. ! The components of ψ do not have to satisfy any auxiliary condition. At any given time they are independent functions of x. Beginning with the first of these requirements, by imposing the condition [γ µ , pˆν ] = γ µ pˆν − pˆν γ µ = 0, (and symmetrizing) ! " 1 ν µ ν µ 2 (γ pˆν + m) (γ pˆµ − m) ψ = {γ , γ } pˆν pˆµ − m ψ = 0 , 2 the latter recovers the Klein-Gordon equation if we define the elements γ µ such that they obey the anticommutation relation,5 {γ ν , γ µ } ≡ γ ν γ µ + γ µ γ ν = 2g µν – thus γ µ , and therefore ψ, can not be scalar. Then, from the expansion of ˆ − m)ψ = i∂t ψ − γ 0 γ · p ˆ ψ − mγ 0 ψ = 0, the Dirac Eq. (15.5), γ 0 (γ 0 pˆ0 − γ · p equation can be brought to the form ˆ i∂t ψ = Hψ,
ˆ =α·p ˆ + βm , H
(15.6)
where the elements of the vector α = γ 0 γ and β = γ 0 obey the commutation relations, {αi , αj } = 2δij ,
β 2 = 1,
{αi , β} = 0 .
(15.7)
ˆ is Hermitian if, and only if, α† = α, and β † = β. Expressed in terms of H † γ, this requirement translates to the condition (γ 0 γ)† ≡ γ † γ 0 = γ 0 γ, and † γ 0 = γ 0 . Altogether, we thus obtain the defining properties of Dirac’s γ matrices, γ µ† = γ 0 γ µ γ 0 ,
{γ µ , γ ν } = 2g µν .
(15.8)
Given that space-time is four-dimensional, the matrices γ must have dimension of at least 4 × 4, which means that ψ has at least four components. It is not, however, a 4-vector; it does not transform like xµ under Lorentz transformations. It is called a spinor, or more correctly, a bispinor with special Lorentz transformations which we will shall discuss presently. ! Info. An explicit representation of the γ matrices which most easily captures the non-relativistic limit is the following, ! " ! " I2 0 0 σ γ0 = , γ= , (15.9) 0 −I2 −σ 0
where σ denote the familiar 2 × 2 Pauli spin matrices which satisfy the relations, σi σj = δij + i$ijk σk , σ † = σ. The latter is known in the literature as the DiracPauli representation. We will adopt the particular representation, ! " ! " ! " 0 1 0 −i 1 0 σ1 = , σ2 = , σ3 = . 1 0 i 0 0 −1 Note that with this definition, the matrices α and β take the form, ! " ! " 0 σ I2 0 α= , β= . σ 0 0 −I2
5
Note that, in some of the literature, you will see the convention [ , ]+ for the anticommutator.
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15.2. DIRAC EQUATION
15.2.1
174
Density and Current
Turning to the second of the requirements placed on the Dirac equation, we now seek the probability density ρ = j 0 . Since ψ is a complex spinor, ρ has to be of the form ψ † M ψ in order to be real and positive. Applying hermitian conjugation to the Dirac equation, we obtain µ← − [(γ µ pˆµ − m)ψ]† = ψ † (−iγ † ∂ µ − m) = 0 ,
← − where ψ † ∂ µ ≡ (∂µ ψ)† . Making use of (15.8), and defining ψ¯ ≡ ψ † γ 0 , the − ¯ ← Dirac equation takes the form ψ(i + ∂ + m) = 0, where we have introduced the Feynman ‘slash’ notation + a ≡ aµ γ µ . Combined with Eq. (15.5) (i.e. − → (i + ∂ − m)ψ = 0), we obtain / 0 ) µ * ← − − → ¯ ψ = 0. ψ¯ + ∂ + + ∂ ψ = ∂µ ψγ
From this result and the continuity relation (15.4) we can identify ¯ µψ , j µ = ψγ
(15.10)
(or, equivalently, (ρ, j) = (ψ † ψ, ψ † αψ)) as the 4-current. In particular, the density ρ = j 0 = ψ † ψ is, as required, positive definite.
15.2.2
Relativistic Covariance
To complete our derivation, we must verify that the Dirac equation remains invariant under Lorentz transformations. More precisely, if a wavefunction ψ(x) obeys the Dirac equation in one frame, its counterpart ψ $ (x$ ) in a Lorentz transformed frame x$ = Λx, must obey the Dirac equation, ) µ $ * iγ ∂µ − m ψ $ (x$ ) = 0 . (15.11)
In order that an observer in the second frame can reconstruct ψ $ from ψ there must exist a local transformation between the wavefunctions. Taking this relation to be linear, we therefore must have, ψ $ (x$ ) = S(Λ)ψ(x) , where S(Λ) represents a non-singular 4 × 4 matrix. Now, using the identity, ∂xν ∂ −1 )ν ∂ = (Λ−1 )ν ∂ , the Dirac equation (15.11) ∂µ$ ≡ ∂x∂" µ = ∂x " µ ∂xν = (Λ µ ∂xν µ ν in the transformed frame takes the form, ) µ −1 ν * iγ (Λ ) µ ∂ν − m S(Λ)ψ(x) = 0 . The latter is compatible with the Dirac equation in the original frame if S(Λ)γ ν S −1 (Λ) = γ µ (Λ−1 )νµ .
(15.12)
To define an explicit form for S(Λ) we must now draw upon some of the defining properties of the Lorentz group discussed earlier. For an infinitesimal proper Lorentz transformation we have Λνµ = g νµ + ω νµ and (Λ−1 )νµ = g νµ − ω νµ + · · ·, where the matrix ωµν is antisymmetric and g νµ ≡ δ νµ . Correspondingly, by Taylor expansion in ω, we can define i S(Λ) = I − Σµν ω µν + · · · , 4 Advanced Quantum Physics
i S −1 (Λ) = I + Σµν ω µν + · · · , 4
15.2. DIRAC EQUATION
175
where the matrices Σµν are also antisymmetric in µν. To first order in ω, Eq. (15.12) yields (a somewhat unrewarding exercise!) * ) (15.13) [γ ν , Σαβ ] = 2i g να γβ − g νβ γα . The latter is satisfied by the set of matrices (another exercise!)6 Σαβ =
i [γα , γβ ] . 2
(15.14)
In summary, if ψ(x) obeys the Dirac equation in one frame, the wavefunction can be obtained in the Lorentz transformed frame by applying the transformation ψ $ (x$ ) = S(Λ)ψ(Λ−1 x$ ). Let us now consider the physical consequences of this Lorentz covariance.
15.2.3
Angular momentum and spin
To explore the physical manifestations of Lorentz covariance, it is instructive to consider the class of spatial rotations. For an anticlockwise spatial rotation by an infinitesimal angle θ about a fixed axis n, x )→ x$ = x − θx × n. In terms of the “Lorentz transformation”, Λ, one has x$i = [Λx]i ≡ xi − ωij xj where ωij = $ijk nk θ, and the remaining elements Λµ0 = Λ0µ = 0. Applied to the argument of the wavefunction we obtain a familiar result,7 ψ(x) = ψ(Λ−1 x$ ) = ψ(x$0 , x$ + x$ × nθ) = (1 − θn · x$ × ∇ + · · ·)ψ(x$ ) ˆ + · · ·)ψ(x$ ), = (1 − iθn · L
ˆ =x ˆ×p ˆ represents the non-relativistic angular momentum operator. where L Formally, the angular momentum operators represent the generators of spatial rotations.8 However, we have seen above that Lorentz covariance demands that the transformed wavefunction be multiplied by S(Λ). Using the definition of ωij above, one finds that i S(Λ) ≡ S(I + ω) = I − $ijk nk Σij θ + · · · 4 Then drawing on the Dirac/Pauli representation, " ! "2 1! i i i 0 σi 0 σj Σij = [γi , γj ] = , = − [σi , σj ] ⊗ I2 = $ijk σk ⊗ I2 , −σ 0 −σ 0 2 2 2 i j one obtains S(Λ) = I − in · Sθ + · · · ,
1 S= 2
!
σ 0
0 σ
"
.
Combining both contributions, we thus obtain ˆ + · · ·)ψ(x$ ) , ψ $ (x$ ) = S(Λ)ψ(Λ−1 x$ ) = (1 − iθn · J 6 Since finite transformations are of the form S(Λ) = exp[−(i/4)Σαβ ω αβ ], one may show that S(Λ) is unitary for spatial rotations, while it is Hermitian for Lorentz boosts. 7 ˆ (θ) = exp(−iθn · Recall that spatial rotataions are generated by the unitary operator, U ˆ L). 8 ˆ For finite transformations, the generator takes the form exp[−iθn · L].
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15.3. FREE PARTICLE SOLUTION OF THE DIRAC EQUATION ˆ=L ˆ + S can be identified as a total effective angular momentum of where J the particle being made up of the orbital component, together with an intrinsic contribution known as spin. The latter is characterised by the defining condition: [Si , Sj ] = i$ijk Sk ,
(Si )2 =
1 4
for each i .
(15.15)
Therefore, in contrast to non-relativistic quantum mechanics, the concept of spin does not need to be grafted onto the Schr¨odinger equation, but emerges naturally from the fundamental invariance of the Dirac equation under Lorentz transformations. As a corollary, we can say that the Dirac equation is a relativistic wave equation for particles of spin 1/2.
15.2.4
Parity
So far, our discussion of the covariance properties of the Dirac equation have only dealt with the subgroup of proper orthochronous Lorentz transformations, L↑+ – i.e. those that can be reached from Λ = I by a sequence of infinitesimal transformations. Taking the parity operation into account, relativistic covariance demands S −1 (P )γ 0 S(P ) = γ 0 ,
S −1 (P )γ i S(P ) = −γ i .
This is achieved if S(P ) = γ 0 eiφ , where φ denotes some arbitrary phase. Taking into account the fact that P 2 = I, φ = 0 or π, and we find ψ $ (x$ ) = S(P )ψ(x) = ηγ 0 ψ(P −1 x$ ) = ηγ 0 ψ(ct$ , −x$ ) ,
(15.16)
where η = ±1 represents the intrinsic parity of the particle.
15.3
Free Particle Solution of the Dirac Equation
Having laid the foundation we will now apply the Dirac equation to the problem of a free relativistic quantum particle. For a free particle, the plane wave ψ(x) = exp[−ip · x]u(p) , 3 with energy E ≡ p0 = ± p2 + m2 will be a solution of the Dirac equation if the components of the spinor u(p) are chosen to satisfy the equation (+ p − m)u(p) = 0. Evidently, as with the Klein-Gordon equation, we see that the Dirac equation therefore admits negative as well as positive energy solutions! Soon, having attached a physical significance to the former, we will see that it is convenient to reverse the sign of p for the negative energy solutions. However, for now, let us continue without worrying about the dilemma posed by the negative energy states. In the Dirac-Pauli block representation, ! 0 " p − m −σ · p µ γ pµ − m = . σ · p −p0 − m Thus, defining the spin elements u(p) = (ξ, η), where ξ and η represent twocomponent spinors, we find the conditions, (p0 − m)ξ = σ · p η and σ · p ξ =
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15.3. FREE PARTICLE SOLUTION OF THE DIRAC EQUATION ·p (p0 + m)η. With (p0 )2 = p2 + m2 , these equations are consistent if η = pσ 0 +m ξ. We therefore obtain the bispinor solution χ(r) u(r) (p) = N (p) σ · p (r) , χ p0 + m
where χ(r) represents any pair of orthogonal two-component vectors, and N (p) is the normalisation. Concerning the choice of χ(r) , in many situations, the most convenient basis is the eigenbasis of helicity – eigenstates of the component of spin resolved in the direction of motion, S·
p (±) σ p (±) 1 χ ≡ · χ = ± χ(±) , |p| 2 |p| 2
ˆ3 , χ(+) = (1, 0) and χ(−) = (0, 1). Then, for the positive e.g., for p = p3 e energy states, the two spinor plane wave solutions can be written in the form χ(±) ψp(±) (x) = N (p)e−ip·x |p| (±) ± 0 χ p +m
Thus, according to the discussion above, the Dirac equation for a free particle admits four solutions, two states with positive energy, and two with negative.
15.3.1
Klein paradox: anti-particles
While the Dirac equation has been shown to have positive definite density, as with the Klein-Gordon equation, it still exhibits negative energy states! To make sense of these states it is illuminating to consider the scattering of a plane wave from a potential step. To be precise, consider a beam of relativistic particles with unit amplitude, energy E, momentum pˆ e3 , and spin ↑ (i.e. χ = (1, 0)), incident upon a potential V (x) = V θ(x3 ) (see figure). At the potential barrier, spin is conserved, while a component of the beam with amplitude r is reflected (with energy E and momentum −pˆ e3 ), and a ˆ3 . component t is transmitted with energy E $ = E − V and momentum p$ e According to the energy-momentum invariant, conservation of energy across the interface dictates that E 2 = p2 + m2 and E $ 2 = p$2 + m2 . Being first order, the boundary conditions on the Dirac equation require only continuity of ψ (cf. the Schr¨odinger equation). Therefore, matching ψ at the step, we obtain the relations 1 1 1 0 p + r 0p = t p0" , − " E+m E+m E +m 0 0 0 from which we find 1 +r = t, and these equations lead to t=
2 , 1+ζ
p (1 − r) p0 +m
1+r 1 = , 1−r ζ
=
p" t. p"0 +m
r=
Setting ζ =
p" (E+m) p (E " +m) ,
1−ζ . 1+ζ
To interpret these solutions, let us consider the current associated with the reflected and transmitted components. Making use of the equation for the current density, j = ψ † αψ, and using the Dirac/Pauli representation wherein ! "! " ! " I2 σ3 σ3 α3 = γ0 γ3 = = , −I2 −σ3 σ3 Advanced Quantum Physics
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15.3. FREE PARTICLE SOLUTION OF THE DIRAC EQUATION ˆ3 -direction is given by the current along e ! " σ3 j3 = ψ † ψ, σ3
j1 = j2 = 0 .
Therefore, up to an overall constant of normalisation, the current densities are given by (i)
j3 =
p0
2p , +m
(t)
j3 =
2(p$ + p$∗ ) 2 |t| , p$0 + m
(r)
j3 = −
2p |r|2 . p0 + m
From these relations we obtain (t)
(p$ + p$∗ ) p0 + m 4 1 = (ζ + ζ ∗ ) 2p p$0 + m |1 + ζ|2 2 4 4 4 1 − ζ 42 2 4 = −|r| = − 44 1+ζ4
j3
= |t|2
(i)
j3
(r)
j3
(i)
j3
from which current conservation can be confirmed: (r)
1+
j3
(i)
j3
(t)
=
|1 + ζ|2 − |1 − ζ|2 2(ζ + ζ ∗ ) j = = 3(i) . 2 2 |1 + ζ| |1 + ζ| j3
Interpreting these results, it is convenient to separate our consideration into three distinct regimes of energy: ! E $ ≡ (E − V ) > m: In this case, from the Klein-Gordon condition (the energy-momentum invariant) p$2 ≡ E $2 − m2 > 0, and (taking p$ > 0 – i.e. beam propagates to the right) ζ > 0 and real. From this result (r) (i) we find |j3 | < |j3 | – as expected, within this interval of energy, a component of the beam is transmitted and the remainder is reflected (cf. non-relativistic quantum mechanics). ! −m < E $ < m: In this case p$2 ≡ E $2 − m2 < 0 and p$ is purely imaginary. From this result it follows that ζ is also pure imaginary and (r) (i) |j3 | = |j3 |. In this regime the under barrier solutions are evanescent and quickly decay to the right of the barrier. All of the beam is reflected (cf. non-relativistic quantum mechanics). ! E $ < −m: Finally, in this case p$2 ≡ E $2 −m2 > 0 and, depending on the (r) (i) sign of p$ , j3 can be greater or less than j3 . But the solution has the " " form e−i(p x−E t) . Since we presume the beam to be propagating to the right, we require E $ < 0 and p$ > 0. From this result it follows that ζ < 0 (r) (i) and we are drawn to the surprising conclusion that |j3 | > |j3 | – more current is reflected that is incident! Since we have already confirmed (t) current conservation, we can deduce that j3 < 0. It is as if a beam of particles were incident from the right. The resolution of this last seeming unphysical result, known as the Klein paradox,9 in fact gives a natural interpretation of the negative energy solutions that plague both the Dirac and Klein-Gordon equations: Dirac particles are fermionic in nature. If we regard the vacuum as comprised of a filled Fermi sea of negative energy states or antiparticles (of negative charge), the Klein Paradox can be resolved as the stimulated emission of particle/antiparticle 9
Indeed one would reach the same conclusion were one to examine the Klein-Gordon equation.
