Advanced quantum mechanics

Introduction S G

State vectors

in re a lity

Stern Gerlach experiment Ù

z

S B

to d e te c to r

N

o v e n

Based on the measurements one can evaluate the z-components Sz of the angular momentum of the atoms and find out that • for the upper distribution Sz = h ¯ /2.

c o llim a to r

• for the lower distribution Sz = −¯h/2.

In the Stern Gerlach experiment • silver atoms are heated in an oven, from which they escape through a narrow slit, • the atoms pass through a collimator and enter an inhomogenous magnetic field, we assume the field to be uniform in the xy-plane and to vary in the z-direction, • a detector measures the intensity of the electrons emerging from the magnetic field as a function of z. We know that • 46 of the 47 electrons of a silver atom form a spherically symmetric shell and the angular momentum of the electron outside the shell is zero, so the magnetic moment due to the orbital motion of the electrons is zero, • the magnetic moment of an electron is cS, where S is the spin of an electron, • the spins of electrons cancel pairwise, • thus the magnetic moment µ of an silver atom is almost solely due to the spin of a single electron, i.e. µ = cS, • the potential energy of a magnetic moment in the magnetic field B is −µ · B, so the force acting in the z-direction on the silver atoms is

In quantum mechanics we say that the atoms are in the angular momentum states ¯h/2 and −¯h/2. The state vector is a mathematical tool used to represent the states. Atoms reaching the detector can be described, for example, by the ket-vectors |Sz ; ↑i and |Sz ; ↓i. Associated with the ket-vectors there are dual bra-vectors hSz ; ↑ | and hSz ; ↓ |. State vectors are assumed • to be a complete description of the described system, • to form a linear (Hilbert) space, so the associated mathematics is the theory of (infinite dimensional) linear spaces. When the atoms leave the oven there is no reason to expect the angular momentum of each atom to be oriented along the z-axis. Since the state vectors form a linear space also the superposition c↑ |Sz ; ↑i + c↓ |Sz ; ↓i is a state vector which obviously describes an atom with angular momentum along both positive and negative z-axis. The magnet in the Stern Gerlach experiment can be thought as an apparatus measuring the z-component of the angular momentum. We saw that after the measurement the atoms are in a definite angular momentum state, i.e. in the measurement the state c↑ |Sz ; ↑i + c↓ |Sz ; ↓i

collapses either to the state |Sz ; ↑i or to the state |Sz ; ↓i. A generalization leads us to the measuring postulates of quantum mechanics: Postulate 1 Every measurable quantity is associated So the measurement of the intensity tells how the with a Hermitean operator whose eigenvectors form a z-component the angular momentum of the silver atoms passing through the magnetic field is distributed. Because complete basis (of a Hilbert space), the atoms emerging from the oven are randomly oriented and we would expect the intensity to behave as shown below. Postulate 2 In a measurement the system makes a transition to an eigenstate of the corresponding operator and the result is the eigenvalue associated with that eigenvector. S G If A is a measurable quantity and A the corresponding Hermitean operator then an arbitrary state |αi can be described as the superposition c la s s ic a lly X In reality the beam is observed to split into two |αi = ca′ |a′ i, components. ′ a Fz = µz

∂Bz . ∂z

where the vectors |a′ i satisfy

Note The matrix representation is not unique, but

A|a′ i = a′ |a′ i. The measuring event A can be depicted symbolically as A

|αi −→ |a′ i. In the Stern Gerlach experiment the measurable quantity is the z-component of the spin. We denote the measuring z and the corresponding Hermitean event by SGˆ operator by Sz . We get ¯ h |Sz ; ↑i 2 ¯h − |Sz ; ↓i 2 c↑ |Sz ; ↑i + c↓ |Sz ; ↓i

Sz |Sz ; ↑i

=

Sz |Sz ; ↓i

=

|Sz ; αi

=

|Sz ; αi

SGˆ z −→

|Sz ; ↑i or

|Sz ; αi

SGˆ z −→

|Sz ; ↓i.

depends on the basis. In the case of our example we get the 2 × 2-matrix representation   ¯h 1 0 Sz 7→ , 0 −1 2 when we use the set {|Sz ; ↑i, |Sz ; ↓i} as the basis. The base vectors map then to the unit vectors   1 |Sz ; ↑i 7→ 0   0 |Sz ; ↓i 7→ 1 of the two dimensional Euclidean space. Although the matrix representations are not unique they are related in a rather simple way. Namely, we know that Theorem 1 If both of the basis {|a′ i} and {|b′ i} are orthonormalized and complete then there exists a unitary operator U so that |b1 i = U |a1 i, |b2 i = U |a2 i, |b3 i = U |a3 i, . . .

′

Because the vectors |a i in the relation

If now X is the representation of an operaor A in the basis {|a′ i} the representation X ′ in the basis {|b′ i} is obtained by the similarity transformation

A|a′ i = a′ |a′ i are eigenvectors of an Hermitean operator they are orthognal with each other. We also suppose that they are normalized, i.e. ha′ |a′′ i = δa′ a′′ . Due to the completeness of the vector set they satisfy X |a′ iha′ | = 1, a′

where 1 stands for the identity operator. This property is called the closure. Using the orthonormality the coefficients in the superposition X |αi = ca′ |a′ i a′

can be written as the scalar product ca′ = ha′ |αi. An arbitrary linear operator B can in turn be written with the help of a complete basis {|a′ i} as B=

X

a′ ,a′′

|a′ iha′ |B|a′′ iha′′ |.

where T is the representation of the base transformation operator U in the basis {|a′ i}. Due to the unitarity of the operator U the matrix T is a unitary matrix. Since • an abstract state vector, excluding an arbitrary phase factor, uniquely describes the physical system, • the states can be written as superpositions of different base sets, and so the abstract operators can take different matrix forms, the physics must be contained in the invariant propertices of these matrices. We know that Theorem 2 If T is a unitary matrix, then the matrices X and T † XT have the same trace and the same eigenvalues. The same theorem is valid also for operators when the trace is defined as X trA = ha′ |A|a′ i. a′

Since

Abstract operators can be represented as matrices: |a1 i |a2 i  ha1 | ha1 |B|a1 i ha1 |B|a2 i ha2 |   ha2 |B|a1 i ha2 |B|a2 i B 7→ ha3 |  ha3 |B|a1 i ha3 |B|a2 i  .. .. .. . . .

X ′ = T † XT,

|a3 i ha1 |B|a3 i ha2 |B|a3 i ha3 |B|a3 i .. .

...  ... ...  . ...  .. .

• quite obviously operators and matrices representing them have the same trace and the same eigenvalues, • due to the postulates 1 and 2 corresponding to a measurable quantity there exists an Hermitean operator and the measuring results are eigenvalues of this operator,

the results of measurements are independent on the particular representation and, in addition, every measuring event corresponding to an operator reachable by a similarity transformation, gives the same results. Which one of the possible eigenvalues will be the result of a measurement is clarified by Postulate 3 If A is the Hermitean operator corresponding to the measurement A, {|a′ i} the eigenvectors of A associated with the eigenvalues {a′ }, then the probability for the result a′ is |ca′ |2 when the system to be measured is in the state X |αi = ca′ |a′ i.

|Sz ; ↑i

S

S G z Ù

S

z z

S G z

S

z

Ù

|Sz ; ↑i = c↑↑ |Sx ; ↑i + c↑↓ |Sx ; ↓i. For the other component we have correspondingly |Sz ; ↓i = c↓↑ |Sx ; ↑i + c↓↓ |Sx ; ↓i. When the intensities are equal the coeffiecients satisfy 1 |c↑↑ | = |c↑↓ | = √ 2 1 |c↓↑ | = |c↓↓ | = √ 2

|Sz ; ↓i =

• the postulate can also be interpreted so that the quantities |ca′ |2 tell the probability for the system being in the state |a′ i, • the physical meaning of the matrix element hα|A|αi is then the expectation value (average) of the measurement and • the normalization condition hα|αi = 1 says that the system is in one of the states |a′ i. Instead of measuring the spin z-component of the atoms with spin polarized along the z-axis we let this polarized beam go through the SGˆ x experiment. The result is exactly like in a single SGˆ z experiment: the beam is again splitted into two components of equal intensity, this time, however, in the x-direction.

S G z Ù

S

S z z

S G x Ù

S

S

SGˆ x −→

x x

1 1 √ |Sx ; ↑i + √ |Sx ; ↓i 2 2   1 1 eiδ1 √ |Sx ; ↑i − √ |Sx ; ↓i . 2 2

ˆ , nor for that There is nothing special in the direction x matter, in any other direction. We could equally well let the beam pass through a SGˆ y experiment, from which we could deduce the relations 1 1 |Sz ; ↑i = √ |Sy ; ↑i + √ |Sy ; ↓i 2 2   1 1 iδ2 √ |Sy ; ↑i − √ |Sy ; ↓i , |Sz ; ↓i = e 2 2 or we could first do the SGˆ x experiment and then the SGˆ y experiment which would give us |Sx ; ↑i = |Sx ; ↓i =

eiδ3 √ |Sy ; ↑i + 2 eiδ3 √ |Sy ; ↑i − 2

eiδ4 √ |Sy ; ↓i 2 eiδ4 √ |Sy ; ↓i. 2

In other words 1 |hSy ; ↑ |Sx ; ↑i| = |hSy ; ↓ |Sx ; ↑i| = √ 2 1 |hSy ; ↑ |Sx ; ↓i| = |hSy ; ↓ |Sx ; ↓i| = √ . 2 We can now deduce that the unknown phase factors must satisfy δ2 − δ1 = π/2 or − π/2. A common choice is δ1 = 0, so we get, for example,

So, we have performed the experiment |Sz ; ↑i

according to the postulate 3. Excluding a phase factor, our postulates determine the transformation coefficients. When we also take into account the orthogonality of the state vectors |Sz ; ↑i and |Sz ; ↓i we can write |Sz ; ↑i =

If we now let the polarized beam to pass through a new SGˆ z experiment we see that the beam from the latter experiment does not split any more. According to the postulate this result can be predicted exactly. We see that

|Sx ; ↓i.

Again the analysis of the experiment gives Sx = h ¯ /2 and Sx = −¯h/2 as the x-components of the angular momenta. We can thus deduce that the state |Sz ; ↑i is, in fact, the superposition

a′

Only if the system already before the measurement is in a definite eigenstate the result can be predicted exactly. For example, in the Stern Gerlach experiment SGˆ z we can block the emerging lower beam so that the spins of the remaining atoms are oriented along the positive z-axis. We say that the system is prepared to the state |Sz ; ↑i.

SGˆ x −→

|Sx ; ↑i or

|Sz ; ↑i = |Sz ; ↓i =

1 √ |Sx ; ↑i + 2 1 √ |Sx ; ↑i − 2

1 √ |Sx ; ↓i 2 1 √ |Sx ; ↓i. 2

Thinking like in classical mechanics, we would expect both the z- and x-components of the spin of the atoms in z and SGˆ x the upper beam passed through the SGˆ experiments to be Sx,z = h ¯ /2. On the other hand, we can reverse the relations above and get 1 1 |Sx ; ↑i = √ |Sz ; ↑i + √ |Sz ; ↓i, 2 2 so the spin state parallel to the positive x-axis is actually a superposition of the spin states parallel to the positive and negative z-axis. A Stern Gerlach experiment confirms this.

S G z Ù

S

S z

z

S G x Ù

S S

x

S G z Ù

x

S

S

z z

After the last SGˆ z measurement we see the beam splitting again into two equally intensive componenents. The experiment tells us that there are quantitities which cannot be measured simultaneously. In this case it is impossible to determine simultaneously both the z- and x-components of the spin. Measuring the one causes the atom to go to a state where both possible results of the other are present. We know that Theorem 3 Commuting operators have common eigenvectors. When we measure the quantity associated with an operator A the system goes to an eigenstate |a′ i of A. If now B commutes with A, i.e. [A, B] = 0, then |a′ i is also an eigenstate of B. When we measure the quantity associated with the operator B while the system is already in an eigenstate of B we get as the result the corresponding eigenvalue of B. So, in this case we can measure both quantities simultaneously. On the other hand, Sx and Sz cannot be measured simultaneously, so we can deduce that [Sx , Sz ] 6= 0. So, in our example a single Stern Gerlach experiment gives as much information as possible (as far as only the spin is concerned), consecutive Stern Gerlach experiments cannot reveal anything new. In general, if we are interested in quantities associated with commuting operators, the states must be characterized by eigenvalues of all these operators. In many cases quantum mechanical problems can be reduced to the tasks to find the set of all possible commuting operators (and their eigenvalues). Once this set is found the states can be classified completely using the eigenvalues of the operators.

Translations The previous discrete spectrum state vector formalism can be generalized also to continuos cases, in practice, by replacing • summations with integrations • Kronecker’s δ-function with Dirac’s δ-function. A typical continuous case is the measurement of position: • the operator x corresponding to the measurement of the x-coordinate of the position is Hermitean, • the eigenvalues {x0 } of x are real, • the eigenvectors {|x0 i} form a complete basis. So, we have x|x0 i = x0 |x0 i Z ∞ 1 = dx0 |x0 ihx0 | −∞ Z ∞ |αi = dx0 |x0 ihx0 |αi, −∞

where |αi is an arbitrary state vector. The quantity hx0 |αi is called a wave function and is usually written down using the function notation hx0 |αi = ψα (x0 ). Obviously, looking at the expansion Z ∞ |αi = dx0 |x0 ihx0 |αi, −∞

The effect of an infinitesimal translation on an arbitrary state can be seen by expanding it using position eigenstates: Z 00 00 |αi −→ T (dx )|αi = T (dx ) d3 x0 |x0 ihx0 |αi Z = d3 x0 |x0 + dx00 ihx0 |αi Z = d3 x0 |x0 ihx0 − dx00 |αi, because x0 is an ordinary integration variable. To construct T (dx0 ) explicitely we need extra constraints: 1. it is natural to require that it preserves the normalization (i.e. the conservation of probability) of the state vectors: hα|αi = hα|T † (dx0 )T (dx0 )|αi. This is satisfied if T (dx0 )is unitary, i.e. T † (dx0 )T (dx0 ) = 1. 2. we require that two consecutive translations are equivalent to a single combined transformation: T (dx0 )T (dx00 ) = T (dx0 + dx00 ). 3. the translation to the opposite direction is equivalent to the inverse of the original translation: T (−dx0 ) = T −1 (dx0 ). 4. we end up with the identity operator when dx0 → 0:

the quantity |ψα (x0 )|2 dx0 can be interpreted according to lim T (dx0 ) = 1. dx0 →0 the postulate 3 as the probability for the state being 0 0 0 0 localized in the neighborhood (x , x + dx ) of the point x . It is easy to see that the operator The position can be generalized to three dimension. We 0 denote by |x i the simultaneous eigenvector of the T (dx0 ) = 1 − iK · dx0 , operators x, y and z, i.e. where the components Kx , Ky and Kz of the vector K |x0 i ≡ |x0 , y 0 , z 0 i are Hermitean operators, satisfies all four conditions. x|x0 i = x0 |x0 i, y|x0 i = y 0 |x0 i, z|x0 i = z 0 |x0 i. Using the definition T (dx0 )|x0 i = |x0 + dx0 i

The exsistence of the common eigenvector requires commutativity of the corresponding operators:

we can show that [xi , xj ] = 0. Let us suppose that the state of the system is localized at the point x0 . We consider an operation which transforms this state to another state, this time localized at the point x0 + dx0 , all other observables keeping their values. This operation is called an infinitesimal translation. The corresponding operator is denoted by T (dx0 ):

[x, T (dx0 )] = dx0 . Substituting the explicit reprersentation T (dx0 ) = 1 − iK · dx0 it is now easy to prove the commutation relation [xi , Kj ] = iδij .

T (dx0 )|x0 i = |x0 + dx0 i. The equations The state vector on the right hand side is again an eigenstate of the position operator. Quite obviously, the vector |x0 i is not an eigenstate of the operator T (dx0 ).

T (dx0 ) = 1 − iK · dx0 T (dx0 )|x0 i = |x0 + dx0 i

can be considered as the definition of K. One would expect the operator K to have something to do with the momentum. It is, however, not quite the momentum, because its dimension is 1/length. Writing p=h ¯K we get an operator p, with dimension of momentum. We take this as the definition of the momemtum. The commutation relation [xi , Kj ] = iδij

So, we have derived the canonical commutation relations or fundamental commutation relations [xi , xj ] = 0,

[pi , pj ] = 0,

[xi , pj ] = i¯hδij .

Recall, that the projection of the state |αi along the state vector |x0 i was called the wave function and was denoted like ψα (x0 ). Since the vectors |x0 i form a complete basis the scalar product between the states |αi and |βi can be written with the help of the wave functions as Z Z hβ|αi = dx0 hβ|x0 ihx0 |αi = dx0 ψβ∗ (x0 )ψα (x0 ),

can now be written in a familiar form like i.e. hβ|αi tells how much the wave functions overlap. If |a0 i is an eigenstate of A we define the corresponding eigenfunction ua0 (x0 ) like

[xi , pj ] = i¯ hδij . Finite translations 0

Consider translation of the distance ∆x along the x-axis: ˆ )|x0 i = |x0 + ∆x0 x ˆ i. T (∆x0 x We construct this translation combining infinitesimal translations of distance ∆x0 /N letting N → ∞: N ipx ∆x0 ˆ ) = lim 1 − T (∆x0 x N →∞ N¯ h   0 ipx ∆x = exp − . h ¯ 

It is relatively easy to show that the translation operators satisfy ˆ ), T (∆x0 x ˆ )] = 0, [T (∆y 0 y so it follows that [py , px ] = 0. Generally [pi , pj ] = 0.

An arbitrary wave function ψα (x0 ) can be expanded using eigenfunctions as X ca0 ua0 (x0 ). ψα (x0 ) = a0

The matrix element hβ|A|αi of an operator A can also be expressed with the help of eigenfunctions like Z Z hβ|A|αi = dx0 dx00 hβ|x0 ihx0 |A|x00 ihx00 |αi Z Z 0 = dx dx00 ψβ∗ (x0 )hx0 |A|x00 iψα (x00 ). To apply this formula we have to evaluate the matrix elements hx0 |A|x00 i, which in general are functions of the two variables x0 and x00 . When A depends only on the position operator x, A = f (x),

This commutation relation tells that it is possible to construct a state vector which is a simultaneous eigenvector of all components of the momentum operator, i.e. there exists a vector |p0 i ≡ |p0x , p0y , p0z i,

the calculations are much simpler: Z hβ|f (x)|αi = dx0 ψβ∗ (x0 )f (x0 )ψα (x0 ). Note f (x) on the left hand side is an operator while f (x0 )

on the right hand side is an ordinary number.

so that px |p0 i = p0x |p0 i,

ua0 (x0 ) = hx0 |a0 i.

py |p0 i = p0y |p0 i,

pz |p0 i = p0z |p0 i.

The effect of the translation T (dx0 ) on an eigenstate of the momentum operator is     ip0 · dx0 ip · dx0 0 0 0 |p i = 1 − |p0 i. T (dx )|p i = 1 − h ¯ h ¯ The state |p0 i is thus an eigenstate of T (dx0 ): a result, which we could have predicted because [p, T (dx0 )] = 0. 0

Note The eigenvalues of T (dx ) are complex because it is not Hermitean.

Momentum operator p in position basis {|x0 i}

For simplicity we consider the one dimensional case. According to the equation Z T (dx00 )|αi = T (dx00 ) d3 x0 |x0 ihx0 |αi Z = d3 x0 |x0 + dx00 ihx0 |αi Z = d3 x0 |x0 ihx0 − dx00 |αi we can write   ip dx00 1− |αi ¯h

Z =

dx0 T (dx00 )|x0 ihx0 |αi

Z

dx0 |x0 ihx0 − dx00 |αi   Z ∂ = dx0 |x0 i hx0 |αi − dx00 0 hx0 |αi . ∂x =

In the last step we have expanded hx0 − dx00 |αi as Taylor series. Comparing both sides of the equation we see that   Z ∂ p|αi = dx0 |x0 i −i¯ h 0 hx0 |αi , ∂x or, taking scalar product with a position eigenstate on both sides, ∂ hx0 |p|αi = −i¯ h 0 hx0 |αi. ∂x In particular, if we choose |αi = |x0 i we get hx0 |p|x00 i = −i¯ h

∂ δ(x0 − x00 ). ∂x0

Taking scalar product with an arbitrary state vector |βi on both sides of   Z ∂ p|αi = dx0 |x0 i −i¯ h 0 hx0 |αi ∂x we get the important relation   Z ∂ hβ|p|αi = dx0 ψβ∗ (x0 ) −i¯h 0 ψα (x0 ). ∂x Just like the position eigenvalues also the momentum eigenvalues p0 comprise a continuum. Analogically we can define the momentum space wave function as hp0 |αi = φα (p0 ). We can move between the momentum and configuration space representations with help of the relations Z ψα (x0 ) = hx0 |αi = dp0 hx0 |p0 ihp0 |αi Z 0 0 φα (p ) = hp |αi = dx0 hp0 |x0 ihx0 |αi. The transformation function hx0 |p0 i can be evaluated by substituting a momentum eigenvector |p0 i for |αi into hx0 |p|αi = −i¯ h

∂ hx0 |αi. ∂x0

Then

∂ hx0 |p0 i. ∂x0 The solution of this differential equation is  0 0 ip x , hx0 |p0 i = C exp ¯h hx0 |p|p0 i = p0 hx0 |p0 i = −i¯ h

where the normalization factor C can be determined from the identity Z hx0 |x00 i = dp0 hx0 |p0 ihp0 |x00 i.

Here the left hand side is simply δ(x0 − x00 ) and the integration of the left hand side gives 2π¯h|C|2 δ(x0 − x00 ). Thus the transformation function is  0 0 ip x 1 0 0 , exp hx |p i = √ ¯h 2π¯h and the relations 0

0

ψα (x )

= hx |αi =

φα (p0 )

= hp0 |αi

=

Z

dp0 hx0 |p0 ihp0 |αi

Z

dx0 hp0 |x0 ihx0 |αi.

can be written as familiar Fourier transforms  0 0 Z  ip x 1 √ dp0 exp ψα (x0 ) = φα (p0 ) ¯h 2π¯h Z    ip0 x0 1 √ dx0 exp − φα (p0 ) = ψα (x0 ). ¯h 2π¯h

Time evolution operator In quantum mechanics • unlike position, time is not an observable. • there is no Hermitean operator whose eigenvalues were the time of the system. • time appears only as a parameter, not as a measurable quantity. So, contradictory to teachings of the relativity theory, time and position are not on equal standing. In relativistic quantum field theories the equality is restored by degrading also the position down to the parameter level. We consider a system which at the moment t0 is in the state |αi. When time goes on there is no reason to expect it to remain in this state. We suppose that at a later moment t the system is described by the state |α, t0 ; ti,

(t > t0 ),

where the parameter t0 reminds us that exactly at that moment the system was in the state |αi. Since the time is a continuous parameter we obviously have lim |α, t0 ; ti = |αi,

t→t0

and can use the shorter notation

external constant magnetic field parallel to the z-axis, it will precess in the xy-plane: the probability for the result x oscillates between 0 and 1 ¯h/2 in the measurement SGˆ as a function of time. In any case, the probability for the result ¯h/2 or −¯h/2 remains always as the constant 1. Generalizing, it is natural to require that X X |ca0 (t0 )|2 = |ca0 (t)|2 . a0

In other words, the normalization of the states does not depend on time: hα, t0 |α, t0 i = hα, t0 ; t|α, t0 ; ti = hα, t0 |U † (t, t0 )U(t, t0 )|α, t0 i. This is satisfied if we require U(t, t0 ) to be unitary, i.e. U † (t, t0 )U(t, t0 ) = 1. 2. Composition property

The evolution from the time t0 to a later time t2 should be equivalent to the evolution from the initial time t0 to an intermediate time t1 followed by the evolution from t1 to the final time t2 , i.e. U(t2 , t0 ) = U(t2 , t1 )U(t1 , t0 ),

|α, t0 i

|α, t0 ; t0 + dti = U(t0 + dt, t0 )|α, t0 i. Due to the continuity condition

evolution −→ |α, t0 ; ti.

We work like we did with translations. We define the time evolution operator U(t, t0 ):

(t2 > t1 > t0 ).

