Qwmtum Mechanics A Self-Conmined COUTSe
Volwne 1
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·Quantum Mechanics A Self-Contained Course David ATKINSON Mahouton Norbert HOUNKONNOU Volume 1
5P
Rinton Press
Quantum Mechanics A Self-Contained Course This book is the frrst in a series of three volumes devoted to quantum mechanics and to quantum field theory. It is self-contained: the only prerequisites are a knowledge of integral calculus and partial differential equations, as well as Newton's mechanics of point masses. New subjects are developed in requisite detail at the various points where they are needed. The mathematics is at the same time explicit but kept in check, so that the reader does not get bogged down in annoying generalizations that might distract him or her from the physics. Each chapter is complemented by ten problems, one hundred in all, that are graded in difficulty, and the student is strongly advised to try them all before looking at the full solutions in the third volume.
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Quantum Mechanics A Self-Contained Course David ATKINSON Mahouton Norbert HOUNKONNOU
Volume 1
Rinton Press, Inc.
© 2001 Rinton Press, Inc. 12 Castleton Road Princeton, New Jersey 08540 USA
All right reserved. No part of this book covered by the copyright hereon may be reproduced or used in any form or by any means-graphic, electronic, or mechanical, including photocopying, recording, taping, or information storage and retrieval system - without permission of the publisher.
Idea for cover design: Jeanne Peijnenburg Published by Rinton Press, Inc. Printed in the United State of America
ISBN 1-58949-022-3
Preface
This book is the first in a series of three volumes devoted respectively to (1) N onrelativistic Quantum Mechanics (2) Relativistic Quantum Field Theory (3) Problems and Solutions, in which the 200 exercises of the first two volumes are solved and discussed. These volumes are self-contained in the sense that the only prerequisites are a knowledge of integral calculus and partial differential equations, as well as Newton's mechanics of point masses. It is not necessary for the reader to know anything about the theory of distributions, Hilbert space or the special functions of mathematical physics. Nor is any knowledge of Lagrangian and Hamiltonian formalism assumed, for all these subjects are developed in requisite detail at the various points where they are needed. On the other hand, it would be helpful for the student to have followed an introductory course on quantum mechanics, to have some notion of what a wave-function is, and to have struggled with the Schrodinger equation in one spatial dimension. In this first volume, we have not made any attempt to cover all subjects of physical interest, but have made a choice appropriate for a course based on two hours of lectures and two hours of problem sessions during one undergraduate semester. Each chapter of this volume is complemented by ten problems, and the student is strongly advised to try them all by himself or herself before looking at our solutions in Volume 3. Physics cannot be learned just by reading or listening, but only by thinking, writing and worrying. The idea of writing a self-contained course originated with one of us (MNH). While the other (DA) was lecturing at the Institute of Mathematics and Physical Sciences of the National University of Benin in February 2001, MNH proposed both the adaptation and extension of the lecture notes that DA had written as an aid to students at the Institute for Theoretical Physics, University of Groningen. v
Preface
vi
The idea was to write a three-part manual of instruction, as much as possible without essential reference to other works. Moreover, MNH suggested contacting Wei Chen of Rinton Press; and this proved indeed to be a felicitous initiative. Despite the plethora of books of instruction on these subjects, we hope that our work will fill a need, and we believe our approach to be pedagogically sound, given its attention to mathematical detail combined with physics, which resulted in the integration of the appropriate mathematical tools at the points in the text where they are needed. The mathematics is at the same time explicit but kept in check, so that the reader does not get bogged down in annoying generalizations that might distract him or her from the physics. At a number of points recourse is taken to the computer to solve transcendental equations; in most cases a simple program is given in Mathematica (which is a trademark of Wolfram Research Inc.), but for readers without access to this system, we indicate where the numerical results can be checked by means of a scientific calculator. We wish to acknowledge the help we have had from reading the books on quantum mechanics that are listed in the bibliography; and in particular we thank Professor Lambrecht Kok, in Groningen, for encouraging us to use some exercises from the book published by Jan Visser and himself, and we are grateful to Professor Jean Pestieau of Louvain-Ia-Neuve for sending us some exercises. We further thank the Office for International Relations, University of Groningen, especially its director, Madeleine Gardeur, for supporting and stimulating our cooperation. The drawings on the covers of the volumes are ambiguous representations. Volume 1: a duck, or is it a rabbit? Volume 2: a vase, or are there two faces? Volume 3: a young, or is it an old woman? The simultaneous existence of two pictures is perhaps the closest metaphor we can find to the fundamental mystery of quantum mechanics, namely the linear superposition of two aspects of reality, each of which separately can be pictured, and whose combination can barely be comprehended by the eye of introspection, but wpich can be apprehe~ded by the power of mathematical language. August 2001,
David Atkinson (The Netherlands) Mahouton Norbert Hounkonnou (Republique du Benin).
Contents
Preface
v
Chapter 1 Transition from Classical to Quantum Mechanics 1.1 Canonical Transformation 1.2 Canonical Momentum 1.3 Poisson Brackets . . . . . 1.4 Hilbert Space . . . . . . . 1.5 Dirac's Transition to Quantum Mechanics 1.6 Dirac Delta Function. 1.7 Momentum Operator. 1.8 The Hamiltonian . . . 1.9 Schrodinger Equation 1.10 Dirac's Views 1.11 Exercises . . . . . . .
1 2 4 6 9 13 14 19 22 24 26 27
Chapter 2 Three-Dimensional Harmonic Oscillator 2.1 Creation and Annihilation Operators. 2.2 Hermite Polynomials . . . . 2.3 Isotropic Oscillator. . . . . 2.4 Many Harmonic Oscillators 2.5 Exercises . . . . . . . . . .
31 32 36 38 40 41
Chapter 3 Orbital Angular Momentum 3.1 Diagonalization of L2 and L3 3.2 Spherical Harmonics 3.3 Zeeman Effect 3.4 Exercises . . . . . .
43 44 50 60 61
vii
viii
Contents
Chapter 4 Central Potential 4.1 Asymptotic Solution for T ---t 0 4.2 Asymptotic Solution for T ---t 00 4.3 3D Oscillator in Polars . . . 4.4 Square-Well Potential . . . . . 4.5 Spherical Bessel Functions . . . 4.6 Vibrational-Rotational Spectra 4.7 Exercises . . . . . . . . . . . .
65 65 68
Chapter 5 Hydrogen Atom and Charmed Quark 5.1 Radial Equation for Coulomb Potential 5.2 Radial Equation for Linear Potential 5.3 WKB Approximation 5.4 Exercises . . . . . . . . . . . . . . .
85 85
Chapter 6 Spin and Addition of Angular Momenta 6.1 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Addition of Angular Momenta . . . . . . . . . . . . . 6.3 Combination of Spin and Orbital Angular Momentum 6.4 Positronium............... 6.5 Exercises . . . . . . . . . . . . . . . . 6.6 Table of Clebsch-Gordan Coefficients.
69 71 75 80 82
91 95 97 99 100 101 105 107
109 112
Chapter 7 Approximate Methods 7.1 Perturbation Theory .. 7.2 Variational Method . . . 7.3 Anharmonic Oscillator . 7.4 Stark Effect . . . . . 7.5 Near Degeneracy . . 7.6 A mmonia Molecule. 7.7 Exercises . . . . . .
113 113 116 118 120 123 124 125
Chapter 8 Scattering Theory 8.1 Lippmann-Schwinger Equation 8.2 Scattering Cross-section .. 8.3 Born Approximation . . . . . 8.4 Unitarity and Phase Shifts .. 8.5 Phase Shifts and Resonances 8.6 Neutron-Proton Scattering 8.7 Exercises . . . . . . . . . . .
129
129 133 135 137 140 141 145
Contents
ix
Chapter 9 Atomic Physics 9.1 Two-Electron System . . . . . . . 9.2 Exchange Term . . . . . . . . . . . 9.3 Hartree and Hartree-Fock Methods 9.4 Periodic Table . . . . . . 9.5 Hydrogenic Molecular Ion 9.6 Hydrogen Molecule. 9.7 Exercises
147 147 154 157 159 162 171 172
Appendix A Completeness of Eigenvectors A.1 Hermitian Operators . . . . . . . . . . . . . A.2 Fundamental Theorem of Linear Equations A.3 Fundamental Theorem of Algebra .. A.4 Diagonalization of a Hermitian Matrix A.5 Commuting Hermitian Matrices . A.6 Infinite Dimensional Spaces A.7 Exercises
175 175 176 181 184 187 188 191
Bibliography
193
Index
194
Chapter 1
Transition from Classical to Quantum Mechanics
Consider a point particle of mass, m, in a conservative force-field: that is, suppose that there exists a scalar potential energy, V(f'), that depends only on the position, f', of the particle, such that the force is given by
(1.1) Generally, it is convenient to express vectors in terms of Cartesian components,
f' = (Xl, X2, X3):
av
Fi.= --8 .
(1.2)
Xi
Newton's second law can then be written
~ [mx.] dt
= _ 8V 8Xi '
t
(1.3)
where dots mean time-derivatives. The non-relativistic kinetic energy of the particle is
(1.4) the convention being made that a repeated index (here i) is to be summed from 1 to 3. Accordingly,
~ [:~l =
- ;:, .
(1.5)
The same equation (1.5) holds for N particles in a conservative force-field. We write the kinetic energy in the form
(1.6) 1
2
Transition from Classical to Quantum Mechanics
where we sum over the Cartesian coordinates of all N particles, with the convention that ffi3i-2 = ffi3i-l = ffi3i is the mass of the i th particle, the x, y, z coordinates of which are X3i-2, X3i-l and X3i respectively. The potential energy may include particle-particle interactions, so long as these are conservative: 3N
V
3N
= 2: Vi + ! 2: Vij . if:. j
i
1.1
(1.7)
Canonical Transformation
We shall now perform a (possibly explicitly time-dependent) transformation of the 3N Cartesian coordinates, Xi, to a set of 3N independent coordinates, qn. A simple example would be the polar coordinates of the N particles, but much more exotic possibilities exist, as we shall see. The purpose is eventually to obtain coordinate-independent equations of motion. By differentiating Eq.(1.6) with respect to qn and to qn, we find
(1.8)
aT aqn
. aXi
= ffiiXi oqn .
(1.9)
Since the qn are functions only of the Xi and t, it follows that the Xi may be written as functions of the qn and t (but not explicitly of the qn), so that
. aXi . Xi = aqk qk
aXi
+ at .
(1.10)
N ow the only dependence on the variables qk is in the explicit occurrence of qk on the right-hand side, so
aXi aqn
aXi aqn·
(1.11)
Hence
(1.12) and by differentiating with respect to time, d [aT] dt aqn
(1.13)
Canonical Transformation
3
By using Eq.(1.3), we can write the first term on the right-hand side as (1.14) since V does not depend on the Xi explicitly. The second term on the right of Eq.(1.13) is equal to 8Tj8qn, as can be seen from Eq.(1.8). Therefore (1.15) where the Lagrangian is defined by L = T - V. This is quite different from the total energy, E = T + V. Since V does not depend explicitly on Xi, neither does it depend explicitly on qn, which means that the partial qn-derivative of T is equal to that of L. Hence Eq.(1.15) can be rewritten as follows: (1.16) which is the Euler-Lagrange equation. The action is defined to be the time-integral of the Lagrangian: (1.17) Hamilton's variational principle states that, if we vary the functions q(t) and q(t), for tl < t < t 2 , but in such a way that q(td and q(t2) remain fixed at their physical values, then, of all the possible values that q(t) and q(t) can have, those values that make S extremal satisfy the Euler-Lagrange equations. We shall now prove this statement.
(1.18) By an integration by parts, we find (1.19) where the integrated terms vanish, since gq is constrained to be zero at tl and t2. Since is otherwise arbitrary, it follows that the expression between square brackets must vanish.
oq
4
1.2
Transition from Classical to Quantum Mechanics
Canonical Momentum
In terms of Cartesian components, the Lagrangian of our system of N particles, interacting with one another, and with an external, conservative field, can be written (1.20) where V may depend explicitly on t (for example, a time-dependent external potential), but not on x. Then clearly aL
-a. = Xi
.
(1.21)
m(i)x(i) ,
where "there is no summation over the bracketed index, (i). We recognize the right-hand side of Eq.(1.21) as a component of the linear momentum, and we now define the generalized, or canonical momentum, by aL
(1.22)
Pn=.. aqn
It is important to realize that the canonical momenta are not necessarily momenta in the Cartesian sense: for example, in terms of spherical polar coordinates, the "momentum" conjugate to the angle, (), isthe angular momentum. The rate of change of the Lagrangian can be written
dL dt
=
aL. aqn qn
aL .. + aqn qn
aL + at .
(1.23)
From the Euler-Lagrange equation,
aL aqn
=
d [aL dt aqn
1'
(1.24)
so that Eq.(1.23) can be rewritten (1.25) The Hamiltonian is defined by
H =Pnqn - L,
(1.26)
and we use Eq.(1.22) to rewrite Eq.(1.25) in the form dH
dt
aL
+ Ft
=
o.
(1.27)
5
Canonical Momentum
If L does not depend explicitly on the time, the second term above is absent, and hence the Hamiltonian is constant in time. Moreover, if the potential energy is conservative, Eq.(1.21) holds, so in terms of the Xi we find
(1.28) which means that the Hamiltonian is equal to the total energy, which is timeindependent. Note that two conditions are necessary for these conclusions to be true: (1) L must not explicitly depend on the time, and (2) V must not explicitly depend on the x. From Eq.(1.26) we deduce
. dPn dH = qn
+ Pn d·qn -
aL dqn - -a. aL d·qn - {) aLdt. -a qn qn t
(1.29)
By using the definition (1.22), we see that the second and the fourth terms above cancel. Moreover, the same definition, combined with the Euler-Lagrange equation, implies that
aL . -=Pn, aqn
(1.30)
so that (1.31) The form of this increment suggests that we consider H as a function of qn, Pn and t (and not independently of tin). Then the following partial derivatives can be read off from Eq.(1.31):
aL at
aH qn=apn
(1.32)
aH Pn=-aqn
(1.33)
aH at
dH dt
.
(1.34)
These are the Hamilton equations, which constitute an alternative to the EulerLagrange system. One of the advantages of Hamilton's approach is that one can readily define more general transformations of the variables. The set {qn, Pn}, may be replaced by another set, {Qn, Pn }, in which the Q and the P can both be functions of
6
Transition from Classical to Quantum Mechanics
all the q and the p, and possibly of t. If the Hamilton equations, Eq.(1.32)Eq.(1.34), remain valid in terms of the new variables (i.e., the transformation. is such that these equations, with qn and Pn replaced respectively by Qn and Pn , are true), then we speak of a canonical transformation. The new variables are just as acceptable as the old ones. Pn is called the momentum canonically conjugate to Qn.
1.3
Poisson Brackets
Now we introduce the Poisson bracket of two arbitrary functions of the canonical variables, say F and G: (1.35) The time-derivative of the arbitrary function, F, can be written
dF
dt
(1.36) The total time-derivative of a function that does not depend explicitly on the time is given by the Poisson bracket of that function with the Hamiltonian: we say that the Hamiltonian generates time-translations. In particular, qn = {qn,H},
(1.37) . These equations are impressively symmetrical between the q's and the p's, and they replace the first two of the Hamilton equations, (1.32) and (1.33). It is easy to see that {qm,Pn} = 8mn ,
(1.38)
where the Kronecker delta, 8mn , is defined to be 1 if m = n, and a otherwise. Dirac's recipe for the intuitive leap from classical to quantum mechanics is to represent dynamical variables by linear operators that act on a suitable space of vectors, and Poisson brackets by commutators (multiplied by a constant).
Poisson Brackets
7
With this interpretation, the dynamical equations (1.37), as well the condition £q,(1.SS), have the same form in classical and in quantum mechanics. The reagonableness of Dirac's recipe stems from the fact that both Poisson brackets and commutators satisfy the same algebra: whether [B, C] stands for a Poisson bracket of two functions of classical canonical coordinates, or the commutator, BC - CB, of two operators acting on a vector space, the following two identities hold:
[B, C] = -[C, B] , [A, [B, C]l
(1.39)
+ [B, [C, All + [C, [A, Bll =
(1.40)
0,
which are respectively called the antisymmetry condition and the Jacobi identity. It is obvious that the antisymmetry condition is satisfied by both commutators and Poisson brackets, and it is easy to see that the Jacobi identity is satisfied by commutators:
[A,[B, CJl
ABC-ACB-BCA+CBA
[B,[C, All
BCA - BAC - CAB + ACB
[C,[A, Bll
CAB - CBA - ABC +BAC.
(1.41)
On adding these three right sides together, we see that the six terms with a plus sign cancel against the six terms with a minus sign. However, it is not obvious that the Poisson brackets satisfy the Jacobi identity, so we shall give here a detailed proof of this fact. Proof of the Jacobi Identity Eq.(l.40) for Poisson Brackets Let (k be one of the qi or one of the Pi. Then
8 8(k {B, C}
8 B 8C -----2
~~~
8 2 B 8C
- +8B -
~~~
{~:. ' c } + { B, ~~ }
82C
~~~~
.
- 8B - -8-C2
~~~
(1.42)
We may write (1.43) where DB is the linear differential operator DB
=
8B 8 8B 8 8qi 8Pi - 8Pi 8qi -
8 ~ (3k 8(k .
(1.44)
8
Transition from Classical to Quantum Mechanics
Here ~k runs over the q's and the p's. The important point is that the functions 13k depend on B, but not on C. Similarly, (1.45) where (1.46) with Ok independent of C. Consider
{A, {B, C}} (1.47)
Clearly
(1.48) since the second-order derivatives cancel. This can be rewritten in terms of the qi and the Pi in the form (1.49) once more with implicit summation. Since we have proved that the functions 'Yk and 6k depend only on A andB, but not on 0, we can evaluate them by substituting simple functions for C on both sides of Eq.(1.49). With C = qi, we find "Yi = - { A, -BB}
BPi
B + {BA} B, = --{A,B}, BPi
BPi
(1.50)
where we have used Eq.(1.39) and Eq.(1.42). Similarly, with C = Pi, we have (1.51 )
Hilbert Space
9
On substituting Eq.(1.50) and Eq.(1.51) into Eq.(1.49), we have, for general C,
{A, {B, C}}
{B, {A, C}} -
8 } 8C { 8Pk {A,B} 8qk
+
{8
8qk
{A,B}
-{C,{A,B}}.
} 8C
8Pk
(1.52)
Finally, with use of Eq.(1.39), we can write the Jacobi Identity for Classical Poisson Brackets in the form
{A, {B,C}}
+ {B, {C, A}} + {C,{A,B}} = O.
(1.53)
END OF PROOF
1.4
Hilbert Space
,;Ve have already indicated that part of Dirac's recipe for making the transition from classical to quantum mechanics consists in representing dynamical variables by linear operators that act on a suitable space of vectors. In fact, such a space can be defined quite abstractly. A linear vector space, 1£, is a set of elements,1/;, 4>, ... , called vectors, such that if1/; E 1£ and 4> E 1£, then for any complex numbers e and d, e'l/J + d4> E 1£. Note that, by choosing 4> = 1/; and e = -d, we see that the null vector, usually just written 0, necessarily belongs to 1£. A scalar product of two vectors, S(1/;,4», is a complex number that satisfies the following axioms:
(1) [S( 'l/J, 4»]* = S( 4>, 1/;) (2) S( 'l/J, e4> + dX) = eS(1/;,4» + dS( 'l/J, X) (3) (a) S(4), 4» > 0 if 4> -I- 0 (b) S(4), 4» = 0 if 4> = O. An example of a linear vector space is the set of alII x n dimensional matrices of the form
(1.54)
10
Transition from Classical to Quantum Mechanics
In matrix algebra nomenclature, such column matrices are indeed usually called vectors. A scalar product on this space is
(1.55)
with implicit summation over i. . A linear functional :F on a vector space is a mapping from the space onto the complex numbers: that is, for any ¢i E 1i, we have F(¢i) E C, where C is the set of all complex numbers, and moreover (1.56) An example of such a linear functional can be constructed, given a fixed vector, 'ljJ, as in Eq.(1.55). Let us rewrite it
The point of putting the subscript 'ljJ on F 7jJ (¢) is to stress the idea that 'ljJ is to be thought of as a fixed parameter that specifies the functional :F7jJ, whilst ¢ is a variable that ranges over the whole vector space. It is obvious that this functional is linear, i.e., it satisfies Eq.(1.56). For every vector 'ljJ E 1i there is a corresponding linear functional F7jJ on 1i. The converse is also true: for every linear functional :F7jJ, there exists a vector 'ljJ E 1i, such that S( 'ljJ, ¢) = F7jJ (¢) this is called the Riesz Theorem.
Proof of the Riesz Theorem Given a linear functional F7jJ and an orthonormal basis, {en}, for which S ( em, en) = 8mn , construct the vector
n
cf., Appendix A.4. This does the job, for
n
Hilbert Space
11
so S(7/;,
n
Now since {em} is a basis,
can be expanded on it, say
and therefore S('lj;,
~F~ (8 (en,~>mem) en) L L amF1j;(S(e em)e n ,
m
n
m
n
n
n )
n
END OF PROOF
Dirac introduced a way of distinguishing notationally between column vectors like Eq.(1.54), which he called kets and wrote 1
S( 7/;,
or more briefly :F1j; = ('lj; I. The dual of 1£ is the space of all linear functionals on 1£, which, because of the Riesz theorem, is isomorphic to 1£ itself. Indeed for every F1j; ('lj;1 belonging to the dual space, there is a corresponding ket in 1£ that satisfies
=
12
Transition from Classical to Quantum Mechanics
A linear vector space, equipped as above with a scalar product, is called a Hilbert space if it is complete. A complete space is one that contains all of its limit points. More technically, a Cauchy sequence of vectors, l4>i), is a sequence for which, for any E > 0, :3p :
II ¢i -
4>i II <
E,
Vi > p and j > p,
(1.57)
where the norm here is defined by
II ¢ 11=
J (4)14>) .
If all Cauchy sequences in a space 1£ are convergent, then there exists for any sequence satisfying Eq.(1.57) a vector 14» E 1£ such that 3p:
II 4>i - ¢ 11<
E,
Vi
> p.
The space is then complete and as such is a Hilbert space. This is automatically the case for the finite dimensional spaces that we have been visualizing so far, as in Eq.(1.54); but there is no need to limit our considerations to a finite number of dimensions - indeed it is essential in quantum mechanics often to work with an infinite number of dimensions. If such an infinite-dimensional normed space is not complete it it is called a prehilbert space, but it may be converted into a genuine Hilbert space by adjoining all of its limit points. An linear operator, A, is defined to be a mapping of a vector space (or part of it, the domain of the operator) into the vector space (or part of it, the range of the operator). Thus if'lj; is a vector in the domain of the operator A, then A'lj; is a vector in its range. The Hermitian conjugate of A, namely At, is defined by
S( ¢, A'lj;)
[S('lj;,At¢)r S(At4>, 'lj;),
(1.58)
for all 'lj; in the domain, and all ¢ in the range of A. It follows that (1.59)
Proof of Eq.{1.59) From the definition of the Hermitian conjugate, for suitable ¢ and 'lj;, S( 4>, AB'lj;) = S( (AB) t ¢, 'lj;) .
(1.60)
However, since B'lj; is itself a vector, we also have
S( ¢, AB'lj;) = S( At ¢, B'lj;) ,
(1.61 )
Dirac's Transition to Quantum Mechanics
13
and since At ¢ is also a vector in the space, (1.62) On combining Eq.(1.60) and Eq.(1.62), we conclude that (AB)t
= BtAt.
END OF PROOF
An operator that is equal to its Hermitian conjugate, A = At, is called a Hermitian operator.
1.5
Dirac's Transition to Quantum Mechanics
Following Dirac's lead, we replace classical observables by Hermitian operators that act on a Hilbert space (but see the reservations made in Sect. A6), and Poisson brackets by commutators, times a constant. In particular, the classical relation Eq.(1.38) is to be replaced by the quantum relation (1.63) where qm is the operator corresponding to the mth Cartesian component of the position of the particle, Pn to the nth component of its momentum. In Eq.(1.63), /\, is some universal constant to be determined. Since the Hermitian conjugate of a product of operators, AB, is Bt At, and all operators representing 0 bservables are Hermitian, it follows that the Hermitian conjugate of this is (1.64) so that /\, = -/\,* i.e., /\, must be purely imaginary. It is conventionally set equal to l/(iTi), so that Eq.(1.63) can be written in the standard form (1.65) where Ti is a real, universal constant that has to be determined by experiment. It is in fact equal to where h is Planck's constant, introduced in the year 1900 to avoid the ultraviolet catastrophe of classical radiation theory. We begin with the motion of a single particle in one dimension, described by a pair of canonical variables, the position and the momentum, satisfying
2:'
qp - pq
= iTi.
(1.66)
Let Ix) be an eigenvector of q corresponding to eigenvalue x,
qlx) = xix).
(1.67)
14
Transition from Classical to Quantum Mechanics
The Hermitian conjugate of this relation is
(xlq = x* (xl.
(1.68)
It must be stressed that Ix) is not allowed to be the null vector in Hilbert space,
for the above equations would then be trivial. According to the third axiom of scalar products in a Hilbert space (see above), we know that
(xix) > 0,
(1.69)
the value zero being excluded. By using Eq.(1.67) and Eq.(1.68) successively, we see that (xlqlx) = x(xlx) = x* (xix) and so, because of Eq.(1.69), we conclude that x = x*, i.e., x is real. This proof works in general: the eigenvalues of any Hermitian operator are all real, and this is one of the reasons that physical observables are required to be represented by Hermitian operators. Suppose now that y ¥- x, and (1.70) We know that y is real, and clearly
(xlqly) = x(xly) = y(xly) ,
(1. 71)
(x - y)(xly) = 0,
(1. 72)
which means that
and that implies (xly) = 0, i.e., eigenvectors corresponding to different eigenvalues are mutually orthogonal. This is again a general result: eigenvectors of a Hermitian operator belonging to different eigenvalues are mutually orthogonal. Let us solve Eq.(1. 72) as an equation for x, at a fixed value of y. In a space of continuous functions, the only possibility for (xIY) would be zero, since it is zero for all x ¥- y; but this would mean (xix) = 0 and that implies that Ix) is the null vector, which we have excluded. To find a nontrivial solution, we have to consider generalized functions, usually called distributions, to which we now turn.
1.6
Dirac Delta Function
In this section we will develop the theory of the Dirac delta function, or more properly the delta distribution, along the lines of the distribution theory of Laurent Schwartz. This abstract approach is not difficult, and it is much more satisfactory than the conventional expositions given in many quantum mechanics books.
15
Dirac Delta Function
A functional is a mapping from a function space onto C, the set of complex numbers. For example, the integral
i:
=G
dxg(x)
is a representative of the mapping of a space of functions, g, onto their integrals, taken over the real line. Of course, this only makes sense if 9 is integrable. A space that we will use below is that of all infinitely differentiable functions of compact support; that is, all functions, g( x), that can be differentiated an arbitrary number of times, and are such that, for each g(x), there is an X min and an X max , such that
g(x) = 0
unless
xmin
<
X
<
X max •
The support [x min , x max ] will not be the same for each g(x), and in general, for any x there will be some g's that do not vanish there. For this space, the Schwartz space, S, the above integral is well-defined, and so the functional is also defined. The distribution 8 is defined, with respect to the space S, to be the functional that maps any function g(x) E S onto the number g(O), i.e., onto its value at x = O. That is all there is to it! Here S is an example of a space of test functions: one is free to choose other spaces, for example the space of all functions that are merely continuous, on which the definition of the mapping
8:
g(x) -----t g(O)
(1.73)
still makes good sense. A distribution like 8 is defined in terms of a space of test functions. Change that space and you change the distribution! The mapping (1. 73) is usually expressed
i:
dx g(x)8(x)
= g(O) ,
(1.74)
where 8(x) is written as if it were a function with argument x: but this is emphatically not the case! In order for an integrable function to be able to . -. . 'sample" tne test IUnCtIOn, g~X), at JUSt one POInt, x = U, It WOUlO nave to vamsn at all points x i= 0, but be such that
L:ox
dx8(x)
=1
for all Xmin < 0 and Xmax > O. Intuitively, if 8(x) were a function, it would have to be infinitely sharply peaked at x = O. There are no such functions! That, however, is no problem: Eq.(1.74) is merely a conventional way of writing Eq.(1.73), no more and certainly no less.
Transition from Classical to Quantum Mechanics
16
Making the change of variables
x -------+ x - y in Eq.(1.74), we obtain
i:
g(x) = g(x - y)
dx g(x)8(x - y)
g(y)
= g(O)
= g(y) .
(1.75)
Since g{x) E S => g(x) E S, we can effectively drop the caret in Eq.{1.75): the space of all g's is the same as the space of all g's, and 8(x - y) is the distribution that maps a function g(x) onto the value g(y). Note that 8 (x) is even. We can prove that by making the change of variables
x -------+ - x
g( -x) = g{x)
g(O)
= g(O)
in Eq.(1.74). This gives
[ : dxg(x)o(-x)
= g(O).
(1.76)
The distribution 8( -x) yields exactly the same result as 8(x) itself; since a distribution is defined as a mapping, and the mappings are the same, it follows that
8( -x)
= 8(x).
We can therefore also write
[ : dx g(x)o(y - x)
= g(y).
(1.77)
In order to obtain a useful representation for the delta function, we recall that the complex Fourier transform of g(x) is
g(k) = [ : dx eikxg(x) ,
(1. 78)
the inverse being 1 g(y) = -2 7r
1
00
.
. dk e-tkYg(k).
(1.79)
-00
These formulae are certainly valid if g(x) E S, indeed, it is enough for g{x) to be a piecewise continuous function, according to the theory of Fourier analysis.
17
Dirac Delta Function
Substituting the first equation above into the second, we obtain ~ 1
g(y)
271"
1
dk e- tky .
00
-00
1
00
dx e'tkxg(x) .
-00
1
00 -1 dxg(x) 271" ~oo
1
dk etk(x-y). .
00
(1.80)
-00
On comparing this with Eq.(1.75), we see that
~ foo dk eik(x-y) . 271" J-oo
8(x - y) =
(1.81)
It should be clear that this integral does not make sense as an integral of a function, yielding another function: it has to be understood in the sense of distribution theory, that is, it is merely a shorthand way of expressing Eq.(1. 78) and Eq.(1. 79). Of course, an intuitive, gut-feeling interpretation of Eq.(1.81) is possible: it says that, unless x = y, the integral (1.81) oscillates itself to death! When x = y the integral gives infinity, so what could it be but the Dirac delta function!? The interchange in integration orders that we perpetrated in Eq.(1.80) is strictly a formality: the golden rule in distribution theory is that formulae are only to be understood in the usual sense as functions after multiplication by a test junction, followed by integration. On taking the derivative of Eq.(1.79), we find
.1
,
~
and hence -g' (0)
2~
.
00
9 (y) = - 271"
dkke-tkYg(k);
(1.82)
-00
I:
dk kg(k)
~ foo dk k foo dx eikxg(x) 271" J -00 J-oo -1 271"
1
I:
00
dxg(x)
-00
1
00
-00
a eikx ax
dk -
dx g(x)8'(x)
where
8'(x)=~!£ foo 271" dx
J
-00
dkeikX=!£8(x). dx
(1.83)
18
Transition from Classical to Quantum Mechanics
In other words, the distribution 8' is the following mapping:
8' :
g(x)
~
(1.84)
-g'(O).
Higher derivatives of the delta function can be similarly defined. The Heaviside step function may be defined by
O(x) =
i~ dyJ(y).
As a (discontinuous) function, it is clear that, for x < 0, 8(x) = 0, and for x > 0, 8(x) = 1, whereas it is not defined - as a function - for x = O. Considering the step function rather as a distribution, we have
i:
i:
dx g(x)8(x)
10
00
dyJ(y)
{O
dx g(x)
dx g(x) .
For any test function with support in (-00,0), this yields zero. The derivative of the step-function is the delta-function: d
dx 8(x) = 8(x). That this formal manipulation yields a correct result can be checked by playing with test functions:
roo
d
- J-oo dx8(x) dxg(x)
10o
d
00
-
dx -g(x) dx
g(O) , which effects the required proof. Note that, in the partial integration above, the boundary terms have been left out, since the test functions, g( x) E S, being of compact support, vanish at infinity. The same proof applies for any space of differentiable test functions that tend to zero as x ~ ±oo. In many elementary treatments, the delta function is introduced as a 'limit', such as
8(x) = lim
1 r::;;:;.
e--+O V 1T€
exp
(x2) - €
.
(1.85)
19
Momentum Operator
This does not exist as a limit in the usual sense: we are told to use it only 'under an integral'. In the language of test functions, we write lim
e~O
y
1 r::;;:;: 1f'E:
1
00
_
dx 9 (x) exp ( - -x2)
00
lim 1'-
e~O y7r
g(O)
1
00
c
dy g(y..;€) exp (_y2)
-00
roo
vii J
-00
dyexp (_y2)
g(O). Notice that taking the limit c -+ 0 in g(yye) to get g(O) is perfectly legitimate, since the test function has compact support. Two other representations for the delta function are
8(x)
=
c 2 + c
(1.86)
8(x) = lim ..!:. sinNx N~oo 7r x
(1.87)
lim..!:.
e~O 7r X
2
and
as can be easily checked by means of test function manipulations.
1.7
Momentum Operator
In the space of tempered distributions, a solution of Eq.(1. 72) is
(xJy) = 8(x - y),
(1.88)
or indeed any multiple of it. The above amounts to a choice of (continuum) normalization. For let g( x) be a continuous test function. Then
L:
dxg(a:)(a: - y)J(a: - y)
= I(a: - y)g(a:)l.~y = o.
As a distribution on this space of test functions, (x - y)8(x - y) = 0, which means that Eq.(1.88) satisfies Eq.(1.72). From Eq.(1.66), Eq.(1.71) and Eq.(1.88), we deduce
(x - y)(xJpJy) = (xJqp - pqJy) = i1i(xJy) = i1i8(x - y).
(1.89)
Transition from Classical to Quantum Mechanics
20
A solution of this equation, in the space of tempered distributions, is (1.90) Proof that Eq.(1.90) satisfies Eq.(1.89) Let g(x) be a test function belonging to the Schwartz space, S. Then
!
dxg(x)(x - y) ~ 8(x - y)
-! !
dX 9 (x)(x- y ):x 8(x- y )
dx8(x - y) :x [g(x)(x - y)]
[g'(x)(x - y)
+ g(x)]x=y
= g(y).
(1.91)
So (x - y) ~8(x - y) amounts to the same mapping as 8(x - y), i.e., it is the same distribution. END OF PROOF
A physical system is described by a state vector I,¢). This state vector is abstract, in the sense that it is an element of a linear vector space - the same Hilbert space that contains the eigenvectors Ix) of q. The wave function of the state in the configuration representation is defined to be
,¢(x) = (xl'¢) . The wave function is a complex number. The probability of finding the particle in the interval a < x < b, given that the state vector is I'¢), is postulated to be dx 1,p(x)1 2 .
[
Consider the space spanned by the eigenvectors Ix) of q. That is, consider the set of all vectors that can be written in the form
I¢) =
!
dy ¢(y)IY) ,
where ¢(y) is a complex number. Then Eq.(1.88) implies
(xl¢) =
!
dy¢(y)8(y - x) = ¢(x) .
(1.92)
21
Momentum Operator
Inserting this into Eq. (1. 92), we find
14» = and since
1
dyly) (yl4» ;
14» is an arbitrary vector in the space, we may express this in the form
1
(1.93)
dyly)(yl = 1,
where the right side here means the unit operator on the space spanned by the eigenvectors Ix) . Consider the matrix element
(xlpl1fJ)
1 in 1 (~ -in 1
dy (xlpIY) (yl1fJ)
=
dy
o(x - y)) (yl,p)
dyJ(x - y)
(~ (YI1f1))
-in~ (xl1fJ) 8x (1.94)
-in :x 1fJ(x) ,
where Eq.(1.90) has been used. The result Eq.(1.94) is important: it is sometimes expressed by saying that the operator p is represented by -in More accurately, we can say that, in the configuration representation, 11fJ) is realized by the differentiable function 1fJ (x), the position operator q by the real number x, and the momentum operator p by the differential operator -in We now construct the vector
tx'
tx .
Ip) =
_1_1 21f'n
dyly) eipyln.
(1.95)
From Eq.(1.90) we see that
(xlplp)
_1_/
dy (xlply)eiPyln
- in
8 dye lPY In -c>(x - y)
21f'n
1 . 1 -'--I 21f'n
- in 21f'n
27rn
8y
8 elPY . In dyc>(x - y)8y
dy (xly) eipyln = p(xlp)·
(1.96)
22
Transition from Classical to Quantum Mechanics
Using Eq.(1.93) and restricting p to the space spanned by the vectors Ix), we have
pip) = pip), i.e., the vector defined in Eq.(1.95) is an eigenvector of the momentum operator p with eigenvalue p. Clearly
(xlp) = _1_ / dy (xly)e iPY / Ti = _1_ 21('n
21('n
J
dy 8(x - y) eipy / Ti = _1_eiPX/Ti, (1.97) 21('n
which indicates that the transformation from Ix) to Ip) is precisely a Fourier transformation.