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15.3. FREE PARTICLE SOLUTION OF THE DIRAC EQUATION Figure 15.2: The photograph shows a
small part of a complicated high energy neutrino event produced in the Fermilab bubble chamber filled with a neon hydrogen mixture. A positron (red) emerging from an electron-positron pair, produced by a gamma ray, curves round through about 180 degrees. Then it seems to change charge: it begins to curve in the opposite direction (blue). What has happened is that the positron has run head-on into a (moreor-less from the point of view of particle physics) stationary electron – transferring all its momentum. This tells us that the mass of the positron equals the mass of the electron.
pairs, the particles moving off towards x3 = −∞ and the antiparticles towards x3 = ∞. What about energy conservation? One might worry that the energy for these pairs is coming from nowhere. However, the electrostatic energy recovered by the antiparticle when its created is sufficient to outweigh the rest mass energy of the particle and antiparticle pair (remember that a repulsive potential for particles is attractive for antiparticles). Taking into account the fact that the minimum energy to create a particle/antiparticle pair is twice the rest mass energy 2 × m, the regime where stimulated emission is seen to occur can be understood. Negative energy states: With this conclusion, it is appropriate to revisit the definition of the free particle plane wave 3 state. In particular, for energies E < 0, it is more sensible to set p0 = + (p2 + m2 ), and redefine the plane wave solution as ψ(x) = v(p)eip·x , where the spinor satisfies the condition (+ p + m)v(p) = 0. Accordingly we find, 5 σ · p (r) 6 χ (r) v (p) = N (p) p0 + m . χ(r) So, to conclude, two relativistic wave equations have been proposed. The first of these, the Klein-Gordon equation was dismissed on the grounds that it exhibited negative probability densities and negative energy states. By contrast, the states of the Dirac equation were found to exhibit a positive definite probability density, and the negative energy states were argued to have a natural interpretation in terms of antiparticles: the vacuum state does not correspond to all states unoccupied but to a state in which all the negative energy states are occupied – the negative energy states are filled up by a Fermi sea of negative energy Fermi particles. For electron degrees of freedom, if a positive energy state is occupied we observe it as a (positive energy) electron of charge q = −e. If a negative energy state is unoccupied we observe it as a (positive energy) antiparticle of charge q = +e, a positron, the antiparticle of the electron. If a very energetic electron interacts with the sea causing a transition from a negative energy state to positive one (by communicating an energy of at least 2m) this is observed as the production of a pair of particles, an electron and a positron from the vacuum (pair production) (see Fig. 15.2). However, the interpretation attached to the negative energy states provides grounds to reconsider the status of the Klein-Gordon equation. Evidently, the Advanced Quantum Physics
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15.4. QUANTIZATION OF RELATIVISTIC FIELDS Dirac equation is not a relativistic wave equation for a single particle. If it were, pair production would not appear. Instead, the interpretation above forces us to consider the wavefunction of the Dirac equation as a quantum field able to host any number of particles – cf. the continuum theory of the quantum harmonic chain. In the next section, we will find that the consideration of the wavefunction as a field revives the Klein-Gordon equation as a theory of scalar (interger spin) particles.
15.4
Quantization of relativistic fields
15.4.1
Info: Scalar field: Klein-Gordon equation revisited
Previously, the Klein-Gordon equation was abandoned as a candidate for a relativistic theory on the basis that (i) it admitted negative energy solutions, and (ii) that the probability density associated with the wavefunction was not positive definite. Yet, having associated the negative energy solutions of the Dirac equation with antiparticles, and identified ψ as a quantum field, it is appropriate that we revisit the Klein-Gordon equation and attempt to revive it as a theory of relativistic particles of spin zero. If φ were a classical field, the Klein-Gordon equation would represent the equation of motion associated with the Lagrangian density (exercise) L=
1 1 ∂µ φ ∂ µ φ − m2 φ2 , 2 2
(cf. our discussion of the low energy modes of the classical harmonic chain and the Maxwell field of the waveguide in chapter 11). Defining the canonical momentum ˙ π(x) = ∂φ˙ L(x) = φ(x) ≡ ∂0 φ(x), the corresponding Hamiltonian density takes the form , 1+ 2 H = π φ˙ − L = π + (∇φ)2 + m2 φ2 . 2
Evidently, the Hamiltonian density is explicitly positive definite. Thus, the scalar field is not plagued by the negative energy problem which beset the single-particle theory. Similarly, the quantization of the classical field will lead to a theory in which the states have positive energy. Following on from our discussion of the harmonic chain in chapter 11, we are already equipped to quantise the classical field theory. However, there we worked explicitly in the Schr¨odinger representation, in which the dynamics was contained within the time-dependent wavefunction ψ(t), and the operators were time-independent. Alternatively, one may implement quantum mechanics in a representation where the time dependence is transferred to the operators instead of the wavefunction — the Heisenberg representation. In this representation, the Schr¨odinger state vector ψS (t) is related to the Heisenberg state vector ψH by the relation, ˆ
ψS (t) = e−iHt ψH ,
ψH = ψS (0) .
ˆ S are related to the Heisenberg operators O ˆ H (t) by Similarly, Schr¨odinger operators O ˆ ˆ ˆ ˆ H (t) = eiHt O OS e−iHt .
ˆ S |ψS 4 and 3ψ ! |O ˆ H |ψH 4 are One can easily check that the matrix elements 3ψS! |O H equivalent in the two representations, and which to use in non-relativistic quantum mechanics is largely a matter of taste and convenience. However, in relativistic quantum field theory, the Heisenberg representation is often preferable to the Schr¨odinger representation. The main reason for this is that by using the former, the Lorentz covariance of the field operators is made manifest.
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15.4. QUANTIZATION OF RELATIVISTIC FIELDS In the Heisenberg representation, the quantisation of the fields is still enforced by ˆ but in this case, we promoting the classical fields to operators, π )→ π ˆ and φ )→ φ, impose the equal time commutation relations, -
. ˆ t), π φ(x, ˆ (x! , t) = iδ 3 (x − x! ),
-
. ˆ t), φ(x ˆ ! , t) = [ˆ φ(x, π (x, t), π ˆ (x! , t)] = 0 ,
ˆ In doing so, the Hamiltonian density takes the form with π ˆ = ∂0 φ. . ˆ 2 + m2 φˆ2 . ˆ=1 π ˆ 2 + (∇φ) H 2
To see the connection between the quantized field and particles we need to Fourier transform the field operators to obtain the normal modes of the Hamiltonian, 7 d4 k ˆ ˆ φ(x) = φ(k)e−ik·x . (2π)4
ˆ However the form of the Fourier elements φ(k) is constrained by the following conˆ ditions. Firstly to maintain Hermiticity of the field operator φ(x) we must choose † ˆ ˆ Fourier coefficients such that φ (k) = φ(−k). Secondly, to ensure that the field opˆ ˆ erator φ(x) obeys the Klein-Gordon equation,10 we require φ(k) ∼ 2πδ(k 2 − m2 ). Taking these conditions together, we require ) * ˆ φ(k) = 2πδ(k 2 − m2 ) θ(k 0 )a(k) + θ(−k 0 )a† (−k) ,
√ where k 0 ≡ ωk ≡ + k2 + m2 , and a(k) represent the operator valued Fourier coefficients. Rearranging the momentum integration, we obtain the Lorentz covariant expansion 7 + , d4 k ˆ φ(x) = 2πδ(k 2 − m2 )θ(k 0 ) a(k)e−ik·x + a† (k)eik·x . 4 (2π)
Integrating over k 0 , and making use of the identity 7 7 d4 k d4 k 2 2 0 2πδ(k − m )θ(k ) = δ(k02 − ωk2 )θ(k 0 ) 4 (2π) (2π)3 7 7 d4 k d4 k 1 0 = δ [(k − ω )(k + ω )] θ(k ) = [δ(k0 − ωk ) + δ(k0 + ωk )] θ(k 0 ) 0 k 0 k 3 (2π) (2π)3 2k0 7 7 7 d3 k dk0 d3 k 0 = δ(k − ω )θ(k ) = , 0 k 3 (2π) 2k0 (2π)3 2ωk one obtains ˆ φ(x) =
7
* d3 k ) a(k)e−ik·x + a† (k)eik·x . (2π)3 2ωk
More compactly, making use of the orthonormality of the basis 7 ↔ 1 fk = 3 e−ik·x , fk∗ (x)i ∂ 0 fk! (x)d3 x = δ 3 (k − k! ), 3 (2π) 2ωk ↔
where A ∂0 B ≡ A∂t B − (∂t A)B, we obtain 7 + , d3 k ˆ 3 φ(x) = a(k)fk (x) + a† (k)fk∗ (x) . (2π)3 2ωk 10
Note that the field operators obey the equation of motion, π(x, ˙ t) = −
∂H = ∇2 φ − m2 φ . ∂φ(x, t)
˙ one finds (∂ 2 + m2 )φ = 0. Together with the relation π = φ,
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15.4. QUANTIZATION OF RELATIVISTIC FIELDS Similarly, ˆ π ˆ (x) ≡ ∂0 φ(x) =
7
3
d3 k (2π)3 2ω
k
+ , iωk −a(k)fk (x) + a† (k)fk∗ (x) .
Making use of the orthogonality relations, the latter can be inverted to give 7 7 3 3 ↔ ↔ ˆ ˆ a(k) = (2π)3 2ωk d3 xfk∗ (x)i ∂0 φ(x), a† (k) = (2π)3 2ωk d3 xφ(x)i ∂0 fk (x) , or, equivalently, 7 / 0 ˆ a(k) = d3 x ωk φ(x) − iˆ π (x) e−ik·x ,
a (k) = †
7
With these definitions, it is left as an exercise to show + , a(k), a† (k! ) = (2π)3 2ωk δ 3 (k − k! ),
/ 0 ˆ d3 x ωk φ(x) + iˆ π (x) eik·x .
+ , [a(k), a(k! )] = a† (k), a† (k! ) = 0 .
The field operators a† and a can therefore be identified as operators that create and annihilate bosonic particles. Although it would be tempting to adopt a different normalisation wherein [a, a† ] = 1 (as is done in many texts), we chose to adopt the convention above where the covariance of the normalisation is manifest. Using this representation, the Hamiltonian is brought to the diagonal form 7 , d3 k ωk + † ˆ = H a (k)a(k) + a(k)a† (k) , 3 (2π) 2ωk 2
a result which can be confirmed by direct substitution. Defining the vacuum state |Ω4 as the state which is annhiliated by a(k), a single particle state is obtained by operating the creation operator on the vacuum, |k4 = a† (k)|Ω4 .
Then 3k! |k4 = 3Ω|a(k! )a† (k)|Ω4 = 3Ω|[a(k! ), a† (k)]|Ω4 = (2π)3 2ωk δ 3 (k! − k). Manyparticle states are defined by |k1 · · · kn 4 = a† (k1 ) · · · a† (kn )|Ω4 where the bosonic statistics of the particles is assured by the commutation relations. Associated with these field operators, one can define the total particle number operator 7 d3 k ˆ = 3 N a† (k)a(k) . (2π)3 2ωk Similarly, the total energy-momentum operator for the system is given by 7 d3 k µ ˆ 3 P = k µ a† (k)a(k) . (2π)3 2ωk
The time component Pˆ 0 of this result can be compared with the Hamiltonian above. In fact, commuting the field operators, the latter is seen to differ from Pˆ 0 by an infinite 8 constant, d3 kωk /2. Yet, had we simply normal ordered11 the operators from the outset, this problem would not have arisen. We therefore discard this infinite constant.
15.4.2
Info: Charged Scalar Field
A generalization of the analysis above to the complex scalar field leads to the Lagrangian, L= 11
1 1 ∂µ φ∂ µ φ¯ − m2 |φ|2 . 2 2
Recall that normal ordering entails the construction of an operator with all the annihilation operators moved to the right and creation operators moved to the left.
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15.4. QUANTIZATION OF RELATIVISTIC FIELDS The latter can √be interpreted as the superposition of two independent scalar fields φ = (φ1 +iφ2 )/ 2, where, for each (real) component φ†r (x) = φr (x). (In fact, we could as easily consider a field with n components.) In this case, the canonical quantisation of the classical fields is achieved by defining (exercise) 7 + , d3 k ˆ 3 φ(x) = a(k)fk (x) + b† (k)fk∗ (x) . 3 (2π) 2ωk (similarly φ† (x)) where both a and b obey bosonic commutation relations, + , + , a(k), a† (k! ) = b(k), b† (k! ) = (2π)3 2ωk δ 3 (k − k! ), + , [a(k), a(k! )] = [b(k), b(k! )] = a(k), b† (k! ) = [a(k), b(k! )] = 0 .
With this definition, the total number operator is given by 7 + † , d3 k ˆ = ˆa + N ˆb , 3 N a (k)a(k) + b† (k)b(k) ≡ N 3 (2π) 2ωk while the energy-momentum operator is defined by 7 + , d3 k 3 k µ a† (k)a(k) + b† (k)b(k) . Pˆ µ = 3 (2π) 2ωk
Thus the complex scalar field has the interpretation of creating different sorts of particles, corresponding to operators a† and b† . To understand the physical interpretation of this difference, let us consider the corresponding charge density operator, 8 ↔ ˆj0 = φˆ† (x)i ∂ 0 φ(x). Once normal ordered, the total charge Q = d3 xj0 (x) is given by 7 + † , d3 k ˆa − N ˆb . ˆ= 3 a (k)a(k) − b† (k)b(k) = N Q 3 (2π) 2ωk From this result we can interpret the particles as carrying an electric charge, equal in magnitude, and opposite in sign. The complex scalar field is a theory of charged particles. The negative density that plagued the Klein-Gordon field is simply a manifestation of particles with negative charge.
15.4.3
Info: Dirac Field
The quantisation of the Klein-Gordon field leads to a theory of relativistic spin zero particles which obey boson statistics. From the quantisation of the Dirac field, we expect a theory of Fermionic spin 1/2 particles. Following on from our consideration of the Klein-Gordon theory, we introduce the Lagrangian density associated with the Dirac equation (exercise) L = ψ¯ (iγ µ ∂µ − m) ψ , ↔
¯ 1 iγ µ ∂ µ −m)ψ). With this definition, the corresponding (or, equivalently, L = ψ( 2 ¯ 0 = iψ † . From the Lagrangian density, canonical momentum is given by ∂ψ˙ L = iψγ we thus obtain the Hamiltonian density, H = ψ¯ (−iγ · ∇ + m) ψ ,
¯ 0 ∂0 ψ = ψ † i∂t ψ. which, making use of the Dirac equation, is equivalent to H = ψiγ For the Dirac theory, we postulate the equal time anticommutation relations 9 : ψα (x, t), iψβ† (x! , t) = iδ 3 (x − x! )δαβ ,
0 (or, equivalently {ψα (x, t), iψ¯β (x! , t)} = γαβ δ 3 (x − x! )), together with
; < {ψα (x, t), ψβ (x! , t)} = ψ¯α (x, t), ψ¯β (x! , t) = 0 .
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15.5. THE LOW ENERGY LIMIT OF THE DIRAC EQUATION Using the general solution of the Dirac equation for a free particle as a basis set, together with the intuition drawn from the study of the complex scalar field, we may with no more ado, introduce the field operators which diagonalise the Hamiltonian density 2 7 . = d3 k ψ(x) = ar (k)u(r) (k)e−ik·x + b†r (k)v (r) (k)eik·x 3 (2π) 2ωk r=1 7 2 . = d3 k - † (r) ik·x (r) −ik·x ¯ ψ(x) = a (k)¯ u (k)e + b (k)¯ v (k)e , r r (2π)3 2ωk r=1
where the annihilation and creation operators also obey the anticommutation relations, ; < ; < ar (k), a†s (k! ) = br (k), b†s (k! ) = (2π)3 2ωk δrs δ 3 (k − k! ) ; < ; < {ar (k), as (k! )} = a†r (k), a†s (k! ) = {br (k), bs (k! )} = b†r (k), b†s (k! ) = 0 .
The latter condition implies the Pauli exclusion principle a† (k)2 = 0. With this definition, a(k)u(k)e−ik·x annilihates a postive energy electron, and b† (k)v(k)eik·x creates a positive energy positron. From these results, making use of the expression for the Hamiltonian density operator, one obtains 2 7 = , + d3 k ˆ = ωk a†r (k)ar (k) − br (k)b†r (k) . H 3 (2π) 2ωk r=1
Were the commutation relations chosen as bosonic, one would conclude the existence of negative energy solutions. However, making use of the anticommutation relations, and dropping the infinite constant (or, rather, normal ordering) one obtains a positive definite result. Expressed as one element of the total energy-momentum operator, one finds Pˆ µ =
2 7 = r=1
, + d3 k k µ a†r (k)ar (k) + b†r (k)br (k) . 3 (2π) 2ωk
Finally, the total charge is given by 7 7 ˆ = ˆj 0 d3 x = d3 xψ † ψ = N ˆa − N ˆb . Q
8 ˆ represents the total number operator. Na = d3 k a† (k)a(k) is the number where N 8 of the particles and Nb = d3 k b† (k)b(k) is the number of antiparticles with opposite charge.