Like in the case of the translation operator we will first look at the infinitesimal evolution

|α, t0 ; t0 i = |α, t0 i. Let’s see, how state vectors evolve when time goes on:

a0

lim |α, t0 ; ti = |αi

t→t0

we have lim U(t0 + dt, t0 ) = 1.

dt→0

|α, t0 ; ti = U(t, t0 )|α, t0 i, which must satisfy physically relevant conditions. 1. Conservation of probability

We expand the state at the moment t0 with the help of the eigenstates of an observable A: X |α, t0 i = ca0 (t0 )|a0 i. a0

At a later moment we get the expansion X |α, t0 ; ti = ca0 (t)|a0 i. a0

In general, we cannot expect the probability for the system being in a specific state |a0 i to remain constant, i.e. in most cases |ca0 (t)| 6= |ca0 (t0 )|. For example, when a spin 21 particle, which at the moment t0 is in the state |Sx ; ↑i, is subjected to an

So, we can assume the deviations of the operator U(t0 + dt, t0 ) from the identity operator to be of the order dt. When we now set U(t0 + dt, t0 ) = 1 − iΩdt, where Ω is a Hermitean operator, we see that it satisfies the composition condition U(t2 , t0 ) = U(t2 , t1 )U(t1 , t0 ),

(t2 > t1 > t0 ),

is unitary and deviates from the identity operator by the term O(dt). The physical meaning of Ω will be revealed when we recall that in classical mechanics the Hamiltonian generates the time evolution. From the definition U(t0 + dt, t0 ) = 1 − iΩdt we see that the dimension of Ω is frequency, so it must be multiplied by a factor before associating it with the Hamiltonian operator H: H = ¯hΩ,

or

iH dt U(t0 + dt, t0 ) = 1 − . h ¯ The factor ¯h here is not necessarily the same as the factor h in the case of translations. It turns out, however, that ¯ in order to recover Newton’s equations of motion in the classical limit both coefficients must be equal. Applying the composition property U(t2 , t0 ) = U(t2 , t1 )U(t1 , t0 ),

(t2 > t1 > t0 )

we get U(t + dt, t0 )

= U(t + dt, t)U(t, t0 )   iH dt U(t, t0 ), = 1− h ¯

where the time difference t − t0 does not need to be infinitesimal. This can be written as   H U(t + dt, t0 ) − U(t, t0 ) = −i dt U(t, t0 ). ¯h Expanding the left hand side as the Taylor series we end up with ∂ i¯h U(t, t0 ) = HU(t, t0 ). ∂t This is the Schr¨ odinger equation of the time evolution operator. Multiplying both sides by the state vector |α, t0 i we get i¯h

∂ U(t, t0 )|α, t0 i = HU(t, t0 )|α, t0 i. ∂t

Since the state |α, t0 i is independent on the time t we can write the Schr¨odinger equation of the state vectors in the form ∂ i¯h |α, t0 ; ti = H|α, t0 ; ti. ∂t In fact, in most cases the state vector Schr¨ odinger equation is unnecessary because all information about the dynamics of the system is contained in the time evolution operator U(t, t0 ). When this operator is known the state of the system at any moment is obtained by applying the definition |α, t0 ; ti = U(t, t0 )|α, t0 i,

(ii) The Hamiltonain H depends on time but the operators H corresponding to different moments of time commute. For example, a spin 12 particle in the magnetic field whose strength varies but direction remains constant as a function of time. A formal solution of the equation i¯h

∂ U(t, t0 ) = HU(t, t0 ) ∂t

is now    Z t i 0 0 dt H(t ) , U(t, t0 ) = exp − ¯h t0 which, again, can be proved by expanding the exponential function as the series. (iii) The operators H evaluated at different moments of time do not commute For example, a spin 21 particle in a magnetic field whose direction changes in the course of time: H is proportional to the term S · B and if now, at the moment t = t1 the magnetic field is parallel to the x-axis and, at the moment t = t2 parallel to the y-axis, then H(t1 ) ∝ BSx and H(t2 ) ∝ BSy , or [H(t1 ), H(t2 )] ∝ B 2 [Sx , Sy ] 6= 0. It can be shown that the formal solution of the Schr¨odinger equation is now U(t, t0 ) = n Z t Z t1 ∞  X −i dt1 dt2 · · · 1+ ¯h t0 t0 n=1 Z tn−1 dtn H(t1 )H(t2 ) · · · H(tn ). t0

This expansion is called the Dyson series. We will assume that our Hamiltonians are time independent until we start working with the so called interaction picture. Suppose that A is an Hermitean operator and [A, H] = 0. Then the eigenstates of A are also eigenstates of H, called energy eigenstates. Denoting corresponding eigenvalues of the Hamiltonian as Ea0 we have

H|a0 i = Ea0 |a0 i. We consider three cases: (i) The Hamiltonian does not depend on time. For example, The time evolution operator can now be written with the a spin 12 particle in a time independent magnetic field help of these eigenstates. Choosing t0 = 0 we get belongs to this category. The solution of the equation     XX iHt iHt 00 00 ∂ = |a iha | exp − |a0 iha0 | exp − i¯h U(t, t0 ) = HU(t, t0 ) h ¯ h ¯ ∂t a0 a00   X iEa0 t is   0 = |a i exp − ha0 |. iH(t − t0 ) h ¯ U(t, t0 ) = exp − a0 h ¯ as can be shown by expanding the exponential function as the Taylor series and differentiating term by term with respect to the time. Another way to get the solution is to compose the finite evolution from the infinitesimal ones:  N   (iH/¯h(t − t0 ) iH(t − t0 ) lim 1 − = exp − . N →∞ N h ¯

Using this form for the time evolution operator we can solve every intial value problem provided that we can expand the initial state with the set {|a0 i}. If, for example, the initial state can be expanded as X X |α, t0 = 0i = |a0 iha0 |αi = ca0 |a0 i, a0

a0

we get 

 iHt |α, t0 = 0; ti = exp − |α, t0 = 0i h ¯   X iEa0 t 0 0 = |a iha |αi exp − . h ¯ 0

in an eigenstate |a0 i of an operator A commuting with the Hamiltonian H. Suppose, we are interested in the expectation value of an operator B which does not necessarily commute either with A or with H. At the moment t the system is in the state |a0 , t0 = 0; ti = U(t, 0)|a0 i.

a

In other words, the expansion coefficients evolve in the course of time as   iEa0 t 0 0 0 ca (t = 0) −→ ca (t) = ca (t = 0) exp − . h ¯ So, the absolute values of the coefficients remain constant. The relative phase between different components will, however, change in the course of time because the oscillation frequencies of different components differ from each other. As a special case we consider an initial state consisting of a single eigenstate: |α, t0 = 0i = |a0 i. At some later moment this state has evolved to the state   iEa0 t 0 . |α, t0 = 0; ti = |a i exp − h ¯ Hence, if the system originally is in an eigenstate of the Hamiltonian H and the operator A it stays there forever. Only the phase factor exp(−iEa0 t/¯ h) can vary. In this sense the observables whose corresponding operators commute with the Hamiltonian, are constants of motion. Observables (or operators) associated with mutually commuting operators are called compatible. As mentioned before, the treatment of a physical problem can in many cases be reduced to the search for a maximal set of compatible operators. If the operators A, B, C, . . . belong to this set, i.e. [A, B] = [B, C] = [A, C] = · · · = 0, and if, furthermore, [A, H] = [B, H] = [C, H] = · · · = 0, that is, also the Hamiltonian is compatible with other operators, then the time evolution operator can be written as   X   iHt iEK 0 t 0 exp − = |K i exp − hK 0 |. ¯h h ¯ 0 K

Here K 0 stands for the collective index:

In this special case we have hBi = ha0 |U † (t, 0)BU(t, 0)|a0 i     iEa0 t iEa0 t 0 B exp − |a0 i = ha | exp ¯h ¯h = ha0 |B|a0 i, that is, the expectation value does not depend on time. For this reason the energy eigenstates are usually called stationary states We now look at the expectation value in a superposition of energy eigenstates, in a non stationary state X |α, t0 = 0i = ca0 |a0 i. a0

It is easy to see, that the expectation value of B is now   XX i(Ea00 − Ea0 )t . c∗a0 ca00 ha0 |B|a00 i exp − hBi = ¯h 00 0 a

a

This time the expectation value consists of terms which oscillate with frequences determind by the Bohr frequency condition ωa00 a0 =

As an application we look at how spin 21 particles behave in a constant magnetic field. When we assume the magnetic moments of the particles to be e¯h/2me c (like electrons), the Hamiltonian is   e H=− S · B. me c ˆ , we have If we choose B k z   eB H=− Sz . me c The operators H and Sz differ only by a constant factor, so they obviously commute and the eigenstates of Sz are also energy eigenstates with energies e¯hB 2me c e¯hB + 2me c

E↑

= −

for state |Sz ; ↑i

E↓

=

for state |Sz ; ↓i.

A|K 0 i = a0 |K 0 i, B|K 0 i = b0 |K 0 i, C|K 0 i = c0 |K 0 i, . . . Thus, the quantum dynamics is completely solved (when H does not depend on time) if we only can find a maximal set of compatible operators commuting also with the Hamiltonian. Let’s now look at the expectation value of an operator. We first assume, that at the moment t = 0 the system is

Ea00 − Ea0 . ¯h

We define the cyclotron frequency ωc so that the energy difference between the states is h ¯ ωc : ωc ≡

|e|B . me c

The Hamiltonian H can now be written as H = ω c Sz , when we assume that e < 0. All information about the evolution of the system is contained in the operator   iωc Sz t . U(t, 0) = exp − h ¯ If at the moment t = 0 the system is in the state |αi = c↑ |Sz ; ↑i + c↓ |Sz ; ↓i, it is easy to see that at the moment t it is in the state   iωc t |α, t0 = 0; ti = c↑ exp − |Sz ; ↑i 2   iωc t +c↓ exp + |Sz ; ↓i. 2 If the initial state happens to be |Sz ; ↑i, meaning that in the previous equation c↑ = 1,

c↓ = 0,

we see that the system will stay in this state at all times. This was to be expected because the state is stationary. We now assume that the initial state is |Sx ; ↑i. From the relation 1 1 |Sx ; ↑i = √ |Sz ; ↑i + √ |Sz ; ↓i 2 2 we see that

Lastly we look at how the statevectors corresponding to different times are correlated. Suppose that at the moment t = 0 the system is described by the state vector |αi, which in the course of time evolves to the state |α, t0 = 0; ti. We define the correlation amplitude C(t) as = hα|α, t0 = 0; ti = hα|U(t, 0)|αi.

C(t)

The absolute value of the correlation amplitude tells us how much the states associated with different moments of time resemble each other. In particular, if the initial state is an energy eigenstate |a0 i, then   iEa0 t , C(t) = exp − ¯h and the absolute value of the correlation amplitude is 1 at all times. When the initial state is a superposition of energy eigenstates we get   X iEa0 t |ca0 |2 exp − C(t) = . ¯h 0 a

When t is relatively large the terms in the sum oscillate rapidly with different frequencies and hence most probably cancel each other. Thus we expect the correlation amplitude decreasing rather rapidly from its initial value 1 at the moment t = 0. We can estimate the value of the expression   X iEa0 t 2 |ca0 | exp − C(t) = ¯h 0 a

1 c↑ = c↓ = √ . 2

For the probabilities that at the moment t the system is in eigenstates of Sx we get |hSx ; ↑ |α, t0 = 0; ti|2

=

|hSx ; ↓ |α, t0 = 0; ti|2

=

ωc t 2 2 ωc t sin . 2 cos2

more concretely when we suppose that the statevectors of the system comprise so many, nearly degenerate, energy eigenvectors that we can think them almost to form a continuum. Then the summation can be replaced by the integration Z X −→ dE ρ(E), ca0 −→ g(E) , 0 a

E≈Ea0

Even if the spin originally were parallel to the positive x-axis a magnetic field parallel to the z-axis makes the direction of the spin to rotate. There is a finite probability for finding the system at some later moment in the state |Sx ; ↓i. The sum of probabilities corresponding to different orientations is 1. It is easy to see that the expectation values of the operator S satisfy   h ¯ hSx i = cos ωc t 2   h ¯ hSy i = sin ωc t 2 hSz i = 0.

where ρ(E) is the density of the energy eigenstates. The expression   X iEa0 t C(t) = |ca0 |2 exp − ¯h 0

Physically this means that the spin precesses in the xy-plane.

In many realistic physical cases |g(E)|2 ρ(E) is concentrated into a small neighborhood (size ∆E) of a

a

can now be written as   Z iEt C(t) = dE |g(E)|2 ρ(E) exp − , ¯h which must satisfy the normalization condition Z dE |g(E)|2 ρ(E) = 1.

point E = E0 . Rewriting the integral representation as   iE0 t C(t) = exp − ¯h   Z i(E − E0 )t × dE |g(E)|2 ρ(E) exp − , h ¯ we see that when t increases the integrand oscillates very rapidly except when the energy interval |E − E0 | is small as compared with h ¯ /t. If the interval, which satisfies |E − E0 | ≈ ¯h/t, is much shorter than ∆E —the interval from which the integral picks up its contribution—, the correlation amplitudes practically vanishes. The characteristic time, after which the absolute value of the correlation amplitude deviates significantly from its initial value 1, is h ¯ . t≈ ∆E Although this equation was derived for a quasi continuous energy spectrum it is also valid for the two state system in our spin precession example: the initial state |Sx ; ↑i starts to lose its identity after the time ≈ 1/ωc = h ¯ /(E↑ − E↓ ) as we can see from the equation |hSx ; ↑ |α, t0 = 0; ti|2 = cos2

ωc t . 2

As a summary we can say that due to the evolution the state vector describing the initial state of the system will not any more describe it after a time interval of order ¯h/∆E. This property is often called the time and energy uncertainty relation. Note, however, that this relation is of completely different character than the uncertainty relation concerning position and momentum because time is not a quantum mechanical observable.

Quantum statistics

Thermodynamics

Density operator:

We define X

ρ≡

σ = −tr(ρ ln ρ).

wi |αi ihαi |

One can show that

i

is

• for a completely stochastic ensemble • Hermitean:

σ = ln N,

ρ† = ρ

when N is the number of the independent states in the system.

• normalized: trρ = 1.

• for a pure ensemble

Density matrix: X

hb00 |ρ|b0 i =

σ = 0.

wi hb00 |αi ihαi |b0 i.

i

Ensemble average: [A]

XX

=

b0

=

hb00 |ρ|b0 ihb0 |A|b00 i

Hence σ measures disorder =⇒ it has something to do with the entropy. The entropy is defined by

b00

S = kσ.

tr(ρA).

In a thermodynamical equilibrium Dynamics

∂ρ = 0, ∂t

|αi i = |αi ; t0 i −→ |αi , t0 ; ti We suppose that the occupation of states is conserved, i.e. wi = constant.

so [ρ, H] = 0 and the operators ρ and H have common eigenstates |ki:

Now ρ(t) =

X

H|ki = Ek |ki ρ|ki = wk |ki.

wi |αi , t0 ; tihαi , t0 ; t|,

i

so ∂ρ i¯h ∂t

 ∂ = wi i¯ |αi , t0 ; ti hαi , t0 ; t| h ∂t i †  X ∂ |αi , t0 ; ti + wi |αi , t0 ; ti −i¯ h ∂t i 

X

Using these eigenstates the density matrix can be represented as X ρ= wk |kihk| k

and σ=−

= Hρ − ρH = −[ρ, H].

Continuum

Example: [A] =

3 0

d x

Z

d3 x00 hx00 |ρ|x0 ihx0 |A|x00 i.

! X

hx00 |ρ|x0 i = hx00 |

wi |αi ihαi | |x0 i

ρkk = wk . In the equilibrium the entropy is at maximum. We maximize σ under conditions P • U = [H] = trρH = k ρkk Ek .

wi ψi (x

Hence δσ

i

=

where the diagonal elements of the density matrix are

• trρ = 1.

Here the density matrix is

X

ρkk ln ρkk ,

k

Like Heisenberg’s equation of motion, but wrong sign! OK, since ρ is not an observable.

Z

X

00

= −

)ψi∗ (x0 ).

δρkk (ln ρkk + 1) = 0

k

δ[H]

i

X

=

X

δρkk Ek = 0

k

Note

hx0 |ρ|x0 i =

X i

wi |ψi (x0 )|2 .

δ(trρ)

=

X k

δρkk = 0.

With the help of Lagrange multipliers we get X δρkk [(ln ρkk + 1) + βEk + γ] = 0, k

so ρkk = e−βEk −γ−1 . The normalization (trρ = 1) gives ρkk =

e−βEk N X e−βEl

(canonical ensemble).

l

It turns out that

1 , kB T where T is the thermodynamical temperature and kB the Boltzmann constant. In statistical mechanics we define the canonical partition function Z: N X Z = tre−βH = e−βEk . β=

k

Now

e−βH . Z The ensemble average can be written as ρ=

[A]


tr e−βH A = trρA = Z "N # X −βEk hk|A|kie =

k N X

. −βEk

e

k

In particular we have N X

U

=

[H] =

Ek e−βEk

k N X

e−βEk

k

∂ = − (ln Z). ∂β Example Electrons in a magnetic field parallel to z axis.

In the basis {|Sz ; ↑i, |Sz ; ↓i} of the eigenstates of the Hamiltonian H = ω c Sz we have

 ρ 7→

e−β¯hωc /2 0 0 eβ¯hωc /2 Z

 ,

where Z = e−β¯hωc /2 + eβ¯hωc /2 . For example the ensemble averages are [Sx ]

= [Sy ] = 0,     h ¯ β¯ hωc [Sz ] = − tanh . 2 2

Angular momentum

We see that 

O(3)

Rx ()Ry () − Ry ()Rx ()

We consider active rotations. 3 × 3 orthogonal matrix R ⇐⇒ rotation inR3 .

1. RRT symmetric ⇒ RRT has 6 independent parameters ⇒ orthogonality condition RRT = 1 gives 6 independent equations ⇒ R has 9 − 6 = 3 free parameters. ˆ (2 angles) by the angle φ ⇒ 3 2. Rotation around n parameters. ˆ vector ⇒ 3 parameters. 3. nφ 3 × 3 orthogonal matrices form a group with respect to the matrix multiplication:

 0 0  0

= Rz (2 ) − 1.

R ←→ D(R), i.e. |αiR = D(R)|αi. We define the angular momentum (J) so that (we are not employing properties of the classical angular momentum x × p)   ˆ J ·n ˆ dφ) = 1 − i dφ D(n, ¯h and require that the rotation operator D • is unitary,

1. R1 R2 is orthogonal if R1 and R2 are orthogonnal.

• is decomposable,

2. R1 (R2 R3 ) = (R1 R2 )R3 , associativity.

• D → 1, when dφ → 0. We see that J must be Hermitean, i.e.

3. ∃ identity I = the unit matrix. 4. if R is orthogonal, then also the inverse matrix R−1 = RT is orthogonal.

J† = J. Moreover, we require that D satisfies the same group properties as R, i.e.

The group is called O(3). Generally rotations do not commute,

Dx ()Dy () − Dy ()Dx () = Dz (2 ) − 1.

R1 R2 6= R2 R1 , so the group is non-Abelian. Rotations around a common axis commute. Rotation around z-axis:   cos φ − sin φ 0 Rz (φ) =  sin φ cos φ 0  0 0 1     x x cos φ − y sin φ Rz  y  =  x sin φ + y cos φ  . z z Infinitesimal rotations up to the order O(2 ): 2 1 − 2 − 0  2 Rz () =    1− 2 0 0 0 1  1 0 0  0 1 − 2 − Rx () =  2 2 0  1 − 2  2 1 − 2 0   0 1 0 Ry () =  2 − 0 1 − 2

−2 0 0

In a Hilbert space we associate

Number of parameters



0 =  2 0

  ,

Since rotations around a common axis commute a finite rotation can be constructed as     N ˆ J ·n φ ˆ D(nφ) = lim 1 − i N →∞ ¯h N   ˆ iJ · nφ = exp − ¯h =

1−i

ˆ ˆ 2 φ2 J · nφ (J · n) − + ···. ¯h 2¯ h2

We apply this up to the order O(2 ): 

! 2 2 iJy  Jy  1− − ¯h 2¯h2 !  2 2 iJy  Jy  iJx  Jx2 2 − 1− − 1 − − ¯h ¯h 2¯h2 2¯h2

iJx  Jx2 2 1− − ¯h 2¯h2



1 1 2 3 2 Jx Jy  + 2 Jy Jx + O( ) ¯h ¯h Jz 2 =1−i − 1. ¯h

=−   ,

We see that   .

[Jx , Jy ] = i¯hJz . Similarly for other components: [Ji , Jj ] = i¯hijk Jk .

We consider:

Now Ry0 (β) = Rz (α)Ry (β)Rz−1 (α) Rz0 (γ) = Ry0 (β)Rz (γ)Ry−1 0 (β),

hJx i ≡ hα|Jx |αi −→ † R hα|Jx |αiR = hα|D z (φ)Jx D z (φ)|αi. We evaluate

so

D†z (φ)Jx Dz (φ) = exp



iJz φ h ¯



 Jx exp −

iJz φ h ¯



R(α, β, γ)

applying the Baker-Hausdorff lemma eiGλ Ae−iGλ =  2 2 i λ A + iλ[G, A] + [G, [G, A]] + · · · 2!  n n i λ + [G, [G, [G, . . . [G, A]]] . . .] + · · · n! where G is Hermitean. So we need the commutators [Jz , Jx ] = i¯ hJy [Jz , [Jz , Jx ]] = i¯ h[Jz , Jy ] = ¯h2 Jx [Jz , [Jz , [Jz , Jx ]]] = ¯h2 [Jz , Jx ] = i¯h3 Jy .. . Substituting into the Baker-Hausdorff lemma we get D†z (φ)Jx Dz (φ) = Jx cos φ − Jy sin φ. Thus the expectation value is hJx i −→ R hα|Jx |αiR = hJx i cos φ − hJy i sin φ. Correspondingly we get for the other components hJy i hJz i

−→ −→

hJy i cos φ + hJx i sin φ hJz i.

We see that the components of the expectation value of the angular momentum operator transform in rotations like a vector in R3 : X hJk i −→ Rkl hJl i. l

Euler angles

1. Rotate the system counterclockwise by the angle α around the z-axis. The y-axis of of the system coordinates rotates then to a new position y 0 . 2. Rotate the system counterclockwise by the angle β around the y 0 -axis. The system z-axis rotates now to a new position z 0 . 3. Rotate the system counterclockwise by the angle γ around the z 0 -axis. Using matrices: R(α, β, γ) ≡ Rz0 (γ)Ry0 (β)Rz (α).

= = = =

Ry0 (β)Rz (γ)Ry−1 0 (β)Ry 0 (β)Rz (α) Ry0 (β)Rz (α)Rz (γ) Rz (α)Ry (β)Rz−1 (α)Rz (α)Rz (γ) Rz (α)Ry (β)Rz (γ).

Correspondingly D(α, β, γ) = Dz (α)Dy (β)Dz (γ).

SU(2)

• the spin returns to its original direction after time t = 2π/ωc .

In the two dimensional space

• the wave vector returns to its original value after time t = 4π/ωc .

{|Sz ; ↑i, |Sz ; ↓i} the spin operators   ¯h Sx = {(|Sz ; ↑ihSz ; ↓ |) + (|Sz ; ↓ihSz ; ↑ |)} 2   i¯h Sy = {−(|Sz ; ↑ihSz ; ↓ |) + (|Sz ; ↓ihSz ; ↑ |)} 2   ¯h Sz = {(|Sz ; ↑ihSz ; ↑ |) − (|Sz ; ↓ihSz ; ↓ |)} 2

Matrix representation

In the basis {|Sz ; ↑i, |Sz ; ↓i} the base vectors are represented as     1 0 |Sz ; ↑i 7→ ≡ χ↑ |Sz ; ↓i 7→ ≡ χ↓ 0 1 hSz ; ↑ |

satisfy the angular momentum commutation relations [Sx , Sy ] = i¯hSz + cyclic permutations. Thus the smallest dimension where these commutation relations can be realized is 2. The state |αi = |Sz ; ↑ihSz ; ↑ |αi + |Sz ; ↓ihSz ; ↓ |αi behaves in the rotation   iSz φ Dz (φ) = exp − h ¯ like

7→ (1, 0) ≡ χ†↑

hSz ; ↓ | 7→ (0, 1) ≡ χ†↓ ,

so an arbitrary state vector is represented as   hSz ; ↑ |αi |αi 7→ hSz ; ↓ |αi hα|

7→ (hα|Sz ; ↑i, hα|Sz ; ↓i).