1.8
The Hamiltonian
Let I¢(t)) be an arbitrary, normalized state vector of a dynamical system. As a result of the dynamics, this vector will in general change with time. The dynamics implies the existence of an evolution operator, U(t), such that
I¢(t))
= U(t)I¢(O)) .
Since Il/J (t)) is normalized at all times,
(¢(O)lut(t)U(t)I¢(O)) = (¢(t)I¢(t))
=
1 = (¢(O)I¢(O)).
This must hold for any I¢ (0) ), so we must have
ut (t)U(t) Thus ut(t)I¢(t))
= 1.
(1.98)
= I¢(O)), and so
(¢ ( t) IU (t) ut (t ) I¢ (t )) = (¢ (0) I¢ (0)) = 1 = (¢ (t) I¢ (t)) ,
and this implies
U(t)ut (t)
= 1.
(1.99)
We express Eq.(1.98)-(1.99) by saying that U is a unitary operator. The generator of time translations, H, is closely connected to U. It satisfies
HI¢(t)) = in
!
I¢(t))
for any state vector I¢(t)). It follows that
HU(t)I¢(O))
=
inU(t)I¢(O)) ,
(1.100)
23
The Hamiltonian
and since this must hold for any state vector I'l/J(O)), we conclude that
HU(t)
=
inU(t).
Because of the unitarity of the evolution operator, this implies
H
=
inU(t)U t (t) ,
and hence
However, since
uut =
1 at all times, we see that
U(t)Ut(t) + U(t)ut(t) = 0, so that the time-translation operator is Hermitian:
(1.101) We shall now show why the operator H, the generator of time-translations, is to be identified as the quantum version of the Hamiltonian. To do this we observe that for any linear operator, p (such as the momentum for instance), we have
(1/1(t) Ipl1/1(t))
('l/J(O) Iut (t)pU( t) I'l/J(O)) ('l/J(O) IPH I'l/J(O)) ,
(1.102)
where PH =
Ut(t)pU(t)
is called the Heisenberg picture version of the operator p. The Heisenberg picture version of the vector 11/1(t)) is simply 11/1(0)). In the Heisenberg picture, all the time-dependence is carried by the operators, and none by the state vectors. It is in this picture that H can be identified as the Hamiltonian. If p does not depend on time, then the time-dependence of PHis wholly borne by the evolution operators:
Since U = -iHUIn, it follows that ut Hermitian. Hence
= iUt H In,
for we have shown that H is
24
Transition from Classical to Quantum Mechanics
Here PH and H H are the Heisenberg picture versions of P and H. Recalling that the commutator divided by in is the quantum analogue of the classical Poisson bracket, we observe that this equation has exactly the form of Hamilton's equations of motion, Eq.(1.37), on condition that we identify H as the quantum mechanical Hamiltonian. In an isolated classical system, the Hamiltonian is time-independent and is equal to the total energy of the system. In quantum theory we suppose also that in an isolated system the Hamiltonian operator H is time-independent, and that its eigenvalues correspond to the possible energies of the system. The fact that H is Hermitian guarantees the reality of these energies.
1. 9
Schrodinger Equation
The generalization of the above to the motion of a particle in 3 dimensions is straightforward. Let qi and Pi be respectively the i th Cartesian components of the position and momentum operators. Then the eigenvector Ix) = IXl)lx2)lx3) of qi, corresponding to the eigenvalue Xi, satisfies (1.103) and the quantization condition is (1.104) A solution of this equation is (1.105) For an arbitrary state vector I
Hence / d3 y (XIPily) (YI'lj1(t))
-in 8~i (xl'lj1(t)) -in 88 'Ij1( t, x) , Xi
(1.106)
25
Schrodinger Equation
where 11jJ(t)) is a state vector describing a physical system at time t. Accordingly, we can say that, in the configuration representation, Pi is represented by the operator -in a~i ' or equivalently
P -t
-ili'~ .
(1.107)
In an isolated classical system, in which the Lagrangian is not an explicit function of the time, the Hamiltonian is equal to the total energy of the system, which is time-independent. In the nonrelativistic mechanics of a particle of mass m, we have
H = p .p 2m
+ V(X') ,
where Vex) is the (conservative) potential. In making the transition to quantum mechanics, we simply replace the Hamiltonian, the momentum and the position by the corresponding linear operators that have these quantities as their eigenvalues, 2
H = ~ 2m
+ V(-+) q,
(1.108)
where p2 = P . p. A physical system of well-defined energy is described by a state vector, /1jJ(t)), that is an eigenvector of the Hamiltonian: d
in dt /1jJ(t))
=
H/1jJ(t))
=
E/1jJ(t)) ,
(1.109)
where E is the total energy of the system. Hence 2
E1jJ(t,x)
= (x/HI1jJ(t)) = (X'I~ + V(q)/1jJ(t)). 2m
(1.110)
To evaluate the right side, note that
!
d 3 y (x IPi 117)(17 /Pi 11jJ( t))
n 2 Jd3y
[~83(x - 17)] ~1jJ(t,17) 8Yi 8Yi
_n 2 V 2 1jJ( t, x) ,
(1.111)
where use has been made of Eq.(1.105)-Eq.(1.106). Since the eigenvalue of qi belonging to (X' / is Xi, and the eigenvalue of any power qi n is xi, the eigenvalue of V(q) is Vex) (at any rate for a Vex) that is Fourier transformable). Hence
Transition from Classical to Quantum Mechanics
26
from Eq.(1.110) we find
in
:t
'ljJ(t, x)
=
E'ljJ(t, x) = -
;~ \l2'ljJ(t, x) + V(x)'ljJ(t, x).
(1.112)
This is the time-dependent Schrodinger equation. Since E is time-independent under the conditions stipulated above, we can solve Eq.(1.109):
-iEt) 1'ljJ(O)). 1'ljJ( t)) = exp ( -nHence
¢(t, i)
(i I¢(t)} = exp (
-~Et) ¢(i) ,
where
'ljJ(x) = 'ljJ(O,x).
(1.113)
Thus the time dependence can be factored out of Eq.(1.112), yielding
E¢(i)
= [- :~ V2 + V(i)] ¢(i),
(1.114)
which is called Schrodinger's time-independent equation.
1.10
Dirac's Views
To conclude this chapter on Dirac's way of making the transition from classical to quantum mechanics, we cite the master himself: We had these equations involving these noncommuting quantities, but to begin with, we had no general interpretation for the equations. That was really a remarkable situation to have in a physical theory. (In any physical theory one usually knows just what one's equations mean before one sets them up.) It was found that this equation gave the particle a spin of haIfa quantum. And also gave it a magnetic moment. It gave just the properties that one needed for an electron. That was really an unexpected bonus for me, completely unexpected. *
*P.A.M. Dirac, 'Directions in Physics', Wiley (1978)
Exercises
1.11
27
Exercises
Problem 1 From the axioms of the theory of linear vector spaces prove: (1) Any linear operator can be written in the form A + iB, where A and B are Hermitian operators. (2) If A and B are Hermitian operators, so is irA, B]. (3) For any vector ¢, S( ¢, At A¢) > o.
Problem 2 Suppose that {en} is an orthonormal basis of a Hilbert space, 1£, and let A be a linear operator on 1£. The trace of A with respect to the basis is defined by
n
(1) Demonstrate that TrA is independent of the choice of the orthonormal basis. (2) If A and B are linear operators on 1£, show that
S(A, B) = Tr(At B) satisfies the axioms of the scalar product.
Problem 3 Suppose that A and B are linear operators on a Hilbert space, 1£, and that (1jJIAI1jJ) = (1jJIBI1jJ) for all1jJ E 1£. Show that necessarily (¢IAI1jJ) = (¢IBI1jJ) for all ¢ E 1£ and 1jJ E 1£. Under these conditions, i.e., when all matrix elements of A and B are equal, we write A = B as an identity of two operators on a given Hilbert space. Problem 4 Consider the following spaces of functions of all1jJ, such that, respectively
H.
o
s
I: I:
dx 11/I(x)12
< 00
dx 11jJ(x)12[1 + Ixln] <
i:
dx1jJ*(x)¢(x) <
00,
00,
for n = 0,1,2, ....
forall¢ E O.
28
Transition from Classical to Quantum Mechanics
Show that
(1) n c 1£ c B. (2) x rf. 1£. oslx. £~ (3) .. iRa; C 0, (4) x 2 cos X (j. 1£ , (5) e- x (j. S.
n
sinx d x ~,
sinx x
E
1£.
x 2 cos x E B.
Problem 5 By introducing a suitable space of test functions, justify the following representations of the Dirac delta distribution: .
1
(a)
8(x) = hm -
(b)
8(x)
~-)-o
1T" X
N
+ €2
.!.. sinNx
lim
=
€
2
-)-00
1T"
X
Problem 6 Let G be the space of complex differentiable test functions, g(x), of compact support, where x is real. It is convenient to extendG slightly to encompass all functions, g, such that g(x) = g(x) + c, where 9 E G and c is any constant. Let us call the extended space G. Let q and p be linear operators on G such that
qg(x)
xg(x) ,
pg(x)
-i-g(x) dx
d
Suppose M to be a linear operator on that
=
G that
-ig'(x). commutes with q and p. Show
(1) q and p are Hermitian on G. (2) M is a constant multiple of the identity operator. Problem 7 Given that [q, p]
= in,
show that
(1) [q-l, p] = -inq-2 (2) [qn, p] = innqn-l n = ±1, ±2, ±3, ... (3) [exp(itq) , p] = -nt exp(itq) . 1 3 (4) [q-2, p] = - ~inq-2 (5) [q ~ ,p] = ~i1iq- ~ Problem 8 In one space dimension, consider the classical Poisson bracket of the square of the coordinate and the square of the conjugate momentum, {q2, p2}.
29
Exercises
(1) Show that the quantum mechanical commutator, [q2, p2], divided by in, does not correspond to that Poisson bracket. (2) How must one rewrite the classical Poisson bracket so that the commutator does so correspond?
Problem 9 Suppose that 'If; (t, i) satisfies the time-dependent Schrodinger equation, with a real, time-independent potential. Define the probability density by
P(t,i) = 1'lf;(t,i)1 2 • (1) Show that
!
pet, x)
+ V . jet, x)
= 0,
where the probability current is defined by
in ,,1.*( +-+ ,,1.( -) J-( t, x-) = - 2m 'f/ t, x-) V 'f/ t, x . +-+
Discover thereby the meaning of the symbol V. (2) Show that, for a square integrable 'If;(t, x),
!I
d3 xP(t,x) = 0,
the integral being over all space. Give a physical interpretation of this result. (3) Evaluate next
! Iv
d3 xP(t,x) ,
in terms of j(t,x), the integral now being over a finite volume, V. Give a physical interpretation of this result.
Problem 10 Two photons fly apart from one another, and are in oppositely oriented circularly polarized states. One strikes a polaroid film with axis parallel to the unit vector ii, the other a polaroid with axis parallel to the unit vector b. Let P++(a, b) be the joint probability that both photons are transmitted through their respective polaroids. Similarly P __ (ii, b) is the probability that both photons are absorbed by the polaroids, P + _ (a, b) is the probability that the photon at the apolaroid is transmitted, while the other is absorbed, and finally P _+ (ii, b) is the probability that the photon at the a polaroid is absorbed, while the other is transmitted.
Transition from Classical to Quantum Mechanics
30
The classical realist assumption is that these probabilities can be separated:
where i and j take on the values + and -, where A signifies the so-called hidden variables, and where peA) is a weight function. The correlation coefficient is defined by
C(a, b) = P++(a, b)
+ P__ (a, b) -
and so we can write
C(a, b) =
P+_(a, b) - P_+(a, b) ,
J
dAp(A)C(a, A)C(b, A) ,
where
and
It is required that
peA) > 0
(a) (b)
J
(c)
-1 < C (
dAp(A) = 1
a, A) ::; 1
and
- 1 < C (b , A) < 1.
The Bell coefficient,
B
= C(il, b) + C(a, b') + C(il', b) -
C(il', b'),
combines four different combinations of polaroid directions. Show the following: (1) The above classical realist assumptions imply (2) Quantum mechanics predicts
..
IBI < 2.
( . . )2 -1.
C(a,b)=2 il·b
(3) The maximum value of the Bell coefficient is 2V2, according to quantum mechanics. (4) The quantum mechanical expression for C(il, b) can be cast into separable form. Which of the classical requirements, (a), (b) or (c) above is violated?
Chapter 2
Three-Dimensional Harmonic Oscillator
VCr),
Consider a particle in a potential field, in three dimensions, for example an atom in a crystal. The potential will typically come from the mean effect of the attractions and repulsions of ail the other atoms, and it will have a minimum, which we can for convenience take to be the origin of coordinates, r = O. The equilibrium position of our atom is then the origin; but in general it will have thermal energy and so it will oscillate about this position. If the potential is analytic near r = 0, we can make a Taylor expansion for small r = (XI, X2, X3): ->
_
V(r) - V(O)
+L 3
j=l
[av] ax' J
+2L 3
1
Xj
0
L av ] 3
j=l k=l
[
2 ax .aXk J
XjXk
+ ....
0
Now the condition for a local minimum means that the first-order partial derivatives vanish; and the constant, V(O), does not contribute to the forces acting on the particle, so one can freely set it to zero. As to the second-order terms, by making a suitable rotation of the coordinate axes, (2.1) we obtain
VCr) = ~ t [a::] x; + .... j==l
8X j
0
i.e., the crossed terms have disappeared inthe new frame of reference. The x, y, Z coordinates are called the principal axes. A condition for a local minimum, i.e., a stable equilibrium, is the positivity of these second-order derivatives. An explicit construction of the canonical transformation Eq.(2.1) can be found in Appendix A.4, where it is proved that any Hermitian matrix can be diagonalized by a suitable unitary transformation. In our case the matrix 31
Three-Dimensional Harmonic Oscillator
32
2
: [ 88 x]
] Xk
0
is a real, symmetric matrix, which is a special case of a Hermitian
matrix, so the appropriate matrix, Ujk , that diagonalizes it is unitary and real, i.e., it is an orthogonal matrix.
2.1
Creation and Annihilation Operators
We now drop the bars - or equivalently suppose that we had already chosen the principal axes as our coordinates - and we write our potential in the form 3
V(i) =
L AjXJ ' j=1
where the Aj are positive constants. The quantum mechanical Hamiltonian is
H= T+ V= t [2~P; + AjqJ] , )=1
where m is the mass of the particle, and qj and Pj are the Cartesian components of the Hermitian operators whose eigenvalues are respectively the position and momentum of the particle. This system is called a (three-dimensional) harmonic oscillator; and, as we have shown, it arises very naturally. The quantization conditions are l.
[qj, qk]
o
2.
(Pj,Pk]
o
3.
[qj, Pk]
(2.2)
from which all the quantum mechanics of the system can be deduced. Note that we no longer distinguish quantum operators by bold-face letters, as we did in the previous chapter. We first rescale the coordinates:
(2.3) Then
(2.4)
Creation and Annihilation Operators
33
and any two Q's commute with each other, and similarly any two P's. Further 3
H = ~
L nWj [pl + Q;] ,
(2.5)
j=l
where
The next step is to introduce the lowering and raising operators:
(2.6) which are sometimes called annihilation and creation operators, respectively. From Eq.(2.4) it follows that l.
[aj, ak]
2.
[a},al]
3.
[aj, at]
o o (2.7)
and from Eq.(2.5) that 3
H =
L nWj[Nj + ~], j=l
where the occupation number operator is defined by N J· -- ata· j J.
(2.8)
The occupation number operators have the important property that their eigenvalues, say nj, are non-negative integers, 0, 1, 2, 3, ... . This means that the eigenvalues of the Hamiltonian, i.e., the possible energy values, are 3
E(nb n2, n3) =
L 1iwj [nj + ~] .
(2.9)
j=l
The energy is thus discrete, and its lowest value is E(O, 0, 0) = ~1i[Wl
+ W2 + W3] ,
which is called the zero-point or ground-state energy.
(2.10)
34
Three-Dimensional Harmonic Oscillator
Proof that the eigenvalues of N j are real and non-negative Let Inj) be an eigenvector of N j , belonging to the eigenvalue nj, i.e.,
Here Inj) is a vector in a Hilbert space, N j is a linear operator, and nj is a number. We do not exclude the possibility that nj = 0, but we do exclude the possibility that Inj) is the null vector, so (nj Inj) > o. Now NJ
=
[a}aj
r
= a}aj
=Nj ,
and so (njINjlnj)t = (njINjlnj), which implies nj(njlnj) = nj(njlnj). Since (njlnj) # 0, it follows that nj = nj so the eigenvalue nj is real. Moreover, nj(njlnj) = (njINjlnj) = (njla}ajlnj)
=
(¢jl¢j)
where I¢j) = ajlnj). Since Inj) # 0 =* (njlnj) > 0, and since (¢jl¢j) any vector I¢j) in a Hilbert space, it follows that nj > O.
> 0 for
END OF PROOF
From Eq.(2.7) and Eq.(2.8) we have
so that
Hence aj Inj) is an eigenvector of N j belonging to the eigenvalue nj - 1, unless ajlnj) = 0, in which case of course it does not count as an eigenvector. Proof of the existence of discrete states By iterating the above procedure, we see that N·a·jn·) J J J Nj[aj]2Inj) N j [aj]3Inj)
(nj - l)ajlnj)
2) [aj]2Inj) (nj - 3) [aj]3Inj) (nj -
(2.11) Thus we construct eigenvalues that become progressively smaller, and hence eventually, if the above sequence were not to stop, we would obtain negative
Creation and Annihilation Operators
35
eigenvalues, contradicting our proof that all eigenvalues of N j must be nonnegative. Therefore the above sequence must stop, before such negative eigenvalues are produced, and this will happen if
for v
= 0,1,2, ... , vo,
but
for then the smallest eigenvalue produced in the above sequence is nj -
VO:
(2.12) Note that Vo is a non-negative integer,
°
being possible. However,
and since
it follows from Eq.(2.12) that (2.13) that is, the eigenvalue nj of N j is necessarily a non-negative integer. END OF PROOF
Thus we have shown the existence of an eigenvector of N j , say 10 j) , belonging to the eigenvalue 0,
and we will normalize it, (OjIOj) = 1. We have also demonstrated the existence of a ladder of states, Inj), which we shall normalize, and they satisfy
(2.14) Introduce the coefficient enj by (2.15) so
(2.16)
36
Three-Dimensional Harmonic Oscillator
From Eq.(2.14) and the fact that the eigenvectors of N j are normalized, we find ICnj 12 = nj. The phase has no significance, and we choose cnj = yfrfj. Now multiply both sides of Eq.(2.15) by a}, obtaining
a}ajlnj) = y'nja}lnj - 1).
(2.17)
From Eq.(2.14) we have
and by iteration this yields
Inj) = [nj!
r
l/2
[a}t j 10j).
(2.18)
Eigenvectors of the Hermitian operator Nj = a}aj that belong to different eigenvalues are necessarily orthogonal, so we have the orthonormality relation (2.19) Since N I , N2 and N3 commute with one another we can construct a vector
that is simultaneously an eigenvector of Nb N 2 , and N 3 • This state is called the ground state if it is unique (or sometimes the vacuum state); but in fact 10) may not be unique. Our oscillating atom may have internal degrees of freedom, for example spin, and for each spin state we will be able to define a corresponding ground state, say 10, s), where s designates the spin state. So long as there is no coupling between the spin degree of freedom and the mean field in which the atom oscillates, all the states 10, s) have the same ground-state energy Eq.(2.10). Very often there is coupling, and then the degeneracy is lifted; but usually this coupling can be treated as a small perturbation by approximate methods. 2.2
Hermite Polynomials
The scaled position operator, Qj, can be represented by the real variable, Xj, and the momentum operator, Pk, by the differential operator, -i 8~k. This representation respects the (scaled) commutation relations Eq.(2.4). Accordingly, the annihilation and creation operators are represented by a·J
a~ J
(2.20)
37
Hermite Polynomials
and so the representation of aj 10)
[Xi +
0 is
=
a~;] "'0 (X"
= 0,
X2 , X 3 )
where 'l/Jo is called the ground state eigenfunction. Eq.(2.21) (aside from normalization) is
'l/JO(X 1,X2,X3 )
(2.21)
The unique solution of
= exp[-~X;]exp[-~X~]exp[-~X5] = exp[-~R2],
(2.22)
where R2 = X; + Xi + Xi. By allowing creation operators to work on this ground state, we obtain
=
'l/Jnln2 n3(XI,X2 ,X3 )
IT 3
[
a ]nj exp[-~X]J
Xj - ax.
j=l
J
3
II exp[- ~X]JHnj (Xj ).
(2.23)
j=l
Here the Hermite polynomial is defined by
dn
Hn(X) = (-It exp[X2] dXn exp[-X2]
n = 0,1,2, ....
(2.24)
Proof of Eq.(2.23) From the definition Eq.(2.24), we have
dn d exp[-X2]Hn(X) = (-It dXn exp[-X2] = - dX exp[-X 2]Hn_1(X). (2.25) Define
so that
-
d~ exp[-~X2]Un_1(X)
expHX2] [X -
d~] Un-leX),
(2.26)
where we have used Eq.(2.25). We find by iteration
Un (X)
=
[X - d~] Un-leX) [X - d~rUo(X)
(2.27)
Three-Dimensional Harmonic Oscillator
38
Since Uo
= exp[-~X2],
we have verified Eq.(2.23).
END OF PROOF
The Hamiltonian Eq.(2.5) is represented by
(2.28) and so
where E(nI, n2, n3) was given in Eq.(2.9). In terms of the original coordinate, r = (XI, X2, X3), this equation reads
which is precisely the Schrodinger equation for our harmonic potential. We have used the notation 'l/Jii (r) - 'l/Jnln2 n 3 (XI, X 2 , X3). Many treatments of the problem start with this equation and deduce the form of its square-integrable solutions. We have reversed that procedure.
2.3
Isotropic Oscillator
=
=
Let us now specialize to the case that Al = A2 A3 A, so that the angular frequencies in the Xl, X2 and X3 directions are the same:
W,
=W2 =W3 _w=
The normalized eigenstate
belongs to the energy eigenvalue
Jg.
39
Isotropic Oscillator
The ground-state energy is accordingly
Eo
= ~1iw.
If the vacuum-state is unique (i.e., there are no extra degrees of freedom like spin), then the degree of degeneracy of a given energy-level, En=(n+~)1iw,
(2.29)
is the number of ways of arranging a triad of non-negative numbers nl, n2, n3, such that nl + n2 + n3 = n. Now nl may be equal to anyone of the n + 1 numbers 0, 1, 2, ... , n, and for a given nl the allowed values of n2 are any of the n - nl + 1 numbers 0, 1, 2, ... , n - nt, the value of n3 then being fixed at n - nl - n2. It follows that the degeneracy of En is n
'"'" ~
(n - nl
+ 1) = (n + 1) 2 -
~n(n +
1) =
Hn + l)(n + 2).
(2.30)
If our atom, influenced by the isotropic harmonic potential arising from the attractions of its crystalline neighbors, is exposed in addition to an external, uniform electric field, {, and it has an electric charge e, then it will experience an extra force e {, which can be derived from the potential er·{ = e£.z = e£r cos 0, w here we have chosen the z-axis to be parallel to {, and we have set £. = 1{ I. The Hamiltonian for this system is H
p2 2m
+ Aq2 + e£q3
}hw (p2 + Q2) + eEQ3V It
mw
'!1iw 2
(p2 + Q2) __e4>._~_ ' 2,,2
(2.31)
where we have introduced the linear transformation (2.32) The system (Q j ' Pk) is a canonical system of coordinates, and the Hamiltonian Eq.(2.31) differs from Eq.(2.5) only by the constant shift that the eigenvalues are
e:r.
We conclude
(2.33) The corresponding bound-state Schrodinger equation is
40
Three-Dimensional Harmonic Oscillator
This shift in the energy levels, which is of quadratic order in the electric field strength, is called the Stark effect, and we will study it again in connection with the hydrogen atom energy levels.
2.4
Many Harmonic Oscillators
The above formalism can be extended with very little effort to the case in which many identical, independent isotropic oscillators coexist. For example, we might consider ions at the lattice-points of a crystal. It is supposed that each ion is acted on by the average harmonic field arising from all the other ions, and that specifically nearest-neighbor interactions can be neglected. Then the Hamiltonian can be written 3N
H
= ~nw 2:[p} + Q;] , j=l
where N is the number of ions in the crystal. Here Pj and Qj are the reduced momentum and position operators, as before, but now j = 1,2,3 refer to the Cartesian coordinates of the first ion, j = 4, 5, 6 to those of the second one, and so on. The canonical commutation relations Eq.(2.4) remain the same, and annihilation and creation operators can be defined as before. The ground-state energy is now ~Nnw, which is very large if, say, N is the Avogadro number! It is convenient to redefine the zero of the energy to coincide with this zero-point energy, and this can be done by simply subtracting ~ Nnw from the Hamiltonian, which accordingly becomes 3N
H = 1iw 2:ajaj. j==l
aJ,
The rest of the formalism can be taken over as well. The creation operators, acting on a state of the crystal, increase its energy by one quantum, namely nw. Indeed, we can say that that a} creates, and aj annihilates a quantum. In quantum field theory we go conceptually even further and regard the corresponding operators as creating and annihilating particles themselves (or more accurately the Hilbert-space state vectors that stand for the particle states in question).
41
Exercises
2.5
Exercises
Problem 1 Prove the recurrence relation for the Hermite polynomial,
where
Deduce a second-order differential equation satisfied by the Hermite polynomial. Work out Ho(x), H 1 (x), H2 (x), H3(X) and H4(X) and sketch them graphically. Problem 2 Two particles, each of mass m, move in a one-dimensional harmonic oscillator well. In addition, there is an interaction, V(fr), which depends only on the separation between the particles,..a; :flr xz. 2!!
t~
~I
-
t~
(1) Show that the Schrodinger equation can be separated into two equations, each of one variable only. (2) Find the energy eigenvalues in the case that \T(\)
;\x 2.
Problem 3 \/( \) :: ~ i-~ Let q and p be the Cartesian coordinates of the position and momentum operators of a particle in a quantum mechanical one-dimensional simple harmonic oscillator. Let In'> be the normalized eillenvector of the_ Hamiltonian belonllinll to tne eIgenvalUe ~ n + ~ ) nw. 1- ne mean, q, ana tne uncertamty, D..q, are q = \ n Iq In) and (~q)2 = (nlq2In) - q2, and similarly for p. Calculate ~q~p for the nth excited state. Comment on the Heisenberg Uncertainty Principle. Problem 4 Calculate the probability of finding a particle in the ground state of an isotropic harmonic oscillator to be in a classically forbidden region. Problem 5 The Thomas-Reiche-Kuhn sum rule for a particle of mass m in one dimension is
Here q is the operator whose eigenvalues give the position of the particle, En is the nth energy level, and In) is the corresponding eigenvector.
(1) Check this sum~~fe. for the simple harmonic oscillator. (2) Prove it for a general potential.
Three-Dimensional Harmonic Oscillator
42
Problem 6 Calculate the energy levels of a 1-D "half" oscillator, defined by the potential Vex) = !mw 2 x 2 for x > 0, and Vex) = 00 for x < O. Which of the eigenvectors of the 3-D isotropic oscillator correspond to those of the half-oscillator? Problem 7 Define N = at a, where at is the Hermitian conjugate of the operator a, and where aa t - ata = l. (1) Prove that a and at have no inverse. (2) Show that the only analytic functions of a and at that commute with N are functions of N.
Problem 8 Let the differential operator a = (x + d/dx)/v'2 act on the space of infinitely differentiable functions, g(x), or, compact support. (1) Calculate the Hermitian conjugate, at, of a. (2) Deduce the commutator of a and at. (3) Prove [ata, at] = at and [ata,a] = -a. Use the language of test-function spaces.
Problem 9 An electron of mass m moves in a two dimensional Cu-O plane in a high-T c superconducting cuprate. Be effective potential is an isotropic oscillator. (1) Express the eigenJlares 'l/Jnl n2 of this system in terms of the eigenstates 'l/Jni ( Xi) of the one-dimensional harmonic oscillator and show that the energy eigenvalues are given by E n1n2 = (nl + n2 + l)hw. (2) Calculate the degeneracy of the nth excited level, i.e., the one with energy En = (n + l)hw. (3) Express the eigenfunctions belonging to the ground state and the first excited state in terms of polar coordinates.
Problem 10 In three dimensions, consider a particle of mass m and potential energy
VCr) where w
=
mw 2 -2-[(1 - r)(x 2 + y2)
+ (1 + r)z2]
> 0 and 0 < r < 1 .
(1) What are the eigenstates and energies of the Hamiltonian? (2) Calculate and discuss, as functions of r, the variation of energy and the degree of degeneracy of the ground state and the first two excited states.
Chapter 3
Orbital Angular Momentum
The orbital angular momentum of our particle is defined by
l
=ijl\p.
(3.1)
In this chapter, it does not matter whether the particle is in a potential field, like the harmonic oscillator of the last chapter, or whether it is free. In subsequent chapters we shall limit attention to central fields, like the Coulomb potential, or the isotropic harmonic oscillator, but for the moment the treatment is completely general. It follows that the components of l satisfy the following Lie algebra: (3.2) with implicit summation over k = 1,2,3. Here the Levi-Civita symbol is defined by 1 Eijk
=
-1
0
1
2
3 } is an even permutation 1, J k 1 2 3 } is an odd permutation if { 1, J k if two or three indices are equal if {
Proof of Eq. (3.2) [q2P3 - q3P2, q3Pl - QlP3] Q2Pl [P3, Q3]
+ QlP2 [Q3, P3]
i1i( QlP2 - Q2Pl) i1iL3,
43
Orbital Angular Momentum
44
where Eq.(2.2) has been used. Clearly the proof goes through in exactly the same way for the other two even permutations, {231} and {312}, while for an odd permutation, we have
[L2' L 1] = - ihL 3 , and similarly for the other two odd permutations, {I, 3, 2} and {3, 2,1}. END OF PROOF
L2 _ L . Land L3 commute with one another: (3.3) Proof of Eq.(3.3)
Li L 3 - L3 L i -L1 L 3L 1 + L1 L 3L 1 L1 [L1' L3] + [L1' L3]L1 -ih(L1L2 + L 2L 1) similarly
L 2[L2, L3] + [L2' L3]L2 ih(L2L1
+ L 1L 2)
and, since L~ commutes with L 3 , we have
END OF PROOF
Incidentally, since [L2, L 3] = 0, it follows that also [L2, L~] [Lr + L~, L~] = 0. Thence we see that
[Li, L~] 3.1
=
0, and hence
= [L~, L;] = [L;, Li] .
Diagonalization of L2 and L3
Because L2 and L3 commute, simultaneous eigenvectors exist * , say
• A proof that one can always construct simultaneous eigenvectors of commuting, Hermitian, finite-dimensional matrices can be found in Appendix A.5.
Diagonalization of L2 and L3
45
(3.4) where m is real, since L3 is hermitian, and {3 is real and non-negative, since L2 is the square of a hermitian operator. For a given f3 > 0,
(3.5) This follows immediately from
({3, mIL 2 1{3, m)
>
({3, mIL~I{3, m) ({3, miL; 1{3, m)
+ ({3, mIL~I{3, m) + ({3, mIL~I{3, m)
so that, from Eq.(3.4),
Define raising and lowering operators
for which
(3.6)
Proof of Eq.(3.6)
[L3, L 1] ± i[L3' L2] iTiL2 ± TiLl ±Ti(L1 ± iL 2) . END OF PROOF
Since L2 commutes with L 3, by symmetry it commutes with L1 and L 2, and so also with L±. Hence
whereas
i.e., L±I{3, m) is an eigenvector of L2 with eigenvalue {3Ti 2 and of L3 with eigenvalue (m±l)Ti, unless it vanishes. Note that the raising and lowering operators, L±, function much as did the raising and lowering operators of the harmonic oscillator.
46
Orbital Angular Momentum
Because of Eq.(3.5), for a given /3 2: 0, there must be a maximum value of m, say mmax, where V7J - 1 < mmax < V7J, such that indeed (3.7) since otherwise the left-hand side of the above would be an eigenvector of L3 belonging to the eigenvalue (mmax + l)n, which is impossible. By repeatedly applying L_ to 1,8, mmax) , one can build up eigenvectors belonging to eigenvalues mmax - 1, mmax - 2, mmax - 3, ... , and so on. However, this process cannot continue ad infinitum, since Eq.(3.5) implies also the existence of a minimum value of m, say mmin' where -J1J < mmin < 1 - .J73, for which (3.8) since otherwise the left-hand side of the above would be an eigenvector of L3 with eigenvalue (mmin - 1)1i, which is impossible. Since mmin can be reached from mmax in a number of integral steps, it follows that N
(3.9)
= mmax - mmin
is an integer. Consider now
(L1 ± iL 2)(L 1 =F iL 2)
Li + L~ =F i[L1' L2] L2 - L~ ± nL3.
(3.10)
Because of Eq.(3.7), we have
o
(/3, mmaxI L - L + 1/3, mmax) (/3 - m~ax - mmax) Ii 2 (,8, mmax 1/3, mmax) ,
and because of Eq.(3.8),
o
(/3, mminI L + L -I/3, mmin) (/3 - m~in
+ mmin) Ii 2 (/3, mmin 1/3, mmin) ,
where we have used Eq.(3.4). It therefore follows that
f3
= mmax(mmax + 1) = mmin(mmin -
1),
(3.11)
since neither (/3, mmaxl/3, mmax) nor (/3, mminl/3, mmin) may vanish. Hence
mmin = -mmax ;
Diagonalization of L2 and L3
47
but then from Eq.(3.9) we see that
N
m tnax
=
2""'
which means that mmax must be an integer, or half an odd integer. It is convenient at this point to adjust the notation a little. Let us write f in place of mmax and If, m) in place of 1,8, m), with j3 = f(f + 1). We have then, instead of Eq.(3.4),
f(f + 1)fi2If, m) mfilf, m) ,
(3.12)
where we have proved so far that f is an integer or a half-integer, and that the permitted values of m are -f, -f + 1, ... , f - 1, f. From Eq.(3.10) we have (3.13) and hence if we set
and require the normalization
(f, mlf, m) = 1 for all m between -f and f, then we find
It is convenient to choose the phase of
cim to be zero, so then
L±lf, m) = fiV(f =f m)(f ± m Note that automatically L+lf,f) Two points must be stressed:
+ 1) If,m ± 1).
(3.14)
= 0 = L_lf, -f).
• The Lie algebra [Eq.(3.2)] has been used exclusively to obtain the above results . • The quantization conditions [Eq.(2.2)] imply the commutation relations Eq.(3.2), but the converse is not true. In the treatment of spin, one only has the equivalent of Eq.(3.2), as we will see later, and then half-integral values of m are allowed, and moreover they occur in nature. For the orbital angular momentum, however, only integral values of m, and hence of f, are allowed. A proof of this fact will now be given.
48
Orbital Angular Momentum
From the commutation relations Eq.(3.2), we know that the only possible eigenvalues of L2 are t(t + 1)n2 , where t is a non-negative integer or half-integer, while the only possible eigenvalues of L 3 , for a given t, are mn, where m must have one of the values -t, -t + 1, ... , t. We now show, by means of a cunning canonical transformation, that the half-integral values are actually excluded. Proof that m must be an integer Define
Q _ qi ±P2 ± y'2 P _
±-
PI
=t=
q2
y'2 .
Then Eq.(2.2) implies
[Q+,Q-J
o
[Q±, P±]
ili
[Q±, P:,=]
0,
(3.15)
i.e., the canonical pairs (Q+,P+) and (Q_,P_) are independent of each other. Moreover, qIP2 - Q2PI
HP; + Q~) -
HP~
+ Q=-).
(3.16)
Note that this looks just like the difference between the Hamiltonians of two simple harmonic oscillators. We exploit this isomorphism in the following treatment, which we give in full. Define annihilation and creation operators
Q± +iP±
v'2h Q± - iP±
v'2h
(3.17)
so that
with all other commutators vanishing. Define the number operators (3.18)
49
Diagonalization of L2 and L3
so that
(3.19) Let 11I±) be a normalized eigenvector of N± with eigenvalue 1I±, i.e., (3.20) Then (3.21) where (3.22) From Eq.(3.21), since (1I±11I±) is positive and (J.L±IJ.L±) is non-negative (it could be the null vector, in which case it would vanish), we see that 1I± are necessarily non-negative real numbers. However, from Eq.(3.19), we find
(a±N± - a±)llI±) (1I± - l)a±llI±),
(3.23)
so that a± 11I±) is an eigenvector of N± with eigenvalue 1I± -1, unless it vanishes, in which case the above equation is trivially true. By repeated applications of the lowering operators, a±, we can generate smaller and smaller eigenvalues of N±, namely 1I± - 1, 1I± - 2, ... , which would eventually reach negative values, which is impossible, as we have just proved. The only way to avoid the contradiction is if there is an eigenvector of smallest eigenvalue, say Illmin ±), for which a± Illmin ±) = o. However this implies N±llImin±) = 0, i.e., lImin± = 0, and so we can write (3.24) where 10) can be taken to be a simultaneous eigenvector of N+ and N_, since these operators commute with one another. From this ground-state 10) we can build up excited states with the help of Eq.(3.19):
N±a110) = (a1N± + a1)10)
= a110)
so that a110) is an eigenvector of N± with eigenvalue unity. By repeated applications of the raising operator, a1, we can build up successively vectors with
Orbital Angular Momentum
50
eigenvalues 2, 3, 4, .... Glancing back at Eq.(3.20), we see that we have established that the eigenvalues v± of the operators N ± are precisely the non-negative integers. From Eq.(3.17), Eq.(3.18) and Eq.(3.15) we find
2~ {(Q± -
N±
iP±)(Q± + iP±)}
2~ {Q~ + pl + i[Q±, P±]} 122
211, {Q ±
+ p± -
11,} .