15.5
The low energy limit of the Dirac equation
To conclude our abridged exploration of the foundations of relativistic quantum mechanics, we turn to the interaction of a relativistic spin 1/2 particle with an electromagnetic field. Suppose that ψ represents a particle of charge q (q = −e for the electron). From non-relativistic quantum mechanics, we expect to obtain the equation describing its interaction with an EM field given by the potential Aµ by the minimal substitution pµ )−→ pµ − qAµ , where A0 ≡ ϕ. Applied to the Dirac equation, we obtain for the interaction of a particle with a given (non-quantized) EM field, [γµ (pµ − qAµ ) − m]ψ = 0, or compactly (+ p − q+A − m)ψ = 0 . Advanced Quantum Physics
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15.5. THE LOW ENERGY LIMIT OF THE DIRAC EQUATION Previously, in chapter 9, we explored the relativistic (fine-structure) corrections to the hydrogen atom. At the time, we alluded to these as the leading relativistic contributions to the Dirac theory. In the following section, we will explore how these corrections are derived. In the Dirac-Pauli representation, ! " ! " 0 σ I2 0 α= , β= . σ 0 0 −I2 we have seen that the plane-wave solution to the Dirac equation for particles can be written in the form ! " χ ψp (x) = N ei(px−Et)/! , cσ ·ˆ p χ mc2 +E where we have restored the parameters ! and c. From this expression, we can see that, at low energies, where |E − mc2 | - mc2 , the second component of the bispinor is smaller than the first by a factor of order v/c. To obtain the non-relavistic limit, we can exploit this asymmetry to develop a perturbative expansion of the coefficients in v/c. Consider then the Dirac equation for a particle moving in a potential (φ, A). Expressed in matrix form, the Dirac equation H = cα · (−i!∇ − e 2 c A) + mc β + eφ is expressed as ! " mc2 + eφ cσ · (−i!∇ − ec A) H= . cσ · (ˆ p − qc A) −mc2 + qφ Defining the bispinor ψ T (x) = (ψa (x), ψb (x)), the Dirac equation translates to the coupled equations, q (mc2 + eφ)ψa + cσ · (ˆ p − A)ψb = Eψa c q cσ · (ˆ p − A)ψa − (mc2 − qφ)ψb = Eψb . c Then, if we define W = E − mc2 , a rearrangement of the second equation obtains ψb =
2mc2
1 q cσ · (ˆ p − A)ψa . + W − qφ c
1 Then, at zeroth order in v/c, we have ψb $ 2mc p − qc A)ψa . Substi2 cσ ·(ˆ tuted into the first equation, we thus obtain the Pauli equation Hnon−rel ψa = W ψa , where
Hnonr el =
1 q .2 σ · (ˆ p − A) + qφ . 2m c
Making use of the Pauli matrix identity σi σj = δij + i$ijk σk , we thus obtain the familiar non-relativistic Schr¨odinger Hamiltonian, Hnon
rel
=
1 q q! (ˆ p − A)2 − σ · (∇ × A) + qφ . 2m c 2mc
As a result, we can identify the spin magnetic moment µS =
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15.5. THE LOW ENERGY LIMIT OF THE DIRAC EQUATION with the gyromagnetic ratio, g = 2. This compares to the measured value of g = 2 × (1.0011596567 ± 0.0000000035), the descrepency form 2 being attributed to small radiative corrections. Let us now consider the expansion to first order in v/c. Here, for simplicity, let us suppose that A = 0. In this case, taking into account the next order term, we obtain ! " 1 V −W ˆ ψa ψb $ 1 + cσ · p 2mc2 2mc2 where V = qφ. Then substituted into the second equation, we obtain 1 2 1 1 2 ˆ) + ˆ )(V − W )(σ · p ˆ ) + V ψa = W ψ a . (σ · p (σ · p 2m 4m2 c2 At this stage, we must be cautious in interpreting ψa as a complete nonrelavistic wavefunction with leading relativistic corrections. To find the true wavefunction, we have to consider the normalization. If we suppose that the original wavefunction is normalized, we can conclude that, 7 7 / 0 d3 xψ † (x, t)ψ(x, t) = d3 x ψa† (x, t)ψa (x, t) + ψb† (x, t)ψb (x, t) 7 7 1 3 † d3 xψa† (x, t)ˆ p2 ψa (x, t) . $ d xψa (x, t)ψa (x, t) + (2mc)2 Therefore, at this order, the normalized wavefunction is set by, ψs = (1 + 1 ˆ 2 )ψa or, inverted, p 8m2 c2 ! " 1 2 ˆ ψa = 1 − p ψs . 8m2 c2 Substituting, then rearranging the equation for ψs , and retaining terms of ˆ non−rel ψs = W ψs , where order (v/c)2 , one ontains (exercise) H ˆ2 ˆ4 p 1 1 ˆ non−rel = p ˆ )V (σ · p ˆ) + V − ˆ2 + p ˆ 2V ) . H − + (σ · p (V p 2m 8m3 c2 4m2 c2 8m2 c2 Then, making use of the identities, ˆ 2 ] = !2 (∇2 V ) + 2i!(∇V ) · p ˆ [V, p ˆ )V = V (σ · p ˆ ) + σ · [ˆ (σ · p p, V ]
ˆ )V (σ · p ˆ) = V p ˆ 2 − i!(∇V ) · p ˆ + !σ · (∇V ) × p ˆ, (σ · p we obtain the final expression (exercise), ˆ2 ˆ4 p ! !2 ˆ non−rel = p ˆ H − + σ · (∇V ) × p + (∇2 V ) . 2 c2 2m 8m3 c2 >4m2 c2 ?@ 8m A > ?@ A spin−orbit coupling
Darwin term
The second term on the right hand side represents the relativistic correction to the kinetic energy, the third term denotes the spin-orbit interaction and the final term is the Darwin term. For atoms, with a central potential, the spin-orbit term can be recast as ˆ S.O. = H
!2 1 !2 1 ˆ. ˆ= σ · (∂r V )r × p (∂r V )σ · L 2 2 4m c r 4m2 c2 r
To address the effects of these relativistic contributions, we refer back to chapter 9. Advanced Quantum Physics
186
Chapter 16
Problem sets Before starting these problems, you might want to revise some of the examples from the Part IB Quantum Physics course. The examples marked with a † are typically more challenging and are the ones to omit if your time is very short, or if you are finding the course difficult. Some of the questions involve a routine piece of bookwork. This is the kind of thing you will have to do in the exam. You are strongly encouraged to do these parts and get feedback in supervisions.
16.1
Problem Set I
1. Quantum mechanics in one-dimension: The following question introduces the concept of a scattering matrix (or S-matrix) in relation to scattering from a potential in one dimension. These concepts will prepare the ground for the study of the three-dimensional scattering problem addressed later in the course. Consider a localized potential in one-dimension (i.e. a potential that is non-zero only over a finite region in space) subject to a beam of quantum particles incident from the left and from the right (see figure). Outside the region of the potential, we know that the wavefunction √ of the particles is described by a plane wave of wavevector k = 2mE/�. The relation between the incoming and outgoing components of the plane wave are specified by a scattering matrix (commonly referred to as the (S-matrix)), i.e. referring to the figure, �� � � � � A C S11 S12 |Ψout � = S|Ψin � ⇒ = S21 S22 D B �a) Consider the action of the probability current operator on a plane wave and hence show that conservation of probability implies that |A|2 + |D|2 = |B|2 + |C|2 . Show that this condition is equivalent to demanding that the S-matrix is unitary, i.e. S † S = �. For matrices which are unitary, the eigenvalues have unit magnitude,1 eiθ – i.e. two scattering phase shifts, in general functions of k, completely describe the scattering in one dimension. For the case of a symmetric potential, V (x) = V (−x), the S-matrix assumes the simple form � � t r S= � r t 1 The proof is as follows: for an eigenvector |v�, such that S|v� = λ|v�, we have the norm �v|S † S|v� = |λ|2 �v|v� = �v|v�, i.e. |λ|2 = 1.
Advanced Quantum Physics
Be−ikx
Ceikx
���� Aeikx
De−ikx
16.1. PROBLEM SET I
188
where r and t are the reflection and transmission amplitudes. �b) Show that unitarity demands that rt∗ + r∗ t = 0 and |r|2 + |t|2 = 1, and hence that |r ± t|2 = 1. Find θ1 and θ2 in terms of r and t. What is the difference in phase between r and t? �c) By matching the boundary conditions, show the elements of the S2 k2 matrix for the scattering of particles of mass m and energy E = �2m γ from a δ-function potential, aV0 δ(x), are given by r = − γ+ik and maV0 ik t = γ+ik , where γ = �2 . Obtain the corresponding scattering phase shifts.
re−ikx
teikx
eikx
V0
2. Operator methods: This problem addresses simple relations that follow from the orthogonality of eigenfunctions and the time-development of wavefunctions. ˆ has two normalized eigenstates |ψ1 � and |ψ2 � which The Hamiltonian H correspond to different eigenvalues E1 and E2 . (a) Show that |ψ1 � and |ψ2 � are orthogonal. ˆ 1 � = |ψ2 � and A|ψ ˆ 2 � = |ψ1 �, calculate (b) For an observable Aˆ where A|ψ the eigenvalues and eigenvectors (which are combinations of |ψ1 � and |ψ2 �).
(c) Assuming that at t = 0 the system is in the state |ψ(t = 0)� = √1 [|ψ1 � − |ψ2 �], find the state of the system |ψ(t)� at time t and 2 show that the probability of the system returning to its initial state is given by P = cos2 [(E1 − E2 )t/2�].
3. Operator methods: This question relates to the probem of coherent or Glauber states. It is included in the problem set as it presents a useful arena in which to practice operator methods. The aim of the problem is to explore properties of coherent states and establish their connection to classical dynamics of the harmonic oscillator. �a) By using the commutation relation [a� a† ] = 1, show that †
†
e−βa aeβa = β + a .
Hint: To prove this result most straightforwardly, consider the β derivative of this expression.
†
Using this result, show that |β� = N eβa |0� is a coherent state, i.e. 2 a|β� = β|β�. Finally, show that the normalization, N = e−|β| /2 . �b) Calculate the expectation values, x0 = �ˆ x� and p0 = �ˆ p�, with 2 2 respect to |β� and, by considering �ˆ x � and �ˆ p �, show that (Δp)2 (Δx)2 = where Δp = pˆ − �ˆ p� (similarly x).
�2 � 4
�c) To determine the coordinate representation to the coherent state, ψ(x) = �x|β�, it is helpful to revert back to the expression for a as a differential operator. Show that the eigenvalue equation a|β� = β|β� translates to the equation, � � � x+ ∂x ψ(x) = βψ(x) . mω Advanced Quantum Physics
Hint: Remember how creation and annihilation operators are related to the phase space operators x ˆ and pˆ. Also, note that the Hermitian conjuate of the eigenvalue equation a|β� = β|β� leads to the relation �β|a† = �β|β ∗ .
16.1. PROBLEM SET I
189
Show that this equation has the solution � � p0 x (x − x0 )2 � +i ψ(x) = N exp − 4(Δx)2 � where x0 and p0 are defined in part (b) above. �d) By expressing |β� in the number basis, show that ωt
|β(t)� = e−i 2 |βe−iωt � . As a result, deduce expressions for x0 (t) and p0 (t) and show they represent solutions to the classical equations of motion. How does the width of the coherent state wavepacket evolve with time?
4. Charged particles in a magnetic field: In lectures, we studied the motion of an electron in a uniform magnetic field working in the Landau gauge, A = (−By� 0� 0). With this gauge choice, the Hamiltonian may be straightforwardly brought to a quantum harmonic oscillator form. In this question, we will address the problem by working the “symmetric” gauge A = (−By/2� Bx/2� 0). The advantage of this gauge choice is that it facilitates the development of the many-particle wavefunction in an aesthetic and useful form. �a) A spinless electron of charge q = −e is confined to the xy-plane and subject to a perpendicular magnetic field B = Bˆ ez . Working in the symmetric gauge A = (−y� x� 0)B/2, show that the electron Hamiltonian is given by � � �2 �2 1 1 1 1 ˆ pˆx − mωy + pˆy + mωx � H= 2m 2 2m 2 where ω =
eB m
denotes the cyclotron frequency.
�b) If units are chosen such that ω = m = � = 1, show that the Hamiltonian can be recast in the dimensionless form � �2 � �2 ˆ = 1 −i∂x − y + 1 −i∂y + x . H 2 2 2 2
�c) Introducting the complex coordinate, z¯ = x + iy and the complex conjugate, z = x − iy, the corresponding derivatives are given by ∂ . From this ∂z¯ = 12 (∂x − i∂y ), and ∂z = 12 (∂x + i∂y ) where ∂z ≡ ∂z definition, confirm that [z� ∂z ] = [¯ z � ∂z¯] = −1� Show that the operators, √ � z� a = 2 ∂z¯ + � 4� � √ z¯ � b = 2 ∂z + 4
[z� ∂z¯] = [¯ z � ∂z ] = 0 . √ � z¯ � 2 −∂z + � 4� � √ z b† = 2 −∂z¯ + � 4 a† =
fulfil the commutation relations [a� a† ] = [b� b† ] = 1 characteristic of creation and annihilation operators. Applied to the Hamiltonian, show that ˆ = a† a + 1 � H 2 Advanced Quantum Physics
Physicists often choose to work in dimensionless units setting � = 1, etc. Think about what this choice entails and, at each stage of the problem, try to infer how dimensionful parameters can be restored.
The choice z = x − iy (as opposed to z = x + iy) is made on purely aesthetic grounds – see the wavefunction below. Also, following convention in this subject, we have chosen to write the complex conjugate as z¯ and not z ∗ .)
16.1. PROBLEM SET I
190
independent of b, i.e. a quantum harmonic oscillator. As a result, we can identify a Landau level Hamiltonian with the level set by the eigenvalue, n of the number operator, n ˆ = a† a. �d) Show that the angular momentum operator takes the form, ˆ z = −(b† b − a† a)� . L ˆ z as −m� (with the Therefore, if we define the eigenvalues of L sign following an accepted convention), the quantum numbers m can take values m = −n� −n + 1� · · · � 0� 1� · · ·. The corresponding (normalized) states of Landau level, n, and angular momentum, m, are given by (b† )m+n (a† )n √ |0� 0� � |n� m� = � (m + n)� n�
where |0� 0� denotes the zero angular momentum ground state of the lowest Landau level. �e) Going back to the definition of the annihilation operator, show that the ground state in the real space representation is given by,2 1 �r|0� 0� = √ e−¯z z/4 . 2π Finally, show that in the lowest Landau level, �r|0� m� = √
1 z m e−¯z z/4 � m 2π2 m�
i.e. aside from the Gaussian factor, the states of the lowest Landau level are given by a polynomial in z – they are said to be analytic functions of z. �f ) The last part of this problem anticipates our studies of the many-particle fermionic system in chapter 8 and should not be attempted until this ground is covered. In this chapter, we will show that the properly antisymmetrized many-particle wavefunction of a non-interacting system of N identical spinless electrons (fermions) is given by the Slater determinant, ψ(r1 � r2 � · · ·) = detij φi (rj ) where φi (r) denote eigenstates of the single-particle problem. Using the identity � � � 1 1 1 ··· � � � � z1 z2 z3 · · · � � � � (zj − zk ) � � z2 z2 z2 · · · � = � � 1 2 3 � . .. . . �� j
known as a Vandemonde determinant, show that the ground state wavefunction of the filled lowest Landau level is given by � � � 1� 2 (zj − zk ) exp − |zi | . ψ(r1 � r2 � · · ·) = 4 j
2
i
In the symmetric gauge, we therefore find that the Landau level states are localized in both x and y directions. This contrasts with what was found from the Landau gauge condition where states were localized along only one direction. Of course, there is no contradiction between these two representations: since the Landau levels have a huge degeneracy, we are at liberty to reconstruct states within the basis.
Advanced Quantum Physics
Note that, for the ground state, we must have a|0� 0� = b|0� 0� = 0.
16.1. PROBLEM SET I
191
Note that, as required, the wavefunction is antisymmetric under the exchange of any two fermions. This expression also emphasizes the exclusion character of the fermionic system with the wavefunction vanishing as any two particles move together.
5. Spin: This question develops the concept of a spinor wavefunction. Using the Pauli matrices, σx , σy and σz , write down the operator corresponding to a component of spin along the axis (θ� φ) in spherical polar coordinates. Show that the eigenvalues of spin in this direction are ±�/2 (as expected), and deduce the corresponding wavefunctions. Hence, infer the wavefunctions for particles whose spins are aligned along the +x, −x, +y and −y directions.
Hint: You can find the relevant spin operator by taking the dot product of the vector σ with a unit vector in the desired direction.
6. Spin: This problem addresses the addition of spin angular momenta. Consider two identical spin 1/2 fermions, and let χ+ (i) represent the state of particle i with spin up, and χ− (i) the state with spin down. Write down the four possible states of the system which have definite exchange symmetry. ˆ 2 = Sˆ+ Sˆ− + Sˆ2 − �Sˆz , where the spin raising and lowering Show that S z operators are given by Sˆ± = Sˆx ± iSˆy . Using this result, or otherwise, show that the four states of definite exchange symmetry are eigenstates ˆ 2 and find the corresponding eigenvalues and hence the total spin of S quantum number for each state. At a given moment, the system is in a state, � � 2 1 ψ= χ+ (1)χ− (2) + χ− (1)χ+ (2) . 3 3 What is the probability of a measurement of the total spin giving the result S = 1?
7. † Spin: This is a challenging problem which addresses several important aspects of the coursework: quantum spin algebra, time-evolution, and spin precession in a field. A Stern-Gerlach apparatus is used as a filter which rejects para-H2 and passes molecules of ortho-H2 (which has resultant nuclear spin one) with spin component +� in the x direction travelling in the y direction. A magnetic field B in the z direction acts over 20mm of path between two such filters in series, and it is found that no molecules of kinetic energy 0.025eV emerge when B = 1.8(n + 1/2) × 10−3 T, where n is an integer. Explain this phenomenon and deduce a value for the magnetic moment of the proton. (Part II 1966)
8. Spin: We have seen that the “ladder operator” formalism provides a framework in which to define and classify the states of the quantum harmonic oscillator. In the following, we will see that the same ladder operator formalism provides a representation of the quantum spin algebra. This representation, known as the Holstein-Primakoff transformation, can be used to Advanced Quantum Physics
Hint: Write the wavefunction corresponding to spin +� in the x direction using as a basis the eigenstates of Sˆz . Then write the time dependence of this wavefunction in the presence of the uniform field B, and find the fraction of the Sx = +� state in this wavefunction at a later time. You can also obtain the same result from a classical precession argument. You may note that J± |j� m� = �[j(j +1)−m(m+1)]1/2 |j� m±1�.
16.1. PROBLEM SET I
192
develop a controlled perturbation theory of spin Hamiltonians. For present purposes, it also gives us an opportunity to practice the operator formalism and spin algebra. According to the quantum spin algebra, the spin operators are defined by the commutation relations, [Sˆi � Sˆj ] = i��ijk Sˆk , where �ijk denotes the antisymmetric tensor. According to the Holstein-Primakoff transformation,3 it is stated that the quantum mechanical spin S operators can be represented by �1/2 � √ a† a − † ˆ S = � 2S a 1 − � 2S
Sˆ+ = (Sˆ− )† �
Sˆz = �(S − a† a) �
where the ladder operators a and a† obey the usual commutation relations, [a� a† ] ≡ aa† − a† a = 1.4 Making use of these relations, show that this definition is indeed consistent with the quantum spin algebra, i.e. [Sˆ+ � Sˆ− ] = 2�Sˆz . Comments: Physically, the ladder operators simply “count” the number of “spin deviations” away from eˆz . For large spin, S – the analogue the semi-classical limit for quantum mechanical spins √ –†the Holstein-Primakoff − transformation affords the expansion, S = � 2S a +O(S −1/2 ) and S + = √ � 2S a + O(S −1/2 ). In this limit, quantum spin models typically become bilinear (i.e. quadratic) in ladder operators and can be “diagonalized” (i.e. solved) in the same manner as the quantum harmonic oscillator Hamiltonian.