The column vector     hSz ; ↑ |αi c↑ χ= ≡ hSz ; ↓ |αi c↓ is called the two component spinor Pauli’s spin matrices

 Dz (φ)|αi =

exp −

iSz φ ¯h

 |αi

= e−iφ/2 |Sz ; ↑ihSz ; ↑ |αi +eiφ/2 |Sz ; ↓ihSz ; ↓ |αi. In particular: Dz (2π)|αi = −|αi. Spin precession

When the Hamiltonian is

Pauli’s spin matrices σi are defined via the relations   ¯h (Sk )ij ≡ (σk )ij , 2 where the matrix elements are evaluated in the basis {|Sz ; ↑i, |Sz ; ↓i}. For example   ¯h S1 = Sx = {(|Sz ; ↑ihSz ; ↓ |) + (|Sz ; ↓ihSz ; ↑ |)}, 2 so

H = ω c Sz the time evolution operator is   iSz ωc t U(t, 0) = exp − = Dz (ωc t). h ¯

hJx i cos φ − hJy i sin φ hJy i cos φ + hJx i sin φ hJz i

one can read that hSx it hSy it hSz it We see that

=

(S1 )12 = (S1 )21

=

or (S1 ) =

Looking at the equations hJx i −→ R hJy i −→ R hJz i −→ R

(S1 )11 = (S1 )22

= hSx it=0 cos ωc t − hSy it=0 sin ωc t = hSy it=0 cos ωc t + hSx it=0 sin ωc t = hSz it=0 .

¯h 2



0 1

1 0

0 ¯h , 2  .

Thus we get       0 1 0 −i 1 0 σ1 = , σ2 = , σ3 = . 1 0 i 0 0 −1 The spin matrices satisfy the anticommutation relations {σi , σj } ≡ σi σj + σj σi = 2δij and the commutation relations [σi , σj ] = 2iijk σk .

Moreover, we see that

Euler’s angles

The spinor rotation matrices corresponding to rotations around z and y axes are  −iα/2  e 0 Dz (α) 7→ 0 eiα/2   cos β/2 − sin β/2 Dy (β) 7→ . sin β/2 cos β/2

σi† = σi , det(σi ) = −1, tr(σi ) = 0. Often the collective vector notation ˆ + σ2 y ˆ + σ3 z ˆ. σ ≡ σ1 x

With the help of Euler’s angles α, β and γ the rotation matrices can be written as

is used for spin matrices. For example we get

1

σ·a ≡

X

D(α, β, γ) 7→ D( 2 ) (α, β, γ) =       β β −e−i(α−γ)/2 sin 2 e−i(α+γ)/2 cos 2      . β β ei(α−γ)/2 sin 2 ei(α+γ)/2 cos 2

ak σk

k

 =

+a3 a1 + ia2

a1 − ia2 −a3

 .

ˆ We seek for the eigenspinor of the matrix σ · n:

and

ˆ = χ. σ · nχ (σ · a)(σ · b)

=

X

σj aj σk bk

Now

j,k

=

X1 2

j,k

=

X

({σj , σk } + [σj , σk ]) aj bk



 sin β cos α ˆ =  sin β sin α  , n cos β

so (δjk + ijki σi )aj bk

j,k

= a · b + iσ · (a × b). A special case of the latter formula is 2

Now     ˆ ˆ iS · nφ iσ · nφ ˆ φ) = exp − D(n, 7→ exp − = h ¯ 2     φ φ ˆ sin 1 cos − iσ · n = 2 2         cos φ2 − inz sin φ2 (−inx − ny ) sin φ2         cos φ2 + inz sin φ2 (−inx + ny ) sin φ2 and the spinors behave in rotations like   ˆ iσ · nφ χ −→ exp − χ. 2 Note the notation σ does not mean that σ would behave

in rotations like a vector, σk −→ R σk . Instead we have Rkl χ† σl χ.

l

ˆ one has For all directions n   ˆ iσ · nφ exp − = −1, 2 φ=2π

sin βe−iα − cos β

 .

ˆ /¯h −iS ·n Dn ˆ (φ) = e

2

X

cos β sin βeiα

ˆ The state where the spin is parallel to the unit vector n, is obviously invariant under rotations

(σ · a) = |a| .

χ† σk χ −→

 ˆ= σ·n

ˆ and thus an eigenstate of the operator S · n. This kind of state can be obtained by rotating the state |Sz ; ↑i 1. angle β around y axis, 2. angle α around z axis, i.e. ˆ · n; ˆ ↑i = S · nD(α, ˆ S · n|S β, 0)|Sz ; ↑i   ¯h D(α, β, 0)|Sz ; ↑i = 2   ¯h ˆ ↑i. = |S · n; 2 Correspondingly for spinors the vector     β −iα/2 cos e 1  2  χ = D( 2 ) (α, β, 0)|Sz ; ↑i =  β sin 2 eiα/2 ˆ is an eigenstate of the matrix σ · n. SU(2)

As a representation of rotations the 2 × 2-matrices ˆ for any n.

ˆ φ/2 ˆ φ) = e−iσ ·n D( 2 ) (n, 1

form obviously a group. These matrices have two characteristic properties:

1. unitarity 

1

D( 2 )

†

 1 −1 = D( 2 ) ,

2. unimodularity 1 ( 2 ) D = 1. A unitary unimodular matrix can be written as   a b U (a, b) = . −b∗ a∗ The unimodularity condition gives 1 = |U | = |a|2 + |b|2 , and we are left with 3 free parameters. The unitarity condition is automatically satisfied because  ∗   a −b a b † U (a, b) U (a, b) = b∗ a −b∗ a∗   |a|2 + |b|2 0 = = 1. 0 |a|2 + |b|2 Matrices U (a, b) form a group since • the matrix U (a1 , b1 )U (a2 , b2 ) = U (a1 a2 − b1 b∗2 , a1 b2 + a∗2 b1 ) is unimodular because |U (a1 a2 − b1 b∗2 , a1 b2 + a∗2 b1 )| = |a1 a2 − b1 b∗2 |2 + |a1 b2 + a∗2 b1 |2 = 1, and thus also unitary. • as a unitary matrix U has the inverse matrix: U −1 (a, b) = U † (a, b) = U (a∗ , −b). • the unit matrix 1 is unitary and unimodular. The group is called SU(2). Comparing with the previous spinor representation 1

ˆ φ) = D( 2 ) (n,      cos φ2 − inz sin φ2    (−inx + ny ) sin φ2

   (−inx − ny ) sin φ2      cos φ2 + inz sin φ2

we see that Re(a) Re(b)

  φ = cos 2   φ = −ny sin 2

  φ Im(a) = −nz sin 2   φ Im(b) = −nx sin . 2

The complex numbers a and b are known as Cayley-Klein’s parameters. Note O(3) and SU(2) are not isomorphic. Example

In O(3): 2π- and 4π-rotations 7→ 1 In SU(2): 2π-rotation 7→ −1 and 4π-rotation 7→ 1. The operations U (a, b) and U (−a, −b) in SU(2) correspond to a single matrix of O(3). The map SU(2) 7→ O(3) is thus 2 to 1. The groups are, however, locally isomorphic.

Angular momentum algebra

• The product of matrices belongs to the group: X (j) (j) (j) Dm00 m (R1 R2 ) = Dm00 m0 (R1 )Dm0 m (R2 ),

It is easy to see that the operator J 2 = Jx Jx + Jy Jy + Jz Jz

m0

where R1 R2 is the combined rotation of the rotations R1 and R2 ,

commutes with the operators Jx , Jy and Jz , [J 2 , Ji ] = 0.

• the inverse operation belongs to the group:

We choose the component Jz and denote the common eigenstate of the operators J 2 and Jz by |j, mi. We know (QM II) that 1 3 J |j, mi = j(j + 1)¯ h |j, mi, j = 0, , 1, , . . . 2 2 Jz |j, mi = m¯h|j, mi, m = −j, −j + 1, . . . , j − 1, j. 2

2

(j)∗

(j)

Dm0 m (R−1 ) = Dmm0 (R). The state vectors |j, mi transform in rotations like X |j, m0 ihj, m0 |D(R)|j, mi D(R)|j, mi = m0

=

We define the ladder operators J+ and J− :

X

(j)

|j, m0 iDm0 m (R).

m0

With the help of the Euler angles

J± ≡ Jx ± iJy .

(j)

Dm0 m (R) =

They satisfy the commutation relations

      iJz α iJy β iJz γ hj, m0 | exp − exp − exp − |j, mi ¯h ¯h ¯h

[J+ , J− ] = 2¯hJz [Jz , J± ] = ±¯ hJ±  2  J , J± = 0.

0

(j)

= e−i(m α+mγ) dm0 m (β), where

We see that

(j) dm0 m (β)

Jz J+ |j, mi = h ¯ J+ Jz |j, mi = (m + 1)¯ hJ+ |j, mi and



iJy β ≡ hj, m | exp − ¯h 0

 |j, mi.

(j)

2

Functions dm0 m can be evaluated using Wigner’s formula

2

J J+ |j, mi = J+ J |j, mi = j(j + 1)¯ hJ+ |j, mi, so we must have J+ |j, mi = c+ |j, m + 1i

k

The factor c+ can be deduced from the normalization condition hj, m|j 0 , m0 i = δjj 0 δmm0 . We end up with J± |j, mi =

p

(j ∓ m)(j ± m + 1)¯ h|j, m ± 1i.

Matrix elements will be hj 0 , m0 |J 2 |j, mi = j(j + 1)¯ h2 δj 0 j δm0 m hj 0 , m0 |Jz |j, mi = m¯ hδj 0 j δm0 m p 0 0 hj , m |J± |j, mi = (j ∓ m)(j ± m + 1)¯ hδj 0 j δm0 ,m±1 . We define Wigner’s function: (j) Dm0 m (R)



ˆ iJ · nφ = hj, m | exp − h ¯ 0

(j)

dm0 m (β) = X 0 (−1)k−m+m

 |j, mi.

p (j + m)!(j − m)!(j + m0 )!(j − m0 )! × (j + m − k)!k!(j − k − m0 )!(k − m + m0 )!  2j−2k+m−m0  2k−m+m0 β β × cos × sin . 2 2 Orbital angular momentum

The components of the classically analogous operator L = x × p satisfy the commutation relations [Li , Lj ] = iijk ¯hLk . Using the spherical coordinates to label the position eigenstates, |x0 i = |r, θ, φi, one can show that ∂ hx0 |αi ∂φ   ∂ ∂ − cot θ cos φ hx0 |αi −i¯h − sin φ ∂θ ∂φ   ∂ ∂ −i¯h cos φ − cot θ sin φ hx0 |αi ∂θ ∂φ   ∂ ∂ ±iφ −i¯he ±i − cot θ hx0 |αi ∂θ ∂φ    1 ∂2 1 ∂ ∂ 2 −¯h + sin θ sin θ ∂θ ∂θ sin2 θ ∂φ2 0 ×hx |αi.

hx0 |Lz |αi = −i¯h

Since   ˆ iJ · nφ [J 2 , D(R)] = [J 2 , exp − ] = 0, h ¯

hx0 |Lx |αi =

we see that D(R) does not chance the j-quantum number, so it cannot have non zero matrix elements between states with different j values. (j) The matrix with matrix elements Dm0 m (R) is the (2j + 1)-dimensional irreducible representation of the rotation operator D(R). (j) The matrices Dm0 m (R) form a group:

hx0 |Ly |αi = hx0 |L± |αi = hx0 |L2 |αi =

We denote the common eigenstate of the operators L2 and Lz by the ket-vector |l, mi, i.e. Lz |l, mi = m¯ h|l, mi L2 |l, mi = l(l + 1)¯ h2 |l, mi. Since R3 can be represented as the direct product

or

s (l) Dm0 (α, β, 0)

=

4π m∗ Yl (θ, φ) (2l + 1)

. β,α

As a special case (l)

(l)

D00 (θ, φ, 0) = d00 (θ) = Pl (cos θ).

R3 = R × Ω, Coupling of angular momenta

where Ω is the surface of the unit sphere (position=distance from the origin and direction) the position eigenstates can be written correspondingly as ˆ |x0 i = |ri|ni.

A1 ⊗ A2 |αi1 ⊗ |βi2 = (A1 |αi1 ) ⊗ (A2 |βi2 )

ˆ form a complete basis on the Here the state vectors |ni surface of the sphere, i.e. Z ˆ n| ˆ = 1. dΩn ˆ |nih We define the spherical harmonic function: ˆ = hn|l, ˆ mi. Ylm (θ, φ) = Ylm (n) ˆ with the equations The scalar product of the vector hn| Lz |l, mi = m¯ h|l, mi 2 L |l, mi = l(l + 1)¯ h2 |l, mi gives −i¯h and 

1 ∂ sin θ ∂θ

∂ m Y (θ, φ) = m¯ hYlm (θ, φ) ∂φ l

 sin θ

∂ ∂θ

 +

 1 ∂2 + l(l + 1) Ylm = 0. sin2 θ ∂φ2

Ylm and D(l)

The state ˆ = |θ, φi |ni is obtained from the state |ˆ z i rotating it first by the angle θ around y-axis and then by the angle φ around z-axis: ˆ = D(R)|ˆ |ni zi = D(α = φ, β = θ, γ = 0)|ˆ zi X = D(φ, θ, 0)|l, mihl, m|ˆ z i. l,m

Ylm ∗ (θ, φ)

=

X

(l) Dm0 m (φ, θ, 0)hl, m|ˆ z i.

m

Now

r hl, m|ˆ zi =

Ylm ∗ (0, φ)

so

r Ylm ∗ (θ, φ)

=

in the product space. Here |αii ∈ Hi . In particular, A1 ⊗ 12 |αi1 ⊗ |βi2 = (A1 |αi1 ) ⊗ |βi2 , where 1i is the identity operator of the space Hi . Correspondingly 11 ⊗ A2 operates only in the subspace H2 of the product space. Usually the subspace of the identity operators, or even the identity operator itself, is not shown, for example A1 ⊗ 12 = A1 ⊗ 1 = A1 . It is easy to verify that operators operating in different subspace commute, i.e. [A1 ⊗ 12 , 11 ⊗ A2 ] = [A1 , A2 ] = 0. In particular we consider two angular momenta J1 and J2 operating in two different Hilbert spaces. They commute: [J1i , J2j ] = 0. The infinitesimal rotation affecting both Hilbert spaces is     ˆ ˆ iJ 1 · nδφ iJ 2 · nδφ 1− ⊗ 1− = ¯h ¯h ˆ i(J 1 ⊗ 1 + 1 ⊗ J 2 ) · nδφ 1− . ¯h The components of the total angular momentum J = J1 ⊗ 1 + 1 ⊗ J2 = J1 + J2 obey the commutation relations [Ji , Jj ] = i¯hijk Jk ,

Furthermore ˆ = hl, m|ni

We consider two Hilbert spaces H1 and H2 . If now Ai is an operator in the space Hi , the notation A1 ⊗ A2 means the operator

=

(2l + 1) δm0 , 4π

(2l + 1) (l) Dm0 (φ, θ, γ = 0) 4π

i.e. J is angular momentum. A finite rotation is constructed analogously:     ˆ ˆ J 1 · nφ J 2 · nφ D1 (R) ⊗ D2 (R) = exp − ⊗ exp − . ¯h h ¯ Base vectors of the whole system We seek in the product space {|j1 m1 i ⊗ |j2 m2 i} for the maximal set of commuting operators. (i) J 21 , J 22 , J1z and J2z .

Their common eigenstates are simply direct products |j1 j2 ; m1 m2 i ≡ |j1 , m1 i ⊗ |j2 , m2 i. If j1 and j2 can be deduced from the context we often denote |m1 m2 i = |j1 j2 ; m1 m2 i.

so the Clebsch-Gordan coefficients satisfy the condition

j1 (j1 + 1)¯ h2 |j1 j2 ; m1 m2 i m1 ¯ h|j1 j2 ; m1 m2 i j2 (j2 + 1)¯ h2 |j1 j2 ; m1 m2 i m2 ¯ h|j1 j2 ; m1 m2 i.

= = = =

we must have m = m1 + m2 ,

The quantum numbers are obtained from the (eigen)equations J 21 |j1 j2 ; m1 m2 i J1z |j1 j2 ; m1 m2 i J 22 |j1 j2 ; m1 m2 i J2z |j1 j2 ; m1 m2 i

is unitary. The elements hj1 j2 ; m1 m2 |j1 j2 ; jmi of the transformation matrix are called Clebsch-Gordan’s coefficients. Since Jz = J1z + J2z ,

hj1 j2 ; m1 m2 |j1 j2 ; jmi = 0, if m 6= m1 + m2 . Further, we must have (QM II) |j1 − j2 | ≤ j ≤ j1 + j2 . It turns out, that the C-G coefficients can be chosen to be real, so the transformation matrix C is in fact orthogonal: X hj1 j2 ; m1 m2 |j1 j2 ; jmihj1 j2 ; m01 m02 |j1 j2 ; jmi

(ii) J 2 , J 21 , J 22 and Jz . Their common eigenstate is denoted as |j1 j2 ; jmi

jm

= δm1 m01 δm2 m02

or shortly |jmi = |j1 j2 ; jmi

X

if the quantum numbers j1 and j2 can be deduced from the context. The quantum numbers are obtained from the equations

m1 m2

J 21 |j1 j2 ; jmi J 22 |j1 j2 ; jmi J 2 |j1 j2 ; jmi Jz |j1 j2 ; jmi

= = = =

j1 (j1 + 1)¯ h2 |j1 j2 ; jmi j2 (j2 + 1)¯ h2 |j1 j2 ; jmi j(j + 1)¯ h2 |j1 j2 ; jmi m¯ h|j1 j2 ; jmi.

Now [J 2 , J1z ] 6= 0,

[J 2 , J2z ] 6= 0,

so we cannot add to the set (i) the operator J 2 , nor to the set (ii) the operators J1z or J2z . Both sets are thus maximal and the corresponding bases complete (and orthonormal), i.e. X X |j1 j2 ; m1 m2 ihj1 j2 ; m1 m2 | = 1 j1 j2 m1 m2

XX

|j1 j2 ; jmihj1 j2 ; jm| =

1.

j1 j2 jm

m1 m2

|j1 j2 ; jmihj1 j2 ; jm| =

= δjj 0 δmm0 . As a special case (j 0 = j and m0 = m = m1 + m2 ) we get the normalization condition X |hj1 j2 ; m1 m2 |j1 j2 ; jmi|2 = 1. m1 m2

Recursion formulas Operating with the ladder operators to the state |j1 j2 ; jmi we get J± |j1 j2 ; jmi = (J1± + J2± )

X

|j1 j2 ; m1 m2 i

m1 m2

×hj1 j2 ; m1 m2 |j1 j2 ; jmi, or p (j ∓ m)(j ± m + 1)|j1 j2 ; j, m ± 1i X X q = (j1 ∓ m01 )(j1 ± m01 + 1) m01 m02

In the subspace where the quantum numbers j1 and j2 are fixed we have the completeness relations X |j1 j2 ; m1 m2 ihj1 j2 ; m1 m2 | = 1 X

hj1 j2 ; m1 m2 |j1 j2 ; jmihj1 j2 ; m1 m2 |j1 j2 ; j 0 m0 i

1.

×|j1 j2 ; m01 ± 1, m02 i q + (j2 ± m02 )(j2 ± m02 + 1)  ×|j1 j2 ; m01 , m02 ± 1i ×hj1 j2 ; m01 m02 |j1 j2 ; jmi.

jm

One can go from the basis (i) to the basis (ii) via the unitary transformation X |j1 j2 ; jmi = |j1 j2 ; m1 m2 ihj1 j2 ; m1 m2 |j1 j2 ; jmi, m1 m2

so also the transformation matrix (C)jm,m1 m2 = hj1 j2 ; m1 m2 |j1 j2 ; jmi

Taking the scalar product on the both sides with the vector hj1 j2 ; m1 m2 | we get p (j ∓ m)(j ± m + 1)hj1 j2 ; m1 m2 |j1 j2 ; j, m ± 1i p = (j1 ∓ m1 + 1)(j1 ± m1 ) ×hj1 j2 ; m1 ∓ 1, m2 |j1 j2 ; jmi p + (j2 ∓ m2 + 1)(j2 ± m2 ) ×hj1 j2 ; m1 , m2 ∓ 1|j1 j2 ; jmi.

The Clebsch-Gordan coefficients are determined uniquely by

Example L + S-coupling.

Now

1. the recursion formulas.

j1 m1

2. the normalization condition X |hj1 j2 ; m1 m2 |j1 j2 ; jmi|2 = 1.

= l = 0, 1, 2, . . . = ml = −l, −l + 1, . . . , l − 1, l 1 = s= 2 1 = ms = ± 2 ( l ± 12 , when l > 0 = 1 when l = 0. 2,

j2

m1 m2

m2

3. the sign conventions, for example 0

j

0

hj1 j2 ; j m |J1z |j1 j2 ; jmi ≥ 0.

m s

Note Due to the sign conventions the order of the

coupling is essential: 1 /2

|j1 j2 ; jmi = ±|j2 j1 ; jmi. Graphical representation of recursion formulas

(m

1

- 1 ,m

2

J

) (m

,m

1

) (m

2

,m

1

+ 1 )

2

(m

,m 1

-1 ) (m 2

1

,m

-

) (m 2

1

+ 1 ,m 2

)

|m1 + m2 | ≤ j.

= j2

+ m 2

1 m1 = ml = m − , 2

m l

m2 = ms =

1 2

and the shorthand notation the J− -recursion gives q (l + 21 + m + 1)(l + 12 − m)hm − 21 , 12 |l + 12 , mi q = (l + m + 12 )(l − m + 21 ) ×hm + 12 , 21 |l + 12 , m + 1i,

A

= j

m 1

= -j1

s hm −

m

1 1 2 , 2 |l

+

1 2 , mi

=

l + m + 12 hm + 12 , 12 |l + 12 , m + 1i. l + m + 32

+ m 2

= j1

1

1

m

l

or

1

2

-

m

m

J

Using the selection rule

Recursion formula in m1 m2 -plane

|m2 | ≤ j2 ,

-

-1 /2

We fix j1 , j2 and j. Then |m1 | ≤ j1 ,

J -

Recursion when j1 = l and j2 = s = 1/2

J +

J

= j

m

(a ) D J

J -

J

+

+

-

B

fo rb id d e n

F

-

= -j2

A

J J

E

2

C

(b ) Using recursion formulas

We see that 1. every C-G coefficient depends on A, 2. the normalization condition determines the absolute value of A, 3. the sign is obtained from the sign conventions.

Applying the same recursion repeatedly we have hm − 21 , 12 |l + 12 , mi s s l + m + 12 l + m + 23 = hm + 23 , 12 |l + 12 , m + 2i l + m + 32 l + m + 25 s s s l + m + 12 l + m + 23 l + m + 25 = l + m + 32 l + m + 25 l + m + 27 hm + 52 , 12 |l + 12 , m + 3i = .. . s l + m + 12 = hl, 21 |l + 12 , l + 12 i. 2l + 1 If j = jmax = j1 + j2 and m = mmax = j1 + j2 one must have |j1 j2 ; jmi = hj1 j2 ; m1 = j1 , m2 = j2 |j1 j2 ; jmi|j1 m1 i|j2 m2 i.

Now the normalization condition

On the other hand we have hj1 j2 ; m1 m2 |D(R)|j1 j2 ; m01 m02 i XX = hj1 j2 ; m1 m2 |j1 j2 ; jmi

|hj1 j2 ; m1 = j1 , m2 = j2 |j1 j2 ; jmi|2 = 1 and the sign convention give

jm j 0 m0

×hj1 j2 ; jm|D(R)|j1 j2 ; j 0 m0 i ×hj1 j2 ; j 0 m0 |j1 j2 ; m01 m02 i XX (j) = hj1 j2 ; m1 m2 |j1 j2 ; jmiDmm0 (R)δjj 0

hj1 j2 ; m1 = j1 , m2 = j2 |j1 j2 ; jmi = 1. Thus, in the case of the spin-orbit coupling, hl, 12 |l + 12 , l + 12 i = 1,

jm j 0 m0

×hj1 j2 ; m01 m02 |j1 j2 ; j 0 m0 i.

or

s hm − 12 , 12 |l + 12 , mi =

l + m + 12 . 2l + 1

We end up with the Clebsch-Gordan series Dm11 m0 (R)Dm22 m0 (R) = 1 X XX 2 hj1 j2 ; m1 m2 |j1 j2 ; jmi j

   

Rotation matrices If D(j1 ) (R) is a rotation matrix in the base {|j1 m1 i|m1 = −j1 , . . . , j1 } and D(j2 ) (R) a rotation matrix in the base {|j2 m2 i|m2 = −j2 , . . . , j2 }, then D(j1 ) (R) ⊗ D(j2 ) (R) is a rotation matrix in the (2j1 + 1) × (2j2 + 1)-dimensional base {|j1 , m1 i ⊗ |j2 , m2 i}. Selecting suitable superpositions of the vectors |j1 , m1 i ⊗ |j2 , m2 i the matrix takes the form like D(j1 ) (R) ⊗ D(j2 ) (R) −→  D(j1 +j2 )   D(j1 +j2 −1)    

0

0 ..