(3.25)
Hence Eq.(3.16) yields
and we have proved that the eigenvalues v± of N± are non-negative integers, so it follows that the eigenvalues of L3 have the form mn, where m must be an integer! Since m must have one of the values -f, -f + 1, -f + 2, ... , f, it follows that f must also be an integer. END OF PROOF
In many treatments of angular momentum that are given in text books, the essential reason that an orbital angular momentum may not have a half-odd integral value is given incorrectly or incompletely. The essential point is that l satisfies more than the Lie algebra Eq.(3.2).
3.2
Spherical Harmonics
In the previous section, we considered orbital angular momentum from an abstract point of view. In this section, we shall continue to use the symbols L3 and L2, but here they refer to the configuration space representations of the corresponding abstract operators. The configuration space representations of eigenvectors are called eigenfunctions. The spherical harmonics are eigenfunctions of L3 and L2: L 3 Yfm((), ¢) = m11,Yfm((),
(3.26) (3.27)
where f is a non-negative integer, and m is an integer satisfying (3.28)
Spherical Harmonics
In addition, the extreme values of m
51
(±.e), are associated with the restrictions (3.29) (3.30)
while, if .e > 0, and
-.e < m < .e -
and, for -f + 1 < m ::;
1,
.e, (3.31)
Finally, the normalization is such that
r Jrr 0 d() sin () J0
27r
d¢ 1Ylm ((), ¢ ) 12 = 1.
(3.32)
The above equations determine, with some redundancy, the spherical harmonics up to a phase-factor that is fixed by convention (see below). We have now to express the angular momentum operators in spherical polar coordinates. Consider first a transformation of Cartesian coordinates, Xl, X2, X3, to some other set of orthogonal coordinates, UI, U2, U3, such that
where the h's are functions of the u's. Let T be an infinitesimal volume. Then the divergence of a vector, can be written
.x,
where S is the surface of the infinitesimal volume, T, and ii is a unit vector normal to this surface. This is the Gaufi theorem. In terms of the components in the UI, U2 and U3 directions, this can be written
where 'permutations' means the two terms obtained by replacing the indices 1,2,3 by respectively 2,3,1 and 3,1,2 . This is, however,
Orbital Angular Momentum
52
When
A is the gradient of a scalar, A =
V'lj;, we find
In the case of spherical polars, we have
so that hI
=1,
h2
=r
,
1
~ sin 0 o'lj; +
h3 = r sin 0 .
Hence we can write V2'lj; =
~ ~r2 o'lj; + r2 or
or
r2 sin 000
00
1 02'lj; r 2 sin2 0 02
The radial partial differential can be simplified by noting 1 0 20'lj; --rr2 or
or
(3.33) and so we have the operator identity 10 2 0 1 0 2 = --r. r2 or or r or2
--r -
We have therefore obtained the following expression for the Laplacian differential operator:
(3.34) Let ea:, ey, and ez be unit vectors parallel to the Cartesian axes, and let er, e() and e
, respectively. Consider a rotation about the z-axis through an angle :
c~s
- sm
sin ) ( cos
~ a: e y
)
•
Spherical Harmonics
53
Here ew is a unit vector in the xy plane, at right angles to e4>. Next consider a rotation by () in the plane defined by ez and ew (see Fig. 3.1 on page 64): (
~r
e (}
)
c~s () - sm ()
= (
sin () ) ( ~ z cos () ew
)
•
Hence
e r e x sin () cos e/> + ey sin () sin e/> + ez cos () e (} ex cos () cos e/> + ey cos () sin e/> - ez sin ()
-e x sin e/> + ey cos ¢
(3.35)
We have the cyclic vector products (3.36) and also (3.37) For spherical polars,
v -- {) __
1{)
__
1
{)
(3.38)
+ e (r;. {)() + e 4> r sin () {)e/> '
= e r {)r
so the angular momentum is represented by
l
=
r Ap
-inrer A V .to { _ -1,,,,
.to { _
-1,/1,
_
{)
erA e (} {)() {)
_
__
1
__
{)}
+ erA e 4> sin () {)e/> 1
{)}
e 4> {)() - e (} sin () {)e/>
(3.39)
'
where we have used Eq.(3.37). With the help of Eq.(3.35) we read off the Cartesian components of the angular momentum:
-in {-sin¢~ {)() -iii { cos c/>
-
cose/>cot(}~} {)e/>
:0 - sin c/> cot 0:c/> }
-in~ {)¢ .
(3.40)
Hence L± = L1 ± iL2
{) + i cot () {)e/> {)} = he ±04>{ ~ ± {)()
•
(3.41 )
Orbital Angular Momentum
54
From Eq.(3.10), we have
and
(3.42) On the other hand
2 L32 - nL3 = - n 2{8 8(p2 - Z.8} 8¢ ; and on adding these equations together, we obtain
2
L =
-n
2{ 1
8. 8 sin 888 sm 8 88
+
1 82 } sin2 8 8¢2 .
(3.43)
Using Eq.(3.34), we can now write (3.44) This last result can also be obtained by squaring Eq.(3.38), but that is rather painful and the above method is easier. The most general solution ofEq.(3.26), in the polar representation Eq.(3.40), namely
IS
(3.45) where Ti (cos 8) is an arbitrary function of 8, which we have written as a function of cos 8 for convenience. Inserting this form into Eq.(3.27), with use of Eq.(3.43), we find
55
Spherical Harmonics
In terms of the variable z = cos (),
(3.46)
this takes the form
d (1 - z2 )-d d [ -d Z z
+ ff( + 1) -
m2 2 1,- z
1Tlm( Z) = 0,
(3.47)
which is called the associated Legendre equation. Given that it is a secondorder ordinary differential equation, it has two independent solutions, and if we only use Eq.(3.26) and Eq.(3.27), there is no convincing way of deciding which particular solution we need. Nevertheless, by using Eq.(3.41), we can pick out the required solution of Eq.(3.47). In configuration space, Eq.(3.14) becomes
e±i~
[±! +
icot 8
:4>1TF'(cos 8lei=~ = J(e Of m)(e ±m + ll1{±1(zl ei(m±1)~
and this reduces to
This is a first-order, homogeneous differential system, and it has a unique solution, up to normalization. It will prove helpful to define (3.48) in terms of which we find
, ddz Rtrn(z) = =fJ(f=fm)(f±m+ l)RFm±l(Z) ' with z = cos (). In particular, we see that
so that
RiAz)
= ci
CF being two z-independent constants. (sin 0) m Rim (cos ()) so that
From Eq.(3.48) we have
= Ti (cos ())
= (sin ()) -m Rem (cos ()) ,
(3.49)
56
Orbital Angular Momentum
and in particular
Ri.-l(cos 0)
(sin 0)21. Ri,-A cos 0) = ci (sin 0)21.
Ril(COS 0) ,
(sin 0) 21 Rt,A cos 0) = ct(sinO)21.
We have, for m
(3.50)
> -£,
(_~) R+
1
.J(£ - m + 1)(£ + m) C
1
dz
(£ - m)! ( d) l+m 2 l (21)!(£ + m)! - dz (1 - z )
by iteration, with use of Eq.(3.50). Similarly, for m
Rim(z)
=
V(l+ m c+
l
(z)
l,m-l
(3.51)
< l,
~ 1)(£ _ m) (~) Rl,m-l(z).
(£ + m)! (2£)!(l- m)!
We have thus two expressions for
.
(~)l-m dz
(
1- z2 l
) .
(3.52)
Tr, namely (3.53)
obtained by iterating m down to -£, and (3.54) obtained by iterating m up to i. Since these expressions must be equal to one another, we can find a relation between the two constants. Evidently c- (£ - m)! (1 _ z2)m
l (£ + m)!
(_~)l+m (1- z2)l = c+ (~)l-m (1- z2)l. dz
l
dz
By equating the coefficients of the highest power of the polynomials on both sides, namely zl+m, we find ct = (-l)fci. Use of this result, and inspection of Eq.(3.53)-(3.54), reveals that (3.55) The Legendre polynomial for integral, non-negative l is defined by 1
P1(z) = 211!
(
d
dz
l )
2
l
(z - 1) ,
(3.56)
57
Spherical Harmonics
and the associated Legendre function of the first kind by
= (1 ~ z2)m/2 (~) m
Pt(z)
Pt(z),
(3.57)
< i. From Eq.(3.53), we have that, for l > 0 and
for integral, non-negative m o <m< l,
(l- m)! m (2l)!2 i l!{l + m)!Pi (z).
(3.58)
On looking back at Eq.(3.47), we see that the associated Legendre functions of the first kind satisfy the second order differential equation d [ -d (1 Z
Z
2
d
)-d Z
+ l(l + 1) -
m2 ]
1- Z
2
m
Pi (z)
= o.
(3.59)
The standard normalization and phase convention are as follows: 1
«() A-.) = (_l)m [2141T + 1 {l - m)!]2 pm( ()) im,p im , (l + m)! i cos e ,
y;
tp
(3.60)
when 0 < m < l, which is consistent with Eq.(3.45) and Eq.(3.53). To cover negative m values, we use Eq.(3.55), which implies 1
«() , A-.) =
l:.T .I im
tp
[2l + 1 (l + m)!]2 41T (l _ m)!
p-m{ ()) im¢ i cos e
,
(3.61)
for -l < m < O. The spherical harmonics have been normalized so that (3.62) Proof of the orthonormality of the spherical harmonics The factor 8mm in Eq.{3.62) is obviously correct, for l
la
h
dq,e;(m-m')¢
= 211"6mm , .
(3.63)
So we can set m = m' in what follows. Multiply the differential equation (3.59) by Plf1'(z) and subtract the result from the expression obtained by interchanging land l'. Integrating the result, we find
[f(f + 1)
~ t(t + 1)]/, dzPt(z)P;'(z)
=
(3.64)
Orbital Angular Momentum
58
A partial integration suffices to show that the right side vanishes: the boundary terms contain a factor 1 - z2, which is zero at the boundaries, and the terms in the integral cancel against one another. Thus when f f:. f' the integral on the left side must vanish. This justifies the factor 8u' in Eq.(3.62). It remains to show that the normalization is correct, i.e., to show that (3.65) for m
~
0, where we have defined
or
{I dz [Pi(z)]2 .
=
i-1
To calculate this integral we use Lemma 1 below, which gives om+1
om _ i
i
- (f-m)(f+m+1) '
and on iterating this up to m = f, we find
of
om+2 i
=
(f - m)(f - m - l)(f
+ m + l)(f + m + 2)
(f+m)! i (f - m)!(2f)! Oi .
(3.66)
From Eq.(3.56) and Eq.(3.57) it is easy to show that pi(z) i
=
(2f)! (1 _ z2)i/2 2if! '
(3.67)
and so it follows that Oi i
= {I
)J2 = i -1 dZ [pi( i z
[(2f)!]2 A 2if! i)
(3.68)
where 2 i
1
Ai
=
/ -1
dz (1 - z)
21.+1 (f!)2
= (2f + 1. )., '
where use has been made of Lemma 2. Hence
om _ i
2 (f+m)! - 2f + 1 (f - m)!
(3.69)
Thus we vindicate Eq.(3.65). The above proof can be immediately extended to m < 0 by means of the equivalence Eq.(3.55). END OF PROOF
59
Spherical Harmonics
Lemma 1
1 1
Cr
-1
cm+l
dz [Pi(Z)J2
=
(£ _
m)(~ + m + 1) .
Proof From the expression (3.57), we have
and this leads to
We integrate both sides of this equation and perform an integration by parts:
The integrated parts are all zero. In particular, m [z [Pi(Z)]2]
~1
= 0, trivially
so if m = 0, and in view of the fact that Pi(±l) = 0 for m > 0 (see Eq.(3.57)). We now use the differential equation (3.59) to simplify the first term under the integral, obtaining
1 1
-1
dz [Pi(z)]2 [£(£ + 1) -
(£ - m)(£ + m
m2
1- z
2-
m
+
m 2z2 1- z
2]
+ l)Cr"
Lemma 2
Ai
-1 =
1
-1
2 i 2i + 1 (£!? dz (1 - z) = (£ ), .
2
Proof The integral can be evaluated by noting that
Ai + Ai-l =
/1
2
dZ(z -1)
-1
i-I 2
1
z =-2£
/1 -1
+1
.
2
i
1
dZ(z -1) =-2£Ai'
Orbital Angular Momentum
60
where a partial integration was performed. Hence Ae
=-
2f 2f + 1 Ae-l
e
=
2ff! (-1) (2f + I)!! Ao
=
e 22e +1(fl)2 (-1) (2f + 1) 1 .
END OF LEMMATA
3.3
Zeeman Effect
If the Hamiltonian of a system commutes with the angular momentum operators,
[I, H]
(3.70)
= 0,
then the angular momentum is conserved, i.e., it is time-independent. This property is shared by the three-dimensional harmonic oscillator and the nonrelativistic model of the hydrogen atom, and indeed it is true for any system with a spherically symmetric potential (Exercise 3.4). On condition that the eigenvector If, m) of L2 and L3 is also an eigenvector of the Hamiltonian, Hlf, m)
= Ee,mlf, m) ,
(3.71)
Ee,m cannot depend on m. This follows from Eq.(3.70), for
Howeyer, from Eq.(3.14}, L± If, m) is proportional to
If, m ± 1), and hence
HL±lf,m) = Ee,m±lL±lf,m) ,
and so Ee,m = E e,m±1, on condition that m < f resp. m > -f. This degeneracy is in general broken if the system is exposed to a magnetic field, whence m is sometimes called the magnetic quantum number. The effect was investigated by Zeeman, and we will explain its simplest form here, returning later to more complicated instances of the effect. Suppose that a particle of mass M and charge e executes a circular trajectory of radius r, under the influence of a magnetic induction, jj. Classically the magnetic force is
.... e .... F = -v /\B, e
and so ~
"""+
e-+
~
e-+
-+
. . ."
-+
F·r = -v /\B·r = - r /\v·B, e e where r is the position and v the velocity of the particle. Introducing the momentum, p = M v, we have
....
F .r
=-
e
Me
..
r /\p. B = -
e........
Me
L .B .
(3.72)
61
Exercises
For circular motion the centripetal force is M v 2 / T, in the direction of ....
-r, so
= - M v 2 = - 2E ,
F .r
E being the kinetic energy of the circulating particle. On combining this with Eq.(3.72), we find
E=;i.B, where
. . _ __e_L f-L -
21v[ c
(3.73)
'
which is interpreted as the magnetic moment of the orbiting particle. In the quantum mechanical treatment of this system, we add to the original Hamiltonian, H, a term j1 . jj to account for the effect of the magnetic field, where j1 is still given by Eq.(3.73), but L is now the quantum operator that represents the angular momentum. Let us treat the case of a constant, uniform magnetic induction, and for convenience we choose the x3-axis parallel to this field. Then the total Hamiltonian is
H' = H -
eB L
2Mc
3,
and the eigenvalues of this operator are given by H I If,m) =
(
El,m - emnB) 2Mc If,m).
(3.74)
Hence the energies, El,m, originally degenerate with respect to m, are split, each level being shifted from its neighbor by the same increment. This equal separation, which is generally observed as a splitting of spectral lines, constitutes the phenomenon called the normal Zeeman effect.
3.4
Exercises
Problem 1 Show that any operator that commutes with two components of the angular momentum operator commutes also with the total angular momentum operator. Problem 2 The position, the momentum and the angular momentum of a particle are represented respectively by the operators if, P and L. Show that
(1)
(2)
L 'p=o=L'if L 1\ L = inL
Orbital Angular Momentum
62
Problem 3 Show that (1)
I /\ if + if /\ I
(2)
[L2, if] =
2ihif -2ih(l/\ if - ihif) =
Problem 4 Given a Hamiltonian
where V depends only on q2 and is such that it ~6sses a Fourier transform, show that [I, H] = o. ~o~~~~(es Problem 5 The wave-function of a particle has the form fer, fJ) cos ¢>. Give the possible results of a measurement of the X3-component of the angular momentum, with the corresponding probabilities. Problem 6 Ii, m} is an eigenket of L2 and L3 with eigenvalues l(l + 1)h2 and mh. Show
Problem 7 A particle in a spherically symmetrical potential is in a state given by the wavefunction
where a and (3 are constants. (1) Take a = f3 = 1. What are the possible results of a measurement of L2? Calculate the corresponding probabilities. What are the possible results and the probabilities associated with measurements of Lx, Ly and L z? (2) Take a = 0 but f3 = 1. Answer the same questions for this case.
Problem 8 Suppose that the wave function 'If;(T') is an eigenfunction of Lx and L y • (1) Does this not contradict the uncertainty principle for noncommuting operators? (2) What are the possible results of a measurement of L z ? (3) Show that 'If; is spherically symmetric.
63
Exercises
Problem 9 Consider the following matrix representations of of the of the angular momentum operator.
X1-
and X2-components
-1,
o (1) Calculate L 3 , using the algebra of angular momentum operators. (2) What is the corresponding value of i!, the angular momentum quantum number? (3) Calculate (L 1), (Lr) and D..L1 = ([L1 - (L 1)]2)1/2. Here () refers to matrix elements taken with respect to a normalized eigenvector of L3 belonging to the eigenvalue 1. (4) Consider the state (I = 1, J2). If is measured in this state and the result + 1 is obtained, what is the state after the measurement? How probable was this result? If L1 is now measured, what are the outcomes and their probabilities? Does this second measurement change the state?
Lr
HI,
Problem 10 The generalization of angular momentum in configuration space from 3 to 4 Euclidean dimensions involves the differential operators Ljk
= -i(x j 8 k
-
Xk8j) ,
where j and k take on the values 1,2,3,4, with 1i = 1. Introduce
(1) Discover the commutation relations between (2) Show that the operators
J
and
K.
each obey the standard commutation relations for an angular momentum, and that they commute with one another.
64
Orbital Angular Momentum
Polar Coordinates: Summary w
rsinO
x
rsinOeos¢
y
r sin 0 sin ¢ = w sin ¢
z
reosO
r2
8
x2
= weos¢
+ y2 + z2
tan- 1
[
= w2
+ z2
[:1
';X'z+ Y'] = tan- 1
.. .. '
.... (" ... ' ..
~.~:~
...... .
-..
. ..
"'-:;'"
"--------p... pp"."ecp .
~
Figure 3.1
:.. ........... -.... .
...........
.... .
..
.. .
'.
.- ..
,...:-:-: ...
... ... ... ... .......... ..,
2
1 {j2
1 {
V = --r + r {)r2 r2
1 {). {) 1 {)2} smO- + - -sin 0 {)O {)O sin2 0 {)¢2
Chapter 4
Central Potential
When the potential depends only on the radial variable r equation,
[-::: V2 + Vir)] .p(T)
=
Ii I, the Schrodinger
= E.p(T) ,
(4.1)
can be written
where use has been made of Eq.(3.44). A particular solution has the form (4.2) and from Eq.(3.27) we find that ui(r) must satisfy (4.3) with.£ a non-negative integer.
4.1
Asymptotic Solution for 'r --+ 0
When r
~
0 and .£
~
1, ui"() r
~
.£(.£ + 1) Ui () 2 r , r 65
Central Potential
66
;2.
on condition that VCr) is less singular than Two independent solutions are r.e+l and r-.e. The second is not of physical interest, since it would lead to (4.4) which diverges for £ ~ 1. This is not compatible with the interpretation of 1-zP(i)1 2 as a probability density. For.f. = 0 the argument is more subtle. In the case that V(r) is analytic at the origin, so that V(O) is well-defined, we may write
Uo" (r)
- -2 k Uo (r ) ,
~
where Ii is a constant. Of the two independent solutions sin kr and cos kr, the second one yields a wave-function
-zP (i)
f'J
cos kr .
(4.5)
r
The reason for excluding this is not because of difficulties with the probability interpretation - the integral (4.4) is finite for .f. = 0 - but rather because Eq.(4.5) does not satisfy the Schrodinger equation at r = O. Indeed, Eq.(4.5) implies
V
2 cos kr - 1 r
21
+v-r
_k2 -zP(i) - 47r8 3 (i). To obtain this result, we have used the following identity in distribution theory: (4.6) This result can be proved by introducing a so-called Green's function, namely
I!
G(r) = (2 1r )3 where
E
eiif ·r
d3 q q2
.,
-ZE
(4.7)
--+ O. Then
(4.8) and
1
G(r) = -4 • 7rr Clearly, Eqs.(4.7)-(4.9) are equivalent to Eq.(4.6).
(4.9)
Asymptotic Solution for r -+ 0
Proof of Eqs.( 4.7)-( 4.9) For any integrable test-function,
fer),
67
Eq.(4.7) implies
(4.10) Here (4.11) is the Fourier transform of fer). Now by the fundamental theorem of Fourier analysis, the inverse of Eq.(4.11) is
fer) = (2~)3 /
d3qe-iifOr
i(in .
We see then from Eq.(4.10) that / d 3 rf(f)V 2 G(r)
= - f(O),
which proves Eq.( 4.8). In polar coordinates we can write Eq.( 4.7) in the form G(r)
=
{OO q2dq rTf dO sin 0 (2Tf d¢ exp(~qr c.os 0)
(2 1 )3 7r
J
1 (2 )2 7r
Jo
1
-(27r)2
J
0
J
0
q -
0
'l,f
{OO dq. 2q2. {I dz eiqrz
loo
1
(21r)2ir
0
q q2
Z€
J-1
e iqr -
e- iqr
dq--:--- - - - - q2 - if iqr
{OO
J
-00
q e iqr
dq (q - iE)(q
+ if)
.
( 4.12)
This integral can be performed by closing the contour of integration in the upper half-plane of the complex variable q. The only pole of the integrand inside the contour is that at q = iE, and its residue yields G(r)
END OF PROOF
=
21ri 'l,E (21r )2ir 2iE 1 41rr
( 4.13)
Central Potential
68
Perhaps the reader feels cheated by the first part of the above proof, on the grounds that it relies on a sophisticated result in Fourier theory. She or he can find a more general, and moreover direct proof of this result in Chapter 8 (see Eq.(8.5) et seq.). When VCr) ~ -;- for small r, as is the case for the Coulomb or the Yukawa potentials, the .e = 0 radial equation looks like
for small r. The most general solution has the form 00
uo(r) = ao [1 -logr]
+ aIr + 2: [av + bv logr] rV, v=2
where ao and al are arbitrary, while av and bll are determined from them by Eq.(4.1). For the same reason as in the case that VCr) is analytic, this does not yield a solution of the three-dimensional Schrodinger equation, unless ao = O. Thus in all cases of interest, the behavior of the radial function as r --+ 0 is as follows: (4.14)
4.2
Asymptotic Solution for r -+
00
For large values of r, Eq.(4.3) becomes /I
ue (r)
2mE
+ -2-ue(r) ~ 0, ;"
on condition that V (r) --+ 0 sufficiently rapidly. In the case that the total energy is negative, we set 2 fi,
2mE
= ---
;,,2
>0
'
so there are two asymptotic solutions, e±KT. Of these, only the negative exponential is acceptable, since the positive one is not square integrable and so could not give rise to a probability density. However, after the regular solution has been chosen at the origin, ue(r) is fixed, apart from normalization, and in general when it is continued from small to large values of r, it will be a combination of both asymptotic solutions:
3D Oscillator in Polars
69
Only for those special values of the energy (and so of K) for which Be = 0 do we have an acceptable bound-state solution. Whether one or more bound states exist depends on the form of the potential, V (r ). When the energy is positive, we set 0
k 2 = 2mE
n2
> ,
and now the continuation of the regular solution to large values of r can be written
where now both solutions are allowed: they are not square integrable, but they are acceptable scattering solutions. With a particular choice of overall normalization, this is rewritten
ue(r)
rv
2ik [e- i (kr-i1r/2)
_
Se(k)e i (kr-i1r/2)]
,
where Se(k) is called the scattering function. It depends of course on the form of V(r). The reader is referred to Chapter 8 for a further study of scattering.
4.3
3D Oscillator in Po lars
Let us reconsider the isotropic harmonic oscillator in three dimensions (refer to Section 2.3). The potential is V(r) = Ar 2 , and with the substitutions p=
ue(r)
=
1/2mA
n
r
2
1+1
1
e-'I P p2- Ve(P) ,
the radial equation Eq.( 4.3) becomes
pvt+ (£ + ~
-
p)v~
+ H2€ -2£- 3)ve = 0,
(4.15)
where
E
€=-
nw
with
w=ff..
Try now the series solution 00
ve(p) =
L akpk. k=O
(4.16)
Central Potential
70
The differential equation Eq.( 4.15) implies 00
L ak [k(k + £ + ~)pk-l + H2€ -
2£ - 3 - 4k)pk] = 0,
k=O
which is equivalent to 00
L
[ak+l(k
+ 1)(k + £ + V -
~ak(4k + 2£ + 3 - 2€)] pk
= o.
k=O
Each coefficient must vanish, so 4k + 2£ + 3 - 2€ ak+l = 4(k + 1)(k + £ + ~) ak
t!'
1
rv
k
+ 1 ak,
(4.17)
as k-+ 00. Hence ak and so in general v.e(p) eP , which is unacceptable. The only exceptions occur when the series Eq.(4.16) terminates, for then u.e(r) is square integrable. This occurs when I"V
I"V
€
= 2k+£+ ~2
for some k = 0,1,2, ... , for then ak+l = 0 and therefore also aj j>k+l. Accordingly,
E
= En =
(n
+ ~ )nw ,
o for
all
(4.18)
with n = 0,1,2, ... , where n = 2k + £. This agrees with Eq.(2.29), which we obtained by treating the isotropic oscillator in Cartesian coordinates. For a given principal quantum number, n, the angular momentum quantum number therefore satisfies £ = n - 2k,
i.e., £ = n, n - 2, n - 4, ... , 1 or O. Thus £ has the same parity (evenness or oddness) as does n. This is a property that we did not appreciate in the Cartesian treatment of Chapter 2. To calculate the degeneracy of each level, we recall that each value of £ is associated with 2£ + 1 values of the azimuthal quantum number m. Hence for £ = n - 2k there are 2n - 4k + 1 degenerate levels. This must be summed over k = 0,1,2, ... , [¥-], since the energy does not depend separately on £, but only on n. Now £ is non-negative and we have terminated the k-sequence with the integral part of ~. The total degeneracy of the n th energy level En is [~ ]
I)2n - 4k k=O
+ 1) =
Hn + 1)(n + 2) ,
71
Square- Well Potential
in agreement with Eq.(2.30). Further aspects of the isotropic oscillator will be brought to light in Exercise 4.10.
4.4
Square-Well Potential
Consider first of all a free particle, I.e., one for which V(r) Schrodinger equation,
-!{\12~(r) = E~(r), 2m
O.
The free
(4.19)
has a solution of the form Eq.( 4.2), with (4.20) where k 2
= 2mE /n?
Set p = kr, so that
d2 f(f + { dp2 +1p2
I)}
u£(r(p)) = O.
(4.21 )
A solution of Eq.(4.21) that vanishes at the origin is
where j£ (p) is called the spherical Bessel function. We shall shortly return to the general case, and to the properties of this function, but for the moment we concentrate on the S-wave, i.e., the case f = 0, for which Eq.(4.21) reduces to
d2 } { dp2 + 1 u£
= 0,
(4.22)
with the solution uo(r) = pjo(p) = sinp = sinkr, vanishing at the origin. Consider next the situation in which V (r) = - Vo (a constant) for r < a, and VCr) = 0 for r > a. If E > 0 and we set
k=
V2m (E + Vo) 1i
'
the radial Schrodinger equation has the form Eq.(4.20), but with k. Accordingly we can write, for r < a,
(4.23)
k in place of (4.24)
Central Potential
72
For r > a, we have the free Schrodinger equation Eq.(4.19)j but the more general solution
Ul(r)
= kr [Ai£(kr) + Bnl(kr)]
(4.25)
is possible, where the spherical Neumann /unctions, nl(p), will also- be defined shortly. For the S-wave, the corresponding solution is pno(p) = - cosp, and Eq.( 4.25) reduces to
uo(r)
= Asin(kr) -
Bcos(kr).
(4.26)
The coefficients A and B have to be determined by matching Ul (r ), and its first derivative, as given by Eq.(4.24) and Eq.(4.25), at the point r = a. A similar situation obtains if E is negative but larger than - Vo, for then E + Vo > 0, and so k is real and the solution Eq.(4.24) is still valid when r < a. However, k is now imaginary, and it is convenient to write Ii
= -ik =V -2mE ,
(4.27)
1i
so that the solution for r
> a can be written
For the S-wave this reads simply
Uo (r)
= A sineilir) -
B cos( ilir)
= iA sinh (lir)
- B cosh( lir) .
(4.28)
Again, A and B have to be determined by matching Ul (r) and its first derivative at the point r = aj but the problem now is that in general the solution will explode exponentially as r -t 00. Only for the special instances (if they exist) in which B = iA will (4.28) reduce to
uo(r)
= -B exp( -lir) ,
( 4.29)
which is an acceptable, square-integrable function. The special cases are boundstate or eigenvalue conditions. We must match the solutions, and their derivatives, at r = a. In the limit € -t 0, U~( a -
€) = u~ (a + €) ,
and these constraints can conveniently be combined into the eigenvalue equation,
(4.30)
73
Square- Well Potential
For the S-wave, this matching condition takes on the form d
-
da log sin ka
d
= da log e -Ita,
which leads to (4.31)
= -kcotka.
K,
However, K, and k are related, according to their definitions Eq.(4.23) and Eq.(4.27), by 2 K,
-22m +k = - 2 Vo,
(4.32)
n
which is a circle in the K, - k plane. In Figure 4.1 we show the first quadrant of this circle for a number of different values of the potential strength, Vo (with a = 1). Also plotted are the curves corresponding to Eq. (4.31). Intersections of the circles and the cotangent curves correspond to energy eigenvalues, i.e., to solutions of the bound-state problem. When the radius of the circle is less than ~, there is no bound state: the potential is in this case too weak to bind even one state. With the radius just equal to ~ (the smallest circle in the figure), there is a zero-energy bound state, and, as Vo is increased, the bound state becomes more deeply bound, until at a radius of a second bound state makes its appearance (this corresponds to the second circle), and so on.
10
3; ,
37T" Figure 4.1
Clearly, if the potential strength is such that
(2N - 1)7T" 2
a
12 Vr
(2N
+ 1)7T"
2
'
where N is a positive integer, then there are precisely N bound states. The smallest value of Vo that permits a bound state is a 7T" (4.33) -yf2mVo = -.
n
2
In this limit, the binding energy, EB = -E, vanishes, as does K,2 = 2mEB/n2. For a very weakly bound state, the binding energy is much less than Vo and Eq.( 4.33) is then approximately satisfied. The deuteron is a weakly bound state of a proton and a neutron, in an S state. We may make a rough model of the nucleus of deuterium, or heavy hydrogen,
74
Central Potential
by supposing that each nucleon (i.e., the proton or the neutron) moves In a spherically symmetric square-well potential induced by strong interaction with the other. The masses of the proton and neutron are approximately equal, namely m n c2 ~ m p c2 ~ 940 MeV ,where 1 MeV = 106 eV, an electron volt (eV) being the energy that an electron acquires when passing through a potential difference of one volt. MeV and eV are units of energy often used in nuclear and atomic physics. The binding energy of the deuteron is measured to be EB = 2.226 MeV, and we suppose this to be much smaller than the depth, Vo, of the square-well potential, so that Eq.( 4.33) may be assumed to be approximately true, i.e., (4.34) However, m is not the mass of the deuteron, nor that of one nucleon, but rather the reduced mass, ~mn' To see this, note that the Hamiltonian of the twoparticle system can be written 2
Pp H = -2mn
2·
+
Pn -2mn
+ V(-qn -
.... ) , qp
where (if p,pp) are the position and momentum operators for the proton, (if n,Pn) the corresponding operators for the neutron, and where V is the mutual strong interaction potential. The masses of the two nucleons have been set equal to one another. Define the center-of-mass and total system coordinates
-
....
if
qn - qp
P
.. ) 2"1 (if n -Pp
Q p
- ) 2"1 (.... qn+qp
Pn +PP'
In terms of these new canonical variables, we find p2
H = 2M
p2
....
+ 2m + V(q),
(4.35)
where M = 2m n , the combined mass of the proton and the neutron, but the reduced mass that appears in the Schrodinger equation is m = ~mn' Accordingly, we can write Eq.( 4.34) as 7r 2(hc)2
4(mn c2 )a 2
•
(4.36)
Now the radius of a small nucleus like the deuteron is about 1 fermi (F) (i.e., 10- 15 m), so we may set a = 1 F. The product hc has the dimension of an energy
75
Spherical Bessel Functions
times a distance, and it is about 200 MeV-F. With the value 940 MeV for the rest energy of one nucleon, we find 7r 2(200) 2 '"" Va = 4 x 940 X 12 '"" 105 MeV. We note that indeed E B < < Va, consistent with the assumption made at the beginning. We can improve on this calculation by writing Eq.(4.31) (with a = 1) as tank k
lic
1
---;==~= =
Jm n c2 EB
I'\,
4.4 F .
(4.37)
Numerical solution yields k = 1.70, as can be checked easily on a mathematical calculator (do not forget to set the angles to radians!). The Mathematica command line
FindRoot[-Tan[k]jk == 4.4, {k, 1.6}] returns {k -> 1.70343}. Thus Va
= EB +
(klic) 2 mnc
2
= 125 MeV.
(4.38)
Further refinements are possible, for example the square-well potential can be replaced by a Yukawa potential,
V(r) = -~ e- cmrr / fi r
,
where m7r is the mass of a 7r-meson or pion (140 MeV), and 9 is the the attractive force engendered by the exchange of a pion between and the neutron. However, the square well gives a rough idea of the the interaction that just manages to bind the nucleons into a stable
4.5
strength of the proton strength of nucleus.
Spherical Bessel Functions
The spherical Bessel and Neumann functions were introduced as special solutions of Eq.(4.21); but so far we have only looked at f = O. To study the general case, consider the expression (4.39)
76
Central Potential
By making various choices for the function 10 (p), we will shortly be able to define not only the spherical Bessel and Neumann functions, but also the spherical Hankel functions, which are convenient for use in bound-state problems. Consider further the differential operators
d
be = dp
f
t d b = -e dp
+-p
f + -. p
(4.40)
It is easy to check that,. for any twice differentiable function, g(p),
f(f-l)} () { _~ 2 gp d 2+
P { _~ dp2
P
+
f(f+1)} () p2 9 P .
(4.41)
b~ and be are ladder operators for the functions fe(p) of Eq.( 4.39), in particular one finds
b~fe-l(p) = fe(p)·
(4.42)
It follows therefore that (4.43) and hence from Eq.(4.41) that (4.44) If the function fe-l(P) is such that
(4.45) then it follows from Eq.( 4.44) that (4.46) However, from Eq.(4.42) this means that (4.47)
77
Spherical Bessel Functions
Since Eq.(4.47) has the same form as Eq.(4.45), with l - 1 replaced by l, it follows by induction that, for any function fo(p) that satisfies
d2
(4.48)
- dp2fo(p) = fo(p) ,
the corresponding function h(p), defined by Eq.( 4.39) with this fo(p), necessarily satisfies d2 { dp2
+1-
l(l + 1) } p2
h(p)
= o.
(4.49)
Note that, whichever solution of Eq.(4.48) is chosen, the recurrence relation
(4.50) holds, this being simply the content of Eq.( 4.42). This recurrence relation may be used to generate higher from lower functions on the ladder. The choice fo (p) = sin p leads to h(p)
= pjf(p)
(4.51)
where the spherical Bessel function is defined by
d)
. ( ) _ ( )f (1 JfP--P -pdp
f
sin-p. p
(4.52)
The choice fo (p) = - cos p leads to
(4.53) w here the spherical Neumann function is defined by nf(p) =
f -( -p) (
1 d - - )
i
pdp
cosp
(4.54)
--. p
To analyze the threshold behavior of jf(p) and ni(p), it is convenient to introduce the variable a = p2, in terms of which (_2)faf / 2 (~ \) f sin Vada) Va(_2)faf/2
(~)\ f da
f
p=o
<Xl
(_2)ia i / 2 ' " (-1)P
L..J
p=i
(-a)P (2p + 1)! ,
p.