9. Addition of angular momenta: The addition of two or more angular momenta is a common problem which will arise this year in atomic, molecular, nuclear and particle physics. It is therefore very important to understand the basic principles. Consider the addition of two angular momenta, �1 = 1 and �2 = 2. According to quantum mechanics, the possible values for the total angular momentum quantum number L range from �1 + �2 to |�1 − �2 |, i.e. 3, 2, 1 in this case. Tabulate the possible values of the corresponding quantum ˆ z ), and show numbers m1 , m2 and M = m1 + m2 (i.e. those relating to L that the values of M correspond to the expected values of L. Repeat for the case �1 = 3, �2 = 1. For the case �1 = 2 and �2 = 1, the state |L = 3� M = 3� can be written down straightforwardly as |�1 = 2� m1 = 2� ⊗ |�2 = 1� m2 = 1�. Use ladder operators to construct explicitly the state |L = 3� M = 2� and then orthogonality to construct the state |L = 2� M = 2� in terms of the |�1 � m1 � ⊗ |�2 � m2 � states. This method can obviously be extended to construct all such states.
10. Addition of angular momenta: Write down the commutation relations between the angular momentum operators Jx , Jy , and Jz (hats not shown�). 3
T. Holstein and H. Primakoff, Field dependence of the intrinsic domain magnetization of a ferromagnet, Phys. Rev. 58, 1098 (1940). 4 If you feel disturbed by the concept of the “square root of an operator”, you should † think of it as defined by a Taylor expansion in the argument, a2sa .
Advanced Quantum Physics
Hint: If you find yourself expanding the square root, you should stop and consider whether there is a simpler method...
16.1. PROBLEM SET I
193
�a) Show that the operators J± = Jx ± iJy act as raising and lowering operators for the z-component of angular momentum, by first calculating the commutator [Jz � J± ]. �b) State the allowed values of the total spin angular momentum for a system of three electrons. �c) The ‘coupled basis’ state |S total spin) is also a state of denoted by | ↑↑↑� ≡ | ↑� ⊗ | spin lowering operator, show
= 3/2� mS = 3/2� (an eigenstate of the ‘uncoupled basis’, which may be ↑� ⊗ | ↑�. By an application of total that
1 |S = 3/2� mS = 1/2� = √ (| ↓↑↑� + | ↑↓↑� + | ↑↑↓�) . 3
Advanced Quantum Physics
You may note that J± |j� m� = �[j(j +1)−m(m+1)]1/2 |j� m±1�.
16.1. PROBLEM SET I
197
Answers: Problem set I 1 1. (a) In one dimension, the current operator is specified by ˆj = 2m (ψ ∗ pˆψ + ∗ ψ(ˆ pψ) ). Applied to the left hand side of the system (outside the region 2 of the potential), where ψ(x) = Aeikx + Be−ikx , we have jleft = !k m (|A| − !k 2 2 2 |B| ); similarly jright = m (|C| − |D| ). Current conservation demands that jleft = jright , i.e. |A|2 + |D|2 = |C|2 + |B|2 – all of the incoming beam is transferred to the outgoing beam, a statement of particle conservation. This conservation of current can be written as "Ψin |Ψin # = "Ψout |Ψout #. Therefore, using the expression for |ψout #, we have
"Ψin |Ψin # = "Ψout |Ψout # = "Ψin !"#$ S † S Ψin # , !
=I
leading to the unitarity condition.
(b) From the unitarity condition, it follows that & % 2 |t| + |r|2 rt∗ + r∗ t S†S = I = rt∗ + r∗ t |t|2 + |r|2 Comparing the matrix elements, we obtained the required result. (c) For the δ-function scattering problem, the wavefunction is given by ' ikx e + re−ikx x < 0 ψ(x) = teikx x>0 For the δ-function potential, the wavefunction must remain continuous as x = 0, and ∂x ψ|+! − ∂x ψ|−! = −
2maV0 ψ(0) . !2
0 This translates to the conditions 1 + r = t, and ik(t − 1 + r) = − 2maV !2 t. As a result, we obtain
t=
ik , 0 ik + maV !2
as required. Using the result from (b), since the potential is symmetric, we have % & t ik + γ k2 −ik + γ γ 2 γ2 2 2 r = ∗ (|t| − 1) = − − 1 = = t −ik + γ k 2 + γ 2 ik + γ k 2 + γ 2 (ik + γ)2 As a result, we obtain the required expression for r.
2.
ˆ 1 # = E1 |ψ1 # and H|ψ ˆ (a) We are told that H|ψ ( ( 2 # = E2 |ψ2 #, where E1 $= E2 . ˆ 2 # = ψ ∗ Hψ ˆ 2 dx = ψ ∗ E2 ψ2 dx = E2 "ψ1 |ψ2 #. Since Therefore "ψ1 |H|ψ 1 1 ˆ is Hermitian, we can write H ) ) ˆ 2 # = (Hψ ˆ 1 )∗ ψ2 dx = (E1 ψ1 )∗ ψ2 dx = E1∗ "ψ1 |ψ2 # = E1 "ψ1 |ψ2 # , "ψ1 |H|ψ since E1 and E2 are real. Therefore, (E1 −E2 )"ψ1 |ψ2 # = 0 and, if E1 $= E2 then "ψ1 |ψ2 # = 0, i.e. |ψ1 # and |ψ2 # are orthogonal. ˆ 1 # = |ψ2 # and A|ψ ˆ 2 # = |ψ1 #, then adding them, A(|ψ ˆ 1 # + |ψ2 #) = (b) If A|ψ ˆ |ψ1 # + |ψ2 # and subtracting, A(|ψ1 # − |ψ2 #) = |ψ2 # − |ψ1 # = −(|ψ1 # − |ψ2 #). Hence we have an eigenvector of a = +1 corresponding to a normalized eigenvector √12 (|ψ1 # + |ψ2 #) and an eigenvalue a = −1 corresponding to eigenvector √12 (|ψ1 # − |ψ2 #).
Advanced Quantum Physics
16.1. PROBLEM SET I
198
ˆ = Eψ = i!∂t ψ, hence (c) The time-dependent Schr¨odinger equation is Hψ ψ(t) = ψ(t = 0)e−iEt/! . Since |ψ1 # and |ψ2 # are eigenstates of the Hamilˆ then we can write, |ψ(t)# = √1 [|ψ1 #e−iE1 t/! − |ψ2 #e−iE2 t/! ]. tonian H 2 * *2 1* * P = |"ψ(t = 0)|ψ(t)#|2 = *["ψ1 | −" ψ2 |][|ψ1 #e−iE1 t/! − |ψ2 #e−iE2 t/! ]* 4 1 = [1 + cos((E1 − E2 )t/!)] = cos2 ((E1 − E2 )t/2!) 2
3. (a) Differentiating the left hand side of the given expression with respect to β, one obtains †
†
e−βa [a, a† ] eβa = 1 . ! "# $ =1
†
†
Integrating, we therefore have that e−βa aeβa = β+“integration constant$$ . By setting β = 0 we can deduce that the “constant” must be a yielding the required result. Using this result, we have that †
†
†
e−βa a|β# = e!−βa"#aeβa$ |0# = (β + a)|0# = β|0# . =β+a
Lastly, to obtain the normalization, we have that ∗
"β|β# = N 2 "0|eβ a |β# = N 2 "0| = N2 2
∞ + (β ∗ a)n |β# n! n=0
∞ + 2 ! (β ∗ β)n "0|β# = N 2 e|β| = 1 . n! n=0
i.e. N = e−|β| /2 as required. (b) For the harmonic oscillator, the creation and annihilation operators are , ! † related to the phase space operators by x ˆ = ˆ= 2mω (a + a ), and p , † −i !mω 2 (a − a ). Therefore, we have ! !mω ∗ "ˆ x# = (β + β ), "ˆ p# = −i (β − β ∗ ) . 2mω 2 Then, using the identity (∆x)2 = "(x − "x#)2 # = "x2 # − "x#2 , we have
! ! "β|(a2 + aa† + a† a + (a† )2 |β# = (1 + (β + β ∗ )2 ), 2mω 2mω !mω ! "ˆ p2 # = − "β|(a2 − aa† − a† a + (a† )2 |β# = − (−1 + (β − β ∗ )2 ) . 2 2mω
"x2 # =
! As a result, we find that ∆x = 2mω and ∆p = !mω 2 leading to the required expression. (c) The equation follows simply from the definion of the operator a and the solution may be checked by substitution. (d) Using the time-evolution of the stationary states, |n(t)# = e−iEn t/! |n(0)#, where En = !ω(n + 1/2), it follows that
|β(t)# = e−iωt/2 e−|β|
2
/2
βn √ e−inωt |n# = e−iωt/2 |e−iωt β# . n!
Therefore, during the time-evolution, the coherent state form is preserved but the centre of mass and momentum follow that of the classical oscillator, x0 (t) = A cos(ϕ + ωt),
p0 (t) = mωA sin(ϕ + ωt) .
The width of the wavepacket remains constant.
Advanced Quantum Physics
16.1. PROBLEM SET I
199
2 ˆ = 1 (ˆ 4. (a) Starting with the electron Hamiltonian H 2m p − qA) , substitution of the expression for the vector potential leads to the required result.
(b) Setting ω = m = ! = 1, we immediately obtained the required dimensionless form of the Hamiltonian. (c) Straightforward substitution of the differential operators leads to the required identities. As a result, we can confirm that [a, a† ] =
2 ([∂z¯, z¯] − [z, ∂z ]) = 1 , 4
and similarly [b, b† ] = 1. In the complex coordinate representation, &2 % &2 i 1 1 −i(∂z + ∂z¯) − (z − z¯) + −(∂z¯ − ∂z ) + (z + z¯) 4 2 4 . 1 1 z¯ / . z/ 1 1 ∂z¯ + + = a† a + = −2∂z¯∂z + z¯z + (z∂z¯ − z¯∂z ) = 2 −∂z + 8 2 4 4 2 2 ˆ =1 H 2
%
(d) The angular momentum operator is given by (! = 1)
i ˆ z = (x × p ˆ )z = −i(x∂y − y∂x ) = − ((z + z¯)(∂z − ∂z¯) − (z − z¯)(∂z + ∂z¯)) L 2 . . z¯ / . z/ z/. z¯ / = −2 −∂z + ∂z¯ + + 2 −∂z¯ + ∂z + = −(a† a − b† b) . 4 4 4 4
(e) In the coordinate representation, the condition a|0, 0# = 0 translates to the equation √ . z/ 2 ∂z¯ + "r|0, 0# = 0 4
We thus obtain the Gaussian expression for the wavefunction. Then, using † m ) the relation |0, m# = (b√m! |0, 0#, we have . z¯ /m 1 −¯zz/4 1 1 √ e "r|0, m# = √ 2m/2 ∂z + =√ z m e−¯zz/4 , 4 2π m! 2π2m m!
as required.
(f ) If we populate the states of the lowest Landau with electrons, starting from states of the lowest angular momentum m, the wavefunction is a 2 Slater determinant involving entries φm (rj ) = zjm e−|zj | /4 . Taking the determinant, and making use of the Vandemonde determinant identity, one obtains the required many-electron wavefunction. ˆ θφ , can be found by forming the 5. The spin operator in the (θ, φ) direction, S ˆ scalar product of the spin operator S with a unit vector in the (θ, φ) direction, (sin θ cos φ, sin θ sin φ, cos θ). Therefore % & ! cos θ e−iφ sin θ ˆ Sθφ = . eiφ sin θ − cos θ 2 We need the eigenvalues of the matrix, i.e. % &% & % & ! cos θ sin θe−iφ u u =λ . v v sin θeiφ − cos θ 2 ˆ θφ are ±!/2, Eliminating u and v, we find λ2 = 1 and hence the eigenvalues of S as expected. Substituting the values λ = ±1 back into the equations relating u and v, we can infer the ratios, uv = e−iφ cot(θ/2) and −e−iφ tan(θ/2). So, in matrix notation, the eigenstates are % & % & cos(θ/2) sin(θ/2) and , eiφ sin(θ/2) −eiφ cos(θ/2)
Advanced Quantum Physics
16.1. PROBLEM SET I
200
for eigenvalues +!/2 and −!/2 respectively. The spin states in the x-direction are obtained by setting θ = π/2, φ = 0, and the spin states in the y-direction are obtained by setting φ = π/4 in these general formulae.
6. The four states are given by φ1 = χ+ (1)χ+ (2), φ2 = χ− (1)χ− (2), φ3 = √1 [χ+ (1)χ− (2) + χ− (1)χ+ (2)], and φ4 = √1 [χ+ (1)χ− (2) − χ− (1)χ+ (2)], where 2 2 φ1 , φ2 and φ3 are symmetric under particle interchange and φ4 is antisymmetric. ˆ 2 − Sˆ2 + !Sˆz , Since Sˆ+ Sˆ− = Sˆx2 + Sˆy2 + i(Sˆy Sˆx − Sˆx Sˆy ) = S z ˆ 2 = Sˆ+ Sˆ− + Sˆ2 − !Sˆz . S z
(16.1)
(1) (2) Noting that Sˆz = Sˆz + Sˆz , and likewise for Sˆ± , and using the basic results for the operation of the spin operators on the single particle states,
1 Sˆz(1) χ± (1) = ± !χ± (1), 2
(1) Sˆ∓ χ± (1) = !χ∓ (1),
(1) Sˆ± χ± (1) = 0 ,
ˆ 2 to each state. For example we can now calculate the effect of applying S ˆ 2 φ1 = !2 [2χ+ (1)χ+ (2) + χ+ (1)χ+ (2) − χ+ (1)χ+ (2)] = 2!2 φ1 , S where the three terms in square brackets correspond to the three operators on the right hand side of Eq. (16.1). Likewise φ2 . Clearly Sˆz φ3 = 0 = Sˆz φ4 , so only the Sˆ+ Sˆ− term need be considered. We find
√ Sˆ+ Sˆ− χ+ (1)χ− (2) = Sˆ+ !χ− (1)χ− (2) = !2 (χ+ (1)χ− (2) + χ− (1)χ+ (2)) 2!2 φ3 . √ Similarly Sˆ+ Sˆ− χ− (1)χ+ (2) = 2!2 φ3 so that ˆ 2 φ3 = 2!2 φ3 , S
ˆ 2 φ4 = 0 . S
Thus, φ1 , φ2 , φ3 all have eigenvalue 2!2 and hence S = 1, while φ4 has S = 0. The state ψ is clearly an eigenstate of Sˆz with eigenvalue 0, and must therefore be a linear combination of φ3 and φ4 . The S = 1 component is thus the φ3 term in the wavefunction with amplitude 2 1 1 1 c3 = "φ3 |ψ# = " χ+ (1)χ− (2) + χ− (1)χ+ (2)| χ+ (1)χ− (2) + χ− (1)χ+ (2)# 3 3 2 2 0 0 = 1/3 + 1/6 The probability of S = 1 is therefore |c3 |2 =
√ 3+2 2 6
= 0.971.
7. Taking as a basis the states of the Sˆz operator, we can deduce the form of the Sˆx operator most easily from the action of the spin raising and lower operators. Noting that, for spin S = 1, Sˆ± |S = 1, m# = ![2 − m(m + 1)]1/2 |1, m ± 1#, we can construct the matrix elements of Sˆ± . The latter are given by 0 1 0 √ † Sˆ+ = 2! 0 0 1 , Sˆ− = Sˆ+ . 0 0 0 Then, using the relation, Sˆx = 12 (Sˆ+ − Sˆ− ), we obtained the matrix elements of the operator and corresponding eigenstates, 0 1 ! Sˆx = √ 1 0 2 0 1
0 1 , 0
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m x = 1 1 1 √ 2 2 1
mx = 0 1 1 √ 0 2 −1
m x = −1 1 1 √ − 2 2 1
16.1. PROBLEM SET I
201
When placed in a magnetic field, B, the molecules will acquire an energy µBmz !, where µ is the magnetic moment of the molecule, which in this case equals twice the magnetic moment of the proton, and mz ! is the eigenvalue of Sˆz . At t = 0, the molecules enter the magnetic field in the mx = 1 state, after which their wavefunction evolves with time in the usual way, i.e. −iµBt −iµBt 1 e√ e 0 0 √ 1 1 ˆ 0 0 0 2 = ψ(t) = e−iµB Sz t/! ψ(0) = 2 . 2 2 iµBt iµBt 0 0 e 1 e
Thus, if µBt = (2n + 1)π, with n an integer, the molecules will be in a pure mx = −1 state, 0 and none will pass the second filter. The time is given by t = L/v = L m/2E, where L = 20 mm and m and E are the mass and energy of the molecules respectively. We thus have % &1/2 (2n + 1)π! 2E = 2.84 10−26 JT−1 µ= BL m
and hence the proton magnetic moment is 1.42 10−26 JT−1 . Note that the result can also be obtained by treating the problem as one of classical precession. The couple = µB = LΩ, where L = ! is the angular momentum and Ω the angular frequency of precession. If Ωt = (2n + 1)π, the molecules have precessed into the mx = −1 state, and the result readily follows. 8. Substituting for the definition of the spin raising and lowering operators using the Holstein-Primakoff transformation, the commutator is obtained as † a+1% &1/2 a #$!" &1/2 % & 1 ˆ+ ˆ− a a a† a a† a † † [ S , S ] = 1 − aa 1 − − a 1 − a 2S!2 2S 2S 2S % & % & a† a a† a a† a† aa a† a = 1− + a† a 1 − − a† a + =1− . 2S 2S 2S S
%
†
With Sˆz = !(S −a† a), we obtain the required commutation relation [Sˆ+ , Sˆ− ] = 2!Sˆz . 9. For the case -1 = 1, -2 = 2, a table of possible ways of forming each value of M = m1 + m2 is shown right. The largest value of M is 3, so the largest value of L must be 3. There must also be a state with M = 2 corresponding to L = 3, but we have two states with M = 2. Therefore there must be a state with L = 2 as well. We need two states with M = 1, one for each of the L = 3, 2 multiplets, but we actually have three states with M = 1, so there must be an L = 1 state as well. All of the M states are now accounted for. For the case -1 = 3, -2 = 1, we can again for a table (see right). Following the same logic as before, we see that states with L = 4, 3, 2 just account for all the states. To construct the states explicitly, we start by writing the L = 3 M = 3 state, since there is only one way of forming M = 3, viz. |3, 3# = |1, 1# ⊗ |2, 2#. We ˆ − , which is simply the sum of the then operate with the lowering operator L lowering operators for the two separate particles. Recalling that: 0 ˆ − |-, m# = -(- + 1) − m(m − 1)!|-, m − 1# , L √ √ √ we obtain 6!|3, 2# = 2!|1, 0# ⊗ |2, 2# + 4!|1, 1 ⊗ |2, 1#, where the first term on the right hand side comes from lowering the 0 - = 1 state and the second from lowering the = 2 state. Hence |3, 2# = 1/3!|1, 0# ⊗ |2, 2# + 0 2/3!|1, 1#⊗|2, 1#. The state |2, 2# must be the orthogonal linear combination, 0 0 i.e. |2, 2# = 2/3!|1, 0# ⊗ |2, 2# − 1/3!|1, 1# ⊗ |2, 1#. Further states could be computed in the same way if required.