. D(|j1 −j2 |)

    .  

One can thus write D(j1 ) ⊗ D(j2 ) = D(j1 +j2 ) ⊕ D(j1 +j2 −1) ⊕ · · · ⊕ D(|j1 −j2 |) . As a check we can calculate the dimensions: (2j1 + 1)(2j2 + 1) = 2(j1 + j2 ) + 1 + 2(j1 + j2 − 1) + 1 + · · · + 2|j1 − j2 | + 1. The matrix elements of the rotation operator satisfy hj1 j2 ; m1 m2 |D(R)|j1 j2 ; m01 m02 i = hj1 m1 |D(R)|j1 m01 ihj2 m2 |D(R)|j2 m02 i (j )

(j )

= Dm11 m0 (R)Dm22 m0 (R). 1

2

(j )

(j )

With the help of the recursion relations, normalization condition and sign convention the rest of the C-G coefficients can be evaluated, too. We get  r r l + m + 12 l − m + 21    |j = l + 21 , mi r 2l + 1 r 2l + 1 =  |j = l − 12 , mi l − m + 21 l + m + 21 − 2l + 1 2l + 1   1 1 |ml = m − 2 , ms = 2 i . |ml = m + 12 , ms = − 12 i

m

m0 (j)

×hj1 j2 ; m01 m02 |j1 j2 ; jm0 iDmm0 (R).

As an application we have Z 1 2 dΩYlm ∗ (θ, φ)Ylm (θ, φ)Ylm (θ, φ) 1 2 s (2l1 + 1)(2l2 + 1) = 4π(2l + 1) ×hl1 l2 ; 00|l1 l2 ; l0ihl1 l2 ; m1 m2 |l1 l2 ; lmi.

3j- 6j- and 9j-symbols

where

The Clebsch-Gordan coefficients obey certain symmetry relations, like hj1 j2 ; m1 m2 |j1 j2 ; jmi = (−1)j1 +j2 −j hj2 j1 ; m2 m1 |j2 j1 ; jmi hj1 j2 ; m1 m2 |j1 j2 ; j3 m3 i s 2j3 + 1 hj2 j3 ; −m2 , m3 |j2 j3 ; j1 m1 i = (−1)j2 +m2 2j1 + 1 hj1 j2 ; m1 m2 |j1 j2 ; j3 m3 i s 2j3 + 1 = (−1)j1 −m1 hj3 j1 ; m3 , −m1 |j3 j1 ; j2 m2 i 2j2 + 1 hj1 j2 ; m1 m2 |j1 j2 ; j3 m3 i = (−1)j1 +j2 −j3 hj1 j2 ; −m1 , −m2 |j1 j2 ; j3 , −m3 i. Note The first relation shows that the coupling order is

essential. We define more symmetric 3j-symbols:   j1 j2 j3 ≡ m1 m2 m3

 δ(j1 j2 j3 ) =

when |j1 − j2 | ≤ j3 ≤ j1 + j2 otherwise.

1, 0,

6j-symbols

Let us couple three angular momenta, j1 , j2 and j3 , to the angular momentum J. There are two ways: 1. first j1 , j2 −→ j12 and then j12 , j3 −→ J. 2. first j2 , j3 −→ j23 and then j23 , j1 −→ J. Let’s choose the first way. The quantum number j12 must satisfy the selection rules |j1 − j2 | |j12 − j3 |

≤ j12 ≤ j1 + j2 ≤ J ≤ j12 + j3 .

The states belonging to different j12 are independent so we must specify the intermediate state j12 . We use the notation |(j1 j2 )j12 j3 ; JM i. Explicitely one has

(−1)j1 −j2 −m2 p hj1 j2 ; m1 m2 |j1 j2 ; j3 , −m3 i. 2j3 + 1

|(j1 j2 )j12 j3 ; JM i XX = |j1 j2 ; j12 m12 i|j3 m3 i m12 m3

They satisfy   j1 j2 j3 m1 m2 m3  j2 j3 = m2 m3  j1 (−1)j1 +j2 +j3 m1  j1 j3 = m1 m3   j1 j2 j3 m1 m2 m3 = (−1)j1 +j2 +j3

×hj12 j3 ; m12 m3 |j12 j3 ; JM i X = |j1 m1 i|j2 m2 i|j3 m3 i m1 m2 m3 m12







×hj1 j2 ; m1 m2 |j1 j2 ; j12 m12 i j3 j1 j2 m3 m1 m2 ×hj12 j3 ; m12 m3 |j12 j3 ; JM i.    j2 j3 j2 j1 j3 Correspondingly the angular momenta coupled in way 2 = m2 m3 m2 m1 m3 satisfy    j2 j3 j2 j1 = |j1 (j2 j3 )j23 ; JM i m2 m3 m2 m1 XX = |j1 m1 i|j2 j3 ; j23 m23 i j1 m1

=

m23 m1



j1 −m1

j2 −m2

j3 −m3

×hj1 j23 ; m1 m23 |j1 j23 ; JM i X = |j1 m1 i|j2 m2 i|j3 m3 i

 .

m1 m2 m3 m23

As an application, we see that the coefficients     3 3 2 2 2 3 2 2 , . 1 1 1 1 −2 −1 2 2

×hj2 j3 ; m2 m3 |j2 j3 ; j23 m23 i ×hj1 j23 ; m1 m23 |j1 j23 ; JM i.

vanish. On the other hand, the orthogonality properties are somewhat more complicated:    XX j1 j2 j3 j1 j2 j3 (2j3 + 1) m1 m2 m3 m01 m02 m3 j3

m3

= δm1 m01 δm2 m02 and X X  j1 m1 m1 m2

=

j2 m2

j3 m3



δj3 j30 δm3 m03 δ(j1 j2 j3 ) p , 2j3 + 1

j1 m1

j2 m2

j30 m03



Both bases are complete so there is a unitary transform between them: X |j1 (j2 j3 )j23 ; JM i = |(j1 j2 )j12 j3 ; JM i j12

×h(j1 j2 )j12 j3 ; JM |j1 (j2 j3 )j23 ; JM i. In the transformation coefficients, recoupling coefficients it is not necessary to show the quantum number M , because Theorem 1 In the transformation X |α; jmi = |β; jmihβ; jm|α; jmi β

the coefficients hβ; jm|α; jmi do not depend on the quantum number m.

Proof: Let us suppose that m < j. Now X |α; j, m + 1i = |β; j, m + 1ihβ; j, m + 1|α; j, m + 1i. β

On the other hand J+ p |α; jmi ¯ (j + m + 1)(j − m) h X = |β; j, m + 1ihβ; jm|α; jmi,

|α; j, m + 1i =

β

so hβ; j, m + 1|α; j, m + 1i = hβ; j, m|α; j, mi The explicit expression for the recoupling coefficients will be h(j1 j2 )j12 j3 ; J|j1 (j2 j3 )j23 ; Ji X hj12 j3 ; JM |j12 j3 ; m12 m3 i = m1 m2 m3 m12 m23

×hj1 j2 ; j12 m12 |j1 j2 ; m1 m2 i ×hj2 j3 ; m2 m3 |j2 j3 ; j23 m23 i ×hj1 j23 ; m1 m23 |j1 j23 ; JM i. We define the more symmetric 6j-symbols:   j1 j2 j12 j3 J j23 (−1)j1 +j2 +j3 +J ≡p (2j12 + 1)(2j23 + 1) ×h(j1 j2 )j12 j3 ; J|j1 (j2 j3 )j23 ; Ji (−1)j1 +j2 +j3 +J =p (2j12 + 1)(2j23 + 1) X × hj1 j2 ; m1 m2 |j1 j2 ; j12 , m1 + m2 i m1 m2

×hj12 j3 ; m1 + m2 , M − m1 − m2 |j12 j3 ; JM i ×hj2 j3 ; m2 , M − m1 − m2 |j2 j3 ; j23 , M − m1 i ×hj1 j23 ; m1 , M − m1 |j1 j23 ; JM i. We can handle analogously the coupling of 4 angular momenta. Transformations from a coupling scheme to another are mediated by the 9j-symbols:    j1 j2 j12  j3 j4 j34   j13 j24 j h(j1 j2 )j12 (j3 j4 )j34 ; j|(j1 j3 )j13 (j2 j4 )j24 ; ji ≡ p . (2j12 + 1)(2j34 + 1)(2j13 + 1)(2j24 + 1

Tensor operators

Applying the Baker-Hausdorff lemma

We have used the vector notation for three component operators for example to express the scalar product, like

eiGλ Ae−iGλ = 

0

0

0

0

Classically a vector is a quantity that under rotations transforms like V ∈ R3 (or ∈ C 3 ), i.e. if R ∈ O(3), then 3 X

we end up with the commutators

Rij Vj .

j=1

[Jj , [Jj , [· · · [Jj , Vi ] · · ·]]].

In quantum mechanics V is a vector operator provided that hV i ∈ C 3 is a vector: R hα|Vi |αiR

= hα|D† (R)Vi D(R)|αi =

3 X

These will be evaluated in turn into operators Vi and Vk (k 6= i, j). A vector operator (V ) is defined so that it satisfies the commutation relation

Rij hα|Vj |αi, [Vi , Jj ] = i¯hijk Vk .

j=1

∀|αi ∈ H, R ∈ O(3). Thus we must have D† (R)Vi D(R) =

X

Rij Vj .

j

Thus the infinitesimal rotations ˆ =1− D(n)

ˆ iJ · n h ¯

We can easily see that for example p, x and J are vector operators. In classical mechanics a quantity which under rotations transforms like XXX · · · Rii0 Rjj 0 Rkk0 · · · Ti0 j 0 k0 ··· , T ijk · · · −→ | {z } j 0 k0 i0 n indeces

is called a Cartesian tensor of the rank n. Example The dyad product of the vectors U and V

satisfy 

   ˆ ˆ iJ · n iJ · n Vi 1 + ¯h h ¯ i ˆ i − Vi J · n) ˆ + O(2 ) = Vi + (J · nV h ¯ X = Rij Vj

1+

Tij = Ui Vj is a tensor of rank 2. Cartesian tensors are reducible, for example the dyad product can be written as

j

or



A + iλ[G, A] + [G, [G, A]] + · · ·  n n i λ + [G, [G, [G, . . . [G, A]]] . . .] + · · · n!

p · x = p x x + p y y + pz z .

Vi0 =

i2 λ2 2!

Ui Vj

X  ˆ = ˆ j. Vi + [Vi , J · n] Rij (n)V i¯h j

Substituting the explicit expressions for rotations, like  2 1 − 2 −  2 R(ˆ z ) =   1 − 2 0 0

infinitesimal

0



 0 , 1

=

(Ui Vj − Uj Vi ) U ·V δij + 3 2   Ui Vj + Uj Vi U ·V + − δij . 2 3

We see that the terms transform under rotations differently: • U 3· V δij is invariant. There is 1 term. (Ui Vj − Uj Vi ) retains its antisymmetry. There are 3 2 terms.   Ui Vj + Uj Vi U · V • − 3 δij retains its symmetry and 2 tracelessness. There are 5 terms. •

we get

 [Vx , Jz ] = Vx − Vy + O(3 ). i¯h Handling similarly the other components we end up with Vx +

[Vi , Jj ] = i¯ hijk Vk . Finite rotation

A finite rotation specified by Euler angles is accomplished by rotating around coordinate axises, so we have to consider such expressions as     iJj φ iJj φ exp Vi exp − . ¯h h ¯

We recognize that the number of terms checks and that the partition might have something to do with the angular momentum since the multiplicities correspond to the multiplicities of the angular momenta l = 0, 1, 2. (k) We define the spherical tensor Tq of rank k so that the ˆ of the spherical function argument n ˆ = hn|lmi ˆ Ylm (n)

is replaced by the vector V :

Under the infinitesimal rotations   ˆ iJ · n ˆ = 1− D(n) ¯h

m=q Tq(k) = Yl=k (V ).

Example The spherical function Y1 :

r r 3 3 z 3 (1) cos θ = 7→ T0 = Vz = 4π 4π r 4π r r   3 x ± iy 3 Vx ± iVy (1) √ = ∓ 7→ T±1 = ∓ √ . 4π 4π 2r 2 r

Y10 Y1±1

Similarly we could construct for example a spherical tensor of rank 2: r r 15 (x ± iy)2 15 (2) ±2 7→ T±2 = (Vx ± iVy )2 . Y2 = 32π 32π r2

a spherical tensor behaves thus like     ˆ ˆ iJ · n iJ · n (k) 1+ Tq 1− ¯h ¯h   k X (k) ˆ iJ · n 0 = Tq0 hkq | 1 + |kqi ¯h 0 q =−k

=

(k)

Tq0 hkq 0 |kqi +

q 0 =−k

(k)

ˆ iTq0 hkq 0 |J · n|kqi,

q 0 =−k

ˆ Tq(k) ] = [J · n,

The tensors are irreducible, i.e. there does not exist any proper subset

X

(k)

ˆ Tq0 hkq 0 |J · n|kqi.

q0

ˆ =z ˆ and x ˆ ± iˆ Choosing n y we get (k)

(k)

[Jz , Tq ] = ¯hqTq

which would remain invariant under rotations. and

Transformation of spherical tensors

Under the rotation R an eigenstate of the direction transforms like

(k)

[J± , Tq ] = ¯h

p (k) (k ∓ q)(k ± q + 1)Tq±1 .

Example Decomposition of the dyad product.

ˆ −→ |n ˆ 0 i = D(R)|ni. ˆ |ni The state vectors |lmi, on the other hand, transform under the rotation R−1 like X (l) D(R−1 )|l, mi = |l, m0 iDm0 m (R−1 ).

We form spherical tensors of rank 1 from the vector operators U and V : U0 = Uz , Ux ± iUy , =∓ √ 2

U±1

m0

So we get ˆ 0) Ylm (n

k X

or

(k) Tq

{Tp(k) , Tp(k) , . . .} ⊂ {Tq(k) |q = −k, . . . , +k}, 1 2

k X

V0 = Vz , V±1 = ∓

Vx ± iVy √ . 2

Now ˆ 0 |lmi = hn|D ˆ † (R)|lmi = hn X (l) −1 0 ˆ ˆ = hn|D(R )|lmi = hn||lm iDm0 m (R−1 ) m0

=

X

0

Ylm

(l) ˆ m0 m (R−1 ) (n)D

=

0

(l)

Tq(1)

=

(2)

∗

ˆ mm0 (R). Ylm (n)D

(2)

T±1

m0

We define a tensor operator Ylm (V ) so that X 0 (l)∗ D† (R)Ylm (V )D(R) = Ylm (V )Dmm0 (R). m0 (k)

Generalizing we define: Tq is a (2k + 1)-component spherical tensor of rank k if and only if D

†

(R)Tq(k) D(R)

=

k X

U ·V U+1 V−1 + U−1 V+1 − U0 V0 = , 3 3 (U × V )q √ , i 2

= −

T±2

m0

X

(0)

T0

(2)

T0

= U±1 V±1 , U±1 V0 + U0 V±1 √ , = 2 U+1 V−1 + 2U0 V0 + U−1 V+1 √ = . 6

In general we have (k ) (k ) Theorem 1 Let Xq1 1 and Zq2 2 be irreducible spherical tensors of rank k1 and k2 . Then XX Tq(k) = hk1 k2 ; q1 q2 |k1 k2 ; kqiXq(k1 1 ) Zq(kq 2 ) q1

(k)∗ (k) Dqq0 (R)Tq0

q 0 =−k

q2

is a (irreducible) spherical tensor of rank k. (k) Proof: We show that Tq transforms like

or equivalently (k)

D(R)Tq D† (R) =

Pk

(k)

(k)

q 0 =−k D q 0 q (R)Tq 0 .

D

†

(R)Tq(k) D(R)

=

k X q 0 =−k

(k)∗

(k)

Dqq0 (R)Tq0 .

Now

Proof: Due to the property [Jz , Tq(k) ] = ¯hqTq(k)

D† (R)Tq(k) D(R) XX = hk1 k2 ; q1 q2 |k1 k2 ; kqi q1

we have

q2

hα0 , j 0 m0 |[Jz , Tq(k) ] − ¯hqTq(k) |α, jmi

×D† (R)Xq(k1 1 ) D(R)D† (R)Zq(k2 2 ) D(R) XXXX = hk1 k2 ; q1 q2 |k1 k2 ; kqi q1

(k )

(k )

(k )

so

q20

q10

q2

= [(m0 − m)¯ h − q¯h] × hα0 , j 0 m0 |Tq(k) |α, jmi = 0,

k00

q1

q 00

q20

q10

q2

q0

×hk1 k2 ; q10 q20 |k1 k2 ; k 00 q 0 i (k00 )

(k )

2

1

where we have substituted the Clebsch-Gordan series expansion (j )

Dm11 m0 (R)Dm22 m0 (R) = 1 X XX 2 hj1 j2 ; m1 m2 |j1 j2 ; jmi m

j

if m0 6= q + m Theorem 3 (Wigner-Eckardt’s theorem) The matrix elements of a tensor operator between eigenstates of the angular momentum satisfy the relation

(k )

×hk1 k2 ; q1 q2 |k1 k2 ; k 00 q 00 iDq0 q00 (R−1 )Xq0 1 Zq0 2 ,

(j )

hα0 , j 0 m0 |Tq(k) |α, jmi = 0,

(k )

×Xq0 1 Dq0 1q1 (R−1 )Zq0 2 Dq0 2q2 (R−1 ) 2 1 X2 XXX X1X X hk1 k2 ; q1 q2 |k1 k2 ; kqi =

m0

hα0 , j 0 m0 |Tq(k) |α, jmi = hjk; mq|jk; j 0 m0 i

hα0 j 0 kT (k) kαji p , 2j + 1

where the reduced matrix element hα0 j 0 kT (k) kαji depends neither on the quantum numbers m, m0 nor on q. (k) Proof: Since Tq is a tensor operator it satisfies the condition p (k) [J± , Tq(k) ] = ¯h (k ∓ q)(k ± q + 1)Tq±1 ,

(j)

×hj1 j2 ; m01 m02 |j1 j2 ; jm0 iDmm0 (R)

so

Taking into account the orthogonality of the Clebsch-Gordan coefficients X hj1 j2 ; m1 m2 |j1 j2 ; jmihj1 j2 ; m1 m2 |j1 j2 ; j 0 m0 i m1 m2

= δjj 0 δmm0 we get D† (R)Tq(k) D(R) XXXXX = δkk00 δqq00 hk1 k2 ; q10 q20 |k1 k2 ; k 00 q 0 i k00

q10

q20

q0

q 00

00

(k )

(k )

(k )

×Dq0 q00 (R−1 )Xq0 1 Zq0 2 , 1

2

which can be rewritten as D† (R)Tq(k) D(R)   X X (k ) (k )  = hk1 k2 ; q10 q20 |k1 k2 ; kq 0 iXq0 1 Zq0 2  q0

1

q10 q20

2

(k)

×Dq0 q (R−1 ) X (k) (k) X (k)∗ (k) = Tq0 Dq0 q (R−1 ) = Dqq0 (R)Tq0 q0

q0

hα0 , j 0 m0 |[J± , Tq(k) ]|α, jmi p (k) =h ¯ (k ∓ q)(k ± q + 1)hα0 , j 0 m0 |Tq±1 |α, jmi. Substituting the matrix elements of the ladder operators we get p (j 0 ± m0 )(j 0 ∓ m0 + 1)hα0 , j 0 , m0 ∓ 1|Tq(k) |α, jmi p = (j ∓ m)(j ± m + 1)hα0 , j 0 , m0 |Tq(k) |α, j, m ± 1i p (k) + (k ∓ q)(k ± q + 1)hα0 , j 0 , m0 |Tq±1 |α, jmi. If we now substituted j 0 → j, m0 → m, j → j1 , m → m1 , k → j2 and q → m2 , we would note that the recursion formula above is exactly like the recursion formula for the Clebsch-Gordan coefficients, p (j ∓ m)(j ± m + 1)hj1 j2 ; m1 m2 |j1 j2 ; j, m ± 1i p = (j1 ∓ m1 + 1)(j1 ± m1 ) ×hj1 j2 ; m1 ∓ 1, m2 |j1 j2 ; jmi p + (j2 ∓ m2 + 1)(j2 ± m2 ) ×hj1 j2 ; m1 , m2 ∓ 1|j1 j2 ; jmi. P Both recursions are of the form j aij xj = 0, or sets of linear homogenous simultaneous equations with the same coefficients aij . So we have two sets of equations X X aij xj = 0, aij yj = 0, j

Matrix elements of tensor operators

Theorem 2 The matrix elements of the tensor operator (k) Tq satisfy hα0 , j 0 m0 |Tq(k) |α, jmi = 0, unless m0 = q + m.

j

one for the matrix elements (xi ) of the tensor operator and the other for the Clebsch-Gordan coefficients (yi ). These sets of equations tell that xj yj = ∀j and k fixed, xk yk

so xj = cyj while c is a proportionality coefficient independent of the indeces j. Thus we see that hα0 , j 0 m0 |Tq(k) |α, jmi = ( constant independent on m, q and m0 ) ×hjk; mq|jk; j 0 m0 i.

The coefficient cjm does not depend either on the quantum number m, because J · V is a scalar operator, so we can write it briefly as cj . Because cj does not depend on the operator V the above equation is valid also when V → J and α0 → α, or hα, jm|J 2 |α, jmi = h ¯ 2 j(j + 1) = cj hαjkJ kαji.

If we write the proportionality coefficient like If we now apply the Wigner-Eckart theorem to the operators Vq and Jq we get

hα0 j 0 kT (k) kαji p 2j + 1 we are through. According to the Wigner-Eckart theorem a matrix element of a tensor operator is a product of two factors, of which • hjk; mq|jk; j 0 m0 i depends only on the geometry, i.e. on the orientation of the system with respect to the z-axis. 0 hα0 jp kT (k) kαji • depends on the dynamics of the 2j + 1 system.

As a special case we have the projection theorem: Theorem 4 Let 1 1 J±1 = ∓ √ (Jx ± iJy ) = ∓ √ J± , 2 2

J0 = J z

be the components of the tensor operator corresponding to the angular momentum. Then hα0 , jm|J · V |α, jmi hjm0 |Jq |jmi. hα , jm |Vq |α, jmi = h2 j(j + 1) ¯ 0

0

Proof: Due to the expansions (0)

U+1 V−1 + U−1 V+1 − U0 V0 U ·V = , 3 3 (U × V )q √ , i 2

T0

= −

Tq(1)

=

(2)

T±2

(2)

T±1

(2)

T0

= U±1 V±1 , U±1 V0 + U0 V±1 √ = , 2 U+1 V−1 + 2U0 V0 + U−1 V+1 √ = 6

we can write hα0 , jm|J · V |α, jmi = hα0 , jm|(J0 V0 − J+1 V−1 − J−1 V+1 )α, jmi = m¯hhα0 , jm|V0 |α, jmi ¯h p +√ (j + m)(j − m + 1)hα0 , j, m − 1|V−1 |α, jmi 2 ¯h p −√ (j − m)(j + m + 1)hα0 , j, m + 1|V+1 |α, jmi 2 = cjm hα0 jkV kαji, where, according to the Wigner-Eckart theorem the coefficient cjm does not depend on α, α0 or V .

hα0 jkV kαji hα0 , jm0 |Vq |α, jmi = . 0 hαjkJ kαji hα, jm |Jq |α, jmi for the ratios of the matrix elements. On the other hand, the right hand side of this equation is hα0 , jm|J · V |α, jmi , hα, jm|J 2 |α, jmi so hα0 , jm0 |Vq |α, jmi =

hα0 , jm|J · V |α, jmi hjm0 |Jq |jmi ¯h2 j(j + 1)

Generalizing one can show that the reduced matrix (k) elements of the irreducible product Tq of two tensor (k2 ) (k1 ) operators, Xq1 and Zq2 , satisfy hα0 j 0 ||T (k) kαji  √ 0 XX k1 = 2k + 1(−1)k+j+j j 00 00 α

0 0

×hα j kX

(k1 )

00 00

j

00 00

k2 j0

kα j ihα j |Z (k2 ) kαji.

k j 00



Symmetry

it follows now [G, H] = 0,

Symmetries, constants of motion and degeneracies Looking at the Lagrange equation of motion   d ∂L ∂L − =0 dt ∂ q˙i ∂qi of classical mechanics one can see that if the Lagrangian L(qi , q˙i ) is invariant under translations, i.e.

so according to the Heisenberg equation of motion dA 1 = [A, H] dt i¯h we have

dG = 0. dt In the Heisenberg formalism the observable G is thus a constant of motion. if H is invariant for example under

L(qi , q˙i ) −→ L(qi + δqi , q˙i ) = L(qi , q˙i ), the momentum pi =

• translations then the momentum is constant of motion.