(p - l)!(2p + 1)!
a P- i .
Central Potential
78
The leading contribution comes from the term p the variable p, we find
=f
in the sum. Reverting to
(4.55) The spherical Neumann function has on the other hand non-integral powers of a in the expansion of (4.56) To get the leading power, it suffices to replace cos y'(i by unity, and since
= (-2)-£ (2f)! -(£+1)/2 ( ~)£_1_ da y'(i f! a , it follows that n£ (p )
rv -
(2f)! ( -p )-£-1 .
--
f!
(4.57)
In view of the threshold behaviors Eq.( 4.55)-( 4.57), and the discussion given in Section 4.1, it will be readily understood why only the spherical Bessel function is allowed as r -+ o. Thus
is an acceptable solution of the free Schrodinger equation. A much more general solution is obtained by superposing solutions: 00
1jJ(f) =
£
L L
A£m(k)j£(kr)Yim(O, ¢;),
£=0 m=-£ where A£m are arbitrary coefficients, constrained only by convergence requirements of the infinite f-series. They may depend on k, but not on r, 0 or ¢;, the essential point being that the Schrodinger equation Eq.(4.1) is linear and does not contain f or m. To consider asymptotic behaviors as p -+ 00, it is convenient to introduce the spherical Hankel function of the first kind, defined by
(4.58)
79
Spherical Bessel Functions
In particular, h~1 ) (p) = -i eip / p, corresponding to the choice 10 (p) Eq.(4.39). Accordingly, the explicit form for this Hankel function is
(1-pdp-d) -p . £
h (1) (p) =
£
. £ -l,(-p)
=
-ie ip in
.
etp
(4.59)
For large values of the argument, p, the leading contribution comes from repeatedly allowing the differential operator to act on e ip , ignoring its effect on 1/ p, since those differentiations produce terms that vanish more quickly than does 1/ p itself. Thus we find, for p -t 00, (1)
i
h£ (p)~-p(-I)
£(
d )
dp
i.
£ iP
i
£ i
(.
i7rf)
e =-p(-l,) eP=-pex p l,P-T
(4.60)
From Eq.( 4.58) we deduce (4.61 )
(4.62) Returning to the discussion of the square-well potential, we recall from Eq.( 4.27) that we had set k = iK for negative energies, where K is positive. In this case,
h (1) (iKT) £
~ - -
1 exp (
-KT -
KT
i7rf)
-
2
.
(4.63)
Thus the Hankel function of the first kind tends to zero exponentially at infinity, and it is the precise combination of spherical Bessel and Neumann functions that is needed for a bound state: for this reason, it is sometimes affectionately called the friendly Hankel function. The Hankel function of the second kind, h?)(p) = j£(p) - in.e(p), is on the contrary unfriendly, in the sense that it explodes exponentially at infinity. The eigenvalue condition Eq.(4.30) reduces to d da log [ j.e(ka - €) ]
=
d da log [Aj.e(ka + €)
+ Bn.e(ka + E)] .
(4.64)
As in the discussion of the S-wave, we have a bound-state condition (i.e., a square-integrable solution) only when B = iA, for only then do the spherical Bessel and Neumann functions combine to form the friendly Hankel function.
80
Central Potential
In this case, Eq.( 4.64) reduces to
! 4.6
log [jl(lca -
E)]
=
d~ log [h~l)(ka + E)]
(4.65)
Vibrational-Rotational Spectra
In this section we shall consider a simple diatomic molecule, for example carbon monoxide, CO, in which the constituent atoms have different masses, say ma and mb. Between the two constituents there is an attractive force, due to the presence of the valence electrons that are shared between the two nuclei. Such a force is described typically by a potential that is positive for small nuclear separations, due to the Coulomb repulsion between the nuclei, becomes negative for intermediate energies, with a well-defined minimum, and tends to zero for large separations, corresponding to the dissociation of the molecule. The Hamiltonian of the molecule can be written 2
H = ~ 2ma
2
+~ + V(I-->qa 2 mb
--> I) , qb
(4.66)
where the potential is a function of the modulus of the difference of the position coordinates of the atoms, ifa and if b. We need to make a canonical transformation to the center-of-mass and the relative coordinates,
Q=
maifa + mbifb ma+mb
P=Pa+ih mbPa - maPb P=-----ma+mb The quantization conditions
with all other commutators vanishing, lead to
with all other commutators vanishing. Thus the center-of-mass and relative coordinates, respectively (Q ,P) and (if,p), are independent canonical variables.
81
Vibrational-Rotational Spectra
The Hamiltonian Eq.(4.66) becomes p2
e2
p2
H=-+--21M 2m ItIl' where
M=ma+mb, and 1
1 ma
1 mb
-=-+-, m
( 4.67)
which shows that, in the treatment of the relative motion of the two atoms, the reduced mass m must be used. Note that the discussion preceding Eq.( 4.35) refers to the special case ma = mb. The motion of the molecule as a whole is free, governed as it is by the kinetic term p2 / (2M) only, while the internal motion of the atoms in the molecule is governed by the Hamiltonian
where we have now specialized to the configuration representation in the relative variables, in which the potential is central. The radial equation has the standard form Eq.(4.3), and we may expand the potential about the point, r = ro, at which it has its minimum:
where we have neglected terms of order (r - ro)3. Here V"(ro) > 0, since V(ro) is the minimum value of V(r). Accordingly, we may write Eq.(4.3) in the approximate form u£"() T
+ { -2m 2[ E -( V ro)] - .:\(T - TO) 2 - f (f + 2 1) } ~
h
U£ (r) =
0,
(4.68)
where A = mV"(To)/h 2 is the coupling parameter of a one-dimensional harmonic oscillator. The energy levels are therefore given by Enl =
V(TO)
1
+ hw(v + 2) +
h 2 f(f + 1) 2 2 mTo
'
where the vibrational angular frequency is determined by the oscillator strength, A. Here v is the oscillator quantum number. A further refinement would be to allow both ro and w to depend on £, since in fact the centrifugal term distorts
82
Central Potential
the form of the attractive potential; indeed, for large .e values there may well be no minimum in the effective potential. The electronic levels are separated by a few electron volts, and transitions between them give rise to photons that are typically in the visible spectrum, but each electronic level is split into equally-spaced vibrational levels, separated by a few millivolts. Transitions between these levels are in the infrared. The vibrational levels are themselves split into sublevels by the rotational degree of freedom, this splitting being not equally spaced, since the centrifugal term is proportional to .e(.e + 1). Transitions between levels at different f-values but the same v-value are in the far infrared. The rotational structure also shows up in the fine structure of the near infrared spectrum, corresponding to transitions say between vibrational level v and v-I, complicated by the fact that the initial state can be in one or another rotational state. Thus the energy difference between the vP and the (v - 1)8 level is slightly different from that between vP and (v - l)D, or between v8 and (v - l)P, and so on. The combination of vibration and rotation thus produces a complex spectrum, which can be used to elucidate the details of the molecular structure.
4.7
Exercises
Problem 1 The spherical Bessel and Neumann functions are respectively
d) pdp
. ( P) -_ ( -p )£ (1 sinp JR. -- £ -
nR.(p)
=
£ -(-p) (
1 d
- -) pdp
p
£
cos p --. p
(1) Prove the recurrence relations j£+l(p)
= -j~(p)
n£+l(p) =
f
+ -iR.(p) p
-n~(p) + ~nR.(p). p
(2) Work out j£(p) and n£(p) for f = 0,1,2,3, and sketch them graphically.
Exercises
83
Problem 2 Let 'ljJ(r) = uo(r)lr be an S-wave solution of the Schrodinger equation for a potential that has bound states, and which is strictly zero for r > a. Define the following function of the total energy:
feE) = u~(a) . u(a) •
Let E B be a bound-state energy. Evaluate f (E B)' Problem 3 Solve the bound state problem for a particle of mass m in the central potential
VCr)
o 00
where a
O
a,
= hi J2m.
:s
(1) List l!J(all the energy levels for which n 3 and f ~ 3. (2) Show that energy levels of the same n are ordered by their f values. (3) Show that energy levels of the same f are ordered by their n values.
Problem 4 Given a central potential, V (r) = - Vo for r < a and V (r) = 0 otherwise, find the minimum value of avo such that there is a P-wave bound state. Problem 5 Suppose that the spherically symmetric potential, V (r ), is equal to the positive constant Vo for r < a, is zero for a < r < b, and is infinite for r > b. Calculate the energies of the ground state and the first excited state, and obtain an approximate expression for the energy splitting of these levels if Vo is very large compared to these energy levels. Problem 6 Show that there cannot be more than one linearly independent bound state wave function for a given E and f, i.e., with the same energy and angular momentum. Problem 7 Given the radial potential energy for a diatomic molecule,
VCr) = V(ro)
+ ~mw2(r -
ro)2
h2
+ 2mr2f(f + 1),
find the radius, rl, at which this potential energy is minimum. Evaluate A and B in the expansion
84
Central Potential
Problem 8 A particle is in a cylindrical potential in three dimensions, defined by V (p) = 0 + x~ < and V(p) = 00 elsewhere. if p =
Jxi
a
(1) Determine the three lowest eigenvalues for states that have P3
L3
=
= o.
(2) Determine the three lowest eigenvalues for states that have P3 =
0 and
o.
Problem 9 A particle in a spherically symmetric potential is described by a wave packet tP = (XIX2 + X2X3 + X3Xl) e- ar2 • What is the probability that a measurement of the square of the orbital angular momentum yields zero? What is the probability that it yields 6h 2 ? If the orbital angular momentum is 2, what are the relative probabilities for m = - 2, -1, 0, 1, 2? Problem 10 With Ii = 1, consider the isotropic harmonic oscillator Hamiltonian in three dimensions,
and the operator
Let Enl.m be the energy level corresponding to principal quantum number n, angular momentum quantum number l, and azimuthal quantum number m,
Hlnlm) = Enl.mlnlm) . (1) Show that [H, U] = 0, (2) Show that (xIUlnll) oc (xln, l + 2, l + 2). Under what restrictions on n and l is this true? (3) Deduce from (1) and (2) that Enl.l. = E n ,I.+2,1.+2, under the same restrictions on nand l as in the previous question. (4) The isotropic oscillator Hamiltonian is invariant under rotations, i.e., orthogonal transformations of the coordinate axes. This leads to the fact that Enl.m is actually independent of m. By expressing Hand U in terms of creation and annihilation operators, discover the larger symmetry of the Hamiltonian that is responsible for the fact that Enl.m is also independent of l.
Chapter 5
Hydrogen Atom and Charmed Quark
In preceding chapters, we have encountered the potential r2 in connection with the harmonic oscillator, and this led to a study of the Hermite polynomials. We have also considered the repulsive, centrifugal potential r- 2 that required the introduction of the spherical Bessel functions. In this chapter we shall study the attractive potentials rand r- 1 , which will lead us to look at Airy functions and Laguerre polynomials. The case of the linear potential will be applied to the problem of the confinement of quarks in mesons, while the Coulomb potential, which we will first study, will be applied to the calculation of the energy levels of the hydrogen atom.
5.1
Radial Equation for Coulomb Potential
The potential energy of the electron in a hydrogen atom, or a (He+, Li++, etc.), is
hydrogen~like
ion
Ze 2 V(r) = - - , r where e is the elementary charge, and Z is the atomic number, i.e., the number of protons in the nucleus. With the notation ~ _ 2mZe 2 ':. -
2
n
and, for negative energies, 2 K
2mE
=---2->0, 11,
85
Hydrogen Atom and Charmed Quark
86
we may write the wave function
(5.1) where ul(r) satisfies
Ul" (r) + ['-:;. - £(£+1)] r2 Ul(r) = '" 2 Ul(r),
(5.2)
and we require solutions with the asymptotic behaviors r l + 1 as r -+ 0, and e- Kr as r -+ 00. Let us define p
=
2"'T
=
.j-BmE
n
)..=~=e2ZJ
n
2",
T
m
(5.3)
2E'
so that the radial Schrodinger equation Eq.(5.2) takes the form
d2ul dp2
+ [~. _ £(£ + 1)] Ul = ~Ul. P
Given the required asymptotics at p
= 0 and p =
(5.4)
4
p2
00,
it is sensible to write
Ul(r) = /+l e - p /2 L(p),
(5.5)
and we find from Eq.(5.4) that
pL"(p) + [2(£ + 1) - p] L'(p) + [A - £ - 1] L(p)
=
O.
(5.6)
Let us try a power series 00
L(p) =
L ajrJ,
(5.7)
j=O
with, say, ao = 1 (arbitrary normalization). From Eq.(5.6) we readily deduce the recurrence relation
j+£+l-A For very large j, this approaches
(5.B)
Radial Equation for Coulomb Potential
87
and so one intuitively expects
L ~J. = e
L(p) '"
P;
j
from Eq.(5.5) this means that
ul(r) '" e+ p / 2 = e KT , up to powers. This behavior is unacceptable, since it leads to a wave function that is not square-integrable, and hence one that is not subject to the probability interpretation. The above sloppy argument can be made rigorous by observing that, for i>A all the aj have the same sign, so no cancellation is possible; and, for any € > 0, it is possible to find a iE such that, for all i > ie ,
-.e,
a"+l _3_
> 1-
~ 2
i +1
aj
Hence aj >ie! - ( 1--€ )j-j. aj. -
i!
2
It is now easy to show that indeed ul(r) blows up exponentially as r -+ 00. Note that analogous reasoning establishes the correctness of the corresponding analysis of the infinite series solution for the isotropic harmonic oscillator (cf., Eq.(4.17) et seq.). The only way to prevent the above disaster is to require L(p) to be a polynomial, which will happen if the series Eq.(5.7) terminates. Then ul(r) behaves like e- KT , up to powers of r, so that it is indeed square-integrable. The series can terminate only if, for some i, say imax,
imax + .e + 1 -
A = 0,
(5.9)
for then, according to Eq.(5.8), ajInax+ 1 = 0, and hence aj = 0 for all i ~ imax+ 1. However, imax is necessarily a non-negative integer (it could be zero); and since A = imax + .e + 1, it follows that A must be an integer, not smaller than .e + 1. Let us write A = n,
(5.10)
which is called the principal quantum number. Writing E = En for the nth energy-level, we have from Eq.(5.3) and Eq.(5.10)
En = -
me 4 Z 2 1 2 2· 21i n
(5.11)
88
Hydrogen Atom and Charmed Quark
Note that Eq.(5.9)-(5.10) imply
f
n
1,2,3, ... ,
R
O,I, ... ,n-l,
-R, -R + 1, ... , R.
m
Note that the energy depends on n only. The ground state, n = 1, R = 0, m = 0 (100 for short) is simple; but the nth energy level has a degeneracy n-l
L(2R + 1) = n 2 • l=O
Properties of Laguerre Polynomials: The Laguerre polynomial is defined by (5.12) where t
= 0,1,2, ... , which indeed is clearly a polynomial in p of order t. Since
by differentiation we have
d dp [ e - P Lt+ 1 (p) ]
By working out left and right sides, we obtain
which reduces to (5.13)
89
Radial Equation for Coulomb Potential
A second recurrence relation can be obtained by using the lemma
d)t+l ( dp pB
=
(d)t+l p dp B
+ (t + 1) (d)t dp B
for any B, which may be easily proved by induction. Now take B
= pt e - p ,
so
By working this out, we obtain our second recurrence relation:
Lt+l(P)
= pL~(p)
+ (t + 1 -
p)Lt(p).
(5.15)
Now differentiate this once: L~+l(P)
= pL~'(p) + (t + 2 -
p)L~(p)
- Lt(p).
Set this expression for L~+l (p) equal to that given in Eq.(5.13). After rearrangement we find the following differential equation for the Laguerre polynomial: pL~'(p)
+ (1 -
p)L~(p)
+ tLt(p) = o.
(5.16)
The associated Laguerre polynomial* is defined by
L;(p) = Up)' L,(p) = Up)' e Up)' [p'eP
P ],
(5.17)
which is a polynomial of order t - s. Now by differentiating Eq.(5.16) s times, we find
pLt'(p) + (1 + s - p)L:'(p) + (t - s)L:(p) = O.
(5.18)
This is called the associated Laguerre equation. END: PROPERTIES OF LAGUERRE POLYNOMIALS
Looking back at Eq.(5.6), with the replacement Eq.(5.10), we find
pL"(p) + [2(£ + 1) - p] L'(p)
+ [n - £ - 1] L(p) = 0,
(5.19)
*Two definitions are used in the literature. Schiff and Ballentine agree with our definition, while Merzbacher and Gasiorowicz use L:(p) to mean what for us would be (_1)8 L!+t(p).
90
Hydrogen Atom and Charmed Quark
so we recognize L(p) to be the associated Laguerre polynomial, Eq.(5.17), with the identifications s = 2£ + 1, t = n + £. We conclude that the unnormalized radial wave function can be written
We see from Eq.(5.1) that the wave function corresponding to a given n, £ and m, can be written
'lfJnfm(i) = Cn£mr£e-K,nT L~f:/(2""nr)pr(cosB)eim4>, and Cnfm is a normalization constant chosen so that
J d3 rl'lfJn£mCr) 12 = 1. Here
Z
""n =nao -, where
ao =
1i2 ~ 0.529A me
-2
is the Bohr radius of the hydrogen atom. A means Angstrom, i.e., 10- 10 m. The ground-state wave function of the hydrogen atom (Z = 1), is
and the average value of r in this state is
We may express this by saying that the average radius of the hydrogen atom, in its ground state, is a little less than 10- 10 m. So far we have treated the nucleus as if it were infinitely massive. Let us rectify this. If (ife , pe) and (ijn, pn) are respectively the position and momentum operators for the electron and the nucleus, and these have masses me and m n , we can write the total Hamiltonian as
(pe)2 (pn)2 Z e2 H=--+------2me 2mn lije - ijn I
(5.20)
As in Sect. 4.6, we make a canonical tranformation to the variables describing the motion of the system as a whole, and the relative motion. This leads to
Radial Equation for Linear Potential
91
where j3 is the momentum of the atom, where pis the relative momentum, and ij = ije - ijn is the relative position of the electron with respect to the nucleus. The mass of the atom is
and the reduced mass of the electron is m, given by 1
1
1
-=-+-. m me mn
(5.21)
The first term in the Hamiltonian, p2/(2M), describes the free motion of the atom as a whole, while the remainder describes the motion of the electron with respect to the nucleus. In configuration representation, this is just 1i 2 -_'\7 2 2m
Ze 2 -r
which is what one would have for a fixed nucleus, except that the mass of the electron has been replaced by its reduced mass. Since Eq.(5.21) implies me m------ 1 + me/m n ' and the ratio of the electron and the proton masses is 1/1836, the correction is small (but experimentally measurable in spectral frequencies).
5.2
Radial Equation for Linear Potential
In the early 1960's the quark theory was proposed, according to which the proton, the neutron, and heavier particles (called collectively hyperons), were postulated to be bound states of three quarks. Thus the proton is made of two 'up' quarks and one 'down' quark, with electric charges lei and -llel respectively, e being the charge of the electron. The neutron is made of one up and two down quarks. Some of the hyperons were known to be strange, in that they could be produced copiously, but decayed only slowly. This was attributed to the existence of a quantum number that was playfully dubbed 'strangeness', which can be changed in a weak, but not in a strong interaction. A third, strange quark was introduced to take care of the hyperons; it has charge -llel, like the down quark, but it also carries the strangeness quantum number. Besides the baryons (Le., the nucleons and the hyperons), there are mesons, whose exchange is responsible for the nuclear force. These mesons are made of one quark and one antiquark. Thus the positive pion is du, while its antipartjcle, the negative pion, isud.
i
92
Hydrogen Atom and Charmed Quark
Towards the end of the 1960's it became clear that, while the quark model brought order into much of the particle chaos, it failed to explain the absence of a particular kind of weak process, namely the decay of particles that should disintegrate, according to the quark model as it then was, via a weak current that carries no electric charge. To explain the absence of such a weak neutral current, a daring hypothesis was made in 1970, namely that a fourth quark exists, carrying a new quantum number that was called 'charm'. The charmed quark has charge ~Iel, like the up quark; and it proved possible to use the new symmetry that now existed between the two quark families [(up, down), (charmed, strange)] to eliminate the unwanted weak neutral currents. The postulated fourth quark implied the existence of new baryons and mesons containing the charmed quark. In particular the bound states of the system ee, involving the charmed quark and its antiparticle, should be especially noteworthy, since the lower levels were expected to be stable against strong decay. In 1974 the ground state of this system, called 'charmonium' (cf. positronium, Sect. 6.5), was found experimentally at an energy of 3.097 GeV. Excited states were quickly found, the first one being at 3.686 GeV. The ground state and the first excited state are both expected to have f = 0, the ground state corresponding to principal quantum number n = 1 and the first excited state to n = 2. We propose to solve the Schrodinger equation for the S wave function in order to calculate the mass of the charmed quark. To do this, we have to make a hypothesis about the force that binds e and e together. According to quantum chromodynamics (QCD), the theory of the strong interaction, the strong force does not fall off as the inverse square of the distance of separation, but is rather independent of distance. The reason for this is that the 'gluons' of QCD, whose exchange is responsible for the strong force, interact strongly with each other in such a way that a one-dimensional tube or string is formed as one tries to separate a quark from an antiquark. The field lines do not spread out, as they do in electrodynamics, and so the force remains the same, no matter how far apart the quarks are. A constant force corresponds to a linearly rising, positive potential, so the radial Schrodinger equation for charmonium reads
, , {+ 2m f(f+1)} ul(r) 1i 2 [E - Ar] r2 ul(r) = 0,
(5.22)
where A is the (unknown) strength of the linear potential, and m is the reduced mass of the ee system. This reduced mass is m = m c /2, me being the unknown mass of the charmed quark (the calculation of the reduced mass is as in the analogous case of the deuteron). To study the two lowest bound states, we set
93
Radial Equation for Linear Potential
l . 0, and we change variables as follows:
(5.23) 1
where j3 =
[..\mc/ 1i?] 3.
This yields
d2uo de 2
-
(5.24)
euo = O.
This equation may be solved by Fourier transformation. Set
(5.25) then in p-space the Schrodinger equation (5.24) reads
.duo(p) dp
't
2-
+ P Uo (p) =
(5.26)
0,
with solution
The Fourier transform is
uo(r(e) )
C [ : dp exp i
2C
lp{ + ~p3"
10= dp cos [p{ + ~p3l .
(5.27)
The integral here is a representation of the so-called Airy function. It is known to tend monotonically to zero as -+ 00 and to oscillate for negative having in fact an infinite number of zeros. In the next section, we will introduce the WKB approximation, which can be used to study the Airy function; for the moment we write
e
e,
(5.28) and note that the first two zeros of the Airy function,
are given by
6
=
-2.338
6
=
-4.088.
(5.29)
94
Hydrogen Atom and Charmed Quark
These zeros can be found by integrating the expression Eq.(5.28) numerically; the function exists also in Mathematica. The following command line:
Plot[AiryAi[x], {x, -5, 5}] yields the graph that is shown in Fig. 5.1. We will gain more insight into the nature of the Airy function, as well as estimates for the zeros, when we look at the WKB approximation.
2
Fig. 5.1
Since r = 0 corresponds to u(O) = 0, translates into
~ =
4
Airy Function
-j3E / A, we see that the physical requirement,
which are the bound-state energies. The masses of the bound states of charmonium are accordingly mn = 2mc + En, since the rest masses of the two quarks must be added to the nonrelativistic bound-state energy, En (we use units in which the speed of light is unity). Hence (5.30) The experimental masses, expressed in units of GeV (= 10 9 eV), are ml
= 3.097
m2 =
3.686,
(5.31 )
the zeros being given in Eq.(5.29), and we can solve Eq.(5.30), with n = 1,2, for
95
WKB Approximation
the unknowns .A and mc. We find, for the mass of the charmed quark,
mc
=
m26 -m16 (~
2 1-6
)
~ 1.155GeV.
This value has been used in quark model calculations involving higher angular momenta, and also for charmed mesons (i.e. mesons where the charmed quark is combined, not with its antiparticle, but with 'ii, d or s). Consistent results are obtained.
5.3
WKB Approximation
The semiclassical, approximate treatment of the Schr6dinger equation that was introduced by Wentzel, Kramers and Brillouin (WKB), has been used in the past to obtain approximate solutions to problems of physical interest. Since the widespread introduction of high-speed computers, its popularity has waned, but we will employ it in this section to understand the asymptotic behavior of the Airy function that was defined in the previous section in connection with the cc bound states. Suppose first that the difference between the potential and the total energy, V(r) - E, is positive, and let us define
K(r)
= +V~": [V(r) - E] ,
so the S wave Schr6dinger equation can be written U~ (r) - ~2(r)uo(r) = O.
Now introduce the new quantity, ¢, according to
Uo (r) = e - 4> (r) ¢'
,
= ..j~2 + ¢" ,
(5.32)
which is in fact equivalent to the Schr6dinger equation. As a first approximation to Eq.(5.32), set ¢"(r) = 0, so
4>'
~~,
and now improve the approximation by setting ¢" =
~'
in Eq.(5.32), giving (5.33)
Hudroaen Atom and Charmed Ouark _
96
This equation integrates to
and hence we find (5.34) the missing lower limit of the r'-integration accounting only for a normalization factor. To see how this works out for the cc bound states, we recall that we had scaled the Schrodinger equation, with the linear potential, to yield Eq.(5.24). In our present notation, with e in place of r, and /'\,2 = e, we see from Eq.(5.34) that, for e > 0, (5.35) with arbitrary normalization. For negative e, we set e = lei e±i7r, corresponding to a rotation from positive to negative values of e, the upper over an anticlockwise, the lower over a clockwise arc in the complex e-plane, yielding (5.36) The required value of uo(r), which would be obtained by integrating the Schrodinger equation from positive to negative values of e, is the real part of Eq.(5.36), so (5.37) Expressions Eq.(5.35) and Eq.(5.37) give indeed the correct asymptotic expressions for the Airy function, respectively for e -+ 00 and e -+ -00, up to normalization. The first two zeros of the expression Eq.(5.37) are obtained by firstly setting the argument of the cosine equal to ~ and secondly to 3;. This yields the estimates
97r] 1 = 161 ~ [8
2
2.32
161 '"
[2~'r =
4.08,
which compare well with the zeros of the Airy function that were quoted in Eq.(5.28).
97
Exercises
5.4
Exercises
Problem 1 The associated Laguerre polynomial is defined by
LHp)
= Up)' eP
UJ
[pte-Pl·
(1) Prove the recurrence relation
L:+1(p) = pL:+l(p) (2) Work out Lt(p) for t them graphically.
+ (s + t + 1- p)L:Cp) - SL:-l(p).
= 0,1,2,3 and all integral values of s, and sketch
Problem 2 Is the electron in a hydrogen atom on the average further away from the proton when it is in the 2P orbital than when it is in the 28 orbital? Problem 3 Calculate the wavelength of the 2P -+ 18 transitions in deuterium. Use these masses for the deuteron and the electron: mdc2 = 1875.6 MeV, m e c2 = 0.51100 MeV, a = e2/(hc) = 1/137.036. Problem 4 Do the bound-state eigenfunctions corresponding to the following potentials form a complete set? Motivate your answers. (1) Attractive Coulomb: VCr) = -e 2 /r (2) The above plus a harmonic oscillator: VCr) = Ar2 - e2/r Problem 5 <'1 .. /:c.1~~",,-) Calculate all the nonvanishing matrix elements (OOOlreI2£m) , where Infm) is the normalized hydrogen atom stationary state corresponding to principal quantum number n, angular momentum quantum number f, and magnetic quantum number m, while x = r sin cos ¢ is the x-coordinate, reckoned from the centerof-mass of the proton and the electron. Problem 6 Let (r8) stand for the mean value of r8 in the (nfm) state of the H atom.
e
(1) With ao s
= 1, derive the recurrence relation
+1
_(rS ) n2
-
(2s
+ 1)(r
S-
1)
1 + -s[(2f + 1)2 - s2](r s - 2 )
4
=
o.
(2) Calculate the mean distance of the electron from the proton, and its standard deviation, y' (r2) - (r)2. For a given n, which values of f do the mean and the standard deviation have extremal values?
98
Hydrogen Atom and Charmed Quark
Problem 7 An electron is in the ground state of tritium (H3). A nuclear reaction changes the nucleus to that of He 3. (1) Calculate the probability that the electron remains in the ground state of He 3 • (2) What is the probability that the electron becomes free?
Problem 8 Consider the transformation
where u.e(r) satifies the radial equation for the hydrogen atom. Show that V.e(r) satisfies the radial equation for the isotropic harmonic oscillator. Discuss the relation between the solutions of the Schrodinger equation for the Coulomb and for the oscillator potentials.
Problem 9 The alkali atoms have an electronic structure which resembles that of hydrogen. In particular, the spectral lines and chemical properties are largely determined by one electron. A model for the potential in which this electron moves is V(r)
2
e = --;:-
(
1
+;b)
Calculate the energy levels.
Problem 10 Let if and p be the position and momentum operators of the electron in a hydrogen atom. The Runge-Lenz vector is defined by .... 1 e2 N = - [p' /\ L - L /\ p] - -if, 2m q --+
--+
--+
where L = if /\ P is the angular momentum operator, where q = where H is the Hamiltonian.
1
[if· if] 2, and
(1) Show that [L, H] = 0 = [N ,H], so that both Land N are conserved. (2) Show that the Hamiltonian has a larger symmetry than that of rotations in three dimensions. Identify this larger symmetry. (3) Show that the energy eigenvalues depend on only one quantum number.
Chapter 6
Spin and Addition of Angular Momenta
Consider the f = 1 (or P-wave) states of a system like the hydrogen atom. Since
with m = -1, 0, 1, L3 can be represented by a 3 x 3 diagonal matrix, with matrix elements
Here the principal quantum number, n, which must be greater than 1, is suppressed. On the other hand, we see from Eq.(3.14) that L±i1,m)
= nJf(f + 1) -
m(m ± 1)i1,m ± 1),
with f = 1, and so it follows that L± are represented by the non-diagonal matrix elements
L+'m
=
(1,m'iL+i1,m)
= nJ2 -
m(m + 1)8m',m+l = Lr:: m'.
(6.1)
In explicit form, the matrices representing the f = 1 orbital angular momentum operators are
L+
= nV2
010') ( 0 0 1 000
L3
=n
( 1 0 0) 0
o
0 0
0 -1
.
These matrices satisfy the algebra (cf., Eq.(3.6) and Eq.(3.10)) 99
Spin and Addition of Angular Momenta
100
6.1
Spin
The angular momentum of an electron in a hydrogen atom is not entirely due to its orbital motion. The electron has also an intrinsic angular momentum that is called its spin. To account for the observed doubling of many spectral lines of hydrogen - that is unexplained in our treatment so far - we suppose that the electron can exist in two spin states, called 'up' and 'down'. We write the wave-function, 'lj;, as a two component object,
corresponding to this spin degree of freedom. Let § be the operator corresponding to the spin angular momentum. We suppose that it satisfies the same algebra as the orbital angular momentum, since we shall wish to add the two to give the total angular momentum,
Thus we require
(6.2) Let
Is, m)
be a simultaneous eigenvector of 8 2 and 8 3 , i.e.,
where in place of e we have written s, the spin quantum number. According to the general analysis, we know that s is limited to the possibilities 0, ~, 1, ~, ... , while, for a given s, m = -s, -s + 1, ... ,s.
Since the spin is not an orbital angular momentum and cannot be written in the form q 1\ p, there is no reason now to exclude the half-integral values, as was the case for orbital angular momenta. Indeed, since we want just two states,
101
Addition of Angular Momenta
corresponding to spin 'up' and 'down', (with respect to, say, the z-axis), we must have s = ~, so that there are two possible values for the eigenvalue of 8 3 , namely -~n, and +~n,. In view of the fact that S31~,m)
= mn,I~,m),
with m = -~,~, 8 3 can be represented by a (2 x 2) diagonal matrix, with elements
8;- ,m I
Since s(s
+ 1) =
O,m/IS31~,m)
= mn,8m/,m'
~, we have (compare Eq.(6.1))
S;;/m - (~,m/IS±I~,m) = n,J~ - m(m± 1)8m / ,m±1' Explicitly,
I t is conventional to use the Pauli matrices, -1, )
;
o
"3
=(
~ -1o );
in terms of which
and so we find i
for 6.2
= 1,2,3 .
Addition of Angular Momenta
Consider the problem of adding two independent angular momenta together: -
J
-a
-b
= J +J
.
They might both be spins, or both orbital angular momenta, or, most interestingly, the spin, §, and the orbital angular moment, I, of the same particle. By
102
Spin and Addition of Angular Momenta
independence we mean that all components Jf commute with all components Jj. Hence
+ [Jf, Jj] in,€ijkJk + in,€ijkJZ = [Jf, Jj]
in,€ijkJk,
J J. J
so that
is a genuine angular momentum. We know from Chapter 3 that J2 _ therefore commutes with each component, J i . Consider simultaneous eigenfunctions of J2 and J 3 • There are two more operators that commute with 2
.... a
both J and J 3 , namely J
.... a
.J
.... b
.... b
and J . J . For example
....a ....b ....a .... a 2[J . J ,J . J ] .... a
2 [Jf , J
.... a
b
= 0.
. J ]J i .... b
.... b
In obtaining the first line, we have thrown away J . J , since that commutes with all components of J ,and also J . J ,since any operator commutes with itself. In the second line, we have used the fact that Jf commutes with J a, and ......"a ......"a then that J . J commutes with every component Jf. It is even easier to show that J3 = Jf + J~ commutes with Ja . Ja. ~a
~a
~a
....a
--.a
.... b
.... b
....
....
We can find simultaneous eigenvectors, jja,l;j,m), of J .J ,J .J ,J.J, and J 3 :
+ 1)n,2Ija,l;j,m) l ( l + 1)n,2Ija,jb;j,m) j(j + l)n? Ija, jb; j, m) ja(ja
mn,lja ,jb; j, m) From the general analysis of Chapter 3, we know that the m's and the j's must be integers or half-integers. The eigenvectors
.... a
.... a
.... b
.... b
b
of the mutually commuting operators J . J ,J . J , Jf, and J 3 form an alternative way of spanning the space of states belonging to given values of ja and jb. There must be a linear relation between these two sets of basis vectors:
" I·a I)'a ,)·b.· ,),m )- _~ ) ,).. b,m a,mb)(.a.b ) ,) ,ma,m bl'a ) ,)·b.· ,),m ) .
(6.3)
Addition of Angular Momenta
103
The numerical factors, (ja,jb,ma,mblja,jb;j,m), called Clebsch-Gordan coefficients, have been tabulated for different values of the angular momenta (see below, Sect. 6.6). It should be noted that J . J does not commute with Jf and Jg! On the other hand, by evaluating
(ja,jb,m a,mbIJ3Ija,l;j,m) = (ja,l,ma,mbIJg
+ J~lja,l;j,m)
in two ways, we obtain
m(ja,l,ma,mblja,l;j,m) = (rna +mb)(ja,jb,ma,mblja,l;j,m), which means that m = m a +mb whenever (ja, l, m a , mblja, l; j, m) f=. O. Hence the sum Eq.(6.3) is in fact only over a single variable, say m a , since m b = m-m a is not an independent variable. We know that the m's satisfy -j
<m <j
and so the maximum value of m, which equals j, is ja + jb. There is only one way to make this vector, namely by compounding the vector corresponding to m a = ja and that corresponding to m b = jb. However, the vector corresponding to the eigenvalue (ja + jb - 1)11, of J 3 can be made in two ways: either m a = ja and m b = jb - 1 or m a = ja - 1 and m b = jb. The eigenvalue (ja + jb - 2)11, can be reached in two ways, and so on. The strategy for working out the eigenfunctions of J and J 3 is to begin with the unique vector that corresponds to the eigenvalue (ja + l)11, of J 3 , namely
The vector corresponding to the same eigenvalue of J2, but to the eigenvalue (ja + jb - 1)11, of J3, can be obtained by using the lowering operator, J - I·a J ,).a)l· ) b,).b) ,
(6.4)
aside from normalization. Repeated applications of J _ to the above state generate all 2j + 1 = 2ja + 2jb + 1 eigenvectors of J that correspond to the J3 eigenvalues ranging from - j to +j. As we mentioned, there are two independent eigenvectors of J3 corresponding to the eigenvalue (j - 1)11,. One linear combination is reached by the operation of Eq. (6.4); the orthogonal linear combination is an eigenvector of J2 and J 3 corresponding to the eigenvalues (j - 1)211,2 and (j - 1)11, respectively. Lowering operators applied to this vector generate the family of eigenvectors corresponding to the angular momentum quantum number j - 1.