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M 3 2 1 0 −1 −2 −3
(m1 , m2 ) (1, 2) (1, 1) (1, 0) (1, −1) (1, −2) (0, −2) (−1, −2)
(0, 2) (0, 1) (0, 0) (0, −1) (−1, 1)
(−1, 2) (−1, 1) (−1, 0)
M 4 3 2 1 0 −1 −2 −3 −4
(m1 , m2 ) (3, 1) (3, 0) (3, −1) (2, −1) (1, −1) (0, −1) (−1, −1) (−2, −1) (−3, −1)
(2, 1) (2, 0) (1, 0) (0, 0) (−1, 0) (−2, 0) (−3, 0)
(1, 1) (0, 1) (−1, 1) (−2, 1) (−3, 1)
16.1. PROBLEM SET I 10. The commutation relations are given by [Ji , Jj ] = i!.ijk Jk . (a) First define the eigenstates ψm : Jz ψm = m!ψm . To see if J± ψm is an eigenstate of Jz , we need to look at Jz J± ψm , which is equal to J± Jz ψm − [J± , Jz ]ψm . The required commutator is [J± , Jz ] = [Jx , Jz ] ± i[Jy , Jz ], from the definition of J± . From the basic commutators given at the start, this is [J± , Jz ] = !) − iJy ± −Jx ) = − ± !J± (if treating ± like a number is confusing, do this separately for J+ and J− ). Going back to Jz J± ψm , we can now write this as J± Jz ψm + ±!J± ψm . The first term is just J± m!ψm , so this is (m ± 1)!(J± ψm ). Thus, J± ψm is an eigenstate of Jz , with eigenvalue (m ± 1)!. This establishes the raising and lowering property of J± .
(b) Two electrons would have a total spin of S = 1 or 0. Adding a third spin 1/2 particle creates total spin S = 3/2 or 1/2 from the S = 1 two-particle state. The S = 0 two-particle state becomes S = 1/2 only on adding the third particle, so total S = 3/2 or 1/2 are the only possibilities. (c) The states with well-defined values of m1 , m2 , and m3 for the z spin components of all particles are the ‘uncoupled basis’. Where all particles are ‘spin up’, this state may be written as | ↑↑↑#. This state is also the mS = 3/2 state of total S = 3/2 (there is no other way to get m1 + m2 + m3 = 3/2 in the uncoupled basis). We can therefore write |S = 3/2, mS = 3/2# = | ↑↑↑#. To get from here to |S = 3/2, mS = 1/2#, we need to apply (1) (2) (3) J− = S− + S− + S− . In other words, the total lowering operator is the sum of the lowering operator for each separate spin (reasonably enough) (1) (2) (3) this follows from the definition of J− and Jx = Sx + Sx + Sx , etc. Now, we need to use the given normalization result. This says that 0 J− |S = 3/2, mS = 3/2# = 15/4 − 3/4!|S = 3/2, mS = 1/2# √ = 3!|S = 3/2, mS = 1/2# . Notice that the total quantum number, J, is the same as the overall spin quantum number, S in this case. Therefore |S = 3/2, mS = 1/2# = √ (1/ 3)J− |S = 3/2, mS = 3/2#. Using 0 the given normalization result again for a single state, S− |1/2, 1/2# = 3/4 + 1/4!|1/2, −1/2#. This establishes the required result.
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202
16.2. PROBLEM SET II
16.2
203
Problem Set II
1. Perturbation theory: This question provides a very instructive application of approximation methods in quantum mechanics. The first part of the problem is straightforward and addresses the perturbative series expansion of the anharmonic oscillator. However, the same problem can be used to illustrate the failure of perturbation theory. The second part of the problem demands an application of the WKB method and illustrates the origin of the limitations of the perturbative scheme. Briefly summarize how perturbation theory can be used to obtain approximate values for the energy of a non-degenerate state when an exact solution of the Schr¨odinger equation is unavailable. (a) An anharmonic one-dimensional oscillator for a particle of mass m has potential V (x) = 12 mω 2 x2 + λx4 , where λ > 0 is small. Using perturbation theory, determine the ground state energy to first order in λ. Consider how your result can be obtained using the ladder operator formalism. † (b)
When λ < 0, the anharmonic oscillator provides an instructive example illustrating the failure of perturbation theory. No matter how small is the value of λ, the potential minimum at x = 0 can only be metastable: For x large enough, the potential eventually turns negative providing an escape route for the particle from the harmonic potential. For this tunneling problem, make use of the WKB method to estimate (roughly) the tunneling probability and thereby elucidate the origin of the failure of the perturbation series expansion in λ.
2. Perturbation theory: In addition to relativistic corrections, the Hamiltonian of the hydrogen atom is also perturbed by the finite range of the nucleus. The following problem exploits perturbation theory to explore the scale of such corrections. The fact that the proton is not a point charge influences the energy levels of the hydrogen atom. This problem may be treated (for simplicity) by regarding the proton as a uniformly charged hollow spherical shell of radius b = 5 × 10−16 m. Show that the change in the electrostatic potential energy corresponds to introducing a perturbation % & e2 1 1 (1) ˆ H (x) = − , r < b, 4π&0 r b into the normal Schr¨odinger equation for the hydrogen atom. Using first order perturbation theory, estimate the energy shifts of the hydrogen 2s and 2p states and comment on your findings. Suggest why measurement of such energy shifts is not a good way of studying the proton charge distribution. 3. Perturbation Theory:This problem shows how perturbation theory may be used to estimate the polarizibility of the hydrogen atom in its ground state. (The induced dipole moment in an applied electric field E is α&0 E where α is the polarizibility.) Advanced Quantum Physics
The ground state of the unperturbed oscillator is given by 2 1/4 ψ0 (x) = ( mω exp(− mωx π! ) 2! ).
You may note that !
0
b
dx x(b2 − x2 )1/2 =
b3 . 3
Hint: You can, and should, simplify the integrals considerably by noting that the size of the nucleus is much smaller than the atomic Bohr radius, i.e. b " a0 . You may " note#that, $ 1 ψ2s = 8πa 1 − 2ar 0 e−r/2a0 , 3 0 −r/2a0
√ ψ2p0 = re
32πa50
cos θ,
−r/2a0
√ ψ2p±1 = re
64πa50
e±iφ sin θ.
16.2. PROBLEM SET II Working to second order in field, E, show that induced polarization of 2 ' |#k|z|0$|2 the ground state |0% is given by α = 2e k"=0 Ek −E0 , where Ek is !0 the unperturbed energy of state |k%. Show that the same result may be obtained from the perturbed wavefunction to first order in E and evaluating the expectation value of the induced electric dipole moment. Evaluation of α is tedious, but a useful upper bound may be obtained by noting that Ek ≥ E1 , where E1 is the energy of the first excited. Using 64πa3 this result, show that α ≤ 3 0 . Compare this with the experimental value of α = 8.5 × 10−30 m3 .
204
To derive the matrix element, &0|z 2 |0%, you will need the ground state of the hydrogen atom, |0% = ( πa1 3 )1/2 e−r/a0 . 0
Variational method:The following three problems involve straightforward applications of the variational state analysis. 4. Give an account of the variational method for estimating the ground state energy of a quantum mechanical system. Explain how the method may also be applied to excited states. Use a trial wavefunction of the form, ( A(a2 − x2 ) −a < x < a ψ(x) = , 0 otherwise
to place an upper bound on the ground state energy of the one-dimensional harmonic oscillator with potential V (x) = mω 2 x2 /2 where m is the mass of the particle and ω the oscillator frequency. Compare your answer with the exact result, and comment.
5. By taking a trial wavefunction proportional to exp(−βr) where β is a variational parameter, obtain an upper limit for the ground state energy of the H atom in terms of atomic constants. Comment on your result.
6.
(a) E1 and E2 are the ground state energies of a particle moving in attractive potentials V1 (r) and V2 (r). Using the variational method, show that E1 ≤ E2 if V1 (r) ≤ V2 (r). (b) Consider a particle moving in a one-dimensional attractive potential V (x), i.e. a potential such that V (x) ≤ 0, for all x and V (x) → 0, as |x| → ∞. Use the variational principle with trial function A exp(−λx2 ) to show that the upper bound on the ground state energy is negative, and hence that for any such potential at least one bound state must exist.
7. The following question addresses the constraints imposed by particle indistinguishability on the allowed spin and spatial states of a two-particle quantum mechanical system. Discuss the special considerations which apply to systems of indistinguishable particles in quantum mechanics, giving examples where they lead to observable consequences. Two non-interacting particles of mass m move in one dimension, their positions given by x1 and x2 . The potential is given by ( 0 0<x
Hint: Use the wavefunction of a particle moving in V2 (r) as a trial wavefunction for potential V1 (r).
16.2. PROBLEM SET II Show that the energy of the system is of the form E = (n21 + n22 )ε where n1 and n2 are integers and find an expression for ε. Consider the state with E = 5ε for each of the following three cases: (a) spin-zero particles; (b) spin-1/2 particles in a spin singlet state; (c) spin-1/2 particles in a spin triplet state. In each case, what is the symmetry of the spin and spatial parts of the wavefunction? Hence write down the spatial wavefunction, and sketch the probability density |ψ(x1 , x2 )|2 in the (x1 , x2 ) plane.
Describe qualitatively how the energies of these states would change if the particles carried electric charge and hence interacted with each other (an example of the exchange interaction).
8. Together with the constraints imposed by particle indistingishability, a second feature of many-body problem in quantum mechanics is their typical analytical intractability! In the vast majority of interacting problems, some approximation is necessary. The following question involves an application of perturbation theory to a two particle system. Two identical spin-zero bosons are placed in a one-dimensional square potential well with infinitely high walls; V = 0 for 0 < x < L, otherwise V = ∞. The normalized single-particle energy eigenstates are given by un = (2/L)1/2 sin(nπx/L). (a) Find the wavefunctions and energies for the ground state and the first two excited states of the system. (b) Suppose that the two bosons interact with each other through the perturbative “contact interaction”, ˆ % (x1 , x2 ) = −V0 Lδ(x1 − x2 ) . H Compute the the ground state energy to first order in V0 .
Advanced Quantum Physics
205
16.2. PROBLEM SET II
202
Answers: Problem set II 1. (a) For λ > 0, the first order shift in ground state energy is given by ! ∞ " mω #1/2 ! ∞ 2 3!2 λ ∆E0 = ψ0∗ λx4 ψ0 dx = dxλx4 e−mωx /! = . π! 4m2 ω 2 −∞ −∞ $ ! Alternatively, using the identity x = 2mω (a + a† ), we have &2 ! !0|(a + a† )4 |0" 2mω λ!2 3!2 λ 2 † 2 † † † 2 † = !0|(a (a ) + aa aa + a a a )|0" = . 4m2 ω 2 4m2 ω 2
∆E0 = λ
%
(b) For λ < 0, the situation becomes more subtle. The potential now takes the form of an upturned double well potential with a metastable minimum at zero. Here, as we will see, conventional perturbative approaches fail. However, we can straightforwardly implement the WKB approach to compute the tunneling amplitude from the well created by the perturbation using the relation, ! 1 b ' −S dx 2m(V (x) − E0 ) . t∼e , S= ! a To implement the WKB method, we have to first identify the classical turning points. If λ is small in magnitude, the ground state energy of the unperturbed oscillator is neglible as compared to the barrier height and we may set E0 = !ω/2 % 0. The classical turning points are determined by the equation E0 = V (x) = 12 mω 2 x2 − λx4 . The latter has the solution at x = a % 0 (with more care, we can show ' that it is simply the ' turning point of the harmonic oscillator, x0 = !/mω) and x = b mω 2 /2λ. (The reflection about x = 0 also gives another set of solutions.) We therefore obtain √ ! 2mλ b 2m2 ω 3 S% dx x(b2 − x2 )1/2 = , ! 3!λ (0 )* + b3 /3
which leads to the transmission probability , 4m2 ω 3 2 |t| ∼ exp − . 3!λ
This result exposes the problem with perturbation theory: it assumes that the ground state energy is an analytic function of λ for some sufficiently small region around λ = 0. However, this result shows that the true ground state solution has an essential singularity in λ and the radius of convergence of the perturbation theory vanishes. 2
e 2. For a point-like nucleas, the hydrogen atom has a potential V (r) = − 4π# . A 0r hollow spherical shell will have the same potential for r > b, but V (r) = V (b) for r < b, by Gauss theorem, and thus its effect can be regarded as adding a perturbation, % & 2 1 1 ˆ (1) = e H − , 4π%0 r b
to the Hamiltonian for r < b, and zero for r > b. energy shift induced by the perturbation is % ! b e2 1 1 2 (1) ˆ 4πr dr − ∆E = !ψ|H |ψ" = 8πa30 4π%0 0 r
Advanced Quantum Physics
For the 2s wavefunction, the 1 b
&% &2 r 1− e−r/a0 . 2a0
16.2. PROBLEM SET II
203
Since b ' a0 , the terms involving r/a0 are negligible in the region of integration, so we can simplify the integral to % & ! b 1 1 e2 e2 b2 2 r dr R , R = ∆E = − = . ∞ ∞ 8π%0 a30 0 r b 6a20 8π%0 a0 Likewise for the 2p0 wavefunction, making the same approximation, we obtain &! π ! b % 1 1 e2 b4 4 2 r R∞ . ∆E = − dθ2π sin θ cos θ = 5 128π 2 %0 a0 0 r b 240a40 0 Both energy shifts are very small, but that for the 2p state is much smaller, because the 2p wavefunction vanishes at the origin. This is not a good method to explore the nucleus because other effects, such as spin-orbit interaction and other relativistic corrections would swamp the nuclear size effect. It is more effective for heavy atoms, with larger nuclei and smaller Bohr radii, and especially for “muonic” atoms, where the greater mass of the muon again reduces the Bohr radius.
3. If, without loss of generality, we take the field to lie along z, the perturbation is ˆ $ = −eEz. At first order in perturbation theory, ∆E = −!0|eEz|0" given by H vanishes since the ground state of the hydrogen atom |0" is an eigenstate parity. The leading contribution to ∆E is therefore the second order term ∆E = . |&k|eEz|0'|2 k%=0 E0 −Ek . If the induced dipole moment is d = α%0 E, its energy of interaction with the electric field is given by ∆E = − 12 d · E = − 12 α%0 E 2 . So, by comparing with our perturbation theory result we obtain the required result. An alternative derivation of this result starts from the first order perturbation . theory expression for the perturbed wavefunction: |ψ" = |0" + k%=0 ck |k", where ck = &k|−eEz|0' E0 −Ek . The dipole moment operator for the electron is ez, and its expectation value in this state is (neglecting small terms of order (c2k )), / !ψ|ez|ψ" = !0|ez|0" + [ck !0|ez|k" + c∗k !k|ez|0"] +O(c2k ) = α%0 E ( )* + k%=0 =0 ( )* + 2E
P
k!=0
|"k|ez|0#|2 Ek −E0
from which the value of α follows as before. Since Ek ≥ E1 for all k, we obtain α≤
2e2 / |!k|z|0"|2 2e2 / !0|z|k"!k|z|0" 2e2 !0|z 2 |0" = = , %0 E1 − E0 %0 E 1 − E0 %0 E1 − E0 k%=0
k%=0
. where we have used the completeness relation I = k |k"!k|. Note that the sum now includes the k = 0 term. Using the explicit form for the Hydrogen ground state, |0" = ( πa1 3 )1/2 e−r/a0 , we evaluate the matrix element, 0
!0|z 2 |0" = !0|r2 cos2 θ|0" =
!
π
2π sin θ cos2 θdθ
0
!
0
∞
r2 drr2 e−2r/a0 = a20 .
We also need the energy difference, E1 − E0 = (1 − 14 )R∞ =
which we obtain α ≤ experiment.
64πa30 3
= 9.9 × 10
−30
3 e2 4 8π#0 a0
from
m , a figure that is not too far from 3
4. From trial wavefunction, we can obtained A from the normalization, 1 = 0 ∞ the 0a 2 5 |ψ|2 dx = A2 −a (x4 − 2a2 x2 + a4 )dx = 16 15 A a . Moreover, using the −∞ identity, % & , 2 !2 2 1 ! 1 2 2 2 2 2 4 ˆ Hψ = − ∂ + mω x ψ = A + mω (a x − x ) 2m x 2 m 2
Advanced Quantum Physics
16.2. PROBLEM SET II
204
the expectation value of the Hamiltonian is given by , 2 , ! a ! 1 15 2!2 4mω 2 a2 ˆ !ψ|H|ψ" = A2 (a2 − x2 ) + mω 2 (a2 x2 − x4 ) dx = + . m 2 8 3ma2 105 −a 1/2 ! Minimising with respect to a we obtain a2 = ( 35 2 ) mω . Substituting this ˆ value of a into our expression for$ !ψ|H|ψ", we obtain the upper bound on the 5 ˆ ground state energy, !ψ|H|ψ" = !ω = 0.598!ω, which is greater than the 14
true ground state energy (!ω/2) as expected. 5. From the normalization, 1 = A2
0∞ 0
4πr2 e−2βr dr =
4πA2 4β 3
β3 π . From 2β −βr , we r e
⇒ A2 =
the identity ∇2 ψ = r12 ∂r (r2 ∂r ψ) = r12 ∂r (−βr2 e−βr ) = β 2 e−βr − have , % & ! ∞ !2 2β −2βr e2 −2βr ˆ !ψ|H|ψ" = A2 4πr2 dr − β 2 e−2βr − e − e 2m r 4π%0 r 0 !2 β 2 e2 β = − . 2m 4π%0 2
−1 me Minimising with respect to β we obtain β = 4π# 2 = a0 , which is the inverse 0! 2 m e 2 ˆ = −13.6 eV. This is the of the Bohr radius and thus !ψ|H|ψ" = − 2! 2 ( 4π# ) 0 correct value for the ground state energy of the Hydrogen atom, as expected, because we chose the correct functional form for the trial function.