∂L ∂ q˙i

• rotations then the angular momentum is a constant motion.

is a conserved quantity, i.e. dpi d = dt dt



∂L ∂ q˙i

 = 0.

Let us suppose now that the Hamiltonian is symmetric under the operations S generated by G: S † HS = H [S, H] = 0 [G, H] = 0.

Formulating classical mechanics using the Hamiltonian function H(qi , pi ) the equations of motion take the forms ∂H ∂qi ∂H . ∂pi

p˙i

= −

q˙i

=

Also looking at these one can see that if H is symmetric under the operation

Let |g 0 i be the eigenstates of G, i.e. G|g 0 i = g 0 |g 0 i and let the system at the moment t0 be in the eigenstate |g 0 i of G. Since the time evolution operator is a functional of the Hamiltonian only,

qi −→ qi + δqi there exists a conserved quantity:

U = U [H], so [G, U ] = 0.

p˙i = 0.

At the moment t we then have In quantum mechanics operations of that kind (translations, rotations, . . .) are associated with a unitary symmetry operator. Let S be an arbitrary symmetry operator. We say that the Hamiltonian H is symmetric, if [S, H] = 0, or due to the unitarity of the operator S equivalently

G|g 0 , t0 ; ti = GU (t0 , t)|g 0 i = U (t0 , t)G|g 0 i = g 0 |g 0 , t0 ; ti, or an eigenstate associated with a particular eigenvalue of G remains always an eigenstate belonging to the same eigenvalue. Let us consider now the energy eigenstates |ni, i.e. H|ni = En |ni.

S † HS = H. The matrix elements of the Hamiltonian are then invariant under that operation. In the case of a continuum symmetry we can look at infinitesimal operations i S = 1 − G, h ¯ where the Hermitean operator G is the generator of that symmetry. From the condition S † HS = H

When the Hamiltonian is symmetric under the operations S we have H(S|ni = SH|ni = En S|ni. If now |ni = 6 S|ni, then the energy states En are degenerate. Thus a symmetry is also usually associated with a degeneracy. Let us suppose now that the symmetry operation S can be parametrized with a continuous quantity, say λ: S = S(λ).

When the Hamiltonian is symmetric under these operations all states S(λ)|ni have the same energy. Example Rotations D(R). If [D(R), H] = 0, then [J , H] = 0,

[J 2 , H] = 0.

So there exist simultaneous eigenvectors |n; jmi of the operators H, J 2 ja Jz . Now all rotated states D(R)|n; jmi belong to the same energy eigenvalue. We know that X (j) D(R)|n; jmi = |n; jm0 iDm0 m (R), m0

that is, every rotated state is a superposition of (2j + 1) linearly independent states. The degeneracy is thus (2j + 1)-fold. Example Atomic electron. The potential acting on an electron is of form U = V (r) + VLS L · S. Now [J , H] = 0,

[J 2 , H] = 0,

where J = L + S. The energy levels are thus (2j + 1)-foldly degenerated. Let’s set the atom in magnetic field parallel to the z-axis. The Hamiltonian is then appended by the term Z = cSz . Now [J 2 , Sz ] 6= 0, so the rotation symmetry is broken and the (2j + 1)-fold degeneracy lifted.

Parity

are equivalent:

The parity or space inversion operation converts a right handed coordinate system to left handed: x −→ −x, y −→ −y, z −→ −z. This is a case of a non continuous operation, i.e. the operation cannot be composed of infinitesimal operations. Thus the non continuous operations have no generator. We consider the parity operation, i.e. we let the parity operator π to act on vectors of a Hilbert space and keep the coordinate system fixed:

πT (dx0 ) = T (−dx0 )π. Substituting T (dx0 ) = 1 −

i 0 dx · p, ¯h

we get the condition {π, p} = 0 or π † pπ = −p, or the momentum changes its sign under the parity operation.

|αi −→ π|αi. Angular momentum and parity

Like in all symmetry operations we require that π is unitary, i.e. π † π = 1.

In the case of the orbital angular momentum

Furthermore we require:

one can easily evaluate

†

hα|π xπ|αi = −hα|x|αi ∀|αi. So we must have π † xπ = −x,

L=x×p

π † Lπ

= π † x × pπ = π † xπ × π † pπ = (−x) × (−p) = L,

so the parity and the angular momentum commute:

or πx = −xπ. The operators x ja π anti commute. Let |x0 i be a position eigenstate, i.e. x|x0 i = x0 |x0 i. Then xπ|x0 i = −πx|x0 i = (−x0 )π|x0 i,

[π, L] = 0. In R3 the parity operator is the matrix   −1 0 0 P =  0 −1 0  , 0 0 −1 so quite obviously

and we must have

P R = RP, ∀R ∈ O(3). π|x0 i = eiϕ | − x0 i.

The phase is usually taken to be ϕ = 0, so

We require that the corresponding operators of the Hilbert space satisfy the same condition, i.e. πD(R) = D(R)π.

π|x0 i = | − x0 i. Applying the parity operator again we get

Looking at the infinitesimal rotation

π 2 |x0 i = |x0 i

ˆ = 1 − iJ · n/¯ ˆ h, D(n) we see that

or 2

π = 1. We see that • the eigenvalues of the operator π can be only ±1, • π −1 = π † = π. Momentum and parity

We require that operations • translation followed by space inversion • space inversion followed by translation to the opposite direction

[π, J ] = 0 or π † J π = J , which is equivalent to the transformation of the orbital angular momentum. We see that under • rotations x and J transform similarly, that is, like vectors or tensors of rank 1. • space inversions x is odd and J even. We say that under the parity operation • odd vectors are polar, • even vectors are axial or pseudovectors.

Let us consider such scalar products as p · x and S · x. One can easily see that under rotation these are invariant, scalars. Under the parity operation they transform like π † p · xπ π † S · xπ

= (−p) · (−x) = p · x = S · (−x) = −S · x.

We say that quantities behaving under rotations like scalars, spherical tensors of rank 0, which under the parity operation are

The explicit expression for spherical functions is s (2l + 1)(l − m)! m m m Pl (θ)eimφ , Yl (θ, φ) = (−1) 4π(l + m)! from which as a special case, m = 0, we obtain r 2l + 1 0 Yl (θ, φ) = Pl (cos θ). 4π Depending on the degree l of the Legendre polynomial it is either even or odd:

• even, are (ordinary) scalars, • odd, are pseudoscalars.

Pl (−z) = (−1)l Pl (z). We see that

Wave functions and parity

Let ψ be the wave function of a spinles particle in the state |αi, i.e. ψ(x0 ) = hx0 |αi.

hx0 |π|α, l0i = (−1)l hx0 |α, l0i, so the state vectors obey

Since the position eigenstates satisfy π|x0 i = | − x0 i,

π|α, l0i = (−1)l |α, l0i. Now [π, L± ] = 0

the wave function of the space inverted state is hx0 |π|αi = h−x0 |αi = ψ(−x0 ). Suppose that |αi is a parity eigenstate, i.e. π|αi = ±|αi. The corresponding wave function obeys the the relation

and Lr± |α, l0i ∝ |α, l, ±ri, so the orbital angular momentum states satisfy the relation π|α, lmi = (−1)l |α, lmi. Theorem 1 If [H, π] = 0,

ψ(−x0 ) = hx0 |π|αi = ±hx0 |αi = ±ψ(x0 ), i.e. it is an even or odd function of its argument. Note Not all physically relevant wave function have parity. For example, [p, π] 6= 0, so a momentum eigenstate is not an eigenstate of the parity. The wave function corresponding to an eigenstate of the momentum is the plane wave ψp0 (x0 ) = eip ·x /¯h , 0

0

which is neither even nor odd. Because [π, L] = 0, the eigenstate |α, lmi of the orbital angular momentum (L2 , Lz ) is also an eigenstate of the parity. Now Rα (r)Ylm (θ, φ) = hx0 |α, lmi. In spherical coordinates the transformation x0 −→ −x0 maps to r −→ r θ −→ π − θ (cos θ −→ − cos θ) φ −→ φ + π (eimφ −→ (−1)m eimφ ).

and |ni is an eigenstate of the Hamiltonian H belonging to the nondegenerate eigenvalue En , i.e. H|ni = En |ni, then |ni is also an eigenstate of the parity. Proof: Using the property π 2 = 1 one can easily see that the state 1 (1 ± π)|ni 2 is a parity eigenstate belonging to the eigenvalue ±1. On the other hand, this is also an eigenstate of the Hamiltonia H with the energy En : 1 1 H( (1 ± π)|ni) = En (1 ± π)|ni. 2 2 Since we supposed the state |ni to be non degenerate the states |ni and 21 (1 ± π)|ni must be the same excluding a phase factor, 1 (1 ± π)|ni = eiϕ |ni, 2 so the state |ni is a parity eigen state belonging to the eigenvalue ±1 Example The energy states of a one dimensional harmonic oscillator are non degenerate and the Hamiltonian even, so the wave functions are either even or odd.

Note The nondegeneracy condition is essential. For

p2 example, the Hamiltonian of a free particle, H = 2m , is even but the energy states

Let us suppose that at the moment t0 = 0 the state of the system is |Li. At a later moment, t, the system is descibed by the state vector |L, t0 = 0; ti 1 = √ (e−iES t/¯h |Si + e−iEA t/¯h |Ai) 2 1 −iES t/¯h (|Si + e−i(EA −ES )t/¯h |Ai), =√ e 2

2

H|p0 i =

p0 |p0 i 2m

are not eigenstates of the parity because π|p0 i = | − p0 i.

because now the time evolution operator is simply The condition of the theorem is not valid because the states |p0 i and | − p0 i are degenerate. We can form parity eigenstates √ 1/ 2(|p0 i ± | − p0 i), which are also degenerate energy (but not momentum) eigenstates. The corresponding wave functions

U(t, t0 = 0) = e−iHt/¯h . At the moment t = T /2 = 2π¯h/2(EA − ES ) the system is in the pure |Ri state and at the moment t = T again in its pure initial state |Li. The system oscillates between the states |Li and |Ri at the angular velocity

ψ±p0 (x0 ) = hx0 | ± p0 i = e±ip ·x /¯h 0

0

are neither even nor odd, whereas 0

0

0

0

0

hx |(|p i + | − p i) ∝ cos p · x /¯ h hx0 |(|p0 i − | − p0 i) ∝ sin p0 · x0 /¯ h are. Example One dimensional symmetric double well

ω=

E A − ES . ¯h

When V → ∞, then EA → ES . Then the states |Li and |Ri are degenerate energy eigenstates but not parity eigenstates. A particle which is localized in one of the wells will remain there forever. Its wave function does not, however, obey the same symmetry as the Hamiltonian: we are dealing with a broken symmetry. Selection rules

Suppose that the states |αi and |βi are parity eigenstates: π|αi = α |αi π|βi = β |βi,

S y m m e tric (S )

A n tis y m m e tric (A )

The ground state is the symmetric state |Si and the first excited state the antisymmetric state |Ai: H|Si π|Si H|Ai π|Ai

= = = =

ES |Si |Si EA |Si −|Ai,

where ES < EA . When the potential barrier V between the wells increases the energy difference between the states decreases: lim (EA − ES ) → 0.

where α and β are the parities (±1) of the states. Now hβ|x|αi = hβ|π † πxπ † π|αi = −α β hβ|x|αi, so hβ|x|αi = 0 unless α = −β . Example The intensity of the dipole transition is

proportional to the matrix element of the operator x between the initial and final states. Dipole transitions are thus possible between states which have opposite parity. Example Dipole moment. If [H, π] = 0, then no non degenerate state has dipole moment: hn|x|ni = 0.

V →∞

We form the superpositions |Li = |Ri =

1 √ (|Si + |Ai) 2 1 √ (|Si − |Ai), 2

which are neither energy nor parity eigenstates.

The same holds for any quantity if the corresponding operator o is odd: π † oπ = −o.

Lattice translations

Obviously we have

We consider a particle in the one dimensional periodic potential V (x ± a) = V (x). a

a

a

H|θi = E0 |θi. Furthermore we get τ (a)|θi =

−iθ

=e

a

a

(b ) The Hamiltonian of the system is not in general invariant under translations τ † (l)xτ (l) = x + l,

τ (l)|x0 i = |x0 + li.

However, when l is exactly equal to the period of the lattice a we have τ † (a)V (x)τ (a) = V (x + a) = V (x). Because the operator corresponding to the kinetic energy in the Hamiltonian is translationally invariant the whole Hamiltonian H satisfies the condition τ † (a)Hτ (a) = H, which, due to the unitarity of the translation operator can be written as [H, τ (a)] = 0. The operators H and τ (a) have thus common eigenstates. Note The operator τ (a) is unitary and hence its

eigenvalues need not be real. Let us suppose that the potential barrier between the lattice points is infinitely high. Let |ni be the state localized in the lattice cell n, i.e.

|θi.

Thus every state corresponding to a value of the continuous parameter θ has the same energy, i.e. the ground state of the system infinitely degenerate. Let us suppose further that • |ni is a state localized at the point n so that τ (a)|ni = |n + 1i, • hx0 |ni = 6 0 (but small), when |x0 − na| > a. Due to the translation symmetry the diagonal elements of the Hamiltonian H in the base {|ni} are all equal to eachother: hn|H|ni = E0 . Let us suppose now that hn0 |H|ni = 6 0 only if n0 = n or n0 = n ± 1. We are dealing with the so called tight binding approximation. When we define ∆ = −hn ± 1|H|ni, we can write H|ni = E0 |ni − ∆|n + 1i − ∆|n − 1i, where we have exploited the orthonormality of the basis {|ni}. Thus the state |ni is not an energy eigen state. Let us look again at the trial |θi =

H|ni = E0 |ni, ∀n. Thus the system has countably infinite number of ground states |ni, n = −∞, . . . , ∞. Now τ (a)|ni = |n + 1i, so the state |ni is not an eigenstate of the translation τ (a). Let’s try ∞ X |θi ≡ einθ |ni, −∞

where θ is a real parameter and −π ≤ θ ≤ π.

ei(n−1)θ |ni

n=−∞

hx0 |ni = 6 0 only if x0 ≈ na. Obviously |ni is a stationary state. Because all lattice cells are exactly alike we must have

∞ X

einθ |n + 1i =

n=−∞

(a )

a

∞ X

∞ X

einθ |ni.

n=−∞

Like before we have τ (a)|θi = e−iθ |θi. Furthermore X H einθ |ni X X = E0 einθ |ni − ∆ einθ |n + 1i X −∆ einθ |n − 1i X X = E0 einθ |ni − ∆ (einθ−iθ + einθ+iθ )|ni X = (E0 − 2∆ cos θ) einθ |ni. The earlier degeneracy will be lifted if ∆ 6= 0 and E0 − 2∆ ≤ E ≤ E0 + 2∆.

Bloch’s theorem

Let us consider the wave function hx0 |θi. In the translated state τ (a)|θi the wave function is hx0 |τ (a)|θi = hx0 − a|θi when the operator τ (a) acts on left. When it acts on right we get hx0 |τ (a)|θi = e−iθ hx0 |θi, so we have hx0 − a|θi = hx0 |θie−iθ . This equation can be solved by substituting 0

hx0 |θi = eikx uk (x0 ), when θ = ka and uk (x0 ) is a periodic function with the period a. We have derived a theorem known as the Bloch theorem: Theorem 1 The wave function of the eigenstate |θi of the translation operator τ (a) can be written as the procuct 0 of the plane wave eikx and a function with the period a. Note When deriving the theorem we exploited only the fact that |θi an eigenstate of the operator τ (a) belonging to the eigenvalue eiθ . Thus it is valid for all periodic systems (whether the tight binding approximation holds or not) With the help of the Bloch theorem the dispersion relation of the energy in the tight binding model can be written as E(k) = E0 − 2∆ cos ka, −

π π ≤k≤ . a a

This continuum of the energies is known as the Brillouin zone.

Time reversal (reversal of motion)

The fact that the operator K does not change the base states can be justified like: The state |a0 i represented in the base {|a0 i} maps to the column vector   0  0    then also x(−t) is a solution.  ..   .  At the moment t = 0 let there be a particle at the point    0  0 x(t = 0) with the momentum p(t = 0). Then a particle at  |a i 7→   1 , the same point but with the momentum −p(t = 0)    0  follows the trajectory x(−t).    .  In the quantum mechanical Schr¨ odinger equation  ..    0 h2 2 ¯ ∂ψ = − ∇ + V ψ, i¯h which is unaffected by the complex conjugation. ∂t 2m Note The effect of the operator K depends thus on the due to the first derivative with respect to the time, choice of the basis states. ψ(x, −t) is not a solution eventhough ψ(x, t) were, but If U is a unitary operator then the operator θ = U K is ψ ∗ (x, −t) is. In quantum mechanics the time reversal has antiunitary. obviously something to do with the complex conjugation. Proof: Firstly Let us consider the symmetry operation θ(c1 |αi + c2 |βi) = U K(c1 |αi + c2 |βi) ˜ |αi −→ |˜ αi, |βi −→ |βi. = (c∗1 U K|αi + c∗2 U K|βi) = (c∗1 θ|αi + c∗2 θ|βi), We require that the absolute value of the scalar product is invariant under that operation: so θ is antiliniear. Secondly, expanding the states |αi and |βi in a complete basis {|a0 i} we get ˜ αi| = |hβ|αi|. |hβ|˜ X θ ha0 |αi∗ U K|a0 i |αi −→ |˜ αi = There are two possibilities to satisfy this condition: The Newton equations of motion are invariant under the transformation t −→ −t: if x(t) is a solution of the equation m¨ x = −∇V (x)

a0

˜ αi = hβ|αi, so the corresponding symmetry 1. hβ|˜ operator is unitary, that is hβ|αi −→ hβ|U U |αi = hβ|αi.

˜ αi = hβ|αi∗ = hα|βi, so the symmetry operator 2. hβ|˜ cannot be unitary. We define the antiunitary operator θ so that ˜ αi = hα|βi∗ hβ|˜ θ(c1 |αi + c2 |βi) = c∗1 θ|αi + c∗2 θ|βi, where

=

X

0

ha |αi∗ U |a0 i

a0

†

The symmetries treated earlier have obeyed this condition.

=

X

hα|a0 iU |a0 i

a0

and ˜ = |βi

X

˜ = ha0 |βi∗ U |a0 i ↔ hβ|

a0

X

ha0 |βiha0 |U † .

a0

Thus the scalar product is XX ˜ αi = ha00 |βiha00 |U † U |a0 ihα|a0 i hβ|˜ a00

=

X

a0

hα|a0 iha0 |βi = hα|βi

a0

|αi −→ |˜ αi = θ|αi,

˜ = θ|βi |βi −→ |βi

If the operator satisfies only the last condition it is called antilinear. We define the complex conjugation operator K so that

We present the state |αi in the base {|a0 i}. The effect of the operator K is then X X K |αi = |a0 iha0 |αi −→ |˜ αi = ha0 |αi∗ K|a0 i a0

=

a0

a0 0

∗

0

ha |αi |a i.

The operator θ is thus indeed antiunitary. Let Θ be the time reversal operator. We consider the transformation |αi −→ Θ|αi, where Θ|αi is the time reversed (motion reversed) state. If |αi is the momentum eigenstate |p0 i, we should have

Kc|αi = c∗ K|αi.

X

= hβ|αi∗ .

Θ|p0 i = eiϕ | − p0 i. Let the system be at the moment t = 0 in the state |αi. At a slightly later moment t = δt it is in the state   iH |α, t0 = 0; t = δti = 1 − δt |αi. ¯h

We apply now, at the moment t = 0, the time reversal operator Θ and let the system evolve under the Hamiltonian H. Then at the moment δt the system is in the state   iH δt Θ|αi. 1− ¯h

and

If the motion of the system is invariant under time reversal this state should be the same as

In partcular, for a Hermitean observable A we have

Θ|α, t0 = 0; −δti, i.e. we first look at the state at the earlier moment −δt and then reverse the direction of the momentum p. Mathematically this condition can be expressed as     iH iH 1− δt Θ|αi = Θ 1 − (−δt) |αi. ¯h h ¯ Thus we must have −iHΘ|i = ΘiH|i, where |i stands for an arbitrary state vector. If Θ were linear we would obtain the anticommutator relation HΘ = −ΘH. If now |ni is an energy eigenstate corresponding to the eigenvalue En then, according to the anticommutation rule HΘ|ni = −ΘH|ni = (−En )Θ|ni, and the state Θ|ni is an energy eigenstate corresponding to the eigenvalue −En . Thus most systems (those, whose energy spectrum is not bounded) would not have any ground state. Thus the operator Θ must be antilinear, and, in order to be a symmetry operator, it must be antiunitary. Using the antilinearity for the right hand side of the condition

hβ| ⊗ |αi = hγ|αi = h˜ α|˜ γi = h˜ α|Θ ⊗† |βi = h˜ α|Θ ⊗† Θ−1 Θ|βi ˜ = h˜ α|Θ ⊗† Θ−1 |βi.

˜ hβ|A|αi = h˜ α|ΘAΘ−1 |βi. We say that the observable A is even or odd under time reversal depending on wheter in the equation ΘAΘ−1 = ±A the upper or the lower sign holds. This together with the equation ˜ hβ|A|αi = h˜ α|ΘAΘ−1 |βi imposes certain conditions on the phases of the matrix elements of the operator A between the time reversed states. Namely, they has to satisfy ˜ αi∗ . hβ|A|αi = ±hβ|A|˜ In particular, the expectation value satisfies the condition hα|A|αi = ±h˜ α|A|˜ αi. Example The expectation value of the momentum

operator p. We require that hα|p|αi = −h˜ α|p|˜ αi, so p is odd, or ΘpΘ−1 = −p. The momentum eigenstates satisfy pΘ|p0 i = −ΘpΘ−1 Θ|p0 i = (−p0 )Θ|p0 i,

−iHΘ|i = ΘiH|i

i.e. Θ|p0 i is the momentum eigenstates correponding to the eigenvalue −p0 :

we can write it as ΘiH|i = −iΘH|i. So, we see that the operators commute: ΘH = HΘ.