Spin and Addition of Angular Momenta
104
In an analogous way, the j - 2, j - 3, ... , families can be built up. The process halts when all (2ja + 1)(2jb + 1) independent state vectors have been used up. This occurs when j reaches the value lJa - jb I, since jO-+jb
2:
(2j
+ 1) =
(2ja
+ 1)(2l + 1),
j=ljO-~jbl
which may be easily proved from the formula n
2:(2j
+ 1) =
n(n + 2).
j=l
It follows that the possible values of j are
I'a - J·b I + 1, ... ,J·a·b J. = IJ'a - J·b I,J +J . This is a most important result. It means that, if we add the spin of an electron (8 = ~) to its orbital angular momentum (£ = 0,1,2, ... ), we obtain two possible angular momenta, j = .e + ~ and j = .e - ~, unless .e = 0, in which case there is only one value, j = ~. An interesting application is the addition of two spins, for example those of the proton and the neutron. The two possible values of the total angular momentum are j = ~ + ~ = 1 or j = ~ - ~ = O. The triplet j = 1 states are obtained by applying J ~ twice to the unique state corresponding to m = ~ for the proton and m = ~ for the neutron. This yields successively the m = 0 and m = -1 states of the triplet. The singlet state is obtained as the remaining orthogonal linear combination. The spin-one triplet turns out to have a loosely bound, but stable state, which we studied in Chap. 4. It forms the nucleus of heavy hydrogen or deuterium. We show now how to calculate explicitly the Clebsch-Gordan coefficients for the case ja = 1 and jb = ~. An application is to a P electron in a hydrogen atom (l = 1, s = ~). The unique state with m = ~ is
and to obtain the j
= ~,
m
=~
state we simply apply the lowering operator:
J _I ~, ~) = (J~
+ J~) 11) a 1~ )b •
From the formula
J±lj, m) = nJj(j + 1) - m(m ± 1)lj, m ± 1)
Combination of Spin and Orbital Angular Momentum
105
(cf., Eq. (3.14) ), we find
n h. . fi V2
2
~2 . !.I~ !.) = 1i~lo)al!.)b + h V2 h . ~2 +!.2 . !.11)al_ 22'2 2 2
!.)b
2'
that is to say
I~,~)
=
jilo)al~)b + jiI1)al_ ~)b.
The state I~, - ~) can be obtained by applying J_ again, or with less labor by applying J + to
The result in either case is
The state
jiI1)al_ ~)b _ jilo)al~)b, is orthogonal to I~, ~ ). It can only be the state I~, ~). By using J _ on this, one can obtain
I~, -~) = jil- l)al~)b _ ylilo)al- ~)b. Of course this last state could more easily be obtained as the remaining orthogonal combination. In general, however, for larger values of ja and jb, the second family of states, corresponding to j = ja+ jb -1, must be 'stripped off' by means of the lowering and/ or raising operators, after which one proceeds systematically to the third family, and so on.
6.3
Combination of Spin and Orbital Angular Momentum
Next we generalize the problem of the addition of the spin angular momentum of an electron with its orbital angular momentum in an atom. In the nonrelativistic approximation in which the orbital angular momentum is a well-defined quantum number, we can specify a stationary state by the quantum numbers n, j, I and mj. There are only two possibilities for j, namely I ± ~ (unless of course 1=0). In the case j = l + ~, let us write H L L3 S S3 H LS J J3 In; l, ~; I + ~,m + ~) = amln; I, m)LI~, ~)8
H L L3 S S3 + bmln; I, m + l)LI~, _~)8.
(6.5)
Spin and Addition of Angular Momenta
106
We propose to evaluate am and bm for generall. For brevity we define C(l,m) = Jl(l
+ 1) -
m(m -1),
so that from Eq.(3.14) we have J_ln;l, ~;l
+ ~,m+~) = hC(l + ~,m+ ~)In;l, ~;l + ~,m -~).
However, J_ can be written as L_
+ S_
(6.6)
and we find
+ S_)ln;l,m)LI~, ~)s = hC(l,m)ln;l,m _l)LI~, ~)s + hln;l,m)LI~, _!)s
(L_
(6.7) and
(L_
+ S_ )In; 1, m + l)LI!, _!)s =
hC(l, m + l)ln; l, m)LI!, _!)s .
(6.8)
On inserting Eq.(6.6), Eq.(6.7) and Eq.(6.8) into Eq.(6.5), we find an expression of the form
In; I, !; 1+ ~, m - !) = am-II n; 1, m - 1) L I ~, !) S + bm -11 n; l, m) L I ~, -!) S, ( 6. 9) with am - l
b m-l
=
C(l
+ ~,m +~)
_ am + bmC(I,m + 1) C(l + !, m + ~)
,
so that am
+ bmC(I, m + 1) amC(l, m)
(6.10)
Define dm = bm C (l, m
am
+ 1)
so that the recurrence relation Eq.(6.10) can be written in the simple form dm -
1
= 1 + dm .
In other words, dm = constant - m, and since bl we have dl = 0 and therefore dm=l-m.
=
0, as we see from Eq.(6.5),
Positronium
107
Thus
l- m
bm am
=
C (l, m+ 1)
/ l- m
=
Y1 + m+ 1 .
It follows from Eq.{6.5) that
In;l,!;l+~,m+!) =A{Jl+m+lln;l,m)LI!,!)S +Jl-mln;l,m+l)LI~,-!)S}, where A is a normalization constant that can be easily evaluated as 1/ J2l Accordingly, we have evaluated the coefficients in Eq.{6.5) as follows:
l+m+l 2l + 1
bm
=
V~ 2f+i.
+ l.
(6.11)
In the case j = 1 - ~, we clearly must have
In; l, ~; 1 - ~,m + ~) = bm In; l, m) L I ~, ~) S
- am In;
l, m
+ 1) L I ~, _ ~ ) S
since this is orthonormal to Eq.{6.5). The configuration representation of these results is (with mj
,
= m + ~) (6.12)
and
(6.13) where
Ylm
=
Ylm ((), ¢) is the usual spherical harmonic, and
x- = (
~
)
are the spin eigenfunctions in matrix representation.
6.4
Positronium
Dirac's relativistic equation for the electron, which we shall study in Volume 2 in detail, led its inventor to predict the existence of the positron, the antiparticle of the electron, possessing the same mass and spin, but differing in the sign of its charge. The electron is conventionally assigned a negative charge, -lei, while that of the positron is predicted to be +Iel. Moreover, the magnetic moment of the electron and positron are predicted to be equal in magnitude, but oppositely oriented with respect to the spin direction. Shortly after the theoretical postulation of the positron it was found experimentally, the first particle of antimatter to be identified.
108
Spin and Addition of Angular Momenta
The positron and the electron, being oppositely charged, form an electromagnetic bound state called positronium, which has a level structure resembling that of the hydrogen atom. However, there are some important differences between positronium and hydrogen. One difference is that, since the electron and the positron have the same mass, m, the reduced mass is m/2 (cf. the reduced mass of the nucleons in the deuteron, Sect. 4.4). This is quite different from the hydrogen atom, where the reduced mass of the electron (Sect. 5.1) differs from m by less than one part in a thousand. Nevertheless, the eigenvectors of the Hamiltonian are classified in the same way, with principal quantum number, n = 1,2,3, ... , angular number, f = 0,1, ... , n - 1, and azimuthal number, m = -f, -f + 1, ... , f. The energy depends only on the principal quantum number, being given by
En =
me 4 1 4h n
- - - 2 2·
(6.14)
The spins of the electron and the positron can be combined to give either 8 = 0 or 8 = 1. As we see from the Clebsch-Gordan coefficients for ~ 0 ~ (see below, Sect. 6.6), the 8 = 0 or singlet state is formed as the antisymmetric combination of the spin eigenvectors of the electron and the positron:
The ground and excited states of this system are called parapositronium, and they can be labeled by the symbol n 2S + 1 L J , where n is the principal quantum number, 8 = 0 for parapositronium, L is replaced by 8, P, D, ... , corresponding to f = 0,1,2, ... , and J is the total angular momentum, in this case equal to f, since 8 = o. An excited state of parapositronium decays by emission of photons, for example the transition from the 2 1P 1 state into the 1180 ground state is accompanied by the emission of a photon of wavelength
'Yhich is in the ultraviolet part of the spectrum. What subsequently happens is that the electron and the positron, whose wave functions have a large overlap in the S wave, annihilate each other, giving rise to two photons,
Each photon has the rest energy, mc2 , of one of the particles, so the wavelength
109
Exercises
in this case is
A=
h me
-
= 25
x 10- 13 m,
which is a very hard X-ray, a so-called '"'(-ray. Besides parapositronium there is also orthopositronium, in which the spins of the electron and the positron combine to give a spin one, or triplet state (see the table of Clebsch-Gordan coefficients). The structure of orthopositronium is richer than that of parapositronium, since the total spin, 8 = 1, can be combined with the orbital angular momentum to generate a number of states. Thus for n = 2 and C = 1, we have three distinct values of J, giving rise to 23 Po, 23 PI and 23 P 2 , which have however the same energy (in the nonrelativistic approximation). Excited orthopositronium de-excites into the ground state, 13 8 1 , and again mutual annihilation of matter and antimatter takes place, but in this case the two-photon decay is forbidden, and decay proceeds more slowly to three photons. The average lifetime of orthopositronium is 1.4 x 10- 7 sec., a thousand times longer than the parapositronium lifetime.
6.5
Exercises
Problem 1 Work out the Clebsch-Gordan coefficients for the combination ~ 0 ~. Problem 2 Show that
(1) an arbitrary matrix in two dimensions can be written as a linear superposition of the three Pauli matrices and the unit matrix. (2) no two-dimensional matrix anticommutes with all of the Pauli matrices.
Problem 3 Properties of the Pauli matrices:
(1) Prove TrO'"i = 0,
TrO'"iO'"j
= 28ij ,
TrO'"iO'"jO'"k
= 2i€ijk ;
= a + b . if,
then TrA = 2a and Trif A = 2b. (3) By using these formulas, show that (b . if )(2. if) = b ·2 + i(b /\ 2) . if (4) Write the following explicitly as 2 x 2 matrices: (i) eU1 , (ii) eiU1 , where 0'"1 is the first Pauli matrix. (5) Show that, if n is a unit vector, exp [iB n . if] = cos B + i sin Bn . if
(2) Show that, if A
Problem 4 Define N = btb, where bt is the Hermitian conjugate of the operator b, where bb t + btb = 1, and b2 = O.
110
Spin and Addition of Angular Momenta
(1) (2) (3) (4)
Is b hermitian? Is N hermitian? Show that N 2 = N and find the eigenvalues of N. Find a representation for b in terms of the Pauli matrices.
Problem 5 The spin interaction energy of an electron and a positron in a positronium bound state is proportional to 0"1 ·0"2, the spin matrices corresponding to the two particles. (1) Evaluate the interaction energies in the singlet state, 10,0), and in the triplet state, 11, m) , m = -1,0, l. (2) Suppose now that a weak, uniform magnetic field is applied to this system. Show that the triplet states 11,1) and 11, -1) are still eigenstates of the Hamiltonian, but that 11,0) and 10,0) are not. (3) Obtain the energies of the three levels. Problem 6 What are the spin and total angular momentum states of two deuterons in an arbitrary orbital angular momentum state, L? Problem 7 A particle is known to have spin one. Measurements of the state of the particle yield (81 ) = 0 = (82 ) and (83 ) = a, where 0 :s: a < 1. What is the most general possibility for the state? Problem 8 The spin of a neutron is aligned with the z-axis. What is the probability that (1) a spin measurement at an angle f) to z will yield ~ h? (2) a subsequent measurement (i.e., after a successful measurement as above) along the z-axis will yield a result - ~ h? (3) Suppose now instead that after the successful measurement (a) above, the measurement (b) is made but not recorded. What is now the pro bability that a subsequent measurement of spin along an axis at angle () to z will yield the result ~ h?
Problem 9 Suppose that two identical spin-half particles are created in the singlet state. They fly apart and the spin of one particle is measured in the direction a, the other in the direction b. (1) Calculate the relative frequencies of the coincidences R(up,up), R(up,down), R(down,up) and R(down, down), as a function of the angle between a and b.
111
Exercises
(2) Calculate the correlation coefficient, defined by
C(a, b)
= R(up, up) - R(up, down) - R(down, up)
+ R(down, down).
(3) Given two possible directions, a and a', for one measurement, and two possible directions, band b', for the other, deduce the maximum of the Bell coefficient,
B
=
C(a, b) + C(a', b) + C(a', b') - C(a, b').
(4) Show that this prediction of quantum mechanics is inconsistent with classical local realism. Problem 10 The Greenberger-Horne-Zeilinger (GHZ) state of three identical is defined by IGHZ) =
spin-~
particles
1
J2 [zd z: z: - z;zb z;] ,
zt
a
where is the eigenket of the z-component of the spin operator of particle belonging to eigenvalue ~ Ii (z-spin up), z~ is the eigenket belonging to eigenvalue - ~Ii (z-spin down), and similarly for the particles band c. Show that, if spin measurements are made on the three particles in the x- or y-directions,
(1) the product of three spins in the x-direction is always - ~ Ii3 , (2) the product of two spins in the y-direction and one spin in the x-direction is always ~h3. (3) Consider a prize game for a team of three players, A, B, and C. The players are told that they will be separated from one another and that each will be asked one of two questions, say X or Y, to which each must give one of two allowed answers, namely +1 or -1. Moreover, either (a) all players will be asked the same question X, or (b) one of the three players will be asked X and the other two Y. After having been asked X or Y, no player may communicate with the others until after all three players have given their answers, 1 or -1. To win the game, the players must give answers such that, in case (a) the product of the three answers is -1, whereas in case (b) the product of the three answers is + 1. (a) Show that no classical strategy gives certainty of a win for the team. (b) Show that a quantum strategy, in which each player may take one of the GHZ particles with her, exists for which a win is certain.
Spin and Addition of Angular Momenta
112
6.6
Table of Clebsch-Gordan Coefficients
I ~)al ~)b 11,1) 11,0) 11, -1) 10,0)
I ~)al-;1)b
l-;l)al~)b
1 0 0 1 1 0 v'2 v'2 0 0 0 1 1 0 -72 72 Clebsch-Gordan coefficients for
~
1-;1 )al-;l)b 0 0 1 0 0 ~
To save space, in the following tables we give only one half of the entries: to obtain the missing entries, make the substitution
li,iz) -+
Ii, -iz) & lia)alib)b -+ 1- ia)al- ib)b
For the entries marked at the right with '8', the missing entry is the same as the one given, while for those marked with 'A' one must change the sign.
11)al~)b
I~, ~) I~, ~)
lo)al~)b
11)al-;1 )b 0
1 0 1 0 VI y'3 I~ _1) 0 0 0 2' 2 0 0 0 I~,-~) 1 0 -y'3 I~, ~) VI 112' _1) 0 0 0 2 Clebsch-Gordan coefficients for 10
12,2) 12,1) 12,0) 12, -1) 12, -2) 11,1) 11,0) 11, -1) 10,0)
11)al1)b 1 0 0 0 0 0
11)alo)b 0
10)aI1)b 0
1
1
v'2 0 0 0
v'2 0 0 0
1
v'2 0 0 0
1
-y'2
11)al 0 0 1
v'6
0 0 0
l)b
8jA
8
A ~
lo)alo)b 0 0
8jA
v1
8
0 0 0 0 0
1 0 v'2 0 0 0 1 1 0 0 -73 .J3 Clebsch-Gordan coefficients for 101
0
A 8
Chapter 7
Approximate Methods
It happens all too often that an interesting physical problem is described by a Schrodinger equation that cannot be solved exactly. We must often resort to a numerical solution using a computer program, for example a Runge-Kutta integration. However, such 'brute force' approaches often obscure the physics, since they are usually not conducive to physical understanding. In this chapter we shall introduce two methods that, while they generally result in small numerical errors in the results, do often provide useful insight.
7.1
Perturbation Theory
The general idea is that the full Hamiltonian can be written as follows,
H = Ho
+ >"H1 ,
where the unperturbed Hamiltonian is amenable to exact treatment, but the full Hamiltonian is not. It is supposed that>.. is a small parameter. Let the unperturbed Schrodinger equation be designated
(7.1) In this section, we suppose that l4>n) is simple (i.e., it is not degenerate with another state); but the other states, l4>k) for k =J. n, may well exhibit degeneracy. Later we shall consider the more general situation in which l4>n) is degenerate too. The states span a Hilbert space, a fact that can be symbolized as follows:
I¢n)(¢nl
+L
I
I¢k)(¢kl = 1.
(7.2)
k
Here the sum is over all states except I¢n), the state that we wish to perturb. 113
A ppro:cimate Methods
114
The full Schrodinger equation will be written (7.3) If the perturbation HI is sufficiently well-behaved, the state-function, l7jJn(..\)), will lie in the space spanned by the I¢>n), and so by Eq.(7.2) we can write
l1/Jn(A)) = I¢>n) (¢>nl1/Jn(A))
+~
,
l¢>k)(¢>kl7jJn(A)) .
(7.4)
k
By taking a matrix element of Eq.(7.3) with the selected state (4)nl, we obtain
(7.5) or equivalently (7.6) To evaluate the numerator here, we multiply Eq.(7.4) by (4)nIHb obtaining
(4)nI H II1/Jn(A)) = (¢>nI H II¢>n)(4>nl1/Jn(A))
+ ~'(4)nIHII4>k)(¢>kl1/Jn(A))
.
k
Upon insertion of this expression in Eq.(7.6), we obtain
En(A) = En(0)
+ A(¢>nIHII¢>n) +
A """ (4)nl1/Jn(A)) ~ (4)nI H II4>k)(¢>kl1/Jn(A)).
(7.7)
We take a matrix element of Eq.(7.3) with a state other than (¢>nl, say (¢>kl:
(7.8) which implies
(¢>kl1/Jn(A)) = A (¢>kI H ll7jJn(A)) En(A) - EkO)
(7.9)
On substituting this into the last matrix element in Eq.(7.7), we find
En(A)
= E(O) + A(¢>nI H II4>n) + n
A2 ~'(¢>nIHII¢>k)(¢>kIHll1/Jn(A)). (4)nl1/Jn(A)) k En(A) - EkO) (7.10)
Note that Eq.(7.10) is exact. From Eq.(7.9) we see that the second term on the right-hand side of Eq.(7.4) is of first order in A, since En(A) - EkO) is 0(1), and hence. (7.11)
115
Perturbation Theory
From Eq.(7.6) or Eq.(7.7), (7.12) Inserting Eq.(7.11) and Eq.(7.12) into the right-hand side ofEq.(7.10), we obtain finally En()..) =
E~O) + )..(¢nIHll¢n) + )..2L:' 1(¢ZOI)Hll¢~~t + O()..3). k
En
(7.13)
- Ek
This is the required expression for a perturbed energy level, to second order, in the absence of degeneracy. The above treatment, based on the exact formula Eq.(7.10), is much more attractive than the expansions given in many textbooks. The second-order expression is often very difficult to calculate for interesting problems. An enormous simplification takes place if, for a given Ho and HI, one can find an operator, 0, such that (7.14) In this case
E~O) + )..(¢nIHll
E~O)
+ )..(
En
,
- Ek
(¢nIHll¢k)(¢kIOI¢n)
+ O()..3).
(7.15)
k
Now we can write the double sum by adding and subtracting the missing term corresponding to k = n:
,
L: (¢nIHll¢k)(¢kIOI¢n) k
k
where use has been made of the fact that the vectors I¢k) span the Hilbert space. Hence we find
As a first application of this remarkable formula, consider the case H = Ho+eEH1
,
where eE takes the place of the small parameter).. of Eq.(7.16), and where Ho
=
p2 2m
+ )..q2
Approximate Methods
116
i.e., the three-dimensional, charged, isotropic harmonic oscillator in a uniform electric field, £, as in Eq.(2.31}. Note that ,X itself is being used as a parameter in the unperturbed Hamiltonian, Ho. Since
it follows that Eq.(7.14) is satisfied in this case by
indeed [0, HoJ = HI is an operator identity. To use Eq.(7.16} for calculating the perturbation of the ground state, 1000), we need to obtain some matrix elements. The first one is
but this vanishes, since it is proportional to
see Eq.(2.3} and Eq.(2.6}. Thus the first-order perturbation is zero, and all that is left of Eq.(7.16) is Eo(£)
= ~liw +
ie 2 £2 21i,X (000Iq3P31000)
+ O((e£)3) .
From Eq.(2.3) and Eq.(2.6) again we have Q3P3
iii t 2 2 = liQ3 P3 = 2(a 3 - a3 + 1},
so (000Iq3P31000) = ~ili. We find then
in agreement with the exact result Eq.(2.33). In this case there are no higherorder terms.
7.2
Variational Method
For some problems, first-order perturbation theory is adequate. That is, we simply take Eq.(7.13) up to first order in the coupling, 'x: (7.17)
Variational Method
117
A simple improvement of first-order perturbation theory is the so-called variational method of Ritz, which is generally much easier to implement than is second-order perturbation theory. We therefore introduce the variational method at this point and illustrate it and first"'-order perturbational theory by means of an example. Let {I'l/Jn)} be a complete set of eigenfunctions of a Hamiltonian, H,
(7.18) n
The completeness condition Eq.(7.18) has been formally written as a discrete sum, which would be appropriate for a harmonic oscillator, for example; but for the hydrogen atom Hamiltonian there is, besides the sum over bound states, 00
n-1
l
L L 2:
l'l/Jnlm)(1/;nlml,
n=1l=O m=-l also an integral over appropriately normalized positive-energy continuum states
corresponding to electron-proton scattering. For a free particle, or one in a purely repulsive potential, we have only the positive-energy continuum and no discrete states. The sum in Eq.(7.18) is intended to symbolize all these possibilities in a concise form. Let I1/;) be any normalized state vector. Then the matrix-element
(1/;IHI'I/J)
L (1/; IHI1/;n) ('l/Jn I'I/J) n
n
(7.19) n
Here E1 is the smallest eigenvalue of H, the ground state. Note that essential use was made of the fact that ('l/JI'l/Jn)(1/;nl'I/J) = 1('l/Jnl1/;)12 is non-negative, and that En > E 1 • Hence
(7.20)
118
Approximate Methods
The variational method simply consists in trying different expressions for I'l/J) and choosing the form that gives the smallest ('l/JIHI'l/J): this is then the approximate value of El - it will be exact if and only if I'l/J) is the ground-state vector, otherwise ('l/JIHI'l/J) is an overestimate.
7.3
Anharmonic Oscillator
Consider an atom in a linear molecule, constrained to vibrate in one dimension. A correction to the usual harmonic force might well be expressed by the following Hamiltonian:
H
=
Ho
+ ,Xq4,
(7.21)
where the unperturbed Hamiltonian is p2 2m
mw 2 2 2
Ho= - + - - q . We know that the ground-state wave function of the one-dimensional harmonic oscillator is (xl'l/Jo)
= 'l/Jo(x) = exp ( - mwx2) 2n '
nw,
corresponding to the eigenvalue Eo = ~ the zero-point energy. In first-order perturbation theory, we calculate
6.E1
J~ dx x4 exp (_ mwx 2 )
0 -
00
dxexp -00
foo
Ii (mwx2 ) --li-
,
(7.22)
where we have divided by the squared norm of 'l/Jo(x), since we gave 'l/Jo(x) in a form that was not normalized. By applying two partial integrations to Eq.(7.22), we can easily calculate
so that, to first order,
Eo('x)
3,Xn 2
= ~nw + 4mw 2 2·
Now let us try the variational method. We need to calculate
(7.23)
119
Anharmonic Oscillator
for some suitable trial function, 'IjJ(x) = (xl'IjJ). It is natural to try exp (_~JLX2), where JL is left as a free parameter to be varied in order to minimize the variational estimate of the ground-state energy. We have successively
and to complete the analysis, we minimize the above expression by varying JL. To compare numbers, let us choose ;: = 1 and ~ nw = 1. In these units, the perturbation result reads
The variational result is
which can be minimized either by writing a computer program, or even by hand, solving a cubic equation on the way. The Mathematica program,
v[lam_, mu-l := mu/2 + 1/(2 * mu) FindMinimum[v[l, mu], {mu, I}]
+ 3 * lam/(4 * mu * mu);
returns {1.40332, {mu-> 1.6717}} for)' = l. Some results are given in the following table: ).
0.1 0.2 1.0
Perturbation 0.17500 1.15000 1.75000
Variation 1.06620 1.12061 1.40332
Numerical 1.06529 1.11829 1.39234
Ground-state energy of the anharmonic oscillator
120
Approximate Methods
The column headed 'Numerical', is the result ofa computer calculation that integrates the Schrodinger equation numerically, and may be taken to give the correct answer to 6 significant figures. Note that the variational method gives quite an accurate answer, even for the rather large value A = 1 of the anharmonic coefficient. As a matter of fact, the second (and higher) order perturbative corrections do not give improvements at this value of A, quite the contrary! The reason is that actually the perturbation series diverges, a consequence of the fact that the x4 'correction' to the harmonic potential x 2 is always dominant for sufficiently large x, no matter how small A is. Nevertheless the perturbation series is not useless, since it is in fact asymptotic, and may be resummed (see Exercise 10 at the end of this chapter).
7.4
Stark Effect
Let us now consider the splitting of some of the spectral lines when a source of incandescent hydrogen is exposed to a strong, uniform electrostatic field. If E is the external electric field, the perturbation induced in the hydrogen atom Hamiltonian can be written AHI = eE.r = e£z = e£r cos
e
(7.24)
where £ = lEI, and where we have chosen the z-axis to be parallel to the electric field (cf., Eq.(2.31) et seq. and Eq.(7.16) et seq., where we considered the Stark effect for the harmonic oscillator). The ground state of the hydrogen atom is simple, the normalized wavefunction being
where for convenience we have chosen the unit of length to be equal to the Bohr radius, ao. It is easy to see that the first-order Stark shift of the ground state is zero, SInce
which vanishes because the integrand of the e-integral is odd:
I.
w
dO sin 0 cos 0 = 0 .
The second-order Stark shift is not zero, and it can be calculated by using the trick of Eq.(7.14) (see Exercise 9).
121
Stark Effect
The four degenerate states corresponding to the principal quantum number n = 2 have the following normalized wave functions:
7j;2oo(i)
2-~(47rr-~(2 - r)e- r / 2
7j;211(i)
-2- ~ (87r)- ~ r e- r / 2 sin () e i ¢
7j;210(i)
2- ~ (47r)- ~ r e- r / 2 cos ().
(7.25)
It is necessary to generalize our treatment of perturbation theory to the case that the energy level that one wishes to perturb is degenerate. We shall give the formalism in first order only, treating the general situation, and then applying it to the n = 2 hydrogen states. Suppose that, instead of Eq.(7.1), we have
[Ho - E~O) ll¢~) = 0,
where, for the given n, {I¢~)} is a finite set of N independent eigenstates, which can always be arranged to be mutually orthonormal, i.e.,
The completeness relation Eq.(7.2) is replaced by
L)¢~)(~I + I:'I¢k)(¢kl
= 1,
k
j
where the first sum is over the states belonging to the eigenvalue E~O) and the second is over all remaining states (note that degeneracies in these states are also allowed). Continuum or scattering states can also be included, in which case an integral term is also present. Instead of Eq.(7.4) we now write
l1Pn(A)) =
I:Qjl~) + L:'I¢k)(¢kl1Pn(A)) , k
j
where
The analog of Eq.(7.5) is now
(¢~I[E~O)
+ AHI - En(A)]LQjl
To first order we have then (7.26)
Approximate Methods
122
The system Eq.(7.26) constitutes N homogeneous algebraic equations for the N numbers 0i. The condition for the existence of solutions is (7.27) This equation may be construed as an Nth-order equation for the unknown EnCA). In some cases it will have N distinct solutions: the perturbation has accordingly lifted the degeneracy. If the perturbation, HI, respects some of the symmetry that was responsible for the degeneracy in the first place, however, some of the roots of Eq.(7.27) will be equal, so that the degeneracy is then only partly lifted. Let us apply this formalism to the n = 2 states of the hydrogen atom, Eq.(7.25), with the Stark perturbation Eq.(7.24). It is immediately clear that all diagonal terms (1Pnlm Ir cos BI1Pnlm) vanish, again because the integrand of the B-integral is odd in every case. (1Pnlmlr cos BI1Pnlm/) vanishes if m =1= m', since then the -integral is zero:
(27r
10
d¢exp [i(m' - m)¢]
=0.
The only terms that do not vanish are
(1P200 Ir cos BI1P21O) *
3~7r
!
d3 rrcosBr(2 - r)e- r cosB = -3 .
The first-order result Eq.(7.26) takes on then the matrix form
o 0 3 0
0200 0211 0210 021-1
= -e£
o
0 0 0 3 0 0 0 o 0 0 0
0200 0211 0210 021-1
Evidently the (211) and (21-1) states are not shifted in first order, i.e.,
but the S-waves satisfy
)
(
0200 ) 0210
= 0.
(7.28)
123
Near Degeneracy
The condition for the existence of a solution is the vanishing of the determinant of the above matrix, and this yields the two solutions (7.29) As can be seen from Eq.{7.28), the lower of the two energies (the minus sign above), corresponds to the symmetric superposition, (!'lfJ200) + !'lfJ210))/y'2j while the upper of the two energies corresponds to the antisymmetric superposition, (!'lfJ200) -!'lfJ210))/y'2. Notice that the i-degeneracy has been lifted, but not the m-degeneracy. 7.5
Near Degeneracy
Due to relativistic effects that we have neglected, the 210 and 200 states of hydrogen are not exactly degenerate even before a Stark field is applied. Write H 0!'lfJ200 )
[E~O) - ~] !'lfJ200)
HO!'lfJ210)
[E~O)
+ ~]!'lfJ210) .
Now add a perturbation )"HI (for example the Stark term eEz). Some linear combinations of the 200 and 210 states will be eigenstates of the full Hamiltonian,
Contracting this against the 200 and 210 states, we find (
E~O)
- Ll- E()")
)..('lfJ210!HI !'lfJ200)
)..(t2oo!HI!'lfJ210) ) ( Q200 ) = 0 . E~) + Ll- E{)") Q210
(7.30)
For simplicity, it has been assumed that the diagonal terms ('lfJ2oo !H1!'lfJ200) and ('lfJ210 !HI !'lfJ210) vanish, as is the case for the Stark effect matrix elements. In order for Eq.{7.30) to have a solution, the determinant of the matrix must vanish; and this condition leads to the following two solutions for the perturbed energy levels:
In the case that
~
-+ 0, we recover the result of the degenerate case: E{)..)
= E~O) ± )..! ('lfJ200 !HI !'lfJ210)! ,
Eq.{7.29) being an example. In the other extreme, ~ E{)..)
= E~O) ± Ll + O{)..2) ,
»
)..! ('lfJ2oo !HI !'lfJ210)!,
124
Approximate Methods
i.e., there is no first-order Stark shift of two non-degenerate levels. We see then that the two cases of simple and of degenerate levels are really two extremes of the situation in which two levels that are close together are perturbed. If the perturbation is large compared with the unperturbed spacing, then there is a first-order shift, as in the degenerate case; whereas if the spacing is large compared with the perturbation, each level acts on its own, as it were, and there is only a second-order shift.
7.6
Ammonia Molecule
A related problem is the ammonia molecule, in which two inequivalent configurations that can be visualized are related to one another like a right and a left hand (i.e., they are inequivalent mirror images of one another, enantiomorphs). These visualizable states are however not eigenfunctions of the Hamiltonian. By symmetry, we have
where Hd is real. Moreover, the freedom we have in choosing the phases of the above vectors will be used to ensure that
is real and positive. Eigenstates of the Hamiltonian are formed as superpositions of these states:
[H - E][ar!¢r)
+ al!¢l)] =
0.
After contraction with (¢l! and (¢r!, we obtain
)( :~ )
=0.
We set the determinant to zero in the familiar fashion, and this yields
With the plus sign, we find a r = al; and with the minus sign, we have a r = -al. The energies are accordingly and
125
Exercises
De-excitations are possible from the upper to the lower state. photon is in the microwave region:
v = E+ - E_ h
=
The emitted
2Hc ~ 24 GHz, h
where 1 GHz = 10 9 sec-I. This frequency yields a wavelength of 1.25 cm., and the transition is used in the ammonia maser, the first device based on Microwave Amplification by Stimulated Emission of Radiation, which was made in 1954, the grandfather of modern lasers. It is salutory to realize that the ammonia molecule is an example of a stationary state that we cannot picture with our classical intuitions. It is not correct to say that the molecule flips back and forth between the states designated by IcPl) and IcPr), for this would not be a stationary state. Rather, the two states that we can visualize are coherently superposed in two ways, namely the symmetric and the antisymmetric combinations. This situation calls to mind the Schrodinger cat paradox, in which one is tempted to imagine a superposition of a dead and a live cat! In Schrodinger's own playful words: "Die ~-Funktion des ganzen Systems wiirde das so zum Ausdruck bringen, da:B in ihr die lebende und die tote Katze zu gleichen Teilen gemischt oder verschmiert sind." ("The ~-function of the whole system would express the matter in such a way that in it the live and the dead cat would be mixed up or smeared out in equal portions.") Although the idea of a superposition of macroscopic states (like those of a live and a dead cat!) does not seem to correspond to anything that we observe, this should not mislead us into thinking that such a thing does not happen on the atomic scale. Indeed, we can only really understand the microwave radiation from a maser, and the analogous light from a laser, on the basis of quantum theory, that constructs from two states which are each visualisable ('anschaulich' in the German la:nguage) unvisualizable eigenstates of the Hamiltonian that have slightly different energies.
7.7
Exercises
Problem 1 Let 81 and 82 be the spin operators of two spin ~ particles. Then
is the spin operator for this two-particle system.
126
Approximate Methods
(1) Consider the Hamiltonian Ho = (8; + 8; - 8'D/h.2. Determine the eigenvalues and eigenvectors of this Hamiltonian. (2) Consider the perturbation HI = sIx - S2x. Calculate the eigenvalues of Ho + >"HI in first-order perturbation theory.
Problem 2 Two spin-half particles, 1 en 2, have the unperturbed Hamiltonian
A perturbing Hamiltonian is brought to bear of the form HI
= B(SlxS2x
-
SlyS2y).
(1) Calculate the eigenvalues and the eigenfunctions of Ho. (2) Calculate the exact eigenvalues of Ho + HI. (3) By means of perturbation theory, calculate the first- and the secondorder shifts of the ground state energy of H 0, as a consequence of the perturbation HI. Compare these results with those of (2).
Problem 3 Suppose that a hydrogen atom is exposed to a uniform electric field, f, and a parallel, uniform magnetic induction, jj. Consider the first excited energy level, corresponding to n = 2, which is quadruply degenerate if f = 0 = jj . (1) Show that in general the level is split into four nondegenerate energy levels. (2) For what values of f and jj are there instead only three levels, and what are the degeneracies of these levels? (3) For what values of f and jj are there only two levels, and what are the degeneracies of these levels? In the answers, assume that the fields are strong compared with the fine structure effects, and ignore the complications caused by the electron spin.
Problem 4 An electron is in a superposition of the Coulomb and the 3D oscillator potential: e2
VCr) = -- + >..r2. r
(1) What is the shift of the ground state energy to first order in >... (2) How large must>.. be in order that the ground state have zero energy? (3) What are the first-order shifts of the n = 2 energies?
127
Exercises
Problem 5 Consider the two-dimensional oscillator Hamiltonian,
Ho
1
p2
= -+ -mw 2m 2
2 2
q
'
with the perturbation HI = Aqlq2 . (1) Calculate the energy shift of the ground state in first- and second-order perturbation theory. (2) Calculate the energy shift of the first excited state in first-order perturbation theory. (3) Calculate the first two energy levels exactly. Problem 6 Show that a function 'l/J(x) exists, such that, if V(x)
1
00
-00
dx'l/J*(x)
[
< 0 for all
n,2 d -2 d 2 + V(x) 1'l/J(x) m x 2
<
-00
< x < 00,
o.
Use this result to prove that an attractive potential in one dimension always has a bound state. What can be said in 2 and 3 dimensions? Problem 1 Given a central potential with bound states, use the variational method to show that
where min (Be) is the smallest energy eigenvalue corresponding to angular momentum t. Problem 8 Estimate by means of the variational method the ground-state energy of an electron in a screened Coulomb potential, e2
V(r) = - - e- ar r
•
Take e- br as test function, and if possible use a computer to find the minima. Make a table of the numerical results, showing E(a)j E(O) as a function of a.
128
Approximate Methods
Problem 9 Calculate the Stark shift of the ground-state energy of the hydrogen atom, to second order in eE, where
H = H
°+ e Er cos e,
H o being the hydrogen atom Hamiltonian. Prove first that, in configuration space, [0, H o]e- r = if cos @e-=;r , where 0 = (1 + ~r)r cos e. The Bohr radius has ..t ~t b een set to umty. - ~ t ces9 t. Problem 10 Consider the anharmonic oscillator in one dimension defined by
H(>..) = L
2
2m
+ q2 + >..q4
and set m = ~ 11,2 , so that the ground state energy of the oscillator in the absence of the quartic perturbation is unity, i.e., Eo(O) = 1. (1) Calculate the ground state energy of the oscillator to second order in perturbation theory,
giving explicit expressions for the coefficients al and a2. (2) From this second-order perturbative result, calculate the [1,1] Pade approximant, Eb1 ,11. The [M, N] Pade approximant to a function that has . . a power senes expanSIOn,
is defined by the relation j[N,Ml (>..)