ˆ 1 and H ˆ 2 with ground state 6. (a) Suppose that the two Hamiltonians are H ˆ ˆ 2 ψ2 = E2 ψ2 . Given wavefunctions ψ1 and ψ2 , i.e. H1 ψ1 = E1 ψ1 , and H ˆ1 = H ˆ 2 − V2 (r) + V1 (r) = H ˆ 2 + ∆V (r). From the that V1 ≤ V2 , we have H variational principle, ˆ 1 |ψ2 " = !ψ2 |H ˆ 2 |ψ2 " − !ψ2 |∆V |ψ2 " = E2 + !ψ2 |∆V |ψ2 " ≤ E2 , E1 ≤ !ψ2 |H
where the last inequality follows because ∆V (r) ≤ 0. Thus E2 ≥ E1 . ˆ = − !2 ∂x2 + V (x), and the normalized trial (b) The Hamiltonian is given by H 2m 2 function is given by ψ =$ (2λ/π)1/4 e−λx . Using standard integrals, we 0∞ 2 2 2 ˆ obtain !ψ|H|ψ" = ! λ + 2λ dxV (x)e−2λx = ! λ + I. Minimsing 2m
π
with respect to λ, we obtain, 0=
!2 I I + + + 2m 2λ 2λ
−∞
1
2λ π
2m
!
2
V (x)(−2x2 )e−2λx ,
where the second term arises from differentiating the normalization in I, and the third term from differentiating the integrand. This is an implicit equation for λ and if we solve for I and substitute into the equation from above, we obtain 1 ! 2 !2 !2 2λ ˆ !ψ|H|ψ" = λ+I =− λ + 2λ dxV (x)(2x2 )e−2λx . 2m 2m π This is our upper bound on the ground state energy, and since V (x) ≤ 0, both terms are manifestly negative. Hence the ground state energy is negative, and at least one bound state must exist.
7. A single particle in the potential well has the (unnormalized) wavefunction and !2 π 2 2 energy, ψn (x) = sin(nπx/L) and E = 2mL ≡ %n2 . The wavefunction for a 2n system of two identical particles must be either symmetric or antisymmetric, i.e. ψ(x1 , x2 ) = sin(n1 πx1 /L) sin(n2 πx2 /L) ± sin(n2 πx1 /L) sin(n1 πx2 /L) ,
with energy (n21 + n22 )%. If E = 5%, we must have n1 = 1, and n2 = 2 (or vice versa).
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16.2. PROBLEM SET II
205
(a) Spin-zero particles are bosons and must have a symmetric wavefunction, ψ(x1 , x2 ) = sin(πx1 /L) sin(2πx2 /L) + sin(2πx1 /L) sin(πx2 /L) = 2 sin(πx1 /L) sin(πx2 /L) [cos(πx1 /L) + cos(πx2 /L)] . Clearly, this has zeros when x1 = 0, L, when x2 = 0, L, and when x1 + x2 = L. (b) Spin 1/2 particles are fermions and must have an antisymmetric wavefunction. In the singlet case, the spin wavefunction is antisymmetric, and hence the spatial wavefunction is symmetric, just as in (a).
Symmetric wavefunction
(c) In the triplet case, the spin wavefunction is symmetric, and hence the spatial wavefunction must be antisymmetric, i.e. ψ(x1 , x2 ) = sin(πx1 /L) sin(2πx2 /L) − sin(2πx1 /L) sin(πx2 /L) 2 sin(πx1 /L) sin(πx2 /L) [cos(πx1 /L) − cos(πx2 /L)] .
Clearly, this has zeros when x1 = 0, L, when x2 = 0, L, and when x1 = x2 . If the particles were charged, they would repel each other through the Coulomb interaction. Therefore, in the spin 1/2 case, the triplet state would have the lower energy, because the particles tend to be further apart. This is an example of the exchange interaction, and is a simplified model of what happens in the Helium atom.
Antisymmetric wavefunction
2 2
π ! 2 8. For the single-particle states, E = n2 8ma Since this well is not centred 2 = %n on zero, the single-particle eigenstates are all just proportional to sin(nπx/2a) ≡ |n".
(a) If we write the two-particle states as |n1 , n2 ", the ground state is |1, 1" (E = 2%). The first excited states are |2, 1" and |1, 2" (E = 5%). The second excited state is |2, 2" (E = 8%). The overall wavefunction needs to be symmetric for bosons, which |1, 1" and |2, 2" are already. These therefore pair with a symmetric spin wavefunction, which is always possible, whether or not the bosons have spin zero. For the first excited state, both symmetric √ and antisymmetric combinations are possible: (|2, 1" ± |1, 2")/ 2; these would need to pair with spin wavefunctions that are respectively symmetric and antisymmetric. If S > 0, both are possible; if S = 0, only the symmetric space state is allowed. The (normalized) ground state wavefunction is given by ψ(x1 , x2 ) = !x1 , x2 |1, 1" =
1 sin(πx1 /2a) sin(πx2 /2a) . a
(b) According to first order perturbation theory, the change in the ground state ˆ $ is given by ∆E = !H ˆ $ ", where the expectation value energy caused by H 00 ∗ ˆ$ involves the unperturbed ψ (x 0 0 eigenfunctions, ∆E = 0 1 , x2 )H ψ(x1 , x2 )dx1 dx2 . Using the identity, f (x1 , x2 )δ(x1 − x2 )dx1 dx2 = f (x1 , x2 )dx1 , for any function f , we have ! ! ! 1 2V0 2a 4 2 ∆E = −2aV0 |ψ(x, x)| dx = − sin (πx/2a) = −4V0 sin4 (πy)dy . a 0 0 The sin4 (πy) looks nasty, but written as sin2 (πy)×sin2 (πy), with sin2 (πy) = (1 − cos(2πy))/2, it is easily evaluated and gives ∆E = −3V0 /2.
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16.3. PROBLEM SET III
16.3
210
Problem Set III
1. Atomic structure: The following question begins with a recapitulation of the leading relativistic corrections to the Schr¨odinger Hamiltonian of hydrogenlike atoms. Explain the physical origin of the following terms in the Hamiltonian for one-electron atoms: p2 )2 ˆ 1 = − (ˆ H 8m3 c2 ˆ ·S ˆ ˆ 2 = 1 (∂r V ) L H 2m2 c2 r 2 !2 ˆ 3 = Ze H 4πδ (3) (r) . 4π#0 8(mc)2 Here m denotes the electron mass, V the electrostatic potential generated by the nucleus, and Z the atomic number. The hydrogenic radial wavefunctions have the form, ψn! (r) =
!
Z a0
"3/2
Gn!
!
Zr a0
"
# $ Zr exp − , na0
Explicit evaluation of numerical factors is not required.
where a0 denotes the Bohr radius and Gn! is a polynomial function of its argument. Show that the expectation values of the energies associated with the three terms listed above all have the same dependence on Z. (Part IB Advanced Physics 1993.)
2. Atomic structure: The following problem addresses the electron configuration of multi-electron atoms. In completing this question, you should remind yourself of the physical origin of the Hund’s rules. Determine the possible spectroscopic terms, 2S+1 LJ , for each of the following electron configurations: (2s)(3p), (2p)2 , (3d)2 , (3d)10 , and (3d)9 . Using Hund’s rules, determine the angular momentum quantum numbers of the ground state of Sm, which has electron configuration (4f )6 .
3. Atomic structure: The following problem addresses the interplay between LS and jj coupling. The ground state of Ge has the configuration (4p)2 and spectroscopic term 3 P0 . Explain the meaning of this notation and state the assumptions about atomic structure on which it is based. Derive the allowed states of a (3p)1 (4s)1 configuration in (i) LS coupling and (ii) jj coupling, sketching an energy level diagram in each case. Explain the interactions which give rise to the various energy splittings. For a certain element, the energy levels given right, which are specified relative to the ground state, are known to belong to a (6p)1 (7s)1 configuration. Deduce what you can about the quantum numbers appropriate to each energy level, given their relative magnitude and the fact that only levels (b) and (d) decay to the J = 0 ground state by electric dipole transitions.
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(a) 4.334 eV (b) 4.375 eV (c) 5.975 eV (d) 6.130 eV
16.3. PROBLEM SET III
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4. † Atomic spectra: By drawing upon your knowledge about the properties of multielectron atoms, the following problem requires the decipher of the atomic spectra of sodium. The three groups of lines shown right, whose frequencies are given in units of 1015 Hz, are observed in the emission spectrum of atomic sodium. By identifying the spectral lines with specific states of sodium, make an index of the associated transitions. In doing so, you will find it helpful to address the following questions: (i) What are the appropriate quantum numbers for the ground state and low energy excited states of sodium? (ii) Which energy levels are split by the spin-orbit interaction? (iii) How (qualitatively) does the spin-orbit coupling depend on n? (iv) Which transitions are allowed by the dipole selection rules?
I Doublets with decreasing doublet splitting: 0.50899 0.90782 1.05086 1.11848 0.50847 0.90765 1.05079 1.11845
Draw an energy level diagram, taking the (3s) state as the zero of energy, and answer the following questions:
III Triplets, with two lines sometimes unresolved: 0.36635 0.52756 0.60207 0.64267 0.365833 0.52704 0.60165 0.64215 0.365831
(a) What is the energy difference between the (5p) J = 3/2 and J = 1/2 states in sodium?
II Doublets with constant doublet splitting: 0.26340 0.48713 0.58225 0.63142 0.26288 0.48662 0.58174 0.63090
(b) Estimate the first ionization energy of sodium. (c) What are the relative importances of the &-dependence of the electronelectron energy, and the j-dependence of the spin-orbit energy in the n = 3 and n = 6 shells?
5. Zeeman effect: This problem involves the study of the infleunce of a weak magnetic field on the spectrum of a multielectron atom. Here we are interested in exploring the interplay between LS coupling and the influence of the external field. For an atom characterised by LS coupling, and subject to a weak uniform magnetic field, derive the expression for the Land´e g-factor, g =1+
J(J + 1) − L(L + 1) + S(S + 1) . 2J(J + 1)
In a Zeeman experiment, the 3 S1 →3 P1 emission of an ensemble of such atoms is observed in the presence of a weak magnetic field B. Describe the resulting Zeeman structure of the atomic levels, and indicate which transitions amongst the split levels are allowed in an electric dipole transition. Sketch the form of the line spectrum seen in some general direction before and after the field is applied. Label, in units of µB B where µB is the Bohr magneton, the positions of the components relative to the energy of the unperturbed transition. If the emission is viewed perpendicular to the direction of the magnetic field, how many lines will be observed and what polarization states will they have?
6. Atomic structure: As well as the interaction between the spin and orbital degrees of freedom of the electron which follow from the relativistic corrections, the magnetic field generated by the nuclear magnetic momentum also lead to corrections to the electron Hamiltonian. These corrections are known as hyperfine coupling. The following problem involves exploring the Advanced Quantum Physics
Hint: Consider how the electric field couples to the electric dipole and how this translates to constraints on the allowed change in MJ .
16.3. PROBLEM SET III evolution of the hyperfine spectrum of the hydrogen atom in the presence of a weak magnetic field. The magnetic part of the Hamiltonian for a hydrogen atom in the 1s state, in the presence of a constant magnetic field B along the z axis, may be cast in the form & % ˆ = B µe σz(e) + µp σz(p) + W σ (e) · σ (p) , H
where the superscripts e and p refer to the electron and proton, the vector components of σ are the Pauli spin operators, µe,p are the respective magnetic dipole moments, and W is a constant. (i) Explain the physical origin of each term in the Hamiltonian. (ii) Using as a basis the states | ↑e $ ⊗ | ↑p $, | ↑e $ ⊗ | ↓p $, | ↓e $ ⊗ | ↑p $, ˆ | ↓e $ ⊗ | ↓p $, and neglecting the small term in µp , show that the H may be represented by the matrix b+W 0 0 0 0 b−W 2W 0 , 0 2W −b − W 0 0 0 0 −b + W where b = µB B. (Explain why µp is small in comparison to µe .)
(iii) Determine the energy levels and sketch their evolution as a function of B, labelling them with as much information as possible about the total angular momenta of the states.
7. Molecular structure: This problem concerns the application of the LCAO method to molecular bonding in the H+ 3 ion. Explain what is meant by the Born-Oppenheimer approximation and discuss how molecular wavefunctions can be formed within this approximation by using the Linear Combination of Atomic Orbitals (LCAO). The H+ 3 ion exists as an isosceles triangle (distance d12 = d23 '= d31 ), with the internal bond angle 60◦ ≤ θ ≤ 180◦ . Treating this ion in the LCAO approximation, introducing the 1s basis state wavefunctions, |ψi $ for the ith atom, show that the electron energy levels are solutions of the secular equation - α−E β γβ -- β - = 0, α−E β - γβ β α−E -
ˆ 1 $ = )ψ2 |H|ψ ˆ 2 $ = )ψ3 |H|ψ ˆ 3 $, β = )ψ1 |H|ψ ˆ 2$ = where α = )ψ1 |H|ψ ˆ ˆ ˆ )ψ2 |H|ψ3 $, and βγ = )ψ1 |H|ψ3 $, where H is the Hamiltonian for the ion. In this case, as an approximation, you may ignore the overlap integrals, )ψi |ψj $. Find the energy levels and sketch them in the range 0 ≤ γ ≤ 1. Show that the ground state must be a spin singlet, and predict which value of γ should be most stable. Qualitatively, how would you expect γ to vary with θ, and hence, what would you expect to be the value of θ in the most stable configuration? [Note that the matrix element β is negative.] (Part II 1998)
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212
16.3. PROBLEM SET III
Figure 16.1: Infra-red rotation-vibration band spectrum of HCl measured in Hz. 8. Molecular structure: This problem explores the relative influence of valance bonding in H2 . In the pure valence bonding approximation, the electronic ground state wavefunction of the H2 molecule is given by ψ VB = C[ψa (r1 )ψb (r2 ) + ψb (r1 )ψa (r2 )] , where ψa and ψb are the (real) ground state wavefunctions of the two hydrogen atoms. (i) Express the . normalization constant C in terms of the overlap integral S = d3 r ψa ψb .
(ii) Express ψ VB in terms of the molecular bonding and antibonding orbitals, (σg (1), σg (2)) and (σu∗ (1), σu∗ (2)), defined in lectures. (iii) Hence find a second molecular state ψ⊥ orthogonal to ψ VB and interpret it in terms of covalent and ionic components. (iv) Use the result S = (1 + ρ + ρ2 /3)e−ρ with ρ = R/a0 * 1.6 to estimate the configuration mixing (i.e. the relative contributions of the VB and ionic bond) in H2 .
9. Molecular spectra: This question addresses the features of the rotationvibration spectra of the diatomic molecule HCl. It also provides an exercise in the application of selection rules to radiative transitions. Explain the following features of the rotation-vibration absorption spectrum of HCl shown in Fig. 16.1: (i) the missing peak at the centre, (ii) the double peak structure, (iii) the line spacing and why it is uneven, (iv) the intensity as a function of wave number. Determine the values of as many molecular parameters as you can from the spectrum.