Θ|p0 i = eiϕ | − p0 i. Similarly we can derive for the position operator x the expressions ΘxΘ−1 = x Θ|x0 i = |x0 i

Note We have not defined the Hermitean conjugate of

the antiunitary operator θ nor have we defined the meaning of the expression hβ|θ. That being, we let the time reversal operator Θ to operate always on the right and with the matrix element hβ|Θ|αi we mean the expression (hβ|) · (Θ|αi). Let ⊗ be an arbitrary linear operator. We define

when we impose the physically sensible condition hα|x|αi = h˜ α|x|˜ αi. We consider the basic commutation relations

|γi ≡ ⊗† |βi, so that

[xi , pj ]|i = i¯hδij |i. Now

hβ|⊗ = hγ|

Θ[xi , pj ]Θ−1 Θ|i = Θi¯hδij |i,

from which, using the antilinearity and the time reversal properties of the operators x and p we get [xi , (−pj )]Θ|i = −i¯ hδij Θ|i. We see thus that the commutation rule [xi , pj ]|i = i¯hδij |i remains invariant under the time reversal. Correspondingly, the requirement of the invariance of the commutation rule [Ji , Jj ] = i¯ hijk Jk leads to the condition ΘJ Θ−1 = −J . This agrees with transformation properties of the orbital angular momentum x × p. Wave functions

so the states |ni and Θ|ni have the same energy. Because the state |ni was supposed to be nondegenerate they must represent the same state. The wave function of the state |ni is hx0 |ni and the one of the state Θ|ni correspondingly hx0 |ni∗ . These must be same (or more accurately, they can differ only by a phase factor which does not depend on the coordinate x0 ), i.e. hx0 |ni = hx0 |ni∗ For example the wave function of a nondegenerate groundstate is always real. For a spinles particle in the state |αi we get Z Θ|αi = Θ dx0 hx0 |αi|x0 i Z = dx0 hx0 |αi∗ |x0 i = K|αi,

Z

d3 x0 Θ|x0 ihx0 |αi∗

i.e. the time reversal is equivalent to the complex conjugation. On the other hand, in the momentum space we have Z Θ|αi = d3 p0 | − p0 ihp0 |αi∗ Z = d3 p0 |p0 ih−p0 |αi∗ ,

Z

d3 x0 |x0 ihx0 |αi∗ ,

because

We expand the state |αi with the help of position eigenstates: Z |αi = d3 x0 |x0 ihx0 |αi. Now Θ|αi = =

HΘ|ni = ΘH|ni = En Θ|ni,

so under the time reversal the wave function ψ(x0 ) = hx0 |αi

Θ|p0 i = | − p0 i. The momentum space wave function transform thus under time reversal like φ(p0 ) −→ φ∗ (−p0 ).

transforms like ψ(x0 ) −→ ψ ∗ (x0 ). If in particular we have ψ(x0 ) = R(r)Ylm (θ, φ),

We consider a spin 21 particle the spin of which is oriented ˆ The corresponding state is obtained by rotating along n. the state |Sz ; ↑i: |n; ↑i = e−iSz α/¯h e−iSy β/¯h |Sz ; ↑i,

we see that Ylm (θ, φ) −→ Ylm ∗ (θ, φ) = (−1)m Yl−m (θ, φ). Because Ylm is the wave function belonging to the state |lmi we must have m

Θ|lmi = (−1) |l, −mi. The probability current corresponding to the wave function R(r)Ylm seems to turn clockwise when looked at from the direction of the positive z-axis and m > 0. The probability current of the corresponding time reversed state on the other hand turns counterclockwise because m changes its sign under the operation. The spinles particles obey Theorem 1 If the Hamiltonian H is invariant under the time reversal and the energy eigenstate |ni nondegenerate then the corresponding energy eigenfunction is real (or more generally a real function times a phase factor independent on the coordinate x0 ).

ˆ where α and β are the direction angles of the vector n. Because ΘJ Θ−1 = −J . we see that Θ|n; ↑i = e−iSz α/¯h e−iSy β/¯h Θ|Sz ; ↑i. Furthermore, due to the oddity of the angular momentum, it follows that ¯h Jz Θ|Sz ; ↑i = − Θ|Sz ; ↑i, 2 so we must have Θ|Sz ; ↑i = η|Sz ; ↓i, where η is an arbitrary phase factor. So we get Θ|n; ↑i = η|n; ↓i.

On the other hand we have

Thus we must have

|n; ↓i = e−iαSz /¯h e−i(π+β)Sy /¯h |Sz ; ↑i,

Θ2 = (−1)2j .

so

Often one chooses −iSz α/¯ h −iSy β/¯ h

η|n; ↓i = Θ|n; ↑i = e e Θ|Sz ; ↑i −iαSz /¯ h −i(π+β)Sy /¯ h = ηe e |Sz ; ↑i. Writing

Θ|jmi = i2m |j, −mi.

Spherical tensors

Θ = U K, U unitary and recalling that the complex conjugation K has no effect on the base states we see that   2Sy K. Θ = ηe−iπSy /¯h K = −iη h ¯

Let us suppose that the operator A is either even or odd, i.e. ΘAΘ(−1) = ±A. We saw that then we have hα|A|αi = ±h˜ α|A|˜ αi.

Now In an eigenstate of the angular momentum we have thus

e−iπSy /¯h |Sz ; ↑i = +|Sz ; ↓i e−iπSy /¯h |Sz ; ↓i = −|Sz ; ↑i, so the effect of the time reversal on a general spin is

hα, jm|A|α, jmi = ±hα, j, −m|A|α, j, −mi. 1 2

state

Θ(c↑ |Sz ; ↑i + c↓ |Sz ; ↓i) = +ηc∗↑ |Sz ; ↓i − ηc∗↓ |Sz ; ↑i. Applying the operator Θ once again we get Θ2 (c↑ |Sz ; ↑i + c↓ |Sz ; ↓i) = −|η|2 c↑ |Sz ; ↑i − |η|2 c↓ |Sz ; ↓i = −(c↑ |Sz ; ↑i + c↓ |Sz ; ↓i),

Let now A be a component of a Hermitian spherical tensor: A = Tq(k) . According to the Wigner-Eckart theorem it is sufficient to consider only the component q = 0. We define T (k) to be even/odd under the time reversal if (k)

(k)

ΘTq=0 Θ−1 = ±Tq=0 . Then we have

i.e. for an arbitrary spin orientation we have

(k)

(k)

hα, jm|T0 |α, jmi = ±hα, j, −m|T0 |α, j, −mi.

2

Θ = −1. From the relation Θ|lmi = (−1)m |l, −mi we see that for spinles particles we have

The state |α, j, −mi is obtained by rotating the state |α, jmi: D(0, π, 0)|α, jmi = eiϕ |α, j, −mi. On the other hand, due to the definition of the spherical tensor

Θ2 = 1. D

†

In general, one can show that Θ2 |j half integeri = −|j half integeri Θ2 |j integeri = +|j integeri.

(R)Tq(k) D(R)

=

k X

(k)∗

(k)

Dqq0 (R)Tq0 ,

q 0 =−k

we get (k)

D† (0, π, 0)T0 D(0, π, 0) =

X

Generally we can write

(k)

D0q (0, π, 0)Tq(k) .

q

Θ = ηe−iπJy /¯h K.

Now (k)

D00 (0, π, 0) = Pk (cos π) = (−1)k ,

Now e−2iπJy /¯h |jmi = (−1)2j |jmi, so

so we have (k)

D† (0, π, 0)T0 D(0, π, 0) Θ2 |jmi =



 Θ ηe−iπJy /¯h |jmi

= |η|2 e−2iπJy /¯h |jmi = (−1)2j |jmi.

(k)

= (−1)k T0

+ (q 6= 0 components).

Furthermore (k)

hα, jm|Tq6=0 |α, jmi = 0,

since the m selection rule would require m = m + q. So we get

the Hamiltonian of an electron contains such terms as S · B,

p · A + A · p.

(k)

hα, jm|T0 |α, jmi (k)

= ±hα, jm|D† (0, π, 0)T0 D(0, π, 0)|α, jmi (k)

= ±(−1)k hα, jm|T0 |α, jmi. Note Unlike under other symmetries the invariance of

The magnetic field B is external, independent on the system, so [Θ, B] = 0 ja [Θ, A] = 0. On the other hand, S and p are odd, or

the Hamiltonian under the time reversal [Θ, H] = 0, does not lead to any conservation laws. This is due to the fact that the time evolution operator is not invariant: ΘU (t, t0 ) 6= U (t, t0 )Θ.

Time reversal and degeneracy

Let us suppose that [Θ, H] = 0. Then the energy eigenstates obey H|ni = En |ni HΘ|ni = En Θ|ni. If we now had Θ|ni = eiδ |ni, then, reapplying the time reversal we would obtain Θ2 |ni = e−iδ Θ|ni = |ni, or Θ2 = 1. This is, however, impossible if the system j is half integer, because then Θ2 = −1. In systems of this kind |ni and Θ|ni are degenerate. Example Electon in electromagnetic field If a particle is influenced by an external static electric field V (x) = eφ(x), then clearly the Hamiltonian H=

p2 + V (x) 2m

is invariant under the time reversal: [Θ, H] = 0. If now there are odd number of electrons in the system the total j is half integer. Thus, in a system of this kind there is at least twofold degeneracy, so called Kramers’ degeneracy. In the magnetic field B =∇×A

ΘSΘ−1 = −S ja ΘpΘ−1 = −p, so [Θ, H] 6= 0. We say that magnetic field breaks the time reversal symmetry and lifts the Kramers degeneracy.

Perturbation theory

We write |ni = |n(0) i + λ|n(1) i + λ2 |n(2) i + · · · 2 (2) ∆n = λ∆(1) n + λ ∆n + · · · .

Stationary perturbation methods Let us suppose that • we have solved completely the problem H0 |n

(0)

i=

Because hn(0) |λV − ∆n |ni = 0,

En(0) |n(0) i.

we have, on the other hand

The basis {|n(0) i} is now complete.

∆n = λhn(0) |V |ni.

• the states |n(0) i are non degenerate. • we want to solve the problem

Thus we get

(H0 + λV )|niλ = En(λ) |niλ .

2 (2) λ∆(1) n + λ ∆n + · · · = λhn(0) |V |n(0) i + λ2 hn(0) |V |n(1) i + · · · .

Usualy the index λ is dropped off. When we denote

Equalizing the coefficients of the powers of the parameter λ we get

∆n ≡ En − En(0) ,

(En(0) − H0 )|ni = (λV − ∆n )|ni. (0) Note Because the expression (En − (0) undefined the operator (En − H0 )−1

−1

(0)

H0 ) |n i is is not well defined. So, in the equation above we cannot invert the operator (0) (En − H0 ). Now hn(0) |λV − ∆n |ni = hn(0) |En(0) − H0 |ni,

so in the state (λV − ∆n )|ni there is no component along the state |n(0) i. We define a projection operator as X φn = 1 − |n(0) ihn(0) | = |k (0) ihk (0) |. k6=n

Now 1 En(0)

− H0

φn =

1

X (0) k6=n En

−

(0) Ek

|ni = |n(0) i +

= hn .. .

(0)

|V |n

(N −1)

φn En(0)

− H0

(λV − ∆n )|ni

×(|n(0) i + λ|n(1) i + · · ·). Equalizing the coefficients of the linear λ-terms we get in the first order O(λ) : |n(1) i

the formal solution is of the form

=

φn En(0)

− H0 φn

En(0) − H0

V |n(0) i −

∆(1) n En(0)

− H0

φn |n(0) i

V |n(0) i,

φn (λV − ∆n )|ni, because (0) φn ∆(1) i = 0. n |n

where lim cn (λ) = 1

We substitute |n(1) i into the expression

and

(0) ∆(2) |V |n(1) i, n = hn

cn (λ) = hn(0) |ni. Diverting from the normal procedure we normalize hn(0) |ni = cn (λ) = 1.

i

|n(0) i + λ|n(1) i + λ2 |n(2) i + · · · φn 2 (2) = |n(0) i + (0) (λV − λ∆(1) n − λ ∆n − · · ·) En − H0

=

λ→0

.

for the state vector the power series of the state vector and the energy correction and we get

|ni → |n(0) i,

1

= hn(0) |V |n(0) i = hn(0) |V |n(1) i .. .

We substitute into the expression

Since in the limit λ → 0 we must have

En(0) − H0

(N ) ∆n

O(λ ) : .. .

(λV − ∆n )|ni = φn (λV − ∆n )|ni.

|ni = cn (λ)|n(0) i +

∆n (2) ∆n

N

|k (0) ihk (0) |

and

(1)

O(λ1 ) : O(λ2 ) : .. .

the eigenvalue equation to be solved takes the form

so (0) ∆(2) |V n = hn

φn En(0)

− H0

V |n(0) i.

We substitute this further into the power series of the state vectors and we get for the coeffients of λ2 the condition φn

O(λ2 ) : |n(2) i = −

φn En(0)

− H0

En(0)

− H0

φn

V

En(0)

hn(0) |V |n(0) i

− H0 φn

En(0) − H0

V |n(0) i V |n(0) i.

Likewise we could continue to higher powers of the parameter λ. This method is known as the Rayleigh-Schr¨ odinger perturbation theory. The explicit expression for the second order energy correction will be ∆(2) n

= hn =

(0)

X

|V

φn

V |n

En(0) − H0

hn(0) |V |k (0) ihk (0) |

k,l

=

X

Vnk

k,l6=n

=

X

hk (0) |l(0) i (0)

En(0) − El

|Vnk |2 (0)

k6=n

En(0) − Ek

(0)

i

− H0

|l(0) ihl(0) |V |n(0) i

so that hn(0) |niN = Zn1/2 hn(0) |ni = Zn1/2 . Thus the normalization factor Zn is the probablility for the perturbed state to be in the unperturbed state. The normalization condition N hn|niN

Zn−1

Vln

.

= hn|ni = (hn(0) | + λhn(1) | + λ2 hn(2) | + · · ·) ×(|n(0) i + λ|n(1) i + λ2 |n(2) i + · · ·) = 1 + λ2 hn(1) |n(1) i + O(λ3 ) X |Vkn |2 = 1 + λ2 + O(λ3 ). (0) 2 (0) k6=n (En − Ek )

≡ En − En(0) = λVnn + λ2

Zn ≈ 1 − λ2

|Vnk |2

X

(0)

k6=n

= Zn hn|ni = 1

Up to the order λ2 the probability for the perturbed state to lie in the unperturbed state is thus

Thus, up to the second order we have ∆n

(0)

gives us

φn En(0)

(0)

Perturbation expansions converge if |Vij /(Ei − Ej )| is ”small”. In general, no exact convergence criterion is known. The state |ni is not normalized. We define the normalized state |niN = Zn1/2 |ni,

En(0) − Ek

+ ···.

Correspondingly, up to the second order the state vector is X Vkn |ni = |n(0) i + λ |k (0) i (0) (0) En − Ek k6=n  X X Vkl Vln +λ2 |k (0) i  (0) (0) (0) (0) k6=n l6=n (En − Ek )(En − El ) ! Vnn Vkn − (0) (En(0) − Ek )2 + ···. We see that the perturbation mixes in also other states (than |n(0) i). We see that

X

|Vkn |2

k6=n

(En(0) − Ek )2

(0)

.

The latter term can be interpreted as the probability for the ”leakage” of the system from the state |n(0) i to other states. Example The quadratic Stark effect. We consider hydrogen like atoms, i.e. atoms with one electron outside a closed shell, under external uniform electric field parallel to the positive z-axis. Now H0 =

p2 + V0 (r) and V = −e|E|z. 2m

We suppose that the eigenstates of H0 are non degenerate (not valid for hydrogen). The energy shift due to the external field is ∆k = −e|E|zkk + e2 |E|2

|zkj |2

X

(0) j6=k Ek

(0)

− Ej

+ ···,

• in the 1st order we need only the matrix element Vnn . where • in the 2nd order the energy levels i and j repel each (0) (0) other. Namely, if Ei < Ej , then the contributions of one of these states to the energy corrections of the other are (2)

∆i

(2)

∆j

=

|Vij |2 (0)

Ei =

(0)

− Ej

|Vij |2 (0)

Ej

(0)

− Ei

<0 >0

and the energy levels move apart from each other.

zkj = hk (0) |z|j (0) i. Since we assumed the states |k (0) i to be non degenerate they are eigenstates of the parity. So, according to the parity selection rule the matrix element of zkk vanishes. Indeces k and j are collective indeces standing for the quantum number triplet (n, l, m). According to the Wigner-Eckart theorem we have zkj

= hn0 , l0 m0 |z|n, lmi = hl1; m0|l1; l0 m0 i

hn0 l0 kT (1) knli √ , 2l + 1

where we have written the operator z as the spherical tensor (1) T0 = z. In order to satisfy zkj 6= 0 we must have

(0)

We want to solve the problem

• l0 = l − 1, l, l + 1. From these l0 = l is not suitable due to the parity selection rule. So we get 0



0

hn , l m |z|n, lmi = 0 unless

l0 = l ± 1 . m0 = m

We define the polarizability α so that 1 ∆ = − α|E|2 . 2

where the summing must be extended also over the continuum states. Let us suppose that (0)

(H0 + λV )|li = El |li with the boundary condition X lim |li → hm(0) |l(0) i|m(0) i, λ→0

m∈D

i.e. we are looking for corrections to the degenerated states. With the help of the energy correction we have to solve (0) (ED − H0 )|li = (λV − ∆l )|li. We write again

As a special case we consider the ground state |0(0) i = |1, 00i of hydrogen atom which is non degenerate when we ignore the spin. The perturbation expansion gives ∞ X |hk (0) |z|1, 00i|2 , α = −2e2 (0) (0) k6=0 E0 − Ek

(0)

E0 − Ek ≈ constant, so that X

(0)

Let’s suppose that the energy state ED is g-foldly degenerated: H0 |m(0) i = ED |m(0) i, ∀|m(0) i ∈ D, dimD = g.

• m0 = m and

0

Degeneracy

|li = |l(0) i + λ|l(1) i + λ2 |l(2) i + · · · (1)

∆l

= λ∆l

(2)

+ λ2 ∆l

+ ···,

so we get (0)

(ED − H0 )(|l(0) i + λ|l(1) i + λ2 |l(2) i + · · ·) (1)

(2)

− λ2 ∆l

= (λV − λ∆l ×(|l

(0)

i + λ|l

(1)

− · · ·)

2 (2)

i + λ |l

i + · · ·).

Equalizing the coefficients of the powers of λ we get in the first order (0)

(ED − H0 )|l(1) i (1)

|hk

(0)

2

|z|1, 00i|

=

k6=0

X

|hk

(0)

2

|z|1, 00i|

all k’s

= (V − ∆l )|l(0) i " # X (1) (0) (0) (0) = (V − ∆l ) |m ihm |l i . m∈D

= h1, 00|z 2 |1, 00i. In the spherically symmetric ground state we obviously have 1 hz 2 i = hx2 i = hy 2 i = hr2 i. 3 Using the explicit wave functions we get hz 2 i = a20 .

Taking the scalar product with the vector hm0 recalling that hm0

(0)

(0)

| and

(0)

|(ED − H0 ) = 0,

we end up with the simultaneous eigenvalue equations X (0) (1) Vm0 m hm(0) |l(0) i = ∆l hm0 |l(0) i. m

Now

(1)

(0)

(0)

(0)

(0)

−E0 + Ek ≥ −E0 + E1

2

  e 1 = 1− , 2a0 4

so

8a0 16a30 ≈ 5.3a30 . 2 = 3 3e The exact summation gives α < 2e2 a20

α=

9a30 = 4.5a30 . 2

The energy corrections ∆l From the equation

are obtained as eigenvalues.

(0)

(ED − H0 )|l(1) i = (λV − ∆l )|l(0) i we also see that (1)

hm(0) |V − ∆l |l(0) i = 0 ∀|m(0) i ∈ D. (1)

Thus the vector (V − ∆l )|l(0) i has no components in the subspace D. Defining a projection operator as φD = 1 −

g X m∈D

|m(0) ihm(0) | =

X k6∈D

|k (0) ihk (0) |

we can write (1)

(1)

(V − ∆l )|l(0) i = φD (V − ∆l )|l(0) i = φD V |l(0) i. We get the equation

We put the atom in external electric field parallel to the z-axis: V = −ez|E|. Now z is the q = 0 component of a spherical tensor:

(0)

(1)

(ED − H0 )|l(1) i = φD (λV − ∆l )|l(0) i, (0)

where now the operator (ED − H0 ) can be inverted: φD

|l(1) i =

(0) ED

X

=

− H0

V |l(0) i

|k (0) iVkl

(0) k6∈D ED

−

(0) Ek

z = T0 . According to the parity selection rule the operator V now has nonzero matrix elements only between states with l = 0 and l = 1, and due to the m-selection rule all states must have the same m: 2s 2p, 0  2s 0 × 2p, 0  × 0  V = 2p, 1  0 0 2p, −1 0 0

.

When we again normalize hl(0) |li = 1,

2p, 1 0 0 0 0

2p, −1  0 0  . 0  0

The nonzero matrix elements are

we get from the equation

h2s|V |2p, m = 0i = h2p, m = 0|V |2si = 3ea0 |E|.

(0)

(ED − H0 )|li = (λV − ∆l )|li

The eigenvalues of the perturbation matrix are for the energy shift (1)

∆± = ±3ea0 |E|

∆l = λhl(0) |V |li. and the eigenvectors

We substitute the power series and get

1 |±i = √ (|2s, m = 0i ± |2p, m = 0i). 2

λhl(0) |V (|l(0) i + λ|l(1) i + λ2 |l(2) i + · · ·) (1)

= λ∆l

(2)

+ λ2 ∆l

+ ···.

Note The energy shift is a linear function of the electric

The second order energy correction is now (2)

∆l

= hl(0) |V |l(1) i = hl(0) |V =

X

|Vkl |2 (0)

k6∈D

φD (0) ED

− H0

V |l(0) i

. (0)

Diagonalize the perturbation matrix.

◦

The resulting eigenvalues are first order corrections for the energy shifts. The corresponding eigenvectors are those zeroth order eigenvectors to which the corrected eigenvectors approach when λ → 0.

3

Let the states m ∈ D to be almost degenerate. We write V = V1 + V 2 , where

Identify the degenerated eigenstates. We suppose that their count is g. Construct the g × g-perturbation matrix V .

◦

2

Nearly degenerated states

ED − Ek

Thus the perturbation calculation in a degenerate system proceeds as follows: 1◦

field. The states |±i are not parity eigenstates so it is perfectly possible that they have permanent electric dipole moment hzi = 6 0.

4◦ Evaluate higher order corrections using non degenerate perturbation methods but omit in the summations all contributions coming from the degenerated state vectors of the space D. Example The Stark efect in the hydrogen atom.

The hydrogen 2s (n = 2, l = 0, m = 0) and 2p (n = 2, l = 1, m = −1, 0, 1) states are degenerate. Their energy is (0) ED = −e2 /8a0 .

V1

=

X X

|m(0) ihm(0) |V |m0(0) ihm0(0) |

m∈D m0 ∈D

V2

= V − V1 .

We proceed so that 1. we diagonalize the Hamiltonian H0 + V1 exactly in the basis {|m(0) i} and 2. handle the term V2 like in an ordinary non degenerate perturbation theory. This is possible since hm0(0) |V2 |m(0) i = 0

∀m, m0 ∈ D.

Example Weak periodic potential.

Now H0 =

p2 2m

and for the perturbation V (x) = V (x + a).

We denote the unperturbed eigenstates by the wave vector: h2 k 2 ¯ H0 |ki = |ki, 2m so that h2 k 2 ¯ (0) . Ek = 2m We impose the periodic boundary conditions hx0 |ki = ψk (x0 ) = hx0 + L|ki = ψk (x0 + L),

Correspondingly the energy up to the second order is (2)

|Vn |2

(0) n6=0 Ek

− Ek+nK

(0)

.

Let us suppose now that k≈−

nK . 2

We diagonalize the Hamiltonian in the basis

for the wave function and get

{|ki, |k + nKi},

0 1 2π ψk (x ) = √ eikx , k = n, n ∈ I. L L

0

i.e. we diagonalize the matrix

Because the potential V is periodic it can be represented as the Fourier series ∞ X

V (x) =

X

(0)

Ek = Ek + V 0 +

inKx

e

Vn ,

n=−∞

|ki (0) Ek

|ki |k + nKi

|k + nKi

+ V0

Vn

Vn∗ (0) Ek + V 0

! .

Its eigenvalues are

where

(0)

K = 2π/a is the reciprocal lattice vector. The only nonzero matrix elements are now hk + nK|V |ki = Vn ,

(0)

because

(0)

E + Ek+nK Ek± = V0 + k 2 v !2 u u E (0) − E (0) t k k+nK + |Vn |2 . ± 2 (0)

When |Ek − Ek+nK |  |Vn |, it reduces to two solutions Z

0 0 0 0 1X Vn dx0 e−ik x einKx einkx L n X = Vn δk+nK,k0 .

hk 0 |V |ki =

n

(0)

Ek+

= Ek + V 0

Ek−

= Ek+nK + V0 ,

(0)

which are first order corrected energies. In the limiting case we get

So the potential couples states |ki, |k + Ki, . . . , |k + nKi, . . . .

(0)

lim k→−nK/2

Ek± = EnK/2 + V0 ± |Vn |.

The corresponding energy denominators are (0)

(0)

Brillouin-Wigner perturbation theory

Ek − Ek+nK .