2:~=o bn>..n 1 + 2::=1 cn>..n M+N
2:: an>..n + O(>..M+N+1) ,
n=O
which defines the coefficients bn and Cn uniquely in terms of the first M + N + 1 coefficients an. (3) Make a table of EbO,l] (>..) ,EbO,2](>..) and Eb1,1] (>..), giving 6 significant figures, for each of the cases>.. = 0.1, 0.2, 1.0, and, if you have the computer facilities, write a program to calculate E o(>") to sufficient accuracy to guarantee the correctness of 7 significant figures.
Chapter 8
Scattering Theory
Up to this point we have mainly been interested in bound states, which are described by square-integrable wave functions. In this chapter we introduce the subject of scattering, in which the energies are positive and the scattering solutions are not square integrable.
8.1
Lippmann-Schwinger Equation
Consider solutions of the Schrodinger equation for positive energies, E =
n;.!2, (8.1)
To convert this partial differential equation into an integral equation we define the Green's function,
(8.2) where
E
-+ O. In polar coordinates this is
_1_1 (2 ) 7r
3
_1_
(27r)2 1 --
(27r)2
00
q
0
2d
roo dq
io
1
1 (27r)2ir
00
0
q
111' dB sm. B 1211' dA.. exp( iqrk2 cos .B) 0
If'
0
r i-I 1
q2
q2 - k 2 - if
2
q -
-
'If
dz eiqrz
q2 e iqr _ e- iqr dq-----------------q2 - k 2 - iE iqr
roo
i-oo dq (q 129
q eiqr
k - if')(q + k
+ if')
,
(8.3)
Scattering Theory
130
where E = 2kE' and k = +.vk2. This integral can be performed by closing the contour of integration in the upper half-plane of the complex variable q. The only pole of the integrand inside the contour is that at q = k +iE', and its residue yields 27ri k eikr -Gk(r) (27r)2ir 2k 1 eikr --- . (8.4)
47r r
We now claim that
(V'2
+ k 2 ) Gk(r) = (V'2 + k 2 )
ikr _e_
41Tr
= -8 3 (r).
(8.5)
We shall now give a direct proof of this equation, using distribution theory methods. Note that the special case k = 0 corresponds to Eq.(4.7)-(4.9), and so the following proof, in which no appeal will be made to the Fourier theorem, may be used in place of that to be found in Chapter 4.
Proof of Eq.{8.5) Let g( r) be an infinitely differentiable test function of finite support. Then
by two partial integrations. In spherical polars, this can be written
f /.00 dO
eikr 1 8 2
r 2dr-- 8 2rg(r).
orr r
(8.6)
We have neglected the angular parts of the Laplacian, since they give zero contribution. This follows from the fact that 7r /.27r /.o sin ede 0 d¢
/.00 dr e~kr . {sin1 e 8e8 sin e 8()8 + sin21 e 8¢2 82 } g(r) 0
can be written
/.= where
and
dre'kr
{[~ dA(i') + [
dOB(i')} ,
Lippmann-Schwinger Equation
131
Clearly A and B are zero because of the properties of the test-function space: it is enough that the first derivative of 9 with respect to () is bounded, and that the first derivative of 9 with respect to ¢ is continuous, so that its values at ¢ = 0 and ¢ = 271" are the same. By two partial integrations of Eq.(8.6) with respect to r, we see that
100 o
. 02 dr e'tkr -rg(r)
[ eikr ~rg(i)] 00 _ ik [00 dr eikr ~rg(i)
or2
or
[eikr
Jo
0
(1 + r! -
or
ikr) g(r)[ - k 2 fo~ dreikr rg(r).
Because of the continuity of the test functions at r = 0 and their vanishing as r ---+ 00, it follows that the integrated term above reduces to ~g(O). Thus we have shown that
!
d3 xg(r)V
In other words
2e~r = ! dO [-g(O) _ k21~ r 2dr e~r g(r)]
!
d3 xg(i) [\7 2
eikr
+ k 2 ] -r-
=
-471"g(0) ,
which is equivalent to Eq.(8.5). END OF PROOF
The general solution of Eq.(8.1) satisfies the Lippmann-Schwinger equation:
1/Jk(i) = 1/J2(i) -
~":
!
d3 r'Gk(1f - i'I)V(i')1/Jk(i') ,
(8.7)
where 1/J£(i) is the general solution of the free Schrodinger equation (8.8)
A special solution, corresponding to a plane wave transmitted along the z axis, is obtained by setting 1/Jf(i) = eik .r = eikz , so that in this case 1/Jk(i) = exp (if. i) -
2~ n~
!
d3 r'Gk(1f -
r'I)V(i')~k(i') .
(8.9)
This is the form of the equation that we use in a discussion of a typical experiment, in which a collimated beam of monochromatic particles is directed at a target, often in the form of a small sample of material.
132
Scattering Theory
Proof that Eq.(8.7) and Eq.(8.8) yield a solution of Eq.(8.1) Let 'l/Jk(i) be any solution of Eq.(8.7). Then
o
(V2
+ k2)1/J~(i)
(\7 2
+ k 2 ) {'I/I.(r) + :~
(V 2
+ k2)1/Jk(i) -
{ \7 2
1 ~": 1
d"r'G.(IT - r'I)V(r')'I/I.(r') }
d 3 r'b 3 (i - i')V(i')1/Jk(f")
+ k 2 - :~V(r) } 'I/I.(T).
(8.10)
END OF PROOF
Typically r = iii refers to distances of the order of meters or more, in the laboratory, whereas r' = Ii' I refers to atomic or nuclear dimensions. Hence it makes sense to consider an approximation that is valid when r > > r',
Ii - i'l ~ ry'l -
-,
....
2i· i' fr2 ~ r - ~, r
so that
(8.11) where k' = ki fr. Note that k' is a vector of the same length as k, but it is in the direction of i, whereas of course k itself is parallel to the z-axis. On condition that V(i') tends to zero sufficiently quickly as r' ---+ 00, we find asymptotically
(8.12) where, according to Eq.(8.7), the scattering amplitude is
fk((},4» = -
2: 1d 2
3 r'
e-ik'or'V(i')1/Jk(i').
(8.13)
Here () and 4> are the polar angles of i with respect to the incident wave. The form of Eq.(8.l2) is that of an outgoing spherical wave, superposed on the incident plane wave.
133
Scattering Gross-section
8.2
Scattering Cross-section
The wave function describing a scattered particle satisfies the time-dependent Schrodinger equation,
and for a real potential the Hermitian conjugate of this equation is
1
. a'!f;* * H='IjJ * [ - n2V +-2 +V . -1,n-='IjJ 2m
at
Therefore
(8.14) H
+-
-+
where V V - V . The rate of change in time of the probability of finding the particle within a given volume V is
(8.15) where A is the surface of the volume V and the suffix N indicates the normal component to this surface. The GauE theorem has been used here. Now consider the scattering solution (8.12), which can be rewritten (8.16) since
k'
= ki
and therefore
Ir.
For large
T,
Scattering Theory
134
where
In a moment we are going to average the flux over a small area, ~A = r2 ~O (see Eq.(8.18) below); and we introduce a normalized, infinitely differentiable smearing function, g(r), that has support ~A. We then replace Sk(r) by
and by repeated partial integrations in the z = cos 0 integral, we obtain at least three negative powers of ikr, on condition that fk (0, ¢) is at least thrice differentiable in the variable z. Under this condition, the smeared Bk(r) can be neglected, since it is o(r- 2 ). Accordingly, in leading orders (8.17) Suppose now that N particles per unit area per unit time fall upon a scattering center. The number that pass each second through a small area ~A is
[*
inN +-+] nN [.... - 2m ~A 'ljJk \1 'ljJk N = m ~A k
2] + k' r2Ifk (O, ¢)I N
(8.18)
If ~A is normal to k, the dominant term is the first one, the other being negligible for large r. The flux (the number of particles per unit area per unit time) is accordingly 1i~k in the forward direction. In any other direction, the
term k can be neglected. The reason is that the incident beam is never really plane, for that would imply an infinite lateral extent. Instead it is finite, although enormous compared with atomic scales, but at a macroscopic distance from the target, in a direction specified by the polar angles 0 and ¢, the detection apparatus lies outside the domain of the incident beam. Here only the second term in Eq.(8.18) contributes, and so there the flux is n~klfk(O,¢)12/r2. This flux tends to zero of course as r --+ 00, since the scattered energy is spread out in all directions. However, at radius r the area subtended by an increment of solid angle dO = sin OdOd¢ is r 2dO, this incremental area being normal to the vector k', so that finally the energy that is scattered into the solid angle dO, divided by the incident energy per unit area per unit time, is (8.19)
Born Approximation
135
This is called the differential cross-section. In practice one does not deal with just one scattering center, but with a target full of them. The measured flux of particles at angles ((), ¢) must then be divided by the number of scattering centers that are to be found per unit area of the incident beam. In most atomic, nuclear or high-energy scattering experiments there are three typical scales that are very different from one another: (1) Size of the target particle (perhaps a proton ~ la- 15 m). (2) Area of the target and incident beam (perhaps a square centimeter). (3) Distance of the target from the detection apparatus (often many meters). Under these conditions the approximations we have made are very good. The total cross-section,
a=
I
da dOdO'
(8.20)
is the scattered energy divided by the incident energy per unit area, both for unit time and per scatterer. The cross-section therefore has the dimensions of an area. It is a basic nexus at which the theoretician and the experimentalist meet: the former calculates the scattering amplitude from his theory and predicts the cross-section from Eq.(8.l9), while the latter measures the intensity of the scattered particles and deduces the cross-section. Such measurements serve to verify or falsify a theory and hence to support or undermine it.
8.3
Born Approximation
The Born approximation consists in iterating Eq.(8.7) once to yield (8.21) For an incident plane wave, ¢2(f) = eik .r , this leads to (8.22) cf., Eq.(8.l3). We see that the scattering amplitude in Born approximation is proportional to the Fourier transform of the potential. In the case that the potential is a central one, we use polar coordinates to perform the angular integrals in Eq.(8.22). It is advantageous to choose the z' axis to be parallel to the vector k - k " for then the ¢' integration is trivial, and
Scattering Theory
136
we obtain
_ ~ foo r,2 dr'V(r') (rr sin 8' d8' eilk -k'lr' cos 0'
n Jo
2m .... 2 Ik - k'in
-....
Since
Ik
- k' I =
k
kk - rr
1
00
Jo
.... r'dr'V(r') sin{lk.... - k' Ir'} .
(8.23)
0
= k· /2 - 2 cos 8 = 2k sin
v
~2'
(8.24)
it follows that the amplitude depends only on the scattering angle 8, and, in this approximation, it is real. We have then
fk(8)-1i~ m
e loo r'dr'V(r') sin {2kr'sin-28 } •
ICsm'2
10
. (8.25)
Consider the potential,
V( r ) -- g e -p.r/Ii , r
where J1, is the mass of the 7T'-meson or pion. This is the form guessed by Yukawa, as a model for the strong interaction between protons and neutrons. The integral Eq.(8.25) can be worked out straightforwardly, giving
The differential cross-section is accordingly
Note that n has dropped out of this expression, and that the sign of 9 is immaterial: it makes no difference to the cross-section whether the interaction is attractive or repulsive. The Coulomb potential is the special case J1, = 0, 9 = e2 • One then obtains dO' e4 dO. - 16E2 sin4 !!. ' 2
which is precisely the classical result that was applied to the a-Au scattering measurements by Rutherford. This led him to the postulation of the existence of a pointlike source of positive charge in the atom, the atomic nucleus.
137
Unitarity and Phase Shifts
8.4
U nitarity and Phase Shifts
When the potential is central, the wave function can be written
(8.26) where Dim are constants. If we choose the x3-axis to be parallel to i, we see that all terms with m -# 0 give zero in Eq.(8.13). Thus fk(f), ¢) A(f) does not depend on ¢, and we can write Eq.(8.12) as
(8.27) effectively setting Dim = 0 for m expanded in a Legendre series,
-# o.
The scattering amplitude can now be
00
= 2::(2£ + l)h(k)Pi(cosf)
fk(f)
(8.28)
,
£=0
where P£(z) is the Legendre polynomial, Eq.(3.56). Here h(k) are called the partial-wave amplitudes, and those corresponding to £ = 0,1,2,3, ... are called respectively the S, P, D, F, ... waves. The incident plane wave can be expanded as follows: 00
'ljJ~(i)
=
eikrcos(J
= 2::(2£ + l)i£j£ (kr)P£ (cos f)
,
(8.29)
£=0
and since r is large, we may use the asymptote Eq.(4.61) to obtain
(8.30) Proof of Eq.(8.29) Let us write 00
eipz
= 2:(2£ + l)A£(p)P£(z) ,
(8.31)
£=0
where we wish to evaluate the partial-wave amplitudes A£ of the plane wave. The tool we need is the orthogonality of the Legendre polynomials:
1/1
"2
-1
8kf dzPk(z)P£(z) = 2£ l'
+
(8.32)
Scattering Theory
138
which is the special case m = 0 of Eq.{3.62). From Eq.{8.31) it then follows that
(8.33) where £ partial integrations have been performed, the integrated terms all being zero. Consider
Bl(p)
=
i:
dZ(Z2 - I)le;pz.
Clearly i
r i-I 1
_i2
dz{z2 _ l)f-l ze i p z
!
1
-1
d (z2 - 1)f. ipz _ ~ B ( ) Z £ lpe - 2£ f P ,
(8.34)
where an integration by parts has been performed. Hence Bf
=
2£~..!!-. B f - 1 = 2f£! [~..!!-.]f Bo = 2f+l £! [~..!!-.]f sin p . pdp
pdp
pdp
p
It follows from Eq.(8.33) that ( _ip)f 2f+l£! Bf(p) .)f [1 d]f --=1,Jf sin p .f . ( p, ) ( -lP -pdp p
(8.35)
where the spherical Bessel function was given in Eq.( 4.52). END OF PROOF
We can also expand the wave function in a Legendre series: (8.36)
139
Unitarity and Phase Shifts
cf., Eq.(8.26) with Dim = (2£ + 1)8mo , and we know that ui(r) satisfies the radial Schrodinger equation (4.3), which can be written
"() + {k2 -
ui r
On combining Eq.(8.27) -
£(.e + 2 1) r
-
2m 2 V ()} r 11,
Ui () r -- 0 .
(8.37)
Eq.(8.36), we see that (8.38)
However, we know that the regular solution of Eq.(8.37) behaves like r i +1 for small r. Suppose we normalize ui(r) so that
for r ---t 0, which is always possible, since the Schrodinger equation is a homogeneous differential equation. Now Eq.(8.37) is a differential equation that can be integrated out from small to large r values. Because V(r) is a real potential, the solution ui{r) must remain real, and for r -+ 00 it has the asymptotic form (8.39) where Ci and 6i are necessarily real, since otherwise ui(r) could not be real. This can be rewritten
so that, on comparing with Eq.{8.38), we find (8.40) where the phase shift, 8i {k), is real and the scattering matrix, Si{k), is unitary. The scattering amplitude can be written (8.41) and so the partial-wave series Eq.{8.28) can be written in the standard form
100. fk(()) =
k 2:(2£ + 1) etOl(k) sin 8i(k)Pi(COS ()) . i=O
(8.42)
140
8.5
Scattering Theory
Phase Shifts and Resonances
If V(r) = 0 for r > a, the solution of the free radial Schrodinger equation there can be written as in Eq.( 4.25),
(8.43) where we stress that the constants Ae and Be depend on.e. For large r, we use the asymptotic formulae Eq.(4.61) and Eq.(4.62) to conclude that
ul(r)
~ Alsin (kr - ~l)
However, from Eq.(8.39) we have Ae
_BlCOS (kr _ ~l) .
(8.44)
= Cecos6e and Be = -Cesin6e, so Be Ae
tan6e = - - .
(8.45)
The continuity equation at r = a has the same form as Eq.(4.30), except that it applies here for any positive E, i.e., any real k. We write it in the form (8.46) where it is understood that the radial wave function on the left, ue(a), is to be obtained by integrating the Schrodinger equation from r = 0 out to r = a. Let us define, in terms of that function,
,eek) = ~ (u~(a) _ ~) k ue(a) a Then, with use of Eq.(8.45), we find
(k)
= j~(ka)
,e
- tan6en~(ka) je(ka) - tan6e ne(ka) ,
(8.47)
Il(k)jl(ka) - j~(ka) Il(k)nl(ka) - n~(ka)
(8.48)
from which it follows that tan6l(k)
=
From the small-k behaviors of the spherical Bessel and Neumann functions, Eq.(4.55) and Eq.(4.57), we find tanol(k)
rv
k 2l+1 ,
and if we stipulate that the phase shift begins at zero, OleO) = 0, this will also be the threshold behavior of the phase shift itself. It can happen that tan6l(k) rises through positive values and passes from plus to minus infinity, corresponding to
141
Neutron-Proton Scattering
a zero in the denominator of Eq.(8.48). If such a zero occurs at k = write, for k ~ ko,
~o,
we may
f/2 tan8e(k) ~ ko - k ' so
Se(k) -1 !£(k)
=
=
2ik
e 2i6t -1 2ik
~
1 f/2 ko ko - k - if/2
In the vicinity of this resonance, the scattering amplitude will be dominated by the contribution from this partial wave, so from Eq.(8.28), we read off
A(())
~
(2£ + l)!£(k)Pe(cos()).
Thus the differential cross-section is approximated by du
2
dO. ~ {( 2£ + 1) IIe (k ) IPe (cos ())} ,
(8.49)
and, from Eq.(8.20), the total cross-section is approximately (8.50) Note that u(k) has its maximum value at k = ko and is one half of its maximum at k = ko ± f /2, whence .r is called the width of the resonance, ko its mass. It is an unstable state, somewhat analogous to a bound state. The expression (8.50) is called the Breit-Wigner resonance formula.
8.6
Neutron-Proton Scattering
In the proton-neutron system, there is a spin-1 bound state, the deuteron, as we saw in Chapter 4. This is a triplet state, since the z-component of the spin has three possible values, namely -1, 0 or 1. However, when one combines two spin! states, there is also a spin-O component, as we saw in Chapter 6. There exists therefore both a triplet and a singlet scattering amplitude, and the effective potential is not necessarily the same in the two states. The singlet potential turns out to be too weak to produce a bound state, but it does give rise to a resonance in the S wave, as we shall see. Let us consider the square-well potential, of width a, as in Section 4.4. For r < a, the regular solution of the Schrodinger equation is
Uo (r)
= A sin kr ,
(8.51)
Scattering Theory
142
where k 2 = mn(E+ VO)/h,2, Vo being the depth of the potential well (recall that the reduced mass of the two-nucleon system is half of the nucleon mass, m n ). For r > a we write
uo(r)
= C sin(kr
+ 80) ,
(8.52)
where k 2 = mnE /h,2, this being equivalent to a superposition of sin kr and cos kr. The matching condition at r = a is logsin(ka + 80 ), dda logsinka = ~ da
(8.53)
+ 80 ) = k
(8.54)
which leads to k tan(ka
tan ka.
-2 2 2 We also have the relation k = k 2 + k5 , where we have defined ko = mn Vo/h, , and we recall from the discussion in Section 4 that the condition for a bound state at zero energy is k5 = 1r 2 /4. Eq.(8.54) can be recast in the form
ktanka - ktanka
£
tanuo =
,
(8.55)
k [tan koa - koa] . ko
(8.56)
k
+ k tanka tanka
and as k ---+ 0, we have k ---+ ko, and so tan 80
rv
-
The triplet S-wave scattering length is defined by Q:ot
. £ / = - hm uo k = a - -1 tan koa .
k~O
ko
(8.57)
From our study of the deuteron bound state in Chapter 4, we have K,
=
v'mnEB h,
-
-
= - k cot ka ~
- ko cot koa ,
(8.58)
so that Q:6 = a + 1/ K,. The triplet scattering length is known to be 5.4 F, and we recall from Eq.(4.37) that 1/K, = 4.4 F, so
a
= Q:ot - -1
= 5.4 - 4.4
= 1 F,
(8.59)
/'i,
in agreement with the value we used for the evaluation of the width of the potential well from the deuteron binding energy. A similar treatment will now be given for the singlet amplitude. We will assume the same width of 1 F; but it will turn out that the depth, ¥Os, will
143
Neutron-Proton Scattering
be less than the corresponding quantity for the triplet amplitude. The singlet S-wave scattering length is Q
s_ 1· s:s/k _ tan ksa - a k ' o - - 1m Uo k-tO
s
where k; = mn Vos Ih 2 • The singlet scattering length has been measured to be -23.7 F; and with a = 1 F we find tanks ks
-- = a -
Qo
= 24.7 .
The Mathematica command line FindRoot[Tan[k]/k
= 1.54, we find
returns {k -> 1.54459}. With ks
vt =
== 24.7, {k, 1.5}]
k;h2 = k 2 (hc)2 = 101 MeV. mn s m n c2
This potential strength is indeed just too small to produce a bound state; recall that the strength required to produce a zero-energy bound state is 105 MeV, and the strength in the triplet state that produces the deuteron (with a binding energy of 2.226 MeV) is 125 MeV. The singlet S-wave phase shift is positive and rises rapidly to ~. The resonance is the reason that the singlet scattering length is so much larger in absolute value than the triplet scattering length. The S-wave scattering length, either for the triplet or the singlet state, yields lim fk(O)
k-tO
= lim 60k(k) = -Qo . k-tO
Thus the threshold value of the total cross-section is
a
= k-tO lim
J
dOlfk(O)1 2
= 47rQ~.
For the triplet and the singlet cases, this leads to
= 366 F2
at
47r(5.4)2
as
47r(28.7)2 = 7058 F2 .
Since the triplet amplitude ~ spin one - has three states (83 = -1,0,1), while the singlet amplitude has only one, the total cross-section, averaged over all four states at zero energy, is
144
Scattering Theory
which agrees well with the measured value. The proton-proton scattering amplitudes are very similar to the neutronproton scattering amplitudes; but there is an important difference in the calculation of the cross-sections from the amplitudes. This is due to the fact that the incident and target protons are indistinguishable, so there can be no difference, in the center-of-mass frame, between scattering through an angle of () and 7r - (). Classically one would simply add the cross-sections for these two angles; but in quantum mechanics we must add the amplitudes. A general property of fermions (i.e., particles with half odd integral spins, like the proton, which has spin ~) is that their wave functions, and scattering amplitudes, are antisymmetric under interchange of two identical fermions. This property is part of what is called the spin-and-statistics theorem, which we will explain more fully in Volume 2. The spin part of the wave function of two protons in a triplet state is even under interchange of the particle labels, indeed from the table of Clebsch-Gordan coefficients for the case ~ ® ~ we have
11,1)
I ~)al
11,0)
~ (I ~)al_ ~)b+ 1_ ~)al ~)b)
11, -1)
~)b
I_~)al_~)b.
Since this spin part of the wave function is even under interchange of the particle labels (a and b), the space part of the wave function, which multiplies the spin part, must be antisymmetric, so that there is overall antisymmetry under interchange of the two protons. This is achieved by writing
Ifk(()) - fk(7r - ())1 2 Ifk(())1 2
+ Ifk(7r - ())1 2 - 2Re[fk*(())fk(7r - ())] .
The spin part of the singlet wave function is on the contrary odd,
so that the spin-and-statistics theorem decrees that the space part must be even. In this case we write
+ fk(7r - ())1 2 Ifk(())1 2 + Ifk(7r - ())1 2 + 2Re[fk* (())fk (7r - ())] . Ifk(())
145
Exercises
8.7
Exercises
Problem 1 The Legendre polynomial is defined by
e old 2 e ( ) Pe(Z) - Pe (Z) = 2et'! dz (Z - 1) . (1) Show that Pe+l(Z) = zPe(z) + ze2;11 Pl(z). (2) Demonstrate (t' + 1)Pe+1 (z) - (2t' + l)zPe(z) + t'Pe- 1 (z) = O. (3) Work out Pe(z) for t' = 0,1,2,3, and sketch them graphically.
Problem 2 Calculate the phase shifts, oe(k), for the hard core potential, V(r) = and V(r) = 0 otherwise.
00
if r < a
(1) Simplify the result for the S-wave. (2) Calculate the total cross-section in the limit k--+ O. (3) Calculate the total cross-section in the limit k --+ 00.
Problem 3 Prove the optical theorem for elastic scattering, that is aT
!
da dO. dO.
47r
=
T1mfk(0) .
l ..
Problem 4 .t1.f (~" ~t) When there is inelasticity, the partial wave scattering function, Se(k), can be written 1}e.9R{20e), where 0 < 1}e(k) < 1. Show that the optical theorem is still true, where now aT is the sum of the elastic and inelastic cross-sections. Problem 5 Calculate the scattering amplitude in Born approximation with the following spherically symmetric potentials, V(r):
(1) -g()(a - r).
(2) - ~:: exp ( - ~ ). Problem 6 Calculate the differential scattering cross-section in Born approximation for the scattering of two identical particles that move under a Yukawa interaction:
146
Scattering Theory
Problem 7 Consider two different particles of spin
~.
The interaction potential is
V(r) = W(r)81 ·82, where 8j = !1iiJj and W(r) is a continuous, scalar function. Calculate in Born approximation the ratio of the probability that, after scattering'one finds ~ 1 0'( ~l z _.L~ 1 SIz = 2h, ... s'joz\ given that, before scattering, Slz = 2h, = -S2z· f"'cc~~ I Problem 8 Let cSt(k) and 8t(k) be the f-wave phase shifts corresponding to the central potentials V(r) and V(r), respectively. Prove that sin(cS- t
-
cSt) =
- -2m 2-
h, k
1
00
dr[V(r) - V{r)]ut(r)ut{r) ,
0
where ut(r) and ut{r) are the radial wave functions. If V(r) = V(r) + AW(r), where W{r) is suitably well-behaved, sketch a perturbation method to be used for small A. Problem 9 By considering the differential operators £±
d
= tanh r ± dr '
solve the S-wave Schrodinger equation for the spherically symmetric potential V (r)
h,2
= - - cosh - 2 r . m
Calculate the S-wave phase shift. Are there any resonances and/or bound states? Problem 10 A nuclear physicist measures a scattering process, and finds on analyzing his data that the following phase shifts (in radians) describe his data well: cSO = 0.88779, cSl = -0.56640, c)2 = 0.23447, and cSt = 0 for f > 3. (1) Draw a graph of the differential cross-section against cos (). (2) A research student of the nuclear physicist analyses the same experimental results but she arrives at the following phase shifts: cSo = 0.10930, cS l .. -0.76993, cS2 = 0.23447, again with all the higher phase shifts vanishing. Draw a graph of the differential cross-section against cos (). Has the research student, or the nuclear physicist made a mistake, or are they both correct? Explain. (3) If there is an infinite number of non-zero phase shifts, are they uniquely determined by the differential cross-section?
Chapter 9
Atomic Physics
9.1
Two-Electron System
Consider a negative hydrogen ion, H-, a helium atom, He, or a positive lithium ion, Li+. In all cases, we have to do with two electrons, at positions, say, i 1 and i 2, and a nucleus of charge Ze, where Z = 1,2 or 3. The Hamiltonian operator in configuration space is
where
In the first instance, we concentrate our efforts on the calculation of the ground state .of this system; and the strategy will involve three steps: • Neglect of the electron-electron repulsion, V12 • • Treatment of V12 by first-order perturbation theory. • The variational method.
Neglect of repulsion term We choose a separable wave function, 'ifJ(i b i 2)'
'ifJ(i 1, i 2)
= 'ifJnlilml (i d'lfJn i m (i 2), 2
147
2
2
Atomic Physics
148
for which the energy is
Here En is the nth energy level of the hydrogen atom, except that e2 Z replaces e2 , i.e., eVZ replaces e. Thus
where a = ~: ::::::: 1~7 is the fine-structure constant. Hence the (nb n2) energy level in the present approximation is
122(1-+1) mc.2 n n
E=--Za 2 The ground-state energy (nl = n2
2 1
2 2
= 1) is therefore given by
.!£ = -Z 2 a? mc 2
(9.1)
The ground state of the hydrogen atom is -~a2mc2 = -13.6 eV; it follows that the above energy is 2Z 2 times that, i.e., -27.2Z2 eV. For the helium atom, Z = 2, so E = -108.8 eV, to be compared with the experimental value of -78.975 eV. We conclude that the electron-electron interaction term, which has so far been neglected, must have quite a large effect - at any rate if quantum mechanics is to be successful in describing the helium atom. First-order perturbation Treating V12 as a perturbation, we have in first order
Here 'l/JIOO is the normalized ground-state wave function of hydrogen, after the replacement of e by eVZ, so
'l/J100(r) =
~ ~ exp [-z~l ' V7r (~) ao ao
where 1i2 me2
ao = - - .
(9.2)
149
Two-Electron System
After the scale change ....
2Z ....
i
Pi = -ri, ao we find 6E =
e 2Z
327r2ao
II .
= 1,2,
d3pl d 3p2
IPl-P2I e
-(PI+P2)
.
In performing the P2-integration, at fixed P1, we use polars and choose the z-axis in the direction of Pl ' Hence Ip 1 - P21 = Pt + p~ - 2P1P2 cos 0 and the integrand is independent of the azimuthal angle, cp, which can accordingly be integrated out to give 27r. Moreover, since
J
d ./ 2 2 d oV Pl + P2 - 2P1P2 cos () =
PlP2 sin 0 2 0' P1P2 cos
J P12 + P22 -
it follows that
(7r
sinOdO
J0 J Pt + p~ -
_1_
PlP2
2P1 P2 cos 0
P1
[J
pi
+ p~ -
2PlP2 cos
0] 7r 0
+ P2 -IPl -
P21
P1P2 2 max(p1' P2) . Hence AE -_ e2 Z / d3 Ple - PI 87r ao
o
1°O 0
p~dp2 maX(Pl, P2)
e -P2 .
The angular integrals in the Pl-integration can now be done trivially, giving a factor 47r, and the remaining integrals are elementary: 6E
1 1
00
Z e 2 -2
ao
P12dP1 e -PI
0
{l 0
00 pidp1 e- P1 e 2 -Z ao 0 5 e4 m - Z 28 1i '
1.
00
PI
P22 -dP2 e P1
-P2 +
1.
00
PI
P22 -dP2 e P2
-P2}
P2dp2 e- P2
PI
In proceeding from the first to the second lines above, we reversed the order of the P1 and P2 integrations, in the first piece of the P2 integration only, and then we relabeled the integration variables (Pl f-t P2)' It is convenient to write 6E _ ~Z 2. me2 - 8 a,
(9.3)
Atomic Physics
150
and if we add this to Eq.(9.1) we find
E
+ .6.E =
TinC 2
-z (z - ~) a?
(9.4)
8
For He, Z (Z - ~) = 11 and so this energy is 121 times the H ground-state energy, i.e., -74.8 eV. This is much closer to the experimental value of -78.975 eV, being out by about 5%. To get better agreement, one could in principle go to the second order perturbation theory; but this involves a lot of numerical work. Variational method We recall from Eq.(7.20) that, given a Hamiltonian H and any vector l'if),
where E1 is the smallest eigenvalue of H, the ground state. We shall now apply the method to the helium atom. On comparing Eq.(9.1) and Eq.(9.4) we see that the effect of the electron-electron repulsion is to reduce the effective value of Z. This effect is called screening, and it is physically understandable at a classical level: each electron does not 'see' the bare nuclear charge all the time, since the other electron is sometimes between it and the nucleus, thus reducing its attraction. This classical picture must not be taken too literally, since really we have a quantum system of two indistinguishable particles; but it suggests that we try for I'l/!) the ground-state vector of a twoelectron atom, in which the electron-electron interaction has been omitted, but where the atomic number Z is replaced by a parameter Z, less than Z, to account for the screening. Z will be chosen at the end to minimize ('l/!IHI'l/!) and hence to optimize the estimate of the ground-state energy.
~et ~lOo(r) be the ground-state wave function Eq.(9.2), but with Z replaced by Z. Then 'l/!100 satisfies
- 2] 'l/!100(r) nt.2 [-2Tin - v 2 -Ze= r
1
-2
--Z a 2
2
2 TinC
'l/!100(r)
so that [-
;,,2
2
ze 2 ] -
2Tin V1 -~ 'l/!lOo(rd
e 2 - -Z 1 - 2Q 2TinC2][ (Z- - Z)'l/!lOo(rd
r1
2
and similarly for H2~lOo(r2)' Hence, if we choose
151
Two-Electron System
then
The only new term to be calculated is the first one, and this is elementary:
The last term has the same form as the first-order perturbation: . we can take over the result Eq.(9.3), of course changing Z there to Z. On gathering these results together, we find
(¢IHI¢) = Z [Z - 2Z + ~] a 2mc2 = {(Z -
z+
156 )2 -
(Z -
156
)2}a 2mc 2 ,
and this must be greater than the ground-state energy for all Z. The minimum of the right side evidently occurs when Z = Z - 156' and we obtain El ( 5)2 2 -< a. - - Z - 16 mc2
(9.5)
For Z = 2 we find El < - 77.5 eV, which is better than first-order perturbation theory, being about 2% too high. The variational result can be improved by fine-tuning the trial function. The following parametrization was used by Hylleraas in 1929, when quantum mechanics was still young:
1f(i' 1, i' ,)
= exp [_iT1 :
,,]
.L ajkmh + T2)i(T1 - T2)klf
1 -
i' ,1 m
.
},k,m
The parameters are the coefficients ajkm and Z. With just one term, aooo, the calculation is the one we have just completed. With three terms, Hylleraas obtained -78.98 eV. Notice that the result Eq.(9.5) has the same form as Eq.(9.1), but with Z replaced by Z - 156' This indicates that the effective charge 'seen' by each electron is reduced by 5e/16, as a result of the presence of the other
Atomic Physics
152
electron. For He the effective nuclear charge is thus 27e/16. Even better results are possible, for example with 13 terms in the Hylleraas series (see table below).
Method Oth perturbation 1st perturbation 1 term variation 3 term variation 13 term variation
Formula (Z = 2) -Z 2a 2mc2
-Z(Z - ~)a2mc2 -(Z - 156 )2a2mc2
Experiment
eV. -108.8 -74.8 -77.5 -78.98 -79.015 -78.975
Ground-state energy of the He atom For the hydrogen anion, H-, the effective charge is 11e/16. This gives a ground-state energy for H- of - (~i)2 a 2mc2 = -0.47a 2mc 2 ~ -12.86eV, whereas the ground-state energy of a hydrogen atom, -~a2mc2 ~ -13.60eV, is slightly lower. If this were the last word, it would mean that H- should be unstable into decay into a neutral H-atom and a free electron. However a treatment a la Hylleraas gives -14.34 eV, in good agreement with experiment. In the following table, we summarize the results of a 13-term variation calculation, with the experimental results, for the cases Z = 1,2,3,4. The agreement is truly impressive, the differences being always less than one per cent, attributable in all cases to relativistic effects that have been neglected in our treatment.
HHe Li+ Be++
Variation -14.34 -79.02 -198.1 -371.6
Experiment -14.35 -78.98 -197.1 -370.0
Ground-state energy in e V of H-, He, Li+, Be++
On the next page, output from a FORTRAN program (Schmid, 1987) for His displayed.
153
Two-Electron System
THE GROUND STATE OF THE HYDROGEN ANION BY THE HYLLERAAS METHOD Nuclear charge Z = 1 Energy of H-atom in ground state:
E = -13.60 eV
Energy of H-anion E = -14.294950 eV Expectation values: Kinetic energy Electron-nucleus interaction <W> Electron-electron interaction Hamilton operator
with 3 terms and
z- = 0.769
= 14.303991 eV = -37.522710 eV 8.923769 eV = = -14.294950 eV
Basis states used and corresponding components of the eigenvector:
State no. State no. State no.
1 2 3
J
K
M
0 0 0
0 0 2
0 1 0
Components of eigenvector .408970 .246321 .182189
*************************************************************************** Energy of H-anion E = -14.338050 eV Expectation values: = Kinetic energy = Electron-nucleus interaction <W> = Electron-electron interaction = Hamilton operator
with 13 terms and Z- = 0.680 14.338264 -37.384232 8.707918 -14.338050
eV eV eV eV
Basis states used and corresponding components of the eigenvector:
State no. State no. State no.
1 2 3
J
K
M
0 0 0
0 0 0
0 1 2
Components of eigenvector .499839 .312642 -.029391
Atomic Physics
154
State State State State State State State State State State
9.2
no. no. no. no. no. no. no. no. no. no.
4 5 6 7 8 9 10 11 12 13
0 1 1 1 2 2 3 0 0 1
0 0 0 0 0 0 0 2 2 2
3 0 1 2 0 1 0 0 1 0
-.000122 -.203656 -.038433 .005354 .040781 -.000042 -.002835 .104698 .013612 -.007983
Exchange Term
So far we have only considered the ground state of He, which may be symbolized (ls)2, indicating that each electron is in an n = 1, f = 0 state. We have completely neglected spin; but we will now consider this complication. Let the two electrons be designated A and B. Electron A can have spin 'up' or 'down' with respect to an arbitrary z-axis. Introduce the eigenvectors x~, such that
stxt = ±~nxt and similarly for electron B. The total spin of the two-electron system S is 1 or O. The eigenvector of the total spin operator belonging to S = 1 is
while the eigenvector belonging to S = 0 is Xsinglet =
(X~x~ - X~x!)/V2.