Advanced Quantum Physics
213
16.3. PROBLEM SET III
211
Answers: Problem set III ! ! ˆ 1 |ψ" ∼ r2 drψ ∗ p ˆ 4 ψ ∼ r2 dr|∇2 ψ|2 , consider 1. (a) Using the relation, !ψ| H the change of variables u = Zr/a0 . Since ∂x = aZ0 ∂ux , ∇2 = ( aZ0 )2 ∇2u and ˆ 1 |ψ" ∼ Z 3 !ψ| H
"
u2 du Z3
#
Z a0
$4 % %2 % 2 % %∇u G(u)e−u/n % ∼ Z 4 . 2
2
Ze Ze , so ∂r V = 4π" (b) For the Coulomb interaction, V = − 4π" 2 and 0r 0r # $ " " dr Zr ˆ 2 |ψ" ∼ r2 dr|ψ|2 Z ∼ Z 4 !ψ| H Gn# e−2Zr/na0 . r3 r a0
The integral is independent of Z, as can easily be seen by making a change ˆ 2 |ψ" ∝ Z 4 . of variables u = Zr/a0 , and hence !ψ| H (c) Making the usual substitution u = Zr/a0 , we have δ 3 (r) = ( aZ0 )3 δ 3 (u) and thus the expectation value of this term in the Hamiltonian is # $3 # $3 " " 2 u du Z Z 2 (3) 2 3 ˆ !ψ| H3 |ψ" ∼ r dr Zδ (r)|ψ| ∼ Z δ (u) G2 (u)e−2u/n , Z3 a0 a0 which is proportional to Z 4 again. 2. (2s)(3p) The allowed values of L and S are S = 0, 1, L = 1. Since the electrons are inequivalent, all combinations of L and S are allowed, i.e. 1 P1 , 3 P0,1,2 . 2 (2p) Since the electrons are equivalent, we may take S = 0 (antisymmetric) with symmetric spatial wavefunction L = 0, 2, or alternatively S = 1 (symmetric) with antisymmetric spatial wavefunction L = 1, i.e. 1 S0 , 1 D2 , 3 P0,1,2 . (3d)2 Since the electrons are equivalent, we may take S = 0 (antisymmetric) with symmetric spatial wavefunction L = 0, 2, 4, or alternatively S = 1 (symmetric) with antisymmetric spatial wavefunction L = 1, 3, i.e. 1 S0 , 1 D2 , 1 G4 , 3 P0,1,2 , 3 F2,3,4 . (3d)10 This is a completely filled shell, so L = S = J = 0, i.e. 1 S0 . (3d)9 The shell has just one unoccupied state, so the values of L, S and J are just those for a single electron in the shell, i.e. L = 2, S = 1/2 and the terms are 2 D3/2,5/2 . (4f )6 Applying Hund’s first rule, the maximum S = 3. Being maximal, this spin state is symmetric with respect to interchange of electrons, so the spatial state must be totally antisymmetric. Hence the six electrons must occupy six different m# values out of the seven (2$ + 1) available. As a result, the ML of the atom can take one of seven possible values, ML = ±3, ±2, ±1, 0. As a result, we can deduce that L = 3 is the only possibility. The shell is less than half full, so Hund’s third rule requires J = |L − S| = 0, i.e. 7 F0 . 3. The 3 P0 notation signifies the total angular momentum quantum numbers in the form 2S+1 LJ , hence S = 1, L = 1 and J = 0 in this case. The notation implies that the LS coupling approximation is appropriate, so that L and S are good quantum numbers, which will be valid if the residual Coulomb interaction is much greater than the spin-orbit interaction. Considering the (3p)1 (4s)1 configuration: (i) In LS coupling, the allowed values of S and L are S = 0, 1 and L = 1. According to Hund’s first rule, S = 1 lies lower in energy. The S = 1, L = 1 state can take three values of J = 0, 1, 2, which are split by the spin-orbit interaction, with J = 0 lying lowest in energy by Hund’s rules. The energy levels therefore form a triplet 3 P0,1,2 and a higher lying singlet 1 P1 .
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16.3. PROBLEM SET III
212
(ii) In jj coupling, the spin-orbit interaction makes a larger contribution that the Coulomb term. In this scheme, we must first find the j values for the two electrons, i.e. j = 1/2, 3/2 for the 3p state and j = 1/2 for the 4s. These j values will be separated by the spin-orbit interaction. The smaller Coulomb term will then differentiate between the different values of J which can be formed by combining the values of j, giving the following terms, (j1 , j2 )J = (1/2, 1/2)0,1 ,
(3/2, 1/2)2,1 .
The states thus form two doublets in this case. The given energy levels clearly form two doublets (a,b) and (c,d), so the jj coupling scheme is the better approximation in this case. The electric dipole selection rules (∆J = ±1, 0; but 0 → 0 forbidden) tell us than only the J = 1 levels can decay to the ground state. We also expect the (1/2, 1/2) level to ˆ · S" ˆ ∝ (J(J + 1) − L(L + 1) − S(S + 1)), so this suffices to lie lower, since !L identify the levels: (a) is (1/2, 1/2)0 ; (b) is (1/2, 1/2)1 ; (c) is (3/2, 1/2)2 ; (d) is (3/2, 1/2)1 .
4. Firstly, in answer to the introductory questions: (i) As an alkali atom, sodium has a single electron outside closed shells. The low energy excited states involve excitation of this electron. The appropriate quantum numbers are $, s and j (the quantum numbers for the whole atom being the same as for the unpaired electron). The allowed states for the electron are [3s(ground state), 3p, 3d], [4s, 4p, 4d, 4f ], [5s, 5p, 5d, 5f, 5g], etc. (ii) The spin-orbit interaction splits each level into a doublet according to j = $±1/2, except for the s-states for which j = 1/2 is the only possibility. (iii) The strength of the spin-orbit interaction decreases with n, since the electrons in higher energy levels experience a smaller magnetic field as the nucleus is better screened. (iv) Finally, the dipole selection rules require ∆J = ±1, 0, parity change, and ∆$ = ±1. Since the ns states are singlets with term 2S1/2 while the np states are doublets with terms 2P3/2,1/2 , we can deduce that all the doublets must involve s ↔ p type transitions. Since those in group II involve the same doublet spacing, they must all involve the same p state. They are therefore likely to involve 4s → 3p, 5s → 3p, 6s → 3p and 7s → 3p transitions respectively. By contrast, those in group I are likely to involve p states decaying to the same s state. Noting that the first one has the same splitting as group II, we can deduce that the they are likely to involve 3p → 3s, 4p → 3s, 5p → 3s and 6p → 3s transitions respectively, with the spin-orbit splitting decreasing with n as expected. The nd states have terms 2D5/2,3/2 . However, since the selection rules prohibit transitions between J = 5/2 and J = 1/2, we can deduce that the transitions between d ↔ p states must involve triplets, as observed in group III. Since the splittings are the same as in group II, we can deduce that 3p is involved again. They must be 3d → 3p, 4d → 3p, 5d → 3p and 6d → 3p transitions respectively. We can compare these predictions with the diagram (right) showing the spectra of sodium (known as a Grotrian diagram). (a) The 5p5/2 and 5p3/2 states are involved in the transitions of frequency 1.05086 and 1.05079 × 1015 Hz, which differ by 7 × 1010 Hz. The energy splitting is thus 7 × 1010 h = 4.6 × 10−23 J = 0.29 meV.
Advanced Quantum Physics
The Grotrian diagram for sodium showing (some) of the dipole allowed transitions. The two sodium D-lines are emphasized.
16.3. PROBLEM SET III
213
(b) We expect the sodium energy levels to converge to the hydrogen levels for large n, and thus to scale like 1/n2 . To test this, note that the ratio of the energy differences (6s − 5s)/(7s − 6s) = 1.9, compared with the 1 1 1 1 − 36 )/( 36 − 49 ) = 1.65, which is not too bad. The energy expected ratio ( 25 difference between the 7s state and the ionisation energy may be estimated as the energy difference between 7s and 6s (Planck’s constant h times 1 1 1 1 0.049×1015 Hz) times ( 49 −∞ )/( 36 − 49 ), which yields h×0.135×1015 Hz. We add this to the energy difference between 7s and 3s, inferred fron the sum of the 0.63142 × 1015 Hz (7s → 3p) and 0.50899 × 1015 Hz (3p → 3s) transitions to obtain h × 1.27 × 1015 Hz, i.e. 5.2 eV. You could use other states instead in a similar way. (c) The spin-orbit energy in the p-states can be estimated from the splittings of the corresponding doublets in group I. The Coulomb effect is given by the difference between the p- and s-levels for a given n. The ratios in the n = 3 and n = 6 cases are: n=3: n=6:
0.00052 spin − orbit = = 1.02 10−3 coulomb 0.51 spin − orbit 0.00003 = = 1.07 10−3 coulomb 0.028
i.e. both effects decrease with n at about equal rates.
5. The derivation of the Land´e g-factor is standard bookwork. The 3 S1 level has g = 2 and 3 P1 level has g = 3/2. Bearing in mind the selection rules, ∆MJ = ±1, 0, the shifts in the energy levels are given by, MJ ∆E ∆E MJ 3 1 2µB B 1 µB B 3 3 2 S1 P1 0 0 0 0 −1 −2µB B −1 − 32 µB B
and thus the shifts in the energies of the allowed transitions are as given right. The line therefore splits into seven components. Viewing perpendicular to the magnetic field, all seven lines will be seen. The ∆MJ = 0 lines corespond to dipoles parallel to the field, and thus the light will be plane polarised in the direction of the field; these will have energy shifts of ± 21 µB B and zero. The ∆MJ = ±1 lines corespond to dipoles perpendicular to the field, and thus the light will be plane polarised perpendicular to the field; these will have energy shifts of ± 32 µB B and ±2µB B.
6.
(i) First two terms represent interaction between the magnetic moments of electron and proton respectively with the external field B; the final term represents the spin-spin (hyperfine) interaction between the electron and proton. (ii) Neglecting the term in µp , and making use of the identity σ (e) · σ (p) = (e) (p) (e) (p) (e) (p) σz σz + 12 (σ+ σ− + σ+ σ− ), where # $ # $ # $ 0 1 0 2 0 0 σz = , σ+ = , σ− = . 1 0 0 0 2 0 We therefore have σ (e) · σ (p) | ↑e " ⊗ | ↑p " = | ↑e " ⊗ | ↑p "
σ (e) · σ (p) | ↑e " ⊗ | ↓p " = −| ↑e " ⊗ | ↓p " + 2| ↓e " ⊗ | ↑p " ,
etc. From these results, we can deduce the matrix elements of the Hamiltonian and confirm the expression given in the problem.
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MJ 1→1 1→0 0→1 0→0 0 → −1 −1 → 0 −1 → −1
∆E + 12 µB B +2µB B − 32 µB B 0 + 32 µB B −2µB B − 12 µB B
16.3. PROBLEM SET III
214
(iii) By inspection, the states | ↑e " ⊗ | ↑p " and | ↓e " ⊗ | ↓p " are energy eigenstates with energies W + b and W − b respectively. The other two energy eigenvalues are given by the solutions of % % % b−W −E % 2W % % = 0, % 2W −b − W − E % √ which leads to E = −W ± 4W 2 + b2 . In the case where b . W , i.e. where the external magnetic field is very weak, these reduce to W +b2 /2W and −3W − b2 /2W , so there is a triplet of S = 1 states with energy close to W , and the singlet S = 0 state with energy close to −3W . In the other limit where b / W , i.e. where the external magnetic field is very strong, this expression reduces to ±b, so we have two states with energy close to +b and two close to −b, corresponding to the two possible orientations of the electron spin.
7. Taking the 1s hydrogen wavefunctions as our basis, the matrix elements of the Hamiltonian are defined in the question. Neglecting the overlap integrals, the upper bound E on the energy levels is given, according to the Rayleigh-Ritz variational method, by setting the determinant of the matrix Hij − Eδij to zero. Multiplying out, and factorizing, this condition translates to the relation * + (α − E − γβ) (α − E)2 + γβ(α − E) − 2β 2 = 0 . This has three solutions:
E = α − γβ,
. 1 , α + β γ ± γ2 + 8 . 2
As θ → 60◦ we expect γ → 1, as all the bond lengths become equal. The solutions in this limit are: E = α − β, α − β, and α + 2β. As θ → 180◦ we expect γ → 0, as atoms√1 and 3 become separated. The solutions in this limit are: E = α, and α ± 2β. Noting that β < 0, the lowest energy is α + 2β corresponding to θ = 60◦ , so the stable configuration should be an equilateral triangle. Since this state is non-degenerate, it can only accommodate the two electrons of the H+ 3 ion if they are in an antisymmetric singlet (S = 0) spin state. 8. Defining (with obvious notation) |VB" = C[|a1 " ⊗ |b2 " + |b1 " ⊗ |a2 "], we have: (i) the normalization |VB": !VB|VB" = C 2 [!a1 |a1 "!b2 |b2 " + 2!a1 |b1 "!b2 |a2 " + !b1 |b1 "!a2 |a2 "] = C 2 (2 + 2SS ∗ ), and hence C 2 = 12 (1 + SS ∗ ). (ii) Defining - the normalized bonding and-antibonding orbitals, |g" = (|a" + |b")/ 2(1 + S) and |u" = (|a" − |b")/ 2(1 − S), we have / / / / 1+S 1−S 1+S 1−S |a" = |g" + |u", |b" = |g" − |u" . 2 2 2 2 As a result, we find that |VB" = C[(1 + S)|g1 " ⊗ |g2 " − (1 − S)|u1 " ⊗ |u2 "].
(iii) The state orthogonal to |VB" is then given by |ψ⊥ " = C[(1 − S)|g1 " ⊗ |g2 " + (1 + S)|u1 " ⊗ |u2 "]. Rewriting this in the |a" and |b" basis, after a little algebra we arrive at: |ψ⊥ " =
C(1 + S 2 ) 2CS {|a1 " ⊗ |a2 " + |b1 " ⊗ |b2 "} − {|a1 " ⊗ |b2 " + |b1 " ⊗ |a2 "} , 2 (1 − S ) (1 − S 2 )
of which the first term represents the ionic component and the second the covalent, i.e. |ψ⊥ " =
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(1 + S 2 ) 2S |IB" − |VB" . (1 − S 2 ) (1 − S 2 )
16.3. PROBLEM SET III
215
(iv) Inserting the given value of ρ gives S = 0.697, and hence the ratio of 2 ) 2 IB/VB in the state |ψ⊥ ": IB/VB = [ (1−S 2S ] = 0.14.
9. For the rotation/vibration system, the energy levels have the form E = E0 + (n + 1/2)!ω +
!2 J(J + 1) , 2I
and the selection rules are ∆n = ±1 and ∆J = ±1. Thus, neglecting any change in the moment of inertia, we have the following energy changes: !2 (J + 1) I !2 ∆E = !ω − J I
R branch J → J + 1 J = 0, 1, 2 · · · ∆E = !ω + P branch
J → J − 1 J = 1, 2 · · ·
(i) The above formulae imply equally spaced lines above and below !ω, but the transition at !ω (∆J = 0, the Q-branch) is not permitted by the parity change selection rule. (ii) The double peaks are an isotope effect. Cl has two isotopes 35 Cl and 37 Cl with abundances in the ratio of roughly 3:1. The less abundant 37 Cl will have a larger reduced mass (by about 1 part in 1000) and thus a lower vibration frequency. Thus there are two separate spectra slightly displaced in frequency. (iii) The uneven spacing results from the change in moment of inertia. If !2 /2I ≡ B initially and B + δB finally (with δB < 0), the formulae for the energy changes become: R − branch J → J + 1
P − branch J → J − 1
∆E = !ω + 2B(J + 1) + δB(J + 1)(J + 2) ∆E = !ω − 2BJ + δBJ(J − 1)
So the spacing in the R-branch above !ω decreases with increasing J while that in the P-branch increases with J. (iv) The intensity of absorption depends on the population of the states, which is given by the product of the degeneracy of the rotational wavefunction and the Boltzmann factor, i.e. (2J + 1) exp (−J(J + 1)!2 /2IkB T ), which exhibits first a rise and then a fall with increasing J, as observed. From the central vibrational frequency ω we can-infer the force constant of the bond k: ω = 2πν 1 5.4 × 1014 Hz. Then, ω = k/µ, where µ is the reduced mass (approximately equal the mass of the Hydrogen atom in this case), so we deduce that k = 490 Nm−1 .
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Problem Set IV
1. Operator methods in quantum mechanics: The following problem combines the practice of a number different methodologies covered in this course. Here we address the excitation spectrum of a quantum mechanical spin chain. It provides a counterpart to the vibrational modes of the quantum harmonic chain considered in lectures. Here we will address the problem in two different ways. In strongly correlated solid state systems, the Coulomb interaction can result in electrons becoming localized to the sites of the underlying crystalline lattice – the Mott transition. However, in these insulating materials, the spin degrees of freedom carried by the constituent electrons can remain mobile – spin fluctuations can be exchanged between neighbouring electrons without motion of charge. Such systems are described ˆ ˆ ˆ =! by quantum magnets, H m!=n Jmn Sm · Sn , where exchange couplings Jmn denotes the matrix elements coupling lattice sites m and n. Since these matrix elements decay rapidly with distance, it is often legitimate to restrict attention to neighbouring sites. Although these matrix elements are typically positive (leading to an antiferromagnetic coupling), in the following we will consider them negative leading to ferromagnetism – i.e. neighbouring spins want to lie parallel. Consider then the one-dimensional spin S quantum (Heisenberg) ferromagnet, ˆ = −J H
" m
ˆm · S ˆ m+1 , S
α,S ˆnβ ] = where J > 0, and the spins obey the quantum spin algebra , [Sˆm γ i!δmn "αβγ Sˆm .
(a) Making use of the spin commutation relations and Ehrenfest’s theorem, show that, in the Heisenberg representation, the spins obey the equations of motion, !
ˆm dS ˆ m × (S ˆ m+1 + S ˆ m−1 ) , = JS dt
where we suppose that the boundary conditions are periodic, i.e. Sm+N = Sm . (b) For large spin S, we may take the spin expectation values to be ˆ m $ = Sm . Moreover, if we are defined by their classical values, #S interested in low-energy excitations of the spin chain, only modes of long wavelength contribute (cf. the vibrational modes of the harmonic chain). In this case, we may develop the Taylor series expansion, Sm+1 = Sm +∂Sm + 12 ∂ 2 Sm +· · ·, where the lattice spacing is taken as unity. In doing so, show that the leading contribution to the equations of motion in the gradient expansion is given by, S˙ = JS × ∂ 2 S . (c) Show that the solution √ to this equation is given by S(x, t) = (c cos(kx− ωt), c sin(kx−ωt), S 2 − c2 ) and determine the dispersion relation, ω(k). Sketch a “snapshot” configuration of the spins in the chain. These low energy modes of the quantum spin chain are known as a spin waves and mirror the phonon excitations of the quantum harmonic Advanced Quantum Physics
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chain. Notice that the dispersion relation in this case has a quadratic dependence on k (cf. non-relativitic particles) as opposed to the linear dependence of the phonon spectrum. If you are feeling energetic, you might contemplate the spin wave modes for the antiferromagnetic chain where the dispersion becomes linear. Let us now consider an alternative approach to the spin wave spectrum which follows an operator-based formalism. (d) Defining the spin raising and lowering operators, Sˆ± = Sˆx ± iSˆy , show that the ferromagnetic Heisenberg model can be written as %& "# 1 $ ˆ+ ˆ− − ˆ+ z ˆz ˆ = −J Sm+1 . H Sˆm Sm+1 + Sm Sm+1 + Sˆm 2 m
Then, making use of the Holstein–Primakoff spin representation √ † − = ! 2S a† (1 − am am )1/2 , S ˆ+ = (Sˆ− )† , (defined on page 192), Sˆm m m m 2S z = !(S −a† a ) where [a , a† ] = δ and Sˆm m m m n mn , show that the Hamiltonian can be expanded as a bilinear (i.e. to quadratic order) in the raising and lowering operators, " † ˆ = −JN S 2 + S H (am+1 − a†m )(am+1 − am ) + O(S 0 ) . m
(e) Being bilinear in operators (i.e. quadratic), the Hamiltonian can be diagonalized by discrete Fourier transformation. With periodic boundary conditions, a†m+N = a†m , defining 1 a†k = √ N
N "
eikm a†m ,
a†m
m=1
1 " −ikm † =√ e ak , N k
where the sum on k = 2πn/N , runs over the N integers n = −N/2 + 1, −N/2 + 2 · · · N/2, show that the transformed operators obey the commutation relations, [ak , a†k! ] = δkk! . In the Fourier representation, show that " ˆ = −JN S 2 + H !ωk a†k ak + O(S 0 )
In lectures, we have discussed the Fourier series expansion, 1 " f (x) = √ fk eikx L k ' L 1 f (x)e−ikx fk = √ L 0 In the following, we are dealing with a discrete lattice where we have to consider the discrete Fourier representation, 1 " fn = √ fk eikn N k 1 " fk = √ fn e−ikn . N n
k
where ωk = 2JS(1 − cos k) = 4JS sin2 (k/2) represents the dispersion of the spin excitations. In the long wavelength limit, k → 0, the energy of the excitations vanishes, ωk → JSk 2 . These low-energy excitations, known as spin waves or magnons, describe the elementary spin-wave excitations of the ferromagnet. Taking into account terms at higher order in the parameter 1/S, one finds interactions between the magnons.