We start with the Schr¨odinger equation

In general (0)

(En − H0 )|ni = λV |ni,

(0)

Ek 6= Ek+nK except when

and take on both sides the scalar product with the state |m(0) i, and get

nK . 2 We suppose that the condition k≈−

k 6= −

(0) (En − Em )hm(0) |ni = λhm(0) |V |ni.

nK 2

holds safely. The first order state vectors are then |k (1) i = |ki +

X n6=0

|k + nKi

Vn (0) Ek

(0)

− Ek+nK

,

We correct the state |n(0) i. We write the corrected state |ni in the form X |ni = |m(0) ihm(0) |ni = |n(0) ihn(0) |ni + φn |ni m

= |n(0) i +

X

|m(0) ihm(0) |ni,

m6=n

and the wave functions (1) ψk (x0 )

X 1 0 0 1 V √ ei(k+nK)x (0) n (0) . = √ eikx + L L Ek − Ek+nK n6=0

which has been normalized, like before, hn(0) |ni = 1.

We substitute into this the scalar products hm(0) |ni =

λhm(0) |V |ni (0) En − E m

,

and end up with the fundamental equation of the Brillouin-Wigner method X

|ni = |n(0) i +

|m(0) i

m6=n

λ (0) En − E m

hm(0) |V |ni.

Iteration gives us the series |ni = |n(0) i + λ

X

|m(0) i

m6=n

+ λ2

XX

|l(0) i

m6=n l6=n

1 (0) E n − Em

1 (0)

En − El

hm(0) |V |n(0) i

hl(0) |V |m(0) i

1

hm(0) |V |n(0) i (0) En − Em X XX 1 hk (0) |V |l(0) i + λ3 |k (0) i (0) E − E n m6=n l6=n k6=n k ×

×

1

En − + ···.

(0) El

hl(0) |V |m(0) i

1 (0) E n − Em

hm(0) |V |n(0) i

Note This is not a power series of the parameter λ because the energy denominators (0) (0) En − Em = En(0) − Em + ∆n

depend also on the parameter λ according to the equation 2 (2) ∆n = λ∆(1) n + λ ∆n + · · ·

Time dependent potentials

The interaction picture observables AI are defined so that

We have solved the problem

AI ≡ eiH0 t/¯h AS e−iH0 t/¯h .

H0 |ni = En |ni In particular we have completely and want to solve the eigenstates of the Hamiltonian H = H0 + V (t). Since the Hamiltonian depends on time we have

VI = eiH0 t/¯h V e−iH0 t/¯h . We see that the equation governing the time dpendence of the interaction picture state vectors is

U= 6 e−iHt/¯h ,

i¯h

so a system in a stationary state |ii can in the course of time get components also in other stationary states. Pictures of the time evolution

At the moment t = 0 let the system be in the state X |αi = cn (0)|ni n

and at the moment t in the state X |α, t0 = 0; ti = cn (t)e−iEn t/¯h |ni.

so

∂ |α, t0 ; tiI = VI |α, t0 ; tiI . ∂t If now AS does not depend on time we can derive i¯h

1 dAI = [AI , H0 ], dt i¯h

n

Note The time dependence of the coefficients cn (t) is due

only to the potential V (t). The effect of the Hamiltonian H0 is in the phase factors e−iEn t/¯h . Schr¨ odinger’s picture The evolution of the state vectors is governed by the time evolution operator: |α, t0 = 0; tiS = U(t)|α, t0 = 0i. Heisenberg’s picture The state vectors remain constant, i.e.

∂ |α, t0 ; tiI ∂t  ∂  iH0 t/¯h e |α, t0 ; tiS = i¯h ∂t = −H0 eiH0 t/¯h |α, t0 ; tiS +eiH0 t/¯h (H0 + V )|α, t0 ; tiS = eiH0 t/¯h V e−iH0 t/¯h eiH0 t/¯h |α, t0 ; tiS ,

which in turn resembles the Heisenberg equation of motion provided that in the latter we substitute for the total Hamiltonian H the stationary operator H0 . We expand state vectors in the base {|ni}: X |α, t0 ; tiI = cn (t)|ni. n

If now t0 = 0 so multiplying the previous expansion X |α, t0 = 0; ti = cn (t)e−iEn t/¯h |ni

|α, t0 = 0, tiH = |α, t0 = 0i. On the other hand, the operators depend on time. We can go from the time independent operators of the Schr¨odinger picture to the operators of the Heisenberg picture via the transformation AH (t) = U † (t)AS U(t). If the Hamiltonian does not depend on time then HH (t) = U † (t)HU(t) = H

n

on both sides by the operator e−iH0 t/¯h we get X |α, t0 = 0; tiI = cn (t)|ni, n

i.e. the coefficients cn are equal. We just derived the equation ∂ i¯h |α, t0 ; tiI = VI |α, t0 ; tiI . ∂t From this we get

and

dAH 1 1 = [AH , HH ] = [AH , H]. dt i¯h i¯h Interaction picture The state vectors depend on time as iH0 t/¯ h

|α, t0 ; tiI ≡ e

i¯h

= hn|VI |α, t0 ; tiI X = hn|VI |mihm|α, t0 ; tiI . m

The matrix elements of the operator VI are

|α, t0 ; tiS .

hn|VI |mi = hn|eiH0 t/¯h V e−iH0 t/¯h |mi = Vnm (t)ei(En −Em )t/¯h

At the moment t = 0 we have obviously |iS = |iI = |iH .

∂ hn|α, t0 ; tiI ∂t

.

Because we furthermore have cn (t) = hn|α, t0 ; tiI , we can write the equation governing the time dependence of the superposition coefficients as i¯h

The system oscillates between the states |1i and |2i with the angular velocity s  γ2 (ω − ω21 )2 + . Ω= 4 ¯h2 The amplitude of the oscillations is at its maximum, i.e. we are in a resonance, when

X d cn (t) = Vnm eiωnm t cm (t), dt m

ω ≈ ω21 =

where

En − Em = −ωmn . h ¯ This system of differential equations can be written explicitely in the matrix form   c˙1  c˙2    i¯h  c˙3    .. .   c1 V11 V12 eiω12 t ···   c2  V21 eiω21 t V · · · 22   =  V33 · · ·    c3  .. .. .. .. . . . . ωnm ≡

Example Spin magnetic resonance.

We put a spin

particle into

• time dependent magnetic field rotating in the xy plane, i.e.    . 

H0 = E1 |1ih1| + E2 |2ih2| (E1 < E2 ) and that the time dependent potential is like V (t) = γeiωt |1ih2| + γe−iωt |2ih1|. The matrix elements Vnm are now ∗ = V21 = V22

1 2

• time independent magnetic field parallel to the z axis,

Example Two state systems. Suppose that

V12 V11

= γeiωt = 0,

ˆ + B1 (ˆ ˆ sin ωt) B = B0 z x cos ωt + y when the fields B0 and B1 are constant. Since the magnetic moment of the electron is µ=

so transitions between the states |1i and |2i are possible. The system of differential equations to be solved is = γeiωt eiω12 t c2 = γe−iωt eiω21 t c1 ,

where

E2 − E1 . h ¯ We can see that the solution satisfying the initial conditions c1 (0) = 1, c2 (0) = 0 ω21 = −ω12 =

is |c2 (t)|2

2

|c1 (t)|

=

=

γ 2 /¯ h2 γ 2 /¯h + (ω − ω21 )2 /4 ( ) 2 2 1/2 γ (ω − ω ) 21 × sin2 + t 4 h2 ¯ 2

2

1 − |c2 (t)| .

e S, me c

the Hamiltonian is the sum of the terms   e¯hB0 (|Sz ; ↑ihSz ; ↑ | − |Sz ; ↓ihSz ; ↓ |) H0 = − 2me c   e¯hB1 V (t) = − 2me c  × cos ωt(|Sz ; ↑ihSz ; ↓ | + |Sz ; ↓ihSz ; ↑ |)  + sin ωt(−i|Sz ; ↑ihSz ; ↓ | + i|Sz ; ↓ihSz ; ↑ |) . If e < 0, then E2 = E↑ =

i¯hc˙1 i¯hc˙2

E2 − E 1 . ¯h

|e|¯hB0 |e|¯hB0 > E 1 = E↓ = − . 2me c 2me c

We can thus identify in the above treated two stated system the quantities: |Sz ; ↑i 7→ |2i (higher state) |Sz ; ↓i 7→ |1i (lower state) |e|B0 7 → ω21 me c e¯hB1 − 7→ γ, ω 7→ ω. 2me c Comparing with our earlier discussion on the spin precession we see that • if B1 = 0 and B0 6= 0, the the expectation value hSx,y i rotates in the course of time counterclockwise but the probabilities |c1 |2 and |c2 |2 remain still constant. • if B1 6= 0, the the coefficients |c1 |2 and |c2 |2 are functions of time.

When the resonance condition ω ≈ ω21 holds the probability for the spin flips |Sz ; ↑i ←→ |Sz ; ↓i is very high. Because experimental production of rotating magnetic fields is difficult it is common to use a field oscillating for example along the x axis. This can be divided into components rotating counterclockwise and clockwise: ˆ cos ωt 2B1 x ˆ sin ωt) = B1 (ˆ x cos ωt + y ˆ sin ωt). +B1 (ˆ x cos ωt − y In experiments one usually has B1  1, B0 so

|e|B1 |e|B0 B1 B1 γ = = = ω21  ω21 . ¯h 4me c me c 4B0 4B0

If now the component rotating counterclockwise triggers the resonance condition ω ≈ ω21 , the the transition probability due to this component is |cR (t)|2

γ 2 /¯ h2 γ /¯h + (ω − ω21 )2 /4 ) ( 2 2 1/2 γ (ω − ω ) 21 t × sin2 + 4 h2 ¯ γ  ≈ sin2 t . ¯h

=

2

2

The clockwise rotating component, ω = −ω21 , contributes |cL (t)|2

≈



γ 2 /¯h2 2 γ /¯ h2 + ω21 ( 1/2 ) γ2 2 2 × sin + ω21 t h2 ¯ 2

|cR (t)|2 .

Time dependent perturbation theory

Here hn|UI (t, 0)|ii = cn (t)

In the interaction picture the time evolution operator is determined by the equation

is the same as the superposition coefficient we used before. From the relation binding the interaction and Schr¨odinger picture state vectors we get

|α, t0 ; tiI = UI (t, t0 )|α, t0 ; t0 iI . Since the time evolution of the state vectors is governed by the equation i¯h

∂ |α, t0 ; tiI ∂t

= VI |α, t0 ; tiI = VI UI (t, t0 )|α, t0 ; t0 iI ,

= eiH0 t/¯h |α, t0 ; tiS = eiH0 t/¯h U(t, t0 )|α, t0 ; t0 iS = eiH0 t/¯h U(t, t0 )e−iH0 t0 /¯h |α, t0 ; t0 iI ,

so the time evolution operators of these pictures are obtained with the help of the formula

we see that i¯ h

|α, t0 ; tiI

UI (t, t0 ) = eiH0 t/¯h U(t, t0 )e−iH0 t0 /¯h .

∂UI (t, t0 ) |α, t0 ; t0 iI = VI UI (t, t0 )|α, t0 ; t0 iI . ∂t

The interaction picture time evolution operator satisfies thus the equation

The matrix elements of the operator UI (t, t0 ) can now be calculated from the relation hn|UI (t, t0 )|ii = ei(En t−Ei t0 )/¯h hn|U(t, t0 )|ii.

d i¯h UI (t, t0 ) = VI (t)UI (t, t0 ). dt

We see that

As the initial condition we have obviously UI (t, t0 ) t=t0 = 1.

• the matrix element hn|UI (t, t0 )|ii is not quite the transition amplitude hn|U(t, t0 )|ii, • the transition probabilities satisfy

Integration gives i ¯h

UI (t, t0 ) = 1 −

Z

|hn|UI (t, t0 )|ii|2 = |hn|U(t, t0 )|ii|2 .

t

VI (t0 )UI (t0 , t0 ) dt0 .

t0

Note If the states |a0 i and |b0 i are not eigenstates of H0

then

By iteration we end up with Dyson’s series

|hb0 |UI (t, t0 )|a0 i|2 6= |hb0 |U(t, t0 )|a0 i|2 .

UI (t, t0 ) i =1− ¯h

Z

i ¯h

Z

=1− 

t

t0

i + − ¯h

"

i VI (t ) 1 − h ¯ 0

Z

t0

# 00

00

00

VI (t )UI (t ) dt

dt

t0

t

dt0 VI (t0 )

t0

2 Z

t

dt t0

0

Z

t0

dt00 VI (t0 )VI (t00 )

t0

n Z t Z t0 i 0 +··· + − dt dt00 · · · ¯h t0 t0 Z t(n−1) × dt(n) VI (t0 )VI (t00 ) · · · VI (t(n) ) 

t0

+···. Let us suppose again that we have solved the problem

0

In this case the matrix elements are evaluated by expanding the states |a0 i and |b0 i in the base {|ni} formed by the eigenstates of H0 . Let us suppose now that at the moment t = t0 the system is in the eigenstate |ii of H0 . This state vector can always be multiplied by an arbitrary phase factor, so the Schr¨odinger picture state vector |i, t0 ; t0 iS can be chosen as |i, t0 ; t0 iS = e−iEi t0 /¯h |ii. Then in the interaction picture we have |i, t0 ; t0 iI = |ii. At the moment t this has evolved to the state X |i, t0 ; tiI = UI (t, t0 )|ii = cn (t)|ni, n

H0 |ni = En |ni completely. Let the initial state of the system be |ii at the moment t = t0 = 0, i.e. |α, t0 = 0; t = 0iI = |ii.

so cn (t) = hn|UI (t, t0 )|ii, as we already noted. Now 1. substitute the Dyson series into this

At the moment t this has evolved to the state |i, t0 = 0; tiI

= UI (t, 0)|ii X = |nihn|UI (t, 0)|ii. n

2. expand the coefficient as a power series of the perturbation (1) (2) cn (t) = c(0) n (t) + cn (t) + cn (t) + · · · ,

(k)

3. equalize the terms cn with the perturbation terms of the order k,

1600

4. denote

1200

1400

ei(En −Ei )t/¯h = eiωni t .

1000

800

We get

t=40

600

c(0) n (t) c(1) n (t)

= δni = − = −

c(2) n (t) =

i ¯ h i ¯h

 −

t=20 400

t

Z

hn|VI (t0 )|ii dt0

t0 Z t

200 t=10 0 -1

0

-0.5

0

eiωni t Vni (t0 ) dt0

0.5

1

f

t0

2 X Z

i ¯h

×e

dt0

Z

t0

m iωmi t00

t

t0

t2

0

dt00 eiωnm t Vnm (t0 )

t0 00

Vmi (t ). 1/t

The probability for the transition from the state |ii to the state |ni can be written as (2) 2 Pr(i → n) = |cn (t)|2 = |c(1) n (t) + cn (t) + · · · | .

ω

When t is large then |cn (t)|2 6= 0 only if Fermi’s golden rule

Consider the constant perturbation  0, when t < 0 V (t) = V (time independent) when t ≥ 0.

t≈

If now ∆t is the time the perturbation has been on then transitions are possible only if

switched on at the moment t = 0. At the moment t = 0 let the system be in the pure state |ii. Now c(0) n c(1) n

= c(0) n (0) = δin Z t 0 i = − Vni eiωni t dt0 h ¯ 0 Vni = (1 − eiωni t ). En − Ei

The transition probability to the state |ni is thus 2 |c(1) n |

= =

|Vni |2 (2 − 2 cos ωni t) |En − Ei |2   4|Vni |2 2 (En − Ei )t sin . 2¯h |En − Ei |2

2π 2π¯h = . ω |En − Ei |

∆t∆E ≈ ¯h. Note If the energy is conserved exactly, i.e.

E n = Ei , then

1 |Vni |2 t2 . ¯2 h The transition probability is proportional to the square of the on-time of the perturbation (and not linearly proportional to the time). In general we are interested in transitions in which the initial state |ii is fixed but the final state |ni can be any state satisfying the energy conservation rule 2 |c(1) n (t)| =

En ≈ Ei

The quantity En − E i ω≡ h ¯ is almost continuous because usually the En states form almost a continuum. The transition probability is now 2 |c(1) n | =

|Vni |2 f (ω), h2 ¯

The total probability for such a transition is now Pr(i → f ) X 2 = |c(1) n (t)| n En ≈Ei

Z

2 dEn ρ(En )|c(1) n |   Z |Vni |2 2 (En − Ei )t ρ(En ) dEn . = 4 sin 2¯h |En − Ei |2

=

where f (ω) =

4 sin2 ωt/2 . ω2

0

ρ(E)dE = the number of states between(E, E + dE). Because

0

• if Em in the term eiωnm t differs from En and at the same time from Ei it oscillates rapidly and contributes nothing.

1 sin2 xt = δ(x), t→∞ π tx2 lim

So we can write

we get lim

t→∞



1 2 En − Ei t 2 sin 2¯ h (En − Ei )

= =

En − Ei πt 2δ 2¯h 4¯h πt δ(En − Ei ). 2¯h



wi→f

Consider the potential V (t) = Veiωt + V † e−iωt , which is again assumed to be switched on at the moment t = 0. When t < 0, the system is supposed to be in the state |ii. The first order term is now c(1) n

m

t

dt0

Z

t0

t0

so in the case of the potential  0, when t < 0 V (t) = V (time independent) when t ≥ 0.

= Above

−

i ¯h

2 X m

Z Vnm Vmi 0

t



0

eiωni t dt0

0

# 1 1 − ei(ω+ωni )t 1 − ei(ω−ωni )t † Vni + Vni . ¯h ω + ωni −ω + ωni

ωni =

En − Ei −→ ωni ± ω. ¯h

(1)

When t → ∞, |cn |2 is thus non zero only if

wi→n =

we have 

0

or or

En ≈ Ei − ¯hω En ≈ Ei + ¯hω.

Obviously, if the first term is important the second one does not contribute and vice versa. The energy of a quantum mechanical system is not conserved in these transitions but the ”external” potential either gives (absorption) or takes (stimulated emission) energy to/from the system. Analogically to the constant potential the transition rate will be

t0

00

=

0

† −iωt Vni eiωt + Vni e

0

dt00 eiωnm t Vnm (t0 )

×eiωmi t Vmi (t00 ),

c(2) n

Z t

ωni + ω ≈ 0 ωni − ω ≈ 0

In the second order we got 2 X Z

i ¯h "

This is of the same form as in the case of our earlier step potential, provided that we substitute

Second order corrections

i ¯h

= − =

but then one implicitely assumes that it will be integrated R in the expression dEn ρ(En )wi→n · · ·.

−

En ≈Ei

Harmonic perturbations

Quite often this is also written as   2π wi→n = |Vni |2 δ(En − Ei ), h ¯



.

|ii −→ |mi −→ |ni.

where |Vni |2 is the average of the term |Vni |2 . Note The total transition probability depends linearly on time t. The transition rate w is defined to be the transition probability per unit time. We end up with the Fermi golden rule ! d X (1) 2 |cn (t)| wi→f = dt n   2π = . |Vni |2 ρ(En ) ¯h En ≈Ei

=

2 X Vnm Vmi 2π V + ρ(E ) = ni n ¯h Ei − Em m

In the second order the term Vnm Vmi can be thougth to describe virtual transitions

The transition probability is thus   2π , |Vni |2 ρ(En )t lim Pr(i → f ) = t→∞ h ¯ En ≈Ei

c(2) n (t)

(1)

• the term eiωni t is same as in the coefficient cn , so it contributes only if En ≈ Ei when t → ∞.

Here ρ(E) is the density of states, i.e.

0

dt0 eiωnm t

Z

t0

00

dt00 eiωmi t

0

Z t 0 0 i X Vnm Vmi (eiωni t − eiωnm t )dt0 . ¯h m Em − Ei 0

2π |Vni |2 δ(En − Ei ± ¯hω). ¯h

Energgy shifts and line widths

Thus, up to the second order the coefficient ci is 2  e2ηt i i ηt |Vii |2 2 ci (t) ≈ 1 − Vii e + − ¯hη ¯h 2η X  |Vmi |2 e2ηt i . + − ¯h 2η(Ei − Em + i¯hη)

Evolution of the initial state

We consider the case where the initial and final states are the same. We switch the interaction on slowly: V (t) = eηt V.

m6=i

Here η → 0 at the end. We suppose that in the far past, t = −∞, the system has been in the state |ii. Check If n 6= i, then the perturbation theory gives c(0) n (t)

=

i = − Vni lim t0 →−∞ ¯h

c(1) n (t)

t

ηt0 iωni t0

e

e

dt

t0

≈

2

Note ∆i is not necessarily real.

so in the limit η → 0 we get the Fermi golden rule   2π |Vni |2 δ(En − Ei ). wi→n ≈ h ¯

We interprete this so that the state |ii evolves gradually like |ii −→ ci (t)|ii = e−i∆i t/¯h |ii.

This is equivalent with our previous result. Let now n = i. We get

(1)

ci

i Vii ¯h η  X −i −i |Vmi |2 Vii + , ¯h ¯h Ei − Em + i¯hη 1−

ci (t) = e−i∆i t/¯h .

η hδ(En − Ei ), 2 = πδ(ωni ) = π¯ η + ωni

=

m6=i

The solution satisfying the initial condition ci (0) = 1 is

Now

(0)

i ¯ h

where we have already set eηt → 1. Note We cannot set in the denominator η = 0, because the states Em can form nearly a continuum in the vicinity of Ei . The logarithmic derivative is thus time independent, i.e. of form i c˙i (t) = − ∆i . ci (t) ¯h

and the transition rate correspondingly   ηe2ηt 2|Vni |2 d |cn (t)|2 ≈ . 2 dt η 2 + ωni h2 ¯

ci

−

i |Vii |2 X |Vmi |2 + ¯h η (Ei − Em + i¯hη)

m6=i

Up to the lowest non vanishing order the transition probability is   |Vni |2 e2ηt 2 |cn (t)| ≈ 2 η 2 + ωni h2 ¯

lim

≈

0

eηt+iωni t i . = − Vni ¯h η + iωni

η→0

Vii − c˙i ci

0 Z

For the logarithmic time derivative of the coefficient ci up to the second order in the perturbation V we get

In the Schr¨odinger picture the latter contains also a phase factor, or

1

e−i∆i t/¯h |ii 7→ e−i∆i t/¯h−iEi t/¯h |ii. Z

i = − Vii lim ¯h t0 →−∞

t

0

eηt dt0 = −

t0

i Vii eηt ¯η h

Due to the perturbation the energy levels shift like Ei −→ Ei + ∆i .

(2) ci

 =

i − ¯h

2 X m

Z

t

× lim

t0 →−∞

 =

i − ¯h

We expand now ∆i as the power series in the perturbation:

|Vmi |2 0 iωim t0 +ηt0

Z

t0

dt e t0

2 X

(1)

00

dt00 eiωmi t

t0

t0 →−∞

dt e t0

2

m6=i

0

eiωmi t +ηt i(ωmi − iη)

i e2ηt = − |Vii |2 2 ¯h 2η  X i |Vmi |2 e2ηt + − . ¯h 2η(Ei − Em + i¯hη) 

+ ···.

Comparing with our previous expression

0

0 iωim t0 +ηt0

(2)

+ ∆i

|Vmi |

t

× lim

∆i = ∆i

2

m

Z

+ηt00

∆i = Vii +

X m6=i

|Vmi |2 , Ei − Em + i¯hη

we see that in the first order we have (1)

∆i

= Vii .

This is equivalent with the time independent perturbation theory.

Because the energies Em for almost a continuum we can in the second order term X m6=i

|Vmi |2 Ei − Em + i¯hη

replace the summation with the integration. To handle the limit η → 0 we recall from the function theory that Z ∞ Z ∞ f (z) f (z) lim dz = ℘ dz − iπf (0), →0+ −∞ z + i −∞ z where ℘ stands for the principal value integral. A common shorthand notation for this is 1 1 = ℘ − iπδ(x). →0 x + i x lim

Thus we get X |Vmi |2 E i − Em m6=i X = −π |Vmi |2 δ(Ei − Em ).