Suppose that electron A is in the orbital state given by (nA,fA,m A ), so that the spatial part of the wave function is
..'f'I,A(-) r1
)pmA( ()) im A 4>l = CnAlAmAPlA e -PLUA+1(2 n A +1 P lA cos 1 e ,
where P = nZ;~o. Similar considerations apply to 'ljJB (r2), describing electron B. If the electrons were distinguishable - for example if one were replaced by a muon - one would write 'ljJA (r1 )'ljJB (r2) and that would be the end of the story. However, the two electrons are indistinguishable; and it does not make sense to say that the one at r1 is specified by the quantum numbers (n A , fA, m A ) and the other at r2 is specified by (n B , fB , m B). The space part of the two-electron
155
Exchange Term
wave function can contain 'ljJA (rl )'ljJB (r2), but also 'ljJA (r2 )'ljJB (rl), in which the two electrons are exchanged; and a priori we could have any linear combination of these two products. There is a fundamental law according to which the wave function describing two or more identical bosons is symmetric under exchange of any two particles, while the wave function for identical fermions is antisymmetric under such an interchange. A boson is a particle of integral spin (in units of h), e.g., a pion (8 = 0), a photon (8 = 1) or a graviton (8 = 2). A fermion is a particle whose spin is an integer plus one half, e.g., an electron (8 = ~) or an n- particle (8 = ~). The law can in fact be proved within the framework of relativistic quantum field theory, where it is called the spin-and-statistics theorem (cf., the discussion of proton-proton scattering in the previous chapter). Since the triplet spin vector Xtriplet is even, and the singlet vector Xsinglet is odd under exchange of A and B, it follows from the spin-and-statistics theorem that Xtriplet must be combined with an odd, and Xsinglet with an even spatial wave function, in order that the total wave function be odd under A f-t B exchange. In other words, the two possible total wave functions for the twoelectron system, including spin, are
Wtriplet =
~ {'ljJA(rl )'ljJB (r2) -
'ljJA(r2)'ljJB (rl) } Xtriplet .
(9.7)
In the case that 'ljJA (r) = 'ljJB (r), i.e., the two spatial wave functions are the same, the triplet combination vanishes: this fact is sometimes expressed by saying that no two electrons can have the same quantum numbers, n, f, m, s, where s = ± ~ is the value of the spin, in units of h, along the quantization axis. This is called the Pauli exclusion principle. More accurately - since after all Xtriplet also contains a state in which 8 3 = 0 - we should say that the spin triplet combination is excluded if the spatial wave functions are identical. In our analysis of the ground state of the He atom, both wave functions are 'ljJ100, so we have to do with the singlet state W = 'ljJlOo(rI)'ljJlOo(r2)Xsinglet, which is Eq.(9.6) with A = B, except for the normalization. The spectroscopic notation for this state is l80 , where the superscript is the number of spin-states, namely 28 + 1, the letter gives the total orbital angular momentum (8 for L = 0, P for L = 1, D for L = 2, ... ), and the subscript is the total angular momentum, J, which is a strict constant of motion. The principal quantum numbers are not
Atomic Physics
156
specified in this notation. The notational alternative, (ls)2, indicates that both electrons are in n = 1, f = 0 states. Let us now consider the case that the two electrons are in different spatial states. Treating V12 in first-order perturbation theory, we find the first-order energy shift to be
The perturbation does not involve the spin variables, so these drop out, since X+ X = 1 for either the singlet or the triplet cases. Accordingly 6.E
=
~
JJ~;~~I
{7jJA(f1}t/JB (f2)
V±£
{t/lA(Tl)t/lB(i'i) ±.pA(i'i)t/lB(Tl)}*
± 7jJA(f2)7jJB (f1)} (9.8)
in which the direct term is
and the exchange term is
where the plus sign applies to the singlet case and the minus sign to the triplet case. The direct term V can be written
where
which exhibits its semiclassical form. The exchange term £ does not have such a classical analogue: it is a pure quantum effect arising from the coherence of superpositions of states. Both V and £ are positive - the former manifestly so and the latter in cases of interest. As a consequence, the electron-electron repulsion raises the (negative) energy-level that one would calculate without V12 - an effect that we already found for the (ls)2 ground state. The increase is
157
H artree and H artree-Fock Methods
greater for the singlet, 1) + E, than for the triplet, 1) - E. Although the electronelectron interaction does not depend explicitly on the spin of the electrons, the fact that the energy levels are different for the singlet and the triplet spin combinations means that there is indeed an implicit dependence. We can make this effect explicit by considering ---
---
.... A
S .S = S
---A
.S
---A
+ 25
___ B
.S
___ B
+8
2 {S(S + 1) - ~ 2S(8
___ B
.S
0 + 1) - ~ 0 + I)}
+ 1) - 3 = {
1
-3
for for
8=1
5=0
Hence we can write the first-order perturbation Eq.(9.8) .6.E =
9.3
~ \ 1 + if A.if B) E .
1) -
Hartree and Hartree-Fock Methods
The Schrodinger equation for an atom with Z electrons is
which is impossible to solve analytically, and very difficult to solve numerically, even when Z is quite small. Nevertheless, this 3Z-dimensional partial differential equation, supplemented by the spin degree of freedom, relativistic corrections such as the spin-orbit coupling, and the antisymmetrization condition that is required by the spin-and-statistics theorem applied to fermions, contains the essential physics of atoms and hence also the essence of chemistry. The Hartree approximation consists in the approximation of separability:
1/J (r 1, r 2, •.. , r z)
=
1/J1(r 1) 1/J2 (r 2) ... 1/J z (r z) .
Application of the variational method leads to the following system of equations:
(-:~ \]~ -~:2
+ Vi (Ti) )
.p,(Ti) = .,.p,(T.),
(9.9)
where
(9.10)
158
Atomic Physics
and (9.11) i
This equation has a clear intuitive meaning: the ith electron satisfies a oneparticle Schrodinger equation with a potential energy that is supplied by the nucleus (attractive) and by all the other electrons (repulsive). The effective charge-density of the jth electron, say Pjerj), is evidently el'l/Jj(rj) 12 , which may be interpreted as the average density of electric charge at the point j, due to the presence of the jth electron. Despite the intuitively appealing nature of the Hartree equation (9.9), it must be stressed that it is an approximation. With the additional (Pauli) requirement that no two electrons may be in the same quantum state (including the spin degree of freedom), this equation yields results for energy levels that generally agree with experiment at the level of 10% to 20% only. Solution is effected by the method of successive approximations: first the electron-electron repulsion terms are ignored, so the zeroth-order solution is a product of hydrogen-like wave functions, as in Section 9.1 for the case Z = 2. The repulsive potential terms Vi (r i) can be calculated in first order, and these are then included in the Schrodinger equation, which is solved to give improved values of the 'l/Ji(ri). The procedure is iterated to convergence, yielding the so-called self-consistent solution of the Z-electron Schrodinger equation in the Hartree approximation. This method is clearly better than the perturbation theory that we explained, since the electron repulsion term is not simply treated as a first-order perturbation. In practice a further central field approximation is often made by averaging over directions:
r
This approximation is less important now that we have high-speed computers with large memories than it was in the pioneering days of quantum atomic physics. The Hartree equation (9.9) can be written (-
:~ '\7 2 J;2) ,p,(r, s)
+e 2
L L Id r' 'l/J; (r', s') Ir. . ~ . . '1 'l/Jj (r', s')'l/Ji(r, s)
(9.12)
3
"-I-"
3.,-'t
S
I
r
=
€(IPi(r, s) ,
where the spin quantum number has been included explicitly. According to the spin-and-statistics theorem, interchanging two identical fermions results in
159
Periodic Table
a change in sign of the total wave function. This exchange effect, including the sign-change, is effected by subtracting from the above a term in which 'l/Ji (i, s) and 'l/Jj(if, Sf) are replaced by 'l/Ji(i', Sf) and 'l/Jj(i, s) respectively:
+ e2 ~ -
e2
~L J=I=1.
f
rl='t
L f d3rf'l/J; (if, Sf) Ii ~ ifl 'l/Jj (if, Sf)'l/Ji (i, s) 8'
d3r f'l/Jj(i f , Sf) Ii
~ if I'l/Jj(i, s)'l/Ji(i f , Sf) =
fi'l/Ji(i, s).
8'
This is called the Hartree-Fock equation; it is much more laborious to solve than the Hartree equation, since it is truly an integro-differential equation. The last term gives rise to exchange contributions to the energy levels, analogous to those discussed in the last section for the helium atom. The Hartree-Fock equations give results that are significantly closer to the experimental measurements than are those of the Hartree equation (9.9).
9.4
Periodic Table
We shall now discuss very briefly the periodic structure of the elements, based on the separable Hartree approximation, according to which there are Zone-particle states, labeled by Z different quantum numbers ni, fi' mi and Si, i = 1,2, ... , Z. It is understood that these one-particle states are modified by the electronelectron repulsion terms, and, if we include the Fock improvement of the Hartree method, also by the exchange effects. Nevertheless, the above quantum numbers can still be used as labels. Let n be the largest of the ni for a given atom in its ground state. For a given value of n, the angular momentum quantum number f can take on the values 0,1,2, ... , n - 1. For a given nand f, there are 2f + 1 allowed values of the azimuthal quantum number m, namely -f, -f + 1, ... , f, and for each of these values there are two possible spin values.
f 2(2f + 1)
s
P
0 2
1 6
D 2 10
F 3 14
Orbital Angular Momentum The seven periods in the periodic table of the elements each begin with the S states, containing 2 elements, and then the P states, containing an additional 6 elements, except of course for the first period, for if n = 1 then f = 0 necessarily. As one adds successively more and more positively charged protons to the
160
Atomic Physics
nucleus, and correspondingly more and more electrons outside it, so more and more energy levels are filled up. Period 1 2 3 4
5 6 7
# 2 8 8 18 18 32
S P D 2 2+6 2+6 2 + 6 + 10 2 + 6 + 10 2 + 6 + 10 Incomplete
F
n 1 2 3 4
5 6 7
Shell K L M N 0 P Q
Periodic Table The D states corresponding to n > 3 do not belong to the chemists' period n, but rather to period n + 1. The reason for this displacement is that, for R = 2, the electrons in question have a greater probability to be situated closer to the lower-lying electrons than are the R = 0 electrons - with the same n and they experience more inter-electron repulsion and so are less tightly bound. The result is that the n = 3, R = 2 electrons are more lightly bound than the n = 4, R = 0 and R = 1 electrons: this gives rise to the first transition series of elements, containing useful elements like iron, chromium and copper. A similar displacement occurs for the second transition series, containing silver, and for the third, containing gold. The n = 4, R = 3 electrons are even better shielded than are the R = 2 electrons, and the corresponding elements are displaced by two periods, giving rise to the 14 rare earths in the sixth period. The above language of repulsion and displacement is only a picturesque way of gaining some intuition about what goes on as one successively increases the nuclear charge and adds electrons: it is not strictly true that one particular electron is in a particular shell-rather the wave function for the electrons is an antisymmetric combination corresponding to all the (n, R, m, s) states that are occupied. Nevertheless, it is meaningful to say, for example, that as one increases the nuclear charge of a Co nucleus by one unit and adds an electron to keep the resulting Ni atom neutral, one more 3d state is occupied, and not a 4p state, since this would cost more energy. The latter option exists, but it is an excited state of the nickel atom. The seventh period, containing in particular radium, uranium and plutonium, is incomplete because of nuclear instability: a-decay occurs in some cases, for example Ra-'----tRn+He, which is due to a quantum-mechanical tunneling effect, ;3-decay in others, and more radical fission (in the presence of thermal neutrons) in yet others, for example n+Pu~fragments.
Periodic Table
161
Two empirical observations, the Hund Rules, are helpful in deciding how the shells are successively filled: a: Other things being equal, the state of highest spin has the lowest energy. b: If an incomplete shell is more than half-filled, the lowest energy corresponds to J = IL + 81, otherwise to J = IL - 81. The two most important corrections to the Hartree-Fock calculations are:
(1) Deviations from the Hartree separability Ansatz and the central-field approximation. We shall assign all these electrostatic corrections to the potential VES ' (2) Spin-orbit coupling, a relativistic (Dirac equation) effect. The corre:.. sponding correction will be designated VLs.
When the first term is the more important, and the spin-orbit coupling can be ignored, or treated as an extremely small additional effect, we speak of the LS or Russell-Saunders coupling scheme. Because the spin-orbit coupling is neglected, not only J, but also the total orbital angular momentum, L, and the total spin, 8, are constants of motion. A Russell-Saunders state corresponding to L = 2, J = ~, and 8 = ~, for example, would be written 4D!., where the superscript 4 2 is the multiplicity, 28 + 1. When the spin-orbit term is more important than the electrostatic correction, each electron can be characterized by the quantum numbers ni, li, ji, mji' with a spin-orbit coupling of the form
(9.13)
This is called the jj coupling scheme, and is important for very heavy atoms, where the large value of the central potential makes Eq.(9.13) the more important contribution. In practice, often both VLS and YES are included in a serious calculation: for all but very heavy atoms the initial classification is on the basis of the LS scheme, with spin-orbit splitting being treated as a small additional perturbation, whereas from about Pb onwards in the periodic table, the jj classification is useful, in which the orbital and spin angular momenta for each electron are combined into a total angular momentum.
162
Atomic Physics
LS:
VES»
l = ~ili ....
J
VLS
§ = Li Si ....
VLS»
JJ:
....
....
....
VES ....
J i =Li+ 8 i
J = LiJi
=L +8
LS and
ii Coupling
Finally, in order to impose the correct antisymmetrization of the Z-electron wave function, one works not with a simple product of one-electron functions, but rather with their Slater determinant:
'l/J1 (r 1,81) 'l/J1 (r 2,82)
'l/J2 (r 1, 81) 'l/J2(r 2,82)
'l/Jz(r 1,81) 'l/Jz(r 2,82)
(9.14)
which guarantees the Pauli exclusion principle, since if any two columns are identical, the determinant is zero.
9.5
Hydrogenic Molecular Ion
The simplest 'molecule' is H2 +, i.e., a hydrogen molecule in which one of the two electrons has been removed. We will treat this system first, before considering the neutral hydrogen molecule itself. Let the two protons, of mass m p , be at positions R 1 and R2, and the electron, of mass m, be at position r. The Schrodinger equation in configuration space is
(9.15) where
PI =
-itiVRl
163
Hydrogenic Molecular Ion
We shall not try to solve this 9-dimensional partial differential equation exactly, rather we shall make some approximations, and finally use the variational method to estimate the binding energy of this molecular ion. Firstly, instead of transforming to the centre-of-mass system of the three particles, we will transform to that of the two protons only. The mass of the twoproton system is nearly 4000 times that of the electron, and relativistic effects, which are being neglected, are relatively more important in any case. Such additional small effects can be included afterwards as first-order perturbations. Let us in the first instance remove the electron completely. The two-proton system is described by
[Pr
2mp
Pi· + 2mp + V12
-
1 - -
E 'if;(R 1 ,R 2 )
= o.
(9.16)
We transform from the canonical pairs {R 1, j5 I} and {R 2, j5 2} to the centreof-mass coordinates, as in the discussion of the deuteron in Chapter 4:
and the difference coordinates
In terms of these variables, the Schrodinger equation (9.16) becomes p2 [ 2~~m
p2
+ 2M + V12 -
1
I
-
-
E 'if; (Rcm,R) = 0,
(9.17)
which is called the Born-Oppenheimer approximation. Here
the total mass of the two protons, while 1\([ --
!.m 2 P
is the reduced mass. Here we use M for this reduced mass of the protons, since we reserve m for the electron mass. As expected, the centre-of-mass motion can be factored out, and it is in fact convenient to work with coordinates such that Rcm is identically zero at all times, i.e., the origin is chosen to be the point
164
Atomic Physics
midway between the two protons. The above system, with the centre-of-mass motion factored out,
p2 [ 2M
l ..
+ V12 -
E 'lj;(R)
= 0,
(9.18)
describes the scattering of two protons. For example, a beam of protons from a synchrotron might be incident on a bubble chamber filled with liquid hydrogen. There are of course no bound states, since the potential, e2
V12
=
R'
is purely repulsive. The situation changes dramatically when we add an electron. We have now (9.19) Since
Rem = 0, we have
By factoring out the centre-of-mass coordinates, we have reduced the dimensionality of the system to 6. We make now the reasonable assumption that, if the ion is in a configuration in which the two protons are close to equilibrium - in which their mutual repulsion is balanced by the attraction to the electron - 'lj; will be a much more slowly varying function of Ii than it is of Hence we neglect the proton kinetic term, :;'lj;(R, r) = - 2~ V'~'lj;(R, r), compared with the electron kinetic term. The Schrodinger equation (9.19) becomes
r.
Ii 2 2 [ - 2m V'r -
e2
e2
IT _ i R I - IT + i R I +
] ........ e2 R - E 'lj; (R , r )
= 0.
(9.20)
Here R can be regarded as a parameter: indeed we could have written Eq.(9.20) down immediately on intuitive grounds, since it describes the situation in which the two protons have no dynamics, but are nailed down at a mutual distance R from one another. The trick now is to use the variational method to estimate the ground-state of the ion. If the electron is very close to proton #1, with proton #2 relatively
165
Hydrogenic Molecular Ion
. far away, a reasonable guess for the eigenfunction would be the ground state wave function of the hydrogen atom, in which proton #2 is a distant spectator. This normalized function is
(9.21)
:;;2.
For cleanliness we have set the unit of length to be the Bohr radius, ao = If on the other hand the electron is close to proton #2, the guess would be
(9.22) with proton #1 as the spectator. However, this way of talking is all wrong: we must think quantum mechanically! The two protons are identical, and it makes no sense to talk about being close to proton #1 and not to proton #2. Rather, we must consider symmetric and antisymmetric superpositions of the above states:
Here C± are normalization constants; they are given by
The overlap integral is
I 11
~71"
d3 r exp[ -
IT - 12 R I - IT + 12 R I]
71"
d3 r exp [-
IT -
R- I - r]
roo drr 2 e-
/1
dcos (j exp[-Vr2
2
Jo
T
(9.23)
+ R2 -
2r Rcos (j]
.
-1
In going to the second line, we shifted the integration variable from if to if + ~ R . The cos (j integral can be performed by substituting u = J r2 + R2 - 2r R cos (j, so d cos (j = - ~~. The integral (9.23) becomes
21
R
00
o
drr e- T
l
R
T
+ duu e- U
•
IR-rl
These integrals are elementary and we find
(9.24)
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Atomic Physics
Let us now consider the expectation values of the Hamiltonian in the states
'l/J± : ('l/Jl ± 'l/J2!H!'l/Jl ± 'l/J2) 2 [1 ± ('l/Jl!'l/J2)] ('l/JIIH!'l/Jl) ± ('l/J2!H!'l/Jl) 1 ± ('l/Jl! 'l/J2)
(9.25)
where the symmetry between 'l/Jl and 'l/J2 has been used. We have already evaluated the overlap integral in the denominator. Furthermore, since
!
--+ --+ [ n? 2 d3 r'l/Jl(R,r) --2 'V'r -
R
2 1
-:2 0
2!
2
2
me
-
e2
+R
d3 r
e2 - -;
!
1
e2
--+
if + ~R! e-
3
d r
--+
!r - :2R!
m
1 2 - -0 me2 + -e
e2
--+
-
--+
!r
2
e2
1 ....
+ :2R !
--+--+ + eR ] 1Pl(R,r)
2 [ 'l/Jl( R--+ r--+)] '
2r
(9.26)
IT + R! .
In proceeding to the last line, we shifted the integration variable from T to .... r--+ - 2"1 R • After switching the sign of cos fJ, we handle the integral with the same substitution as before, finding
~ 7r
!
d3r
e-2~ if + R!
=
~ roo rdr e-2r rR+r da, R
Jo
J1R-rl
the integrals now being elementary. We obtain
('l/Jl!H!'l/Jl) = - ~02me2
e2
+ R (1 + R) e- 2R .
(9.27)
The other matrix element in Eq.(9.25) is
I!
-
7r
!J
d3 r
e - r --+ exp [-
if+R!
if + R ! ]
roo rdre- r rR+r dae- u o J1R-rl
,
(9.28)
167
Hydrogenic Molecular Ion
where
(1
is as before. We obtain
(1/12IHI1/1,) =
[-!,,2 mc2+ ~] (1/1,11/12) - .2(1 + R) e-
R .
(9.29)
!:
Since we have set the Bohr radius to unity, we have e 2 = = o:2mc2 . Inserting the results Eq.(9.24), Eq.(9.27) and Eq.(9.29) into Eq.(9.25), we find finally (9.30) where
+ R) e- R 1± (1 + R + ~R2) e- R
1 - ~R2 ± (1
2
f±(R) = =f R e-
R
.
(9.31)
By the variation principle, ('lfJ±IHI'lfJ±) is, for all values of R, an upper bound on the ground state of the ion. Accordingly, we vary R to obtain the smallest possible value for the right side of Eq.(9.30). The following table gives values of f.... (R) as a function of R. obtained from the Mathematica function
f[R_, p_]:
=
-2
* p * Exp[-R] * (1 - (2/3)R * R + p * (1 + R) * Exp[-R])/ (R * (1 + p * (1 + R + R * R/3) * Exp[-R])); R
f+(R)
f_(R)
0.10 1.00 1.50 2.00 2.20 2.40 2.45 2.50 2.55 2.60 3.00 4.00
-18.009060 -.423267 -.009972 .107543 .122985 .129083 .129543 .129659 .129465 .128992 .118165 .073732
-21.634880 -2.090802 -1.145122 -.678293 -.557160 -.460042 -.438838 -.418716 -.399612 -.381467 -.264881 -.110226
Here p = ±1 corresponds to ± in Eq.(9.31). The command line
FindMinimum[-f[R, 1], {R, 2}] returns {-0.129662, {R- > 2.49283}}. The first number is the minimum of - f+(R), i.e., minus the maximum of f+(R). From the table we see that f+(R)
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Atomic Physics
has a maximum of about 0.13 at R ~ 2.5. This maximum of f+(R) corresponds to a minimum of the energy, which indicates that this R value is an estimate of the equilibrium separation of the protons in the hydrogen molecular ion. Since our length unit is the Bohr radius, the calculated equilibrium separation is 1.3 A, compared with the experimental value of 1.06 A. The calculated binding energy is about 0.13 times the ground-state energy of the hydrogen atom (13.6 eV), i.e., 1.77 eV, as compared with 2.8 eV from experiment. As can be seen from the above table, the antisymmetric combination of wave functions does not yield a finite maximum, and this can be understood readily. In this case the wave function at the mid-point between the protons (our origin of coordinates) is zero, and the probability that the electron is between the two protons is much smaller than it is in the symmetric case. Accordingly, the mutual repulsion of the protons is insufficiently compensated by the attraction of the electron, and the minimum energy (the maximum of f_(R)) corresponds to R -+ 00. Notice that the stability of the hydrogenic molecular ion, that we have just explained, is a quintessentially quantum phenomenon. If we had only used 1/11, obtaining then Eq.(9.27) as our upper bound on the bound-state energy, we would have found, as in the antisymmetric case, a minimum value in the limit of infinite R, corresponding to the dissociation of the system into a neutral hydrogen atom and a free proton at infinity. This is what one might intuitively expect, and it is of course a possibility; however, by bringing the second proton up to about an Angstrom of the first, energy is gained. 2 If we could ignore the proton-proton repulsion term ~ in Eq.(9.20), we could imagine bringing the protons together, so that R = 0, and then we would have a helium cation, which is a hydrogen-like ion with nuclear charge Z = 2. This is not so whimsical as it seems, for although a diproton is not stable, a dideuteron is stable. If one brings two deuterons closer and closer together, at first there is a growing Coulomb repulsion, but at very small separation the attractive strong attraction takes over. Thisforce arises from the exchange of 7r-mesons, or pions, between the nucleons, and it is described by the Yukawa potential:
Vy
=-
[m7rIicR]
g R exp -
(9.32)
'
m7r
is the mass of a where g is the strong interaction coupling constant, and pion, which is about 140 MeV / c2 • This nuclear force is attractive and is much stronger at short distances than the repulsive electrostatic force, but it is of short
169
Hydrogenic Molecular Ion
range, because of the exponential term. In fact, we can rewrite Eq.(9.32) as (9.33) where 197 MeV F 4 R 7 -1 " -hc --'" ",1. F m7l"c2 140 MeV r-..J
r-..J
Here F stands for fermi, which is 10- 15 m, the typical nuclear scale of distance (cf., Sect. 4.4). Now since 1 A = 105 F, it is clear that the nuclear attraction wins over the Coulomb repulsion only at separations that are tiny, as compared with typical atomic distances like the Bohr radius of the hydrogen atom. Nevertheless, it is energetically well worth the effort, since the difference in mass between two deuterons and one alpha particle (a nucleus of He 4 ) (x c2 !) is released as energy. This fusion reaction occurs, directly of indirectly, in stars, hydrogen bombs, and in experimental fusion reactors. The ground-state energy of a hydrogen-like atom with nuclear charge Z is _~Z2Q:2mc2, and the corresponding normalized wave function is
-
'!/J(r)
=
Vb--;- exp [-] -Zr
.
(9.34)
After fusion of two deuterons the nuclear charge is 2e, and in that case Eq.(9.34), with Z = 2, should be used to estimate the ground-state energy, not Z = 1, which is what we had in Eq.(9.21) et seq. With the deuterons (or protons) separated by an Angstrom instead of a fermi, which is the actual situation in our hydrogenic molecular ion, it is reasonable to expect that the best effective Z to choose for the trial function Eq.(9.34) should lie between 1 and 2. At distances of many Angstrom, Z = 2 should give the tail of the wave function well, but at distances of less than an Angstrom one would prefer Z = 1. Evidently we should be able to improve the variational calculation by allowing Z to be a parameter that will be adjusted: we expect the optimal value to be closer to 1 than to 2. We replace Eq.(9.21) and Eq.(9.22) by
and insert these functions into Eq.(9.25). It is then easy to see that the overlap integral Eq.(9.24) becomes (9.35)
Atomic Physics
170
where
-
-
R=ZR. The Hamiltonian satisfies
[::, + ZV. + (1 - Z)V. + V2 + V'2] 1"'+) [-~Z-2 a 2me2 + (1- Z)V1 + V2 + V12 ] ItP+) ,
(9.36)
and by symmetry it is easy to see that (tP2IV1ItP1) = (tP2IV2ItP1). Hence
(nt. IHlnt. ) = _!z-2 2 2+v, +(1- Z)(1hIVl l1P1) + (tP1I V2ItP1) + (2 - Z)(tP2I V2ItP1) . Cf'+ Cf'+ 2 a me 12 1 + (tP2ItP1) By scaling arguments, we can see that, to obtain (1P1IV2ItP1) or (tP2IV21.~h) from the corresponding expressions without tildes, we simply multiply by Z and replace R everywhere by R. We find
(tP1IV1ItP1) (tP11lt2ltP1) (tP2IV2ItP1) We conclude that
(tP+IHltP+) {
-~a2me2 x
=
Z2 _ 2Z{(Z - l)R + [1 +_ (Z - l)R ~ (Z -_ ~)R2]_e-R + (1 + R) e- 2R }} . R{l + (1 + R + ~R2) e- R }
It is left as an exercise to the student to write a short program, in Mathematica or another language, in order to minimize numerically the above matrix element with respect to Rand Z. We find the optimal value of Z to be about 1.25, which fits our expectations well enough. The optimal R is 3.125, but this corresponds to R = RI Z ~ 2.5, which is the same value for the separation, in units of the Bohr radius, that we obtained before. The binding energy is now improved to 0.173 times the hydrogen atom ground-state energy. The results in more useful units are as follows:
-
Z 1 1.25 Experiment
-
-
R( Angstrom) 1.3 1.3 1.06
Binding energy (eV) 1.77 2.2 2.8
171
Hydrogen Molecule
The results are satisfactory, and the accuracy can be improved by choosing more complicated trial functions, a la Hylleraas.
9.6
Hydrogen Molecule
The hydrogen molecule, with its two protons and two electrons, is pictured below. The total Hamiltonian for the electrons is now
pi + p~ H =
2m
e2
-
e2
e2
e2
iiI - ~ R I - iiI + ~ R I -- Ir2 - ~ R I - 1T2 +
e2 ~ RI+ R
e2
+ ITI - r21 '
the last term being of course the electron-electron mutual repulsion term. This can be rewritten (9.37) where, for i
= 1,2, (9.38)
HI is the Hamiltonian for an H-
-e anion composed of electron 1 and -e the two protons, while H2 is that composed of electron 2 and the two protons. In each term, each electron is associated with both protons: we thus speak of the e molecular orbital method, and 2 it has much application also to complicated molecules. Notice Hydrogen Molecule 2 that the term e / R in Eq.(9.37) has a negative sign; this is right, since it occurs in HI + H2 twice with a positive sign. A reasonable first approximation is to drop the last two terms in Eq.(9.37): after all, one is attractive and one is repulsive, and the average separation of the electrons from one another might be expected to be very roughly the same as the separation of the two protons from one another, so the contribution of one term to the ground-state energy should largely be canceled by that of the other.
!R
e
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Atomic Physics
With the approximate Hamiltonian
the ground state wave function is simply the spin-singlet combination
'I/J+(R, Ti, i 2) = 'I/J+(R, id'I/J+(R, i2)Xsinglet, where 'I/J+(R, i) is the H- ground-state wave function that we calculated in the last section. The ground-state energy, in this approximation, is just twice that of the H- ion, and hence the binding energy of the hydrogen molecule, being the difference between the energies of two H- ions and two H atoms, is thus twice the binding energy of the H- ion. We calculated 2.2 eV for the latter, so we get 4.4 eV for the binding energy of H2. The experimental value is 4.75 eV, so we seem to have booked a big success with little extra work. The goodness of the agreement is actually fortuitous, because if we use the experimental value of the H- binding energy instead of what we calculated in the previous section, we get 5.6 eV. Moreover, 'improving' the calculation by including the term - eR2 (easy) and the term + ,_TI-T2 e 2 _ , (hard!), does not improve agreement with experiment, rather it makes it significantly worse. This does not mean that quantum mechanics has failed; but it does mean that we have gone about as far as we can with hand calculations and simple programs on the personal computer. To proceed further one needs sophisticated programming, involving complicated trial functions and numerical integration routines. That way lies quantum chemistry, littered with messy calculations but hugely crowned with success.
9.7
Exercises
Problem 1 Consider the following terms: IS, 2p, 3p, 3D, 2D, ID, 4D. (1) What levels may arise from these terms (i.e., which values of spin, orbital angular momentum, and total angular momentum can occur)? Express your findings in the standard form, 2S+1L J . (2) Arrange in order of increasing energy the levels that may arise from the following configurations: ls2p, 2p3p, 3p3d.
Problem 2 Calculate the probability density, p(i) = \'I/J+(R, i)\2, for the electron in the Hi molecular ion, and plot it on the line joining the two protons. Evaluate
II ).
p( R ) / p( kR) and p( 0) / p( k
173
Exercises
Problem 3 Calculate the first and second ionization energies for helium (i.e., the energy required to strip first one, and then the other of the electrons from the atom). Compare your results with the experimental values of 24.85 eV and 54.40 eV. Problem 4 The first few S terms of helium have the following energies, relative to 182 IS, the ground state: 1828 IS : 20.89436; 1828 3S : 20.08735; 1838 IS : 23.23007; 1838 3S : 23.02549 (all in units of the electron volt) . Calculate the contributions to the energy coming from the overlap integrals in the 1828 and the 1838 configurations. Problem 5 Calculate the energy of the ground state of the lithium atom. With the Bohr radius set equal to unity, use the trial functions 2Z;/2 exp( - ZI r) for the Is electrons, and cZ;/2 exp( - Z2r /2)(1 - dZ 2r) for the 2s electron. Here ZI and Z2 are variational parameters, while c and d are determined by considerations of normalization and of orthogonality. Problem 6 Consider the beryllium atom. (1) Write down the Slater determinant for the ground term of this atom, and find an expression for its energy in terms of Coulomb and exchange integrals. (2) Find expressions for the energy in terms of the Hartree-Fock expression for the configuration 18 2 28 2 . (3) Evaluate the expectation value < wI H I w>
Problem 7 Write down the part of the ground-state wave function that represents the pstate electrons in the atom of nitrogen in terms of '¢21m, m = -1,0, 1. Assume that the Hund rules apply. Problem 8 Consider a 3 - d electron in an atom that is in a crystal lattice. The energy level of this electron is fivefold degenerate. The corresponding wave functions are
(ill)
=
xzJ(r)
(iI2) = yzf(r)
(iI3) = xyJ(r)
(1) Show that these are indeed 3d wave functions.
174
Atomic Physics
The influence of the crystal environment on the electron is described by the perturbing (rhombic) potential
V(r) = Ax2 + By2
+ ez 2 .
(2) Calculate the energy shifts induced by V in first order perturbation theory. (3) What are these shifts for a tetragonal field (A = B)? (4) Show that the expectation value of the z-component of the angular momentum always vanishes in the presence of this field (in first order). (5) Is the same true for the x- and y-components?
Problem 9 Carbon dioxide is a linear molecule (OCO) that can combine with an electron to form an ion. Suppose that the electron has energy Eo when it is attached either to one of the oxygen atoms or to the carbon atom. The state of the molecule when the electron is attached to the carbon atom is Ie), while the states corresponding to its being attached to one or other of the oxygen atoms are respectively 101) and 102 ). These three states are not eigenvectors of the Hamiltonian, because there is a small transition probability that the electron jumps from the central carbon atom to one of the oxygen atoms, or vice versa. The probability that a direct transition takes place between one oxygen atom and the other is very small, and may be neglected. (1) Write down a Hamiltonian for this ion. (2) Determine its energy levels and its eigenfunctions. (3) An external electric field is applied, so that the electron's energy when it is attached to one oxygen atom is Eo +<5, and when attached to the other is Eo - <5, while the energy of attachment to the carbon atom remains unchanged. Determine the energy shifts to second order in <5. Problem 10 Six identical atoms are in a stationary state, with an atom at each of the vertices of a regular hexagon. One electron is added to the molecule. Let Ij) , j = 1,2,3,4,5,6, be orthonormal states corresponding to the electron's being attached to atom j. Transition from state Ii) is only possible to a neighboring state, i.e., to state Ik) if k = j ± 1 (modulo 6). Let H be the Hamiltonian of the system, and set UIHlj) = Eo, and UIHlk) = a e i8 if k = j + 1 (modulo 6). (1) Calculate the energy levels of this system in terms of Eo, a and B. (2) Make a sketch of the energy levels when 0 < () < < a < < Eo. (3) What are the eigenstates and the eigenvalues (with their degeneracies) in the case () = O?
Appendix A
Completeness of Eigenvectors
A.l
Hermitian Operators
In quantum mechanics it is assumed that possible states of a physical system are represented by vectors in a Hilbert space, and that this space is spanned by the eigenvectors of any Hermitian operator that represents a physical observable. For example, the eigenvectors l'ljJn) of the Hamiltonian operator,
can be used as a basis for the Hilbert space. Any vector 14» can be expanded in terms of these eigenvectors:
n
Of course, if there is degeneracy, i.e., more than one independent eigenvector belonging to a given eigenvalue, they must all be included in the above expansion. One speaks in general of an eigenspace belonging to a given eigenvalue: the statement is then that the union of all the eigenspaces of any physical operator (i.e., an operator that represents a physical observable) is the whole space. This assumption, which at first sight seems surprising in its audacity, is actually inspired by the fact that, in a finite number N of dimensions, the eigenvectors of any Hermitian matrix do in fact span the whole of the N-dimensional space. In the following sections we will prove this result, which involves showing that any finite-dimensional, Hermitian matrix can be diagonalized by a unitary transformation. Before we get that far, however, we need two fundamental theorems - that of linear equations and that of algebra - and since these theorems, when they are first encountered by the student, are often not proved adequately, we will first provide constructive proofs. 175
176
Completeness of Eigenvectors
Before starting on these proofs, however, it is salutary to realize that they do not generalize to an infinite number of dimensions. Some Hermitian operators do exist on infinite-dimensional, separable Hilbert spaces for which the eigenvectors do not span the whole space. The axiom of the completeness of the eigenvectors of physical operators is therefore an assumption with physical consequences. Many (but not all) of the properties of physical operators in quantum mechanics are shared by finite-dimensional Hermitian matrices, and it is therefore a good idea to acquire a thorough understanding of the latter.