2. The following problem involves some revision of time-dependent perturbation theory and then applies the methodology to the problem of an induced transition in a hydrogen atom. Suppose that a system is prepared in an energy eigenstate ψ0 at time t = 0 when a weak perturbation V (t) is applied. Show that the probability of finding the system in state ψn at time t is given approximately by |cn (t)|2 where ' 1 t # i(En −E0 )t! /! #ψn |V (t# )|ψ0 $ . dt e cn (t) = i! 0 Advanced Quantum Physics
Measurements of the spin-wave dispersion relations for the ferromagnet La0.7 Sr0.3 MnO3 .
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At times t > 0, an electric field Ez = E0 exp −t/τ is applied to a hydrogen atom, initially prepared in its ground state. Working to first order in the electric field, find the probability that, after a long time, t ' τ , the atom is in (i) the 2s state, and (ii) one of the 2p states (state which one?).
3. This problem addresses the question of spontaneous emission in hydrogen. Use the formula for the Einstein A-coefficient derived in class to calculate the lifetime of the 2p state of atomic Hydrogen. Without detailed evaluation of the matrix element, explain why the 3s level of hydrogen has a lifetime roughly 100 times longer than the 2p level. And why is the lifetime of the 2s level very much longer than 2p, by a factor ∼ 108 ? 4. Scattering theory: The following problem revises the derivation of the Born approximation and then applies it to the high energy scattering from an attractive square well potential. Show that the Born approximation yields the following expression for the elastic scattering of a particle of mass m and momentum !k from a spherically symmetric potential V (r), *2 ( ) *' * dσ 2m 2 ** ∞ * , = V (r)r sin (Kr)dr * * dΩ !2 K 0
where K = 2k sin(θ/2) and θ is the angle through which the particle is scattered. Obtain the differential cross-section for scattering from a potential, # −V0 r ≤ a V (r) = 0 r>a
and verify that the scattering is isotropic when the energy of the incident particle or the size of the scatterer is sufficiently low, so that Ka * 1. Obtain an expression for the total cross-section in this limit.
5. Scattering theory: The following problem involves the application of the partial wave scattering method to a simplified model of a nucleus. As a crude model for an effective nuclear potential which binds together protons and neutrons, consider a repulsive radial shell potential, V (r) = !2 2m U0 δ(r − R). (a) Calculate the s-wave scattering phase shift as a function of U0 . (b) Assuming that U0 ' 1/R and U0 ' k, show that if tan(kR) is not close to zero, the phase shift resembles that of a hard sphere. (c) Continuing with these assumptions show that, if tan(kR) is close (but not equal to) zero, resonance is possible (i.e. the cross section reaches its maximum value). Compare the resonance energy with that of a bound state of the spherical shell with an infinitely impenetrable wall. This is an example of a “scattering resonance”, when the incident energy matches a quasi bound-state. Advanced Quantum Physics
You will need the same matrix elements as you computed in the previous problem.
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6. † Relativistic quantum mechanics: The following problem establishes an important relation in the study of relativistic covariance in lectures. In lectures, it was shown that Lorentz covariance of the Dirac equation relies on the condition S(Λ)γ µ S −1 (Λ) = (Λ−1 )µν γ ν . For an infinitesimal proper Lorentz transformation Λµν = g µν +ω µν , S(Λ) = I− 4i Σµν ω µν +· · ·, where ωµν and Σµν are antisymmetric. Show that, to leading order in ω, this condition translates to % $ [γ µ , Σαβ ] = 2i g µα γβ − g µβ γα . Show that the following represents a consistent solution, Σαβ = 2i [γα , γβ ].
7. Relativistic quantum mechanics: In lectures, we used the relativistic covariance of the Dirac equation to deduce the existence of an intrinsic angular momentum known as spin. In the following problem, we can establish the consistency of this construction by showning that the Dirac Hamiltonian ˆ commutes with the total angular momentum, J. ˆ = α·p ˆ + βm commutes with both Verify that the Dirac Hamiltonian H ˆ =L ˆ +S the helicity operator S · p/|p|, and the angular momentum J ˆ =x×p ˆ. where L 8. † Relativistic quantum mechanics: The following problem relates to the vacuum instability and the stimulation of particle/antiparticle pairs. Following on from the discussion of the potential step in lectures, consider the problem of the transmission through a potential barrier of width a and hight eV . Applyly the boundary conditions at the edge of the barrier, show that the transmission probability is given by * * 2 *−2 * # # (1 + ζ ) * * |t | = *cos(p a) − i sin(p a) , 2ζ * 2
where ζ =
p! (p0 +m) , p(p0 ! +m)
#
E ≡ p0 , and E # ≡ p0 = E − eV . Analyse the #
transmission probability in the energy ranges p0 ≡ E # > m, −m < E # < m, and E # < −m. In the third regime, explain why a condition of perfect transmission can be obtained.
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Answers: Problem set IV 1.
(a) From Ehrenfest’s theorem, the equation of motion for the spin is given ˆ ˆ m ]. Making use of the spin commutation relation, we ˆ S by −i! dSdtm = [H, have (summation on repeated spin indicies assumed) −i!
β dSˆm α ˆβ α α γ ˆα α = −J[Sˆm , Sm ](Sˆm+1 + Sˆm−1 ) = −i!J!αβγ Sˆm (Sm+1 + Sˆm−1 ). dt
We thus obtain the required equation of motion. (b) Since Sm+1 +Sm−1 " 2S|x=m +∂ 2 S|x=m and, for classical vectors, S×S = 0, we obtain the required equation of motion. (c) Substituting the √ expression for S(x, t), we find that the equation is solved with ω(k) = Jk 2 S 2 − c2 . The corresponding spin configuration is shown right. (d) Substituting for the spin raising and lowering operators, the identity is clear. Expanding to the spin raising and lowering operators to leading † order in a2Sa about the ferromagnetic ground state (in which all spins are aligned along eˆz , we obtain !" # $% ˆ = −JN S 2 + JS H a†m am + a†m+1 am+1 − a†m am+1 + h.c. + O(S 0 ) , m
where h.c. denotes the Hermitian conjugate. Rearranging, we obtain the required expression for the Hamiltonian.
(e) With the definitions given in the problem, [ak , a†k! ] =
1 ! −ikm+ik! n 1 ! −i(k−k! )m e [am , a†n ] = e = δkk! . & '( ) N m,n N m δmn
Then subtituted into the Hamiltonian, ˆ = −JN S 2 + S H
! 1 ! ! ! ei(k−k )m (eik − 1)(e−ik − 1)a†k ak! N m kk! & '( )
= −JN S + S 2
! k
δkk!
|e − 1|2 a†k ak . ik
From this result we obtain the required dispersion relation.
2. Standard bookwork allows a derivation of the amplitude cn (t). In the present case, with V (t) = eE0 ze−t/τ , the matrix element %ψ2s |z|ψ1s & = 0 since the 1s and 2s wavefunctions both have even parity while z has odd parity. Therefore the probability of finding the atom in the 2s state is identically zero. The matrix elements %ψ2p±1 |z|ψ1s & = 0 since the φ part of the integral will vanishes, * 2π %ψ2p±1 |z|ψ1s & ∼ dφe±iφ = 0 . 0
The only non-zero matrix element is: + ,1/2 + ,1/2 * * 1 1 2 2 −r/a0 −r/2a0 %ψ2p0 |z|ψ1s & = r dr r e e 2π sin θdθ cos2 θ 32πa50 πa30 1 4! 4π 256a0 √ . = √ · · = 4 2πa40 (3/2a0 )5 3 243 2
Taking the limit as t → ∞, the t" integral is given by, * ∞ ! ! 1 dt" e−t /τ ei(E2p −E1s )t /! = , 1/τ − i∆E/! 0
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where ∆E = E2p − E1s = 3R∞ /4. Putting all this together we obtain the probability of being in the 2p0 state after a long time as 1 e2 E02 a20 215 · . 10 2 3 ∆E + !2 /τ 2
|c2p0 (∞)|2 =
3. From the lecture notes, the decay rate for unpolarized light is given by, A=
ω 3 |dkj |2 , 3π!0 c3 !
and the lifetime is thus τ = 1/A. Take for example the 2p0 state of Hydrogen decaying to 1s (the other 2p states must have the same lifetime, but this one depends on the same matrix elements that we computed in in the previous question. Only the z-component of d is non-zero for this transition, (the φ integral yields zero if you compute the matrix elements of x or y) giving, %2p0 | ez |1s& =
256ea0 √ = 6.31 × 10−30 Cm . 243 2
The energy of the emitted photon is !ω =
3 3 me4 R∞ = · 4 4 2(4π!0 )2 !2
ω = 1.56 × 1016 Hz .
⇒
Hence, the lifetime of the state is τ = 1.56 × 10−9 s. The only lower lying state to which 3s can decay is 2p according to the selection rules. We can expect the matrix element %3s| ez |2p& ∼ ea0 on dimensional grounds, and thus not very different from %2p| ez |1s&. The main difference between the lifetimes of the 3s and 2p levels will arise from the difference in ω 3 . For the 3s→2p transition, 1 1 5 !ω = ( − )R∞ = R∞ . 4 9 36 The ratio of the lifetimes is therefore approximately + ,3 τ (3s) 3 36 ∼ · ∼ 150 . τ (2p) 4 5 The only state lying below 2s is 1s, but the decay 2s→1s is not allowed by the electric dipole selection rules. The 2s state is “metastable”. The dominant decay is actually via two-photon emission, a process which can arise through second order perturbation theory, and occurs very slowly. In practice, atoms may well make transitions from 2s to 2p (for example) before decay takes place as a result of collision processes. Alternatively, decay of the 2s state may be induced by the application of an external electric field, which mixes 2s and 2p through the Stark effect. 4. From the lecture notes, the Born Approximation gives, /2 - m .2 //* / dσ / V (r)ei∆·r d3 r/ , = / / 2 dΩ 2π!
where ∆ is the difference beweeen incoming and outgoing wave vectors, of magnitude 2k sin2 (θ/2). In the case where V (r) = V (r), i.e. where the potential is centrally symmetric, it is convenient to take ∆ as the axis of polar coordinates for the purpose of integration, so that ∆ · r = |∆|r cos θ" . The integral thus becomes * * ! i∆·r 3 V (r)e d r = V (r)ei∆r cos θ 2π sin θ" dθ" r2 dr 0 1π * * ! ei∆r cos θ 4π 2 = 2π V (r)r dr V (r)rdr sin(∆r) , = i∆r ∆ 0
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and hence dσ = dΩ
+
2m ∆!2
/2 ,2 / * / / / V (r)rdr sin(∆r)/ . / /
Taking V (r) = −V0 for r ≤ a, and V (r) = 0 otherwise, the integral becomes (integrating by parts), 4a * a 5 23 * a cos(∆r) cos(∆r) + dr −V0 r sin(∆r)dr = −V0 −r ∆ ∆ 0 0 0 V0 = − 2 (sin(∆a) − ∆a cos(∆a)) , ∆ and thus 3 42 dσ 2mV0 = (sin(∆a) − ∆a cos(∆a)) . dΩ !2 ∆ 3 In the low energy limit, ∆ → 0, sin(∆a) − ∆a cos(∆a) ≈ ∆a −
1 (∆a)3 − ∆a(1 − (∆a)2 /2) = (∆a)3 /3 , 3!
and hence dσ = dΩ
+
2mV0 a3 3!2
,2
.
This is independent of ∆ and hence independent of θ, so isotropic, as required. The total cross-section is obtained by integrating over solid angles, which simply involves multiplying by 4π in this case + ,2 2mV0 a3 σtot = 4π . 3!2
5.
(a) When kR - 1, s-wave scattering dominates. In this case, the problem is equivalent to a one-dimensional scattering problem with an infinite wall at the origin and a δ-function repulsive potential at r = R. The wavefunction has the solution, 2 C sin kr r R From the continuity condition on the wavefunction and the derivative, we obtain A sin(kR) = sin(KR + δ0 ) kA cos(kR) − k cos(kR + δ0 ) = U0 sin(kR + δ0 ) . 0) From the first equation, we obtain A = sin(kR+δ which substituted into sin(kR) the second equation, leads to the relation 3 4 k tan(kR) δ0 = tan−1 − kR . k − U0 tan(kR)
The structure is similar to that obtained for the spherical square potential but with different resonant behaviour.
(b) With U0 . 1/R, k, and U0 tan(kR) . k, we obtain the resonance condition k tan(kR) k " " 0, k − U0 tan(kR) −U0 tan(kR)
i.e. δ0 " −kR, the value that it would have for a hard sphere.
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(c) Now supose that tan(kR) is small. In this case, we have a resonance when k − U0 tan(kR) = 0, i.e. tan(kR) = Uk0 - 1, and δ0 =
π π − kR " . 2 2
2 4π The cross-section σ0 = 4π k2 sin δ0 " k2 . The resonance is near tan(kR) = 0, which implies that kR = (2n+1)π/2, the quasi-bound state of the well.
6. Substituting the definition of S(Λ) into the defining condition we obtain + , + , i i i 1 − Σαβ ω αβ γ µ 1 + Σγδ ω γδ = γ µ + [γ µ , Σαβ ] ω αβ + · · · 4 4 4 = (g µν − ω µν ) γ ν . Rearranging the left and right hand sides, we obtain i µ [γ , Σαβ ] ω αβ = −ω βα g µβ γα ≡ ω αβ g µβ γα , 4 from which we obtain the required identity. The latter equation is shown to be consistent with the solution Σαβ = (i/2)[γα , γβ ] by making use of the anticommutation relation of the γ matrices.
7. Using the identity ˆ, S · p ˆ ] = pˆi pˆj [α · p
+
0 [σi , σj ]
[σi , σj ] 0
,
= 2i!ijk pˆi pˆj
+
0 σk
σk 0
,
.
ˆ×p ˆ = 0, we find that the Hamiltonian commutes with the Therefore, since p Helicity operator. Turning to the angular momentum, taking each term separately, ˆ L ˆ i ] = !ijk [α · p ˆ, x [H, ˆj pˆk ] = !ijk (αl pˆl x ˆj pˆk − x ˆj pˆk αl pˆl ) ˆ. = !ijk (−iαl δlj pˆk ) = −iα × p 1 ˆ S] = [α · p ˆ , S] = (αi pˆi σj − σj αi pˆi ) [H, 2 , + 3+ ,4 + , 1 1 0 1 ˆ 0 σ·p σ 0 ˆ , σ] = , = [σ · p ˆ σ·p 0 0 σ 2 2 1 0 + , 0 1 ˆ × σ = −iˆ = −i p p × α. 1 0 ˆ = 0. ˆ J] Putting these terms together we find [H,
8. Applying the plane wave solution of the Dirac equation ψ(p) = e−p·x u(p) (defined in this form for positive and negative energy states) to the two edges of the potential step, we obtain the boundary conditions 1 1 0 −ipa/2 0 ipa/2 +r p e e p − E+m E+m 0 0 1 1 ! 0 ! ip! a/2 0 = t" p! e−ip a/2 + r" e − E !p+m E ! +m 0 0 1 1 1 ! ! 0 ! −ip a/2 0 0 t" p! eip a/2 + r" = t p eipa/2 , e − E !p+m E ! +m E+m 0 0 0
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where the reflection and transmission coefficients are defined in the figure. From these equations we obtain 2e−ipa/2 2reipa/2 teipa/2 teipa/2
!
!
!
!
= t" (1 + ζ)e−ip a/2 + r" (1 − ζ)eip a/2
= t" (1 − ζ)e−ip a/2 + r" (1 + ζ)eip a/2 !
!
= eip a/2 t" + e−ip a/2 r" - ! . ! = ζ eip a/2 t" − e−ip a/2 r" .
Rearranging these equations we obtain r" =
1 2 −i(p−p! )a/2 , !a !a e −ip −1 −ip 1 + ζ µe −µ e
where µ = (1 − ζ)/(1 + ζ). Finally, with this result, we obtain t = e−ipa
1 cos(p" a) − i sin(p" a)(1 + ζ 2 )/2ζ
From this result, we obtain the expression for the transmitted current shown in the question. For energies E " > m, the particles traverse the barrier as a plane wave. In particular, when p" a = nπ there is perfect transmission. For m > E " > −m, p" is imaginary and exchange of particles occurs by resonant tunnelling across the barrier. For energies E " < −m, the Klein paradox regime, p" is real and positive, and there is again perfect transmission when p" a = nπ. Here the transmission is mediated by negative energy states under the barrier.
Advanced Quantum Physics