A.2
Fundamental Theorem of Linear Equations
Consider the system of N homogeneous, linear, algebraic equations
+ al2 x 2 + a13x3 + ... + alNXN a2l X I + a22 X 2 + a23X3 + ... + a2NXN a3l X I + a32 X 2 + a33x3 + ... + a3NXN
0
anXI
0
0
(A.I)
0, which we conventionally write in matrix form an
al2
al3
a21
a22
a23
a31
a32
a33
=
0,
(A.2)
or even more compactly as ax = 0. In this last form, of course, much is left implicit! The determinant of the matrix a is the number defined by
2
(A.3)
Appendix. On page 177, at the. bottom of the page, the two possibilities are alternatives. One should read: 1: EITHER aij = 0 ... 2: OR at least ... On page 178, three lines from the bottom, one should read We now iterate: N ----+ N - 1 :
b ----+ a :
GO TO I!
177
Fundamental Theorem of Linear Equations
two indices im and in are equal. The summations in Eq.(A.3) are all to be taken from 1 to N. We shall need to use a few properties of determinants, which we first introduce and prove. The first property is that, if two rows are interchanged, then the determinant of the new matrix is minus that of the original one. This follows immediately from the definition Eq.(A.3), for if, say, amirn and ani n are interchanged, then an even permutation changes into an odd one, and vice versa, and this yields a minus sign for every term in the sum Eq. (A.3). As a simple corollary, if two rows are identical, then the determinant vanishes, for then the determinant is at the same time equal to plus and minus itself. The determinant of a matrix does not change if one adds a constant multiple of one row to another row. This can be seen as follows: suppose for instance that we add A. times the first row of Eq.(A.2) to the second row. The determinant of the new matrix is
{i'~.} {i'~.} +A
{
{
1
2
3
~l
~2
~3
1
2
3
~l
~2
~3
L { {ili2 i S ... }
} ali, {a2i'
+ Aali, }a3i•
.•.
} ali, a2i,a3i• ...
1
2
3
~l
~2
~3
} ali, ali,a3i, ...
deta. We shall prove the fundamental theorem of linear equations, that det a = 0 is a necessary and sufficient condition that ax = 0 has at least one nontrivial solution (i.e., x =I- 0). To be precise, we restrict the matrix a and the solution x to the field of complex numbers, but aside from that specification, there is no further restriction on the matrix. We show first that, if det a = 0, then ax = 0 has a nontrivial solution. We have one of two possibilities:
1:
aij
= 0 for all i,j E {1,2, ... ,N}, thus Xl, X2, ..• , XN are
arbitrary, so we certainly have a nontrivial solution, indeed N linearly independent ones. 2: at least one aij is nonzero. We then take the liberty of renaming the indices and rearranging the equations, as necessary, so that all, with the new labeling, is this nonzero matrix element. Since the left-hand side of the first line of Eq.(A.l) equals zero, we may add or subtract any multiple of it to any or all of the other equations. Subtract then a21/ all times the first line from line 2, a31/all times the first line from line 3,
178
Completeness of Eigenvectors
and so on, up to aNI/an times the first line from line N. One obtains, in matrix form, an 0 0
al2
al3
alN
Xl
b 22
b 23
X2
b32
b33
b2N b3N
0
bN2
bN3
bNN
XN
X3
= 0,
(A.4)
where bij
-
aij -
ail an
-.-alj
i, j E {2, ... , N}. Note that, although the matrix in Eq.(A.4) is not
the same as that in Eq.(A.2), the determinant of the new matrix is the same as that of the old one, since subtracting multiples of one row from another does not change the value of a matrix's determinant. The first line gives
while the rest can be written
= O. bN2
(A.5)
b N3
Moreover, from Eq.(A.3) and Eq.(A.4), we see that deta = au detb where b is the matrix in Eq.(A.5). Since det a = 0 but au =1= 0, it follows that det b = 0, and so we have reduced the problem to the one with which we started, but with one equation and one variable less! The method we have given is called Gaussian reduction, after its inventor, GauB. We now iterate:
GOTO I! The procedure certainly stops if the dimension of the 'matrix' has been reduced to 1. In that case det a = 0 implies a = 0, so the last variable is arbitrary.
179
Fundamental Theorem of Linear Equations
Alternatively, the procedure may stop prematurely, if all the matrix elements vanish at some dimension, say N' > 2. In this case the original matrix a is said to be of rank N - N', and there are N' linearly independent solutions of the original equation. This concludes the proof that, if det a = 0, then there is a nontrivial solution of the homogeneous equation. We now show the converse, and to do that it is convenient to consider first the inhomogeneous system a1l
a12
al3
alN
Xl
a21
a22
a23
a2N
X2
a31
a32
a33
a3N
X3
dl d2 d3
aNI
aN2
aN3
aNN
XN
dN
(A.6)
or more compactly ax
=d.
(A.7)
Here a and d are known and X is unknown. From the definition Eq.(A.3) we clearly have also
2
3
(A.8)
since the sum here involves precisely the same terms, with the same signs, as the sum in Eq.(A.3). This form is an evaluation by columns instead of by rows. Clearly a determinant with two identical columns must vanish; and we may add any multiple of a column to any or all columns of a matrix without altering the value of its determinant. We may write
L ainnAinn,
deta =
in where 2
is called the cofactor of ainn. The symbols in and (linn indicate that the sum over in and the term ainn are missing. If m =f. n, clearly
L ai in
n 7n
A i n n = 0,
180
Completeness of Eigenvectors
since this determinant is the same as that of a, but with the nth column replaced by the mth column, which latter therefore occurs twice. Hence, replacing the dummy index in by p for elegance, we have
L apnApm = 8
mn det a.
p
If det a
#- 0, we define the inverse of a by a;,~
= Apm / det a
so that
L a;,~apn = 8mn . p
In matrix notation, this reads a-Ia = I, where '1' here means the unit matrix. From Eq.(A.7) we find x = a-lax - a-Id. Thus, on condition that the determinant of a does not vanish, the system of equations Eq.(A.6) has the solution a-Id for any d. The solution is unique, for ifaxi = d and aX2 = d, then Xl - X2 = a-l(axl - aX2) = a-l(d - d) = O. In particular, in the case d = 0, which is the homogeneous system that interests us here, this unique solution is clearly 0, the trivial solution. Hence det a = is a necessary, as well as a sufficient condition that the system Eq.(A.l) has a nontrivial solution. From the definition Eq.(A.3), the cofactor may also be written
°
2
3
since this involves precisely the same terms. Hence
L anpAmp = 8
mn det
a,
p
and so, if det a
#- 0, we have aa- l
--
I
,
(A.9)
i.e., a-l acts as an inverse also from the right, justifying its name. A matrix a has an inverse, i.e., aa- l = I = a-la, if and only if det a #- O. We have just seen how to construct the inverse if the determinant does not vanish; conversely, if the determinant does vanish, we know that there exists a nontrivial x such that ax = 0, and if an inverse were to exist, we could prove that x = a-lax = 0, which contradicts the fact that x is nontrivial. Hence there is indeed no inverse matrix when det a = O.
Fundamental Theorem of Algebra
181
Lastly, we shall prove the useful result that the determinant of the product of two matrices is equal to the product of the determinants of the two matrices, that is, det ac = det a . det c. From Eq.(A.3) we have
{i'~.} ,~ {i'~ }{;'~
=
detac
2
3
{
..
i~
.. } {
2
3 (A.IO)
Now on condition that jr,i2,j3,'" ,jN are all different, we have
2
3
2
3
J2
J3
N } { jr IN II
since if iI, i2, i3, ... ,iN is an even permutation of 1, 2, 3, ... , N, then j1, i2, i3, ... . . . , j N is either an even or an odd permutation of both 1, 2, 3, ... , N, and iI, i2, i3, ... ,iN, whereas if i 1,i2,i3, ... ,iN is an odd permutation of I,2,3, ... ,N, then jllj2,i3, ... ,jN is necessarily an even permutation of either 1,2,3, ... ,N or of iI, i2, i3, ... , iN, and an odd permutation of the other. Hence
detac
{j,~} {"~ X
{
.• } {
1
2
3
...
]I
J2
J3
... IN
J1 1,1
J2 1,2
N
J3 1,3
} a'j, a2ha3;, ... aNjN
IN IN
} Cj, i, Chi, C;,i, ... Cj N
(A.II)
iN .
The result det ac = det a . det c, follows immediately. From this it is clear that det ac = det ca, since both sides are equal to det a . det c .
A.3
Fundamental Theorem of Algebra
The fundamental theorem of algebra states that every Nth order polynomial, P(z), has precisely N roots (i.e., zeros) in the field of complex numbers, although the roots need not all be distinct. It is often proved in text-books by showing that, for a sufficiently large contour,
fd z P'(z) P(z) = 27riN, and then Cauchy's theorem implies that the integrand must have precisely N poles, counting multiplicity. We can well object to this proof on the grounds that
182
Completeness of Eigenvectors
one should not prove any elementary and basic theorem by using a sophisticated result. After all, by the time we reach the blase stage of writing contour integrals for meromorphic functions, might it not be that we have, somewhere along the way, used the very theorem that we are trying to prove, and thus committed the cardinal sin of petitio principii? Is it not possible to produce an elementary proof of this basic result? Indeed it is: Consider then an Nth order polynomial (N > 1) N
L anzn ,
P(z) =
n=O
where it is supposed that aN =J. 0, so this is truly of Nth order, but where some or all of the other coefficients may vanish. For Izl sufficiently large the term aN zN will have a larger modulus than the sum of the rest: indeed we will be able to find a real positive R so great that, on the circle Izl = R, we have IP(z) I > ~ laNIRN. Inside this circle of radius R in the complex z-plane, IP(z)1 must reach its minimum, non-negative value at least at one point, say zo0 We shall suppose that this minimum value, IP(zo)l, is not zero, but is strictly positive. This will lead to a contradiction, allowing us to conclude on the contrary that IP(z)l, and hence P(z) itself, has at least one zero in the circle of radius R. The polynomial can be rewritten in the form N
P(z) =
L cn(z -
(A.12)
zo)n ,
n=O
where the coefficients Cn can easily be computed from the coefficients an, given zo° It can occur that some of the Cn vanish, but certainly CN =J. 0, since P(z) is an Nth order polynomial, and Co = P(zo) =I 0, since IP(zo)1 > by assumption. We shall write z - Zo = r eiB and
°
N
P(z) = Co + Ckrk eikfJ
+
L
cnrn einfJ
n=k+l
Here k could be 1, but it could exceptionally be N, in which case the sum would be absent. Let us write the known complex constants
and choose the phase () of z - Zo, which is at our disposition, to satisfy () _
-
7r
+ cPo - cPl k
.
Fundamental Theorem of Algebra
183
Hence, with z satisfying this constraint, N
P{z) = (Icol - ICklrk) eir/>o
+
L
cnrn einlJ
n=k+l Now for all r such that
< ro
l/k
I~ I
, we can certainly find a real, positive constant,
0,
N
L
cnrn einlJ < or k + 1 ,
n=k+l so in this domain (A.13) For r small but non-zero the right-hand side of the above inequality is less than Icol = IP{zo)l, and this contradicts the assumption that IP{zo)1 > 0 is a minimum of IP{z)1 in the circle Izl < R. The theorem has thus been proved by reductio ad absurdum. The basic trick was to find a direction in the complex plane along which IP{zo)1 could be made smaller, so that a local minimum other than zero is excluded. Since we now know that a minimum occurs at a point where Co = P{zo) = 0, it follows that Eq.{A.12) can be written N-l
P(z) = (z - zo)
L
Cn+l{Z - zo)n ,
n=O
i.e., z - Zo times a polynomial of order N - 1. However, this latter polynomial can be subjected to the same reductio, showing that it too has a zero that can be factored out. The procedure stops when a zeroth order polynomial, i.e., a constant, is left. Thus Eq.(A.12) can finally be written N
P{z)
= CN
II (z -
zn),
n=l
where the zn,.which may.not all be, distinct, are called the.roots mial, being the solutions of the equation
P{z) =
o.
oLth~
polyno
N
184
A.4
Completeness of Eigenvectors
Diagonalization of a Hermitian Matrix
We shall show that any Hermitian matrix can be reduced to diagonal form by a unitary transformation. That is, if
= a,
at
then a matrix
u
exists such that
u tu
=
1 and
a diag =
(A.I4)
utau
is a diagonal matrix. Note that, in a finite number of dimensions, u t u = 1 implies Idet ul = 1, which in turn implies the existence of an inverse of u, which is of course u t. This means that also uu t = 1. A matrix that satisfies u t u = 1 = uu t is said to be unitary. Now Eq.(A.I4) is equivalent to ua diag
= au,
or, written out fully, Uu
Ul2
ulN
Al
0
u21
U22
u2N
0
A2
0 0
uNI
uN2
uNN
0
0
AN
au
al2
alN
Uu
ul2
UIN
a21
a22
a2N
u21
u22
u2N
aNI
aN2
aNN
UNI
UN2
uNN
where the Ai are not necessarily all distinct. For a constructive proof of this statement, consider the matrix equation au a21
A
al2 a22 -
aN!
aN2
A
alN
Uu
a2N
U21
aNN-A
UNI
=
o.
(A.I5)
We know from Sect. A.2 that a necessary and sufficient condition for Eq.(A.15) to have a nontrivial solution is the vanishing of the determinant of the matrix 0" tne left \.e. ditt t~ AI1 ;; 0 where 1 ii l he 'Ioit m,t"llC This 'iletefmiu"l\llt is however an Nth order polY.b.Omla\ ia A, and we know trorn the Sect. A.a ihai 1'1; has N roots, which may however not all be distinct. We call one of these roots Al (we know there is at least one), and we construct by the method of the Sect.
-
185
Diagonalization of a Hermitian Matrix
A.2 a nontrivial, normalized solution for the Un, the first column of the matrix u. Let us now require the remaining Uij, j = 2, 3, ... ,N to be normalized, and to be orthogonal to Uil, and also to one another, i.e., 2::1 UijUik = bjk , for all j, k E {I, N}. This is always possible, and of course there is much freedom in the choice, subject to the constraints. With this construction, U is unitary and U t au has the form .A1
0
0
0
0 0
b22 b32
b23 b33
b2N b3N
0
bN2
bN3
bNN
(A.16)
The zeros in the first column follow from the fact that 2: j .A1 2: j uljuj1
=
2:k uljajkuk1
=
.A1bil; and the zeros in the first row follow from this, and the
fact that Ut au is Hermitian: [u tau] t = Ut au. Hence the first step of the diagonalization has been performed. Let us write the above in shorthand notation as U
t
( .A1
au = (
(A.17)
\. 0
where b is a matrix of order N - 1. By using the property det ac at the end of Sect. A.2, we have
= det ca , proved
det[a - .AI] = det[(a - .AI)uut ] = det[ut(a - .AI)u]. From Eq.(A.17),
det[a - .AI]
= det (
.A1 - .A 0
b _0>.I )
= (>.. -
>')det[b -
>.II ;
where I is now the unit matrix in N -1 dimensions. The right side of Eq.(A.17) is a Hermitian matrix and so it follows that .A1 is real and b is a Hermitian matrix . .A1 is called an eigenvalue of a. By the above procedure we have reduced the eigenvalue problem for a Hermitian matrix in N -dimensions to one in N - 1 dimensions, in the process having 'split off' one eigenvalue. By iterating the procedure we can proceed to split off the next eigenvalue, thereby reducing the dimension again. It is important to stress that the next eigenvalue may possibly not be different from .A1. The method nowhere requires the successive eigenvalues to be different: it is enough that the matrix b be Hermitian to ensure that it has at least one eigenvalue.
186
Completeness of Eigenvectors
At each step of the procedure a unitary transformation of the matrix is made, so that finally, after the original matrix a has been reduced to diagonal form, N - 1 successive unitary transformations have been made, which of course can also be written in retrospect as one unitary transformation, as in Eq.(A.I4). The columns of U are just the eigenvectors of a: U is not unique, for the order of the eigenvalues in a diag can be changed, and in the case of degeneracy the choice of basis in the eigenspace is not unique, but the important point is that there are always N orthonormal eigenvectors. Since the N eigenvectors are mutually orthogonal, they are necessarily independent of one another, and so they span the entire N-dimensional space. This is really the end of the proof; but to do the job a little more formally, consider a basis of orthonormal vectors that obviously spans the space, namely {ej}, where 1 0 0
el =
e2
0 1 0
=
o
eN
=
o
0 0 0
1
Such a set of vectors is called an orthonormal basis. An arbitrary vector x can be expanded in terms of them: (A.I8)
x = LXkek. k
The N eigenvectors of the matrix a that we have constructed can be written
Uj = L Uijei· i
Using the unitarity of the matrix u, we find
ek
= L 6ki ei = L L UkjUijei = L UkjUj . i
i
j
j
Hence
x
= LXk LUkjUj = LCjUj, k
where
j
j
Commuting Hermitian Matrices
187
This concludes the explicit demonstration that an arbitrary vector x can be written as a linear superposition of the N orthonormalized eigenvectors Uj of an arbitrary Hermitian matrix a.
A.5
Commuting Hermitian Matrices
In this section we shall show that if two Hermitian matrices, say a and b, commute with one another, then it is always possible to find simultaneous eigenvectors. Let us first consider the case that .A is a simple (or nondegenerate) eigenvalue of a, i.e., ax = .Ax,
and there is only one independent vector x that satisfies this equation for the given .A. Since ab = ba, it follows that abx
= .Abx,
and hence bx is also an eigenvector of a belonging to the same eigenvalue .A. Since there is only one such eigenvector, it follows that bx must be proportional to x, i.e., x is also an eigenvector of b. When there is degeneracy the situation is a little more complicated. Suppose now that, for the given eigenvalue .A,
for q = 1,2, ... ,m, with m < N, where we can suppose without loss of generality that the m independent eigenvectors x(q) have been orthonormalized: X (p)tx(q) --
..
U pq
.
Again we have abx(q) = .Abx(q) ,
so bx(q) is an eigenvector of a belonging to the eigenvalue.A. However, since there are now m such independent eigenvalues, we can only assert that bx(q) IS some linear combination of them: m
bx(q)
=
L r=l
Bqrx(r) .
188
Completeness of Eigenvectors
The orthonormality of the eigenvectors implies that x(p)tbx(q) -
B
qp'
and from the hermiticity of b we find easily that B;p
= B pq ,
so that Bpq are the elements of an m x m Hermitian matrix. We know that a unitary matrix U can be constructed that diagonalizes B. To be precise, there is an array U pq such that
"U ~ pq B qr
"~ Bdiag pq Uqr
-
q
-
Bdiag Upr . pp
q
It follows then that b" Upq x(q) -- " " U pq B qr x(r) -~ ~~ q
r
q
Bdiag" U pr x(r) . pp ~ r
For each of the m values of p, the linear superposition :L: q Upqx(q) is an eigenvector of the matrix b. If all of the elements B~;ag are distinct, then the degeneracy of the eigenvalue ). of a has been resolved, in the sense that each of the vectors :L: q Upqx(q) can be labeled by a different pair of eigenvalues. In the case however that some of the B~~ag are the same, some degeneracy remains. One can look for a third matrix, c, that commutes with a and b, in order to lift the degeneracy completely, and if that does not work, then a fourth and a fifth matrix, until one eventually has a complete set of commuting matrices such that each simultaneous eigenvector has a unique assignment of eigenvalues. A.6
Infinite Dimensional Spaces
In this section we shall illustrate some of the differences between finite and infinite dimensional spaces, and the operators on them. It would be helpful to work through Chapter 1 before reading this section. The configuration-space representation of the kinetic energy of a particle is p2 h2 2 -=--V, 2m 2m
(A.19)
and it is an important operator in quantum mechanics (see Sect. 1. 7 et seq.). In this section we shall for convenience consider one spatial dimension only, in which V 2 reduces to D2 = d2 / dx 2 , but the general conclusions apply to the three-dimensional case as well.
189
Infinite Dimensional Spaces
Consider a twice-differentiable function, ¢>( x), of the one-dimensional variable, x, that is restricted to the interval -L/2 < x < L/2, where we shall consider separately the cases that L is finite, or that it is infinite. Let D2 be the operation of taking the second derivative, i.e.,
Is D2 Hermitian and does it possess a complete set of eigenvectors? The answers depend on how the infinite-dimensional space on which D2 acts is defined. We shall restrict the space to consist of twice-differentiable functions; and we define a scalar product between two functions, ¢> and 'ljJ, to be S('ljJ,¢» = /L/2 dx'ljJ*(x)¢>(x) , -L/2
see Sect. 1.4. The Hermitian conjugate of D2 is defined by S( 'ljJ, D2t ¢»
=
{S( ¢>, D2'ljJ)} * ,
for all ¢> and 'ljJ in the space. Thus D2 is Hermitian if
However, by two partial integrations we see readily that /
L/2 dx'ljJ*(X)¢>"(X) -L/2
= /L/2
dx'ljJ"*(x)¢>(x)
+ B,
(A.20)
-L/2
in which the boundary term is
The operator D2 is Hermitian only if B vanishes, which will not be the case unless we impose further restrictions on our space. With L finite and with a space of bounded functions that satisfy periodic boundary conditions, i.e., ¢>(L/2) = ¢>( -L/2), D2 is indeed Hermitian. Moreover, the two independent solutions of
(A.21) are ¢>(x) = exp(±vlAx). These solutions satisfy the condition of periodicity if
where n is an integer. Thus the eigenvalues form a discrete (but infinite) set, and the eigenvectors are bounded in modulus. Moreover, it is known from the theory
190
Completeness of Eigenvectors
of Fourier series that any bounded, twice-differentiable, periodic function can be represented by a convergent Fourier series, i.e., the eigenvectors form indeed a complete set: they span the space. Since the number of Fourier coefficients is in general infinite, the vector space describing the system has an infinite number of dimensions. To obtain the above desired result, however, we had to restrict the space by the artificial constraint of periodicity. If we relax this constraint, n2 is not Hermitian. If we let L = 00, which is more natural, and require the functions in the space to be bounded, as well as twice differentiable, then the eigenvalues are continuous, comprising all the negative real numbers, and the eigenvectors are e±iJ.Lx, where J.1 = J A. By the theory of Fourier analysis, we know that the functions e±iJ.Lx are complete, in the sense that any bounded, twice-differentiable function can be written as a Fourier integral, but n 2 is not Hermitian on this space, since the boundary terms in Eq.(A.20) are ill-defined in this case. True, the boundary terms oscillate themselves to death, so we might hope that n2 is 'almost' Hermitian, and shares the good features of such operators. Indeed, the kinetic energy operator, Eq.(A.19), has the desirable property that its eigenvalues are the positive, real numbers, and moreover its eigenvectors span the infinite-dimensional vector space. More can be said about the peculiarities of infinite-dimensional vector spaces; and we will return to the matter in Volume 2. The option chosen in this book is to drop the requirement that all physical observables be strictly represented by Hermitian operators on Hilbert space. The essential requirements are that they have real eigenvalues and that their eigenvectors are complete, and this is certainly true of the kinetic energy operator if it is defined on a suitable vector space. Dirac himself was careful never to limit his observables to those that are Hermitian on a Hilbert space, indeed he wrote in his book explicitly that the bra and ket vectors form a more general space than a Hilbert space. He requires an observable to have real eigenvalues and to have a complete set of eigenvectors; but he adds that this is a physical assumption and that it is often very difficult to decide mathematically whether a particular dynamical quantity is an observable. If one can prove that the corresponding operator is Hermitian, the job is done, of course; but if it is not one can sometimes still check the conditions of reality and completeness, as we did for the kinetic energy. Other options, which we will consider in Volume 2, are (1) Extending the space, so that the kinetic energy, and scattering operators, are truly Hermitian. This involves a 'rigged' Hilbert space. (2) Discarding Hilbert space and working in an algebraic approach with projection operators. Von Neumann handled the difficulty in this way.
191
Exercises
A.7
Exercises
In these exercises, whenever the word 'matrix' or 'operator' is used, this means a square matrix of finite dimension. Problem 1 Prove that the matrix (1 - iA) has an inverse if A is Hermitian. Problem 2 Let I be the unit matrix, and A an arbitrary square matrix. Show (1) det [1 + fA] = 1 + fTr (A) + O(f2), where the trace of the matrix is the sum of the matrix elements on the diagonal, i.e., Tr (A) = I.: j Ajj , (2) det[exp(A)] = exp[Tr (A)] .
Problem 3 A Hermitian operator P is called a projection operator if p2 = P. (1) Prove that the only projection operator with an inverse is the identity operator. (2) Let PI and P 2 be projection operators. Prove that if PI + P 2 is a projection operator, then P I P 2 = P 2 P I = o. Operators satisfying this condition are called orthogonal operators.
Problem 4 Consider the three orthonormal vectors, {I
ei
> H=ll given by
(1) Find the projection operators Pi associated with each of these vectors. (2) Prove that Pi projects onto the line along I ei > . (3) Verify that L~=I Pi = 13 •
Problem 5 Show that the symmetrizing operator, S, defined such that, for any matrix, Yes), depending on a variable, s, SV(s) = ![V(s) + V( -s)], and the antisymmetrizing operator, A, defined such that AV(s) = HV(s) - V( -s)], are projection operators. Problem 6 Prove that two Hermitian matrices commute if there exists a complete orthonormal set of common eigenvectors.
192
Completeness of Eigenvectors
Problem 7 Let s be a complex variable and A a constant square matrix of finite dimension. Define Yes) = exp(sA). Show (1) d~is) = A V = V A (2) {V(s)V( -s)} = 0 and hence that the inverse of the matrix Yes) is
:8
given by V-1(s) = V( -s) (3) Using the above results, show that if A is a Hermitian matrix, then eiA is a unitary matrix.
Problem 8 If A and B are square matrices and A commutes with [A, B], show that, for any
n=1,2, ... (1) [An, [A, Bl] = 0 (2) [An, B] = nAn-l [A, B]] (3) t@tA, B] etA [A, B] Problem 9
t ~f\ . 8 'J ~ ,t t A. t1 .
Let t be a complex variable, and define
J(t) = etA.etB.e-t(A+B) where A and B are square matrices that both commute with [A, BJ. Show
(1) f'(t) = 2tJ(t)[A, B] (2) J(t) = exp {t 2 [A, Bn (3) eA.eB=exp{A+B+[A,B]} Problem 10 Let t be a complex variable, and define
get) = etA B e- tA where A and B are square matrices that do not commute with [A, B]. Show
(1) g'(t) = [A,g(t)] (2) get) = exp[tOA]B, where OA is the linear operator, called the commutator operator, that is defined by nAG = [A, GJ, where G is any square matrix.
(3) eAB e- A = B + [A, B] + HA, [A, B]] + ~ [A, [A, [A, BJ]] + ... Here the (n + l)th term has a factor (n!)-l and contains n nested commutators.
Bibliography
Atkins, P.W., (1983) Molecular Quantum Mechanics, second edition, Clarendon Press, Oxford. Atkins, P.W., (1983) Solutions Manual for Molecular Quantum Mechanics, Clarendon Press, Oxford. Basdevant, J-L., (1986) Mecanique Quantique, Ecole Poly technique, Paris. "'Ballentine, L.E., (1989) Quantum Mechanics, Prentice-Hall, New Jersey. ..,.Dirac, P.A.M., (1958) Quantum Mechanics, fourth edition, Oxford University Press, Oxford . .J Gasiorowicz, S., (1974) Quantum Physics, second edition, John Wiley and Sons Inc., New York. Gol'dman, I.I. and Krivchenkov, V.D., Problems in Quantum Mechanics, Pergamon, London. JGoswami, A., (1992) Quantum Mechanics, Wm. C. Brown, Dubuque. JGreiner, W., (1989) Quantum Mechanics, An Introduction, Springer-Verlag, New York. Haar, D. ter, (1975) Problems in Quantum Mechanics, third edition, Pion Ltd., London. JItzykson, C. and Zuber, J.B., (1980) Quantum Field Theory, McGraw-Hill Inc., New York. Kok, L.P. and Visser, J., (1996) Quantum Mechanics, Problems and Solutions, second edition, Coulomb Press Leyden, Leiden. j Liboff, R.L., (1992) Quantum Mechanics, Addison-Wesley Publishing Company Inc., Reading. Mandl, F., (1992) Quantum Mechanics, John Wiley and Sons, Chichester. McMurry, S.M., (1993) Quantum Mechanics, Addison-Wesley Publishing Company Inc., Reading. ~Merzbacher, E., (1998) Quantum Mechanics, third edition, John Wiley and Sons, New York. JSchiff, L.L, (1968) Quantum Mechanics, third edition, McGraw-Hill, New York. Schmid, E.W., Spitz, G. and Losch, W. (1987) Theoretische Physik mit dem Personal Computer, Springer-Verlag, Berlin . .1 Townsend, J.S., (1992) A Modern Approach to Quantum Mechanics, McGraw-Hill, New York.
193
Index
Action, 3 Addition of angular momenta, 101 Clebsch-Gordan coefficients, 101, 112 spin and orbital parts, 105 two spins, 104 Airy function, 93 Alkali atom, 98 Almost degenerate levels, 123 Ammonia molecule, 124 maser, 125 Amplitude, 132 Angular momentum, 44 eigenvalues, 47 eigenvectors, 57 matrices, 99 operators, 43 in four dimensions, 63 Anharmonic oscillator, 118, 128 Anisotropic oscillator, 32 Annihilation operator, 32, 45, 76 Antisymmetric wave function, 155 Slater determinant, 162 Associated Legendre functions,56 Asymptotic behavior r -+ 0, 65,78 r -+ 00, 68, 79 Asymptotic expansion for Airy function, 96 for Bessel functions, 79 for plane wave, 137 Atomic structure, 147 Axioms of vector spaces, 9, 27
Bell inequality, 29, 110 Beryllium, 173 Bessel functions, 75, 82 Born approximation, 135, 145 Bound states, deuterium, 75 harmonic oscillator, 37 linear potential, 93 square well, 73 Bra vectors, 11 Breit-Wigner resonance, 140 Canonical momentum, 4 Canonical transformation, 2 Carbon dioxide, 174 Carbon monoxide, 80 Cauchy sequence, 12 Center-of-mass coordinates, 74, 81, 90 Central potential, 65 Charmonium, 92 Classical mechanics, 1 Clebsch-Gordan coefficients, 101, 112 Commutation relations, for q and p, 13 for components of I, 43 Compact support, 15 Complete space, 12 Completeness of Eigenvectors, 175, 188 Complex vector space, 9 Configuration representation, 20 Conservative field, 1 Coulomb potential, 85 194
195
Index
Creation operator, 32, 45, 76 Cross-section, 133 optical theorem, 145 identical particles, 144 Cuprate, high T c , 42 Degenerac~
124, 188 for harmonic oscillator, 39 for hydrogen atom, 88 in perturbation theory, 120 for simultaneous eigenvectors, 188 Determinant, 176 Deuterium, 97 Deuteron, 74, 110 Diagonalizaton of matrices, 184 Diatomic molecule, 83 carbon monoxide, 80 hydrogen, 171 Differential cross section, 133 Differential equation for Airy function, 93 for Hermite polynomial, 41 for associated· Laguerre function, 89 for associated Legendre function, 57 Differential operator, 21 Dirac delta function, 14 Dirac notation, 80 Distribution, 14 Domain of operator, 12 Eigenfunctions, 37, 50 Eigenvalues, 13, 186 Eigenvectors, 13, 188 Exchange term, 154 Exclusion principle, 155 Expansion postulate, 175, 190 Evolution operator, 22 Euler-Lagrange equation, 3 Fine structure constant, 148 Flux conservation, 29 Fourier theorem, 67 Functional, 10, 15 Fundamental theorem of linear equations, 176 of algebra, 182
Gaussian reduction, 178 Generator of time translation, 22. epees Horne-Zeilinger state, 111 Green's function, 66, 129 Ground state, 90, 118 of harmonic oscillator, 37 of He atom, 152 ofH atom, 90 of square well, 73 Hamilton equations, 5 Hamiltonian, 4, 22 Hamilton's variation principle, 3 Harmonic oscillator, 32, 69 Hartree-Fock method, 157, 173 Heaviside step function, 18 Heisenberg picture, 23 Heisenberg uncertainty relation, 41, 62 Helium atom, 147 electron-electron repulsion, 147 effective spin-dependent term, 157 Hermite polynomials, 36,41 Hermitian operators, 175 Hermitian conjugate, 12 diagonalization, 184 commuting matrices, 187 Hilbert space, 9 Hund rules, 161, 173 Hydrogen atom, 85 degeneracy of eigenvalues, 88 Hydrogen ion, 162 Hydrogen molecule, 171 Hyperons, 91 Identical particles in He atom, 155 in Ht ion, 165 proton-proton scattering, 144 Ions, 85, 152 helium ionization energies, 173 hydrogen ion, 162 Infinite dimensional spaces, 188 Jacobi identity, 6 proof for commutators, 7 proof for Poisson brackets, 7
196
Kronecker delta, 6 Ket vectors, 11 Ladder operator, 33, 45, 76 Lagrangian, 3 Laguerre polynomials, 88, 97 Laplacian, 25, 52 Legendre functions, 55 Legendre polynomials, 56, 137, 145 Levi-Civita symbol, 43 Linear potential, 91 Linear vector space, 9 linear operators, 12 Lippmann-Schwinger equation, 129 Lowering operator, 33, 45, 76 Maser, 125 Matrix, 176 commuting Hermitian matrices, 187 determinant, 176 determinant of matrix product, 181 diagonalization of matrix, 184 GauE reduction, 178 inverse of matrix, 180 orthonormal basis, 186 permutation, 176 Mean, 41, 63 Molecule, 80, 83, 174 H2 atom, 171 Ht ion, 165 vibrational-rotational states, 80 Momentum operator, 19 Neutron-proton scattering, 141 Nitrogen, 173 Norm, 12 N umber operator, 33 Null vector, 9 Observables, 175, 190 Operators, 12, 22, 23, 32 annihilation, 32 45, 76 creation, 32, 45, 76 hermitian, 13 ladder, 33, 45, 76 linear, 12
Index
lowering, 33, 45, 76 projection, 191 raising, 32, 45, 76 symmetry, 91 unitary, 22 Optical theorem, 145 Orbital angular momentum, 44 Orthonormal basis, 10, 27, 186 Oscillator, 32, 69 Overlap integral, 165 Pauli principle, 155 Pade approximants, 128 Pauli exclusion principle, 155 Pauli matrices, 101, 109, 126 Periodic table, 159 Perturbation theory, 113 first order, 115, 121, 126 second order, 116, 128 Phase shifts, 139, 145 neutron-proton S wave, 141 proton-proton S wave, 144 resonant, 140 threshold behavior, 140 Poisson brackets, 6 Polar coordinates, 51, 64 Position operator, 13 Positronium, 107, 110 Potential Coulomb, 85 simple harmonic, 32, 69 linear, 91 square-well, 71 Yukawa, 75, 146, 168 Potential energy, 2 Projection operator, 191 Proton-neutron scattering, 141 Probability density, 29 Proton-proton scattering, 144 Projection operator, 191 Quantum chromo dynamics, 92 Quadratic potential, 32, 69 Quantization of angular momentum, 44 Quarks, 91
197
Index
Radial equation, 65 Raising operator, 32, 45, 76 Range of operator, 12 Representation of 8-function, 19, 28 Relative motion, 74, 81, 90 Resonance, 140 singlet neutron-proton state, 143 Riesz theorem, 10 Ritz variational principle, 117 Runge-Lenz vector, 98 Scalar product, 9 Scattering, 129 amplitude, 132 Born approximation, 135, 145 cross-section, 133 phase shifts, 139 resonant, 140 scattering length, 142 square well, 74, 141 Schrodinger's cat, 125 Schrodinger equation, 24 asymptotics, 65, 79 center-of-mass coordinates, 74, 81, 90 central potential, 65 charmonium, 92 hydrogen atom, 85 hydrogen ion, 162 hydrogen molecule, 171 simple harmonic oscillator, 38 Schwartz space, 15 Screened Coulomb potential, 127 Screening of charge, 150 Separable potential, 146 Simple harmonic oscillator, 32, 69 in one dimension, 41 in two dimensions, 42, 126 in three dimensions, 32, 42, 126 isotropic oscillator, 41, 69 Singlet state, 104, 143 Slater determinant, 162 S-matrix, 139 phase shifts, 139 resonances, 140 Spectroscopic notation, 161, 172
Spherical Bessel function, 75, 82 Spherical Hankel functions, 78 Spherical Neumann function, 77, 82 Spherical harmonics, 54, 57 Spin, 100 Spin-orbit coupling, 161 Spin-and-statistics theorem, 144, 155 Square integrability, 66, 68 Square well potential, 71 bound states, 72, scattering states, 74, 141 Stark effect, in harmonic oscillator, 39 in hydrogen atom, 120, 128 Step function, 18 Strangeness, 91 Superposition of states, 125 Symmetry operator, 91 Test functions, 15 Thomas-Reiche-Kuhn rule, 41 Triplet state, 104, 142 Tritium, 98 Two-electron system, 147 Uncertainty, 41, 62, 63 Unitary operators, 22 Unitary transformation, 184 Variational method, 117 Vector spaces, 9 Vibrational-rotational spectra, 80 Wave function, 20 Width of resonance, 141 WKB approximation, 95 X-ray photon, 108 Yukawa potential, 75, 146, 168 in Born approximation, 136 Zeeman effect, 60 Zero-point energy, 33