(2)

Re(∆i )

= ℘

(2)

Im(∆i )

m6=i

The right hand side of the latter equation is familiar from the Fermi golden rule, so we can write h i X 2 2π X (2) |Vmi |2 δ(Ei − Em ) = − Im ∆i . wi→m = ¯h h ¯ m6=i

m6=i

The coefficient ci (t) can be written with the help of the energy shift as ci (t) = e−(i/¯h)[Re(∆i )t]+(1/¯h)[Im(∆i )t] . We define

Γi 2 ≡ − Im(∆i ). ¯h h ¯

Then |ci (t)|2 = e2Im(∆i )t/¯h = e−Γi t/¯h . Thus the quantity Γi tells us at which rate the state |ii disappears. The quantity τi =

h ¯ Γi

is thus the average life time of the state |ii. In the Schr¨odinger picture the time evolution is Z h ¯ −iEi t/¯ h f (E)e−iEt/¯h dE, ci (t)e = 2π where the energy spectrum Z f (E) = e−i[Ei +Re(∆i )]t/¯h−Γi t/2¯h eiEt/¯h dt is the Fourier transform of the coefficient ci (t)e−iEi t/¯h . Now |f (E)|2 ∝

1 , {E − [Ei + Re(∆i )]}2 + Γ2i /4

so Γi —or excluding the factor -2, the imaginary part of the energy shift— is the width of the decay line and the real part of the energy shift what is usually called the energy shift. In the case of harmonic perturbations we can repeat the same derivation provided that we substitute Em − Ei 7→ Em − Ei ± ¯hω.

Radiation and matter

Earlier we saw that in the case of the harmonic potential

We handle the interaction of radiation and matter semiclassically: • the radiation field classically,

V (t) = Veiωt + V † e−iωt transitions are possible if ωni + ω ≈ 0 ωni − ω ≈ 0

• the matter quantum mechanically, • OK, if there is large number of photons in the volume ≈ λ3 ,

E B

1 ∂ A c ∂t = ∇ × A. = −

The energy flux —energy/unit area/unit time— is  2  B2 c Emax + max . cU = 2 8π 8π For a monochromatic plane wave we have h i ˆ ·x−iωt + e−i(ω/c)n ˆ ·x+iωt , A = A0 ˆ ei(ω/c)n ˆ and ˆ are the directions of the propagation and where n polarization of the plane wave. Due to the transverse condition ∇·A=0 ˆ The energy flux is then we have ˆ⊥n. 1 ω2 |A0 |2 . 2π c A particle in the radiation field has the mechanical momentum  e e e2 e 2 p− A = p2 − p · A − A · p + 2 A2 c c c c 2 e e = p2 − 2 A · p + 2 A2 , c c since due to the transvers condition cU =

p · A = A · p. The Hamiltonian of an electron in the field is now 1  e 2 H = p − A + eφ(x) 2me c p2 e ≈ + eφ(x) − A · p, 2me me c when we drop off the term |A|2 . Now   e − A·p me c   e =− A0 ˆ · p me c h i ˆ ·x−iωt + e−i(ω/c)n ˆ ·x+iωt . × ei(ω/c)n

En ≈ Ei − ¯hω En ≈ Ei + ¯hω,

or

• in the case of the spontaneous emission we impose a fictive field equivalent with the quantum theory. The vector potential A of the classical radiation field can always be chosen to satisfy the transverse condition: ∇ · A = 0. The electric and magnetic field are obtained from the vector potential as

or or

eiωt −iωt

e

←→ ←→

stimulated emission absorption.

Absorption

In the case of the radiation field,  eA0  i(ω/c)n † ˆ ·x ˆ · p Vni =− e me c ni is the matrix element corresponding to the absorption. The transition rate is then wi→n

=

2π e2 ˆ ·x ˆ · p|ii|2 |A0 |2 |hn|ei(ω/c)n ¯h m2e c2 ×δ(En − Ei − ¯hω).

We should note that • if the final states |ni form a continuum we integrate weighing with the state density ρ(En ). • if the final states |ni are discrete they, nevertheless, are not ground states so that their energy cannot be extremely accurate. • collisions can broaden the energy levels. • the incoming radiation is not usually completely monochromatic. So we write the δ-function as γ  1  . δ(ω − ωni ) = lim γ→0 2π (ω − ωni )2 + γ 2 /4 We define the absorption cross section: σabs =

(energy/unit time) absorbed by the atom (i → n)

.

energy flux of the radiation field

Since in every absorption process the atom absorbs the energy ¯hω, we have σabs

¯ ωwi→n h 1 ω2 2 A 2π c 0 2π e2 ˆ ·x ˆ · p|ii|2 ¯hω |A0 |2 |hn|ei(ω/c)n ¯h m2e c2 = 1 ω2 |A0 |2 2π c ×δ(En − Ei − ¯hω)   4π 2 ¯h e2 ˆ ·x ˆ · p|ii|2 |hn|ei(ω/c)n = m2e ω ¯hc ×δ(En − Ei − ¯hω). =

Here e2 /¯hc is the fine structure constant α ≈ 1/137. In order the absorption to be possible the energy quantum ¯hω of the radiation must be of the order of the energy level spacing: ¯hω ≈

Integration gives Z X σabs (ω) dω = 4π 2 αωni |hn|x|ii|2 . n

The oscillator strength is defined as follows:

Ze2 Ze2 ≈ , (a0 /Z) Ratom

fni ≡

when Z is the atomic number. Now

2me ωni |hn|x|ii|2 . ¯h

One can show that it satisfies so called Thomas-Reiche-Kuhn’s sum rule: X fni = 1.

137Ratom c λ c¯ hRatom ≈ = ≈ ω 2π Z Ze2 or

Z Ratom ≈  1. λ 2π137 We expand the exponential function in the expression for the cross section as the power series ˆ ·x = 1 + i ω n ˆ · x + ···. ei(ω/c)n c

n

We see that  2  Z 4π 2 α¯h e . σabs (ω) dω = = 2π 2 c 2me me c2 This is known as the oscillator sum rule of classical electrodynamics.

Now ω ω Z Z ˆ · xi ≈ Ratom ≈ hn , Ratom = c c 137Ratom 137 so it is usually enough if we keep only the term 1. We have then the so called electric dipole approximation. Thus in the electric dipole approximation

Photoelectric effect

The initial state |ii is atomic but the final state |ni is in the continuum formed by the plane waves |kf i. In the absorption cross section we have now to weigh the function δ(ωni − ω) with the final state density ρ(En ): σabs

4π 2 ¯h ˆ ·x ˆ · p|ii|2 α|hn|ei(ω/c)n m2e ω ×ρ(En )δ(En − Ei − ¯hω).

=

ˆ ·x ˆ · p|ii −→ ˆ · hn|p|ii. hn|ei(ω/c)n We choose

Under the periodic boundary conditions in the L-sided cube we have 0 eikf ·x hx0 |kf i = , L3/2 where 2πni , ni = 0, ±1, ±2, . . . . ki = L When L → ∞, the variable n, defined via the relation

ˆ k x ˆ and n ˆ kz ˆ. Let the states |ni be the solutions of the problem H0 |ni = En |ni, H0 = Because [x, H0 ] =

p2 + eφ(x) 2me

i¯hpx , me

n2 = n2x + n2y + n2z ,

we have

can be considered continuous. Then the volume in the solid angle dΩ bounded by the surfaces n0 = n and n0 = n + dn is n2 dn dΩ. The final state energy is

me hn|[x, H0 ]|ii i¯ h = ime ωni hn|x|ii.

hn|px |ii =

(1)

Since x is a superposition of the spherical tensors T±1 we get the selection rules m0 − m = ±1 |j 0 − j| = 0, 1. If we had ˆ , the same selection rules were valid. • ˆ k y (1)

ˆ , we should have m0 = m, because z = T0 . • ˆ k z In the dipole approximation the absorption cross section is σabs = 4π 2 αωni |hn|x|ii|2 δ(ω − ωni ).

E=

¯ 2 kf2 h ¯h2 n2 (2π)2 = . 2me 2me L2

The number of states with the wave vector kf in the interval (E, E + dE) and in the solid angle is  3
dkf dn L 2 n dΩ dE = k2f dΩ dE dE 2π dE  3 L me = kf dE dΩ. 2π ¯h2 The differential cross section is now 3 dσ 4π 2 α¯h ˆ ·x ˆ · p|ii|2 me kf L . = |hkf |ei(ω/c)n 2 2 dΩ me ω ¯h (2π)3

Example Emission of an electron from the innermost

shell. The wave function of the initial state is approximately like the one of the hydrogen ground state provided we substitute a0 −→ a0 /Z: 0

hx |ii ≈



Z a0

3/2

e−iZr/a0 .

The matrix element is now ˆ ·x ˆ · p|ii hkf |ei(ω/c)n 0 Z e−ikf ·x i(ω/c)n ˆ ·x0 = ˆ · d3 x0 e 3/2 L "  3/2 # Z −Zr/a0 ×(−i¯h∇) e . a0 Integrating by parts and noting that due to the ˆ = 0 we have transversal condition ˆ · n h i ˆ ·x0 = 0. ˆ · ∇ei(ω/c)n We get ˆ ·x ˆ · p|ii hkf |ei(ω/c)n Z ω ¯hˆ · kf ˆ )·x0 ψ 0 d3 x0 ei(kf − c n = atom (x ). 3/2 L Thus the matrix element is proportional to the Fourier transform of the atomic wave function with the respect of the variable ω  ˆ q = kf − n. c As the final result we can write the differential cross section as (ˆ · kf )2 Z 5 dσ 1 = 32e2 kf   . dΩ me cω a50 (Z/a0 )2 + q 2 4 ˆ and n ˆ kz ˆ , the differential cross section can If now ˆ k x be written using the polar angle θ, the azimuthal angle φ and the relations kf 2

(ˆ · kf )

q2

= kf (sin θ cos φ, sin θ sin φ, cos θ) = kf2 sin2 θ cos2 φ  ω 2 ω = kf2 − 2kf cos θ + . c c

Relativistic quantum mechanics Classical fields

Here Li is the linear Lagrangian density. We go to continuum by substituting the limits

L = L(qi , q˙i ) = T − V of classical mechanics does not depend explicitely on time. From the Hamilton variation principle Z t2 δ L(qi , q˙i ) dt = 0, δqi (t) t=t = 0

Now

Z

where

The Hamiltonian function of the Hamiltonian mechanics is X H= pi q˙i − L, i

where the canonically conjugated momentum pi of qi is

t1

  Z t2 Z ∂η dx L η, η, dt ˙ =δ ∂x t    Z 1Z ∂L ∂η ∂L δη + δ = dt dx ∂η ∂(∂η/∂x) ∂x   ∂L ∂η + δ ∂(∂η/∂t) ∂t  Z Z ∂L ∂L ∂ = dt dx + (δη) ∂η ∂(∂η/∂x) ∂x  ∂ ∂L (δη) + ∂(∂η/∂t) ∂t    Z Z ∂L ∂ ∂L = dt dx − ∂η ∂x ∂(∂η/∂x)   ∂L ∂ − δη. ∂t ∂(∂η/∂t)

The equations of motion are now

p˙i

∂H ∂pi ∂H = − . ∂qi =

Many body system

- h

i

(i-1 )a

h

- k (h

)

i-1

ia

i+ 1

)

(i+ 1 )a

h

i-1

- h i

i

h

i+ 1

To get the variation to vanish for all δη one must satisfy the Euler-Lagrange equation ∂ ∂L ∂ ∂L ∂L + − = 0. ∂x ∂(∂η/∂x) ∂t ∂(∂η/∂t) ∂η

We consider N identical particles coupled to eachother by identical parallel springs. The Lagrangian of the system is L = T −V N  1 X 2 mη˙ i − k(ηi+1 − ηi )2 , = 2 i when ηi is the deviation of the particle i from its equilibrium position ia. We write this as N

 1 X 2 mη˙ i − k(ηi+1 − ηi )2 L = 2 i "  2 # N X 1 m 2 ηi+1 − ηi = a η˙ − ka 2 a i a i N X i

aLi .

"  2 # 1 ∂η L= µη˙ 2 − Y . 2 ∂x

In the continuous case the Hamiltonian variation principle takes the form Z t2 δ L dt

∂L . pi = ∂ q˙i

q˙i

L dx,

L=

one can derive the equations of motion   d ∂L ∂L − = 0. dt ∂ q˙i ∂qi

=

∂η ∂x → Y = Young’s modulus. →

1,2

t1

- k (h

→ dx → µ = linear mass density

a m a ηi+1 − ηi a ka

We suppose that the Lagrange function

When

"  2 # ∂η 1 2 L= µη˙ − Y , 2 ∂x

then ∂L ∂η ∂ ∂L ∂x ∂(∂η/∂x) ∂ ∂L ∂t ∂(∂η/∂t)

=

0

∂ ∂η ∂2η = −Y ∂x ∂x ∂x2 2 ∂ η = µ 2. ∂t = −Y

Substituting into the Euler-Lagrange equation we get Y

∂2η ∂2η = 0, 2 −µ ∂x ∂t2

which describes p a wave progating in one dimension with the velocity Y /µ. We define the canonically conjugated momentum π=

∂L ∂ η˙

and the Hamiltonian density H = ηπ ˙ − L. Now π = µη, ˙ so H

"  2 # 1 ∂η = µη˙ 2 − µη˙ 2 − Y 2 ∂x  2 1 2 1 ∂η = . µη˙ + Y 2 2 ∂x

The Lagrangian formalism generalizes easily to three dimensions. The Euler-Lagrange equation takes then the form 3 X ∂ ∂ ∂L ∂L ∂L + − = 0. ∂xk ∂(∂φ/∂xk ) ∂t ∂(∂φ/∂t) ∂φ

k=1

Covariant formulation

We employ the metrics where the components of a four-vector bµ are bµ = (b1 , b2 , b3 , b4 ) = (b, ib0 ). In particular the coordinate four-vector is xµ = (x1 , x2 , x3 , x4 ) = (x, ict). Under Lorentz transformations the coordinate vector transforms according to the equation x0µ = aµν xν . The coefficients of the Lorentz transformation satisfy the orthogonality condition aµν aµλ = δνλ ,

(a−1 )µν = aνµ ,

so that xµ = (a−1 )µν x0ν = aνµ x0ν . We define the four-vector so that under Lorentz transformations it transforms like xµ . Now ∂ ∂xν ∂ ∂ = aµν , 0 = 0 ∂xν ∂xµ ∂xµ ∂xν so the four-gradient ∂/∂xµ is a four-vector. The scalar product of the four-vectors b and c, b · c = bµ cµ =

3 X

bj cj + b4 c4

j=1

= b · c − b0 c0 ,

is invariant under Lorentz transformations. Using the four-vector notation the Euler-Lagrange equation can be written into the compact form   ∂L ∂L ∂ − = 0. ∂xµ ∂(∂φ/∂xµ ) ∂φ We see that the field equation derived from the Lagrangian density L is covariant provided that the Lagrangian density L itself is relativistically scalar (invariant under Lorentz transformations).

Klein-Gordon’s equation

so the solution describes a free field. We include the term

We consider the scalar field φ(x) which, according to its definition, behaves under Lorentz transformation like

Lint = −φρ, where ρ is the (usually position dependent) density of the source. The field equation is now

φ0 (x0 ) = φ(x). Now

φ − µ2 φ = ρ.

L = L(φ, ∂φ/∂xµ ). When we choose

Since we want • the Lagrangian density to be invariant under Lorentz transformations • a linear wave equation, the Lagrangian density can contain only the terms φ2 and

∂φ ∂φ . ∂xµ ∂xµ

One possible form for the Lagrangian density is   1 ∂φ ∂φ 2 2 L=− +µ φ . 2 ∂xµ ∂xµ Substituting this into the Euler-Lagrange equation   ∂L ∂L ∂ − = 0, ∂xµ ∂(∂φ/∂xµ ) ∂φ we get −



∂ ∂xµ

∂φ ∂xµ



+ µ2 φ = 0.

ρ = Gδ(x) and seek for a stationary solution we end up with the equation (∇2 − µ2 )φ = Gδ(x). We substitute φ using its Fourier transform Z 1 ˜ φ(x) = d3 keik·x φ(k), (2π)2/3 where ˜ φ(k) =

Z

1 (2π)3/2

d3 x e−ik·x φ(x).

We end up with the algebraic equation ˜ (−k2 − µ2 )φ(k) =

G (2π)3/2

of the Fourier components. Its solution is ˜ φ(k) =−

1 G . (2π)2/3 k 2 + µ2

Taking the Fourier transform we get the solution

If we employ the notation 1 ∂2 = ∇ − 2 2, c ∂t

φ(x) = −

2

we end up with the Klein-Gordon equation φ − µ2 φ = 0.

G e−µr , 4π r

known as the Yukawa potential. Let’s suppose that the meson field of a nucleon at the point x1 satisfies the equations (∇22 − µ2 )φ = Gδ(x1 − x2 ). Its solution is thus the Yukawa potential.

Heuristic derivation

We substitute into the relativistic energy-momentum relation E 2 − |p|2 c2 = m2 c4

φ(x2 ) = −

G e−µ|x2 −x1 | . 4π |x2 − x1 |

Because the Hamiltonian density was

the operators E 7→ i¯h

∂ , ∂t

pk 7→ −i¯ h

∂ , ∂xk

H = ηπ ˙ − L, the Hamiltonian density of the interaction is

and get  −

∂2 m2 c2 2 2 2 +∇ − c ∂t h2 ¯

When we set µ=

Hint = −Lint

 φ = 0.

mc 1 , [µ] = , h ¯ length

we end up with the Klein-Gordon equation. There are no sources in the Lagrangian density   1 ∂φ ∂φ 2 2 L=− +µ φ 2 ∂xµ ∂xµ

and the total interaction Hamiltonian Z Z 3 Hint = Hint d x = φρ d3 x. We see that the interaction energy of nucleons located at the points x1 and x2 is (1,2)

Hint

=−

G e−µ|x2 −x1 | . 4π |x2 − x1 |

Note Unlike in the Coulomb case, this interaction is

atractive and short ranged. In the reality there are 3 mesons (π + , π 0 , π − ), with different charges but with (almost) equal masses, consistent with the thory. We expand our theory so that we consider two real fields, φ1 and φ2 , for two particles with equal masses. From these we construct the complex fields φ = φ∗

=

Thus we get ∂sµ = 0, ∂xµ  sµ = i

∂φ∗ ∂φ φ − φ∗ ∂xµ ∂xµ

 .

We see that

The Lagrangian density for the free fields can be written using either the complex or real fields:     1 ∂φ1 ∂φ1 1 ∂φ2 ∂φ2 2 2 2 2 L = − + µ φ1 − + µ φ2 2 ∂xµ ∂xµ 2 ∂xµ ∂xµ   ∗ ∂φ ∂φ + µ2 φ∗ φ . = − ∂xµ ∂xµ Considering the fields φ and φ∗ independent we get two Euler-Lagrange equations =

0

=

0,

which can be further written as two Klein-Gordon equations φ∗ − µ2 φ∗ = 0 φ − µ2 φ = 0. We define the first order gauge transformation so that the fields transform under it like φ0 φ∗ 0

δL = 0.

where

φ1 + iφ2 √ 2 φ1 − iφ2 √ . 2

∂L ∂ ∂L − ∂xµ ∂(∂φ/∂xµ ) ∂φ ∂ ∂L ∂L − ∂xµ ∂(∂φ∗ /∂xµ ) ∂φ∗

In a small neighborhood of the solutions φ and φ∗ the Lagrangian density is invarinat so we must have

= eiλ φ = e−iλ φ∗ ,

when λ is a real parameter. Let λ be now an arbitrary, infinitesimally small, number. Then δφ = iλφ δφ∗ = −iλφ∗ . The Lagrangian density transforms then as    ∂L ∂φ ∂L δφ + δ δL = ∂φ ∂(∂φ/∂xµ ) ∂xµ   ∗  ∂L ∗ ∂L ∂φ + δφ + δ ∂φ∗ ∂(∂φ∗ /∂xµ ) ∂xµ    ∂L ∂ ∂L = − δφ ∂φ ∂xµ ∂(∂φ/∂xµ )    ∂L ∂ ∂L + − δφ∗ ∂φ∗ ∂xµ ∂(∂φ∗ /∂xµ )   ∂L ∂L ∂ ∗ + δφ + δφ ∂xµ ∂(∂φ/∂xµ ) ∂(∂φ∗ /∂xµ )  ∗  ∂ ∂φ ∂φ = −iλ φ − φ∗ . ∂xµ ∂xµ ∂xµ

• a complex field φ is associated with a conserved four-vector density sµ , • if we exchange φ ←→ φ∗ , then sµ ←→ −sµ . We interpret this so that • sµ is the charge current density, • φ carries the charge e, • φ∗ carries the charge −e, • the previous real field corresponds to neutral mesons.

Photons

where

We consider a radiation field whose vector potential A satisfies the transversality condition ∇ · A = 0.

ω = |k|c. Now A(x, t) 1 XX =√ (ck,α (t)ˆ(α) eik·x + c∗k,α (t)ˆ(α) e−ik·x ) V k α 1 XX =√ (ck,α (0)ˆ(α) eik·x + c∗k,α (0)ˆ(α) e−ik·x ), V k α

Because the electric and magnetic fields E B

1 ∂ A c ∂t = ∇×A = −

satisfy the free space Maxwell equations ∇·E ∇×E ∇·B ∇×B

=

0

∂B = − ∂t = 0 1 ∂E , = c ∂t

the vector potential satisfies the wave equation ∇2 A −

where we have employed the four-vector notation

1 ∂2A = 0. c2 ∂t2

We write the vector potential at the moment t = 0 as a superposition of the periodically normalized plane waves in an L-sided cube,

k · x = k · x − ωt = k · x − |k|ct. The Hamiltonian function of the classical radiation field is Z 1 H = (|B|2 + |E|2 ) d3 x 2 Z  1  = |∇ × A|2 + |(1/c)(∂A/∂t)|2 d3 x. 2 A straightforward calculation shows that XX H= 2(ω/c)2 c∗k,α ck,α . k α Because the coefficients

uk,α (x) = ˆ(α) eik·x , like:

ck,α (t) = ck,α (0)e−iωt satisfy the equation of motion

A(x, t)|t=0

1 X X = √ (ck,α (0)uk,α (x) V k α=1,2 +c∗k,α (0)u∗k,α (x)).

Here V = L3 and ˆ(α) , α = 1, 2 are real polarization vectors. Due to the transversality condition we have

c¨k,α = −ω 2 ck,α , it would look like the classical radiation field were composed of independent harmonic oscillators. We define the variables Qk,α Pk,α

ˆ(α) · k = 0, so the polarization can chosen so that the vectors (ˆ(1) , ˆ(2) , k/|k|) form a righthanded rectangular coordinate system. The Fourier components uk,α satisfy the orthogonality conditions Z 1 d3 x uk,α · u∗k0 ,α0 = δkk0 δαα0 . V Due to the periodicity conditions the wave vectors can take the values kx , ky , kz = 2nπ/L,

n = ±1, ±2, . . . .

At the moment t 6= 0 the vector potential is obtained simply by setting ck,α (t) c∗k,α (t)

= ck,α (0)e−iωt = c∗k,α (0)eiωt ,

1 (c + c∗k,α ) c k,α iω = − (ck,α − c∗k,α ). c =

With the help of these the Hamiltonian function can be written as XX 1 H= (Pk2 ,α + ω 2 Q2k,α ). 2 k α Since ∂H ∂Qk,α ∂H ∂Pk,α

= −P˙k,α =

+Q˙ k,α ,

the variables Pk,α and Qk,α are canonically conjugated and the Hamiltonian function the sum of the total energies of the corresponding harmonic oscillators. Thus the classical radiation field can be thought to be a collection of independent harmonic oscillators. There

• every oscillator is characterized by the wave vector k and the polarization ˆ(α) ,

Thus we can write a† |nk ,α i = c+ |nk ,α + 1i k,α ak,α |nk ,α i = c− |nk ,α − 1i.

• the dynamic variables of every oscillator are combinations of Fourier coefficients. We quantize these oscillators by postulating that Pk,α and Qk,α are not any more pure numbers but operators which satisfy the canonical commutation rules [Qk,α , Pk ,α0 ] = i¯ hδkk δ [Qk,α , Qk0 ,α0 ] = 0 0

0

[Pk,α , Pk0 ,α0 ]

=

√

a† k,α

=

√

2¯ hω 1 2¯ hω

(ωQk,α + iPk,α ) (ωQk,α − iPk,α ).

It is easy to see that they satisfy the commutation relations [ak,α , a† 0 0 ] k ,α

= δkk0 δαα0

[ak,α , ak0 ,α0 ]

=

[a† , a† 0 0 ] = 0. k,α k ,α

Note In these relations the operators must be evaluated

at the same moment, i.e. [ak,α , a† 0 0 ] stands in fact for k ,α the commutator [ak,α (t), a† 0 0 (t)]. k ,α We further define the Hermitean operator Nk,α = a† ak,α . k,α It is easy to see that

†

[a , Nk0 ,α0 ] k,α

= δkk0 δαα0 ak,α = −δkk0 δαα0 a† . k,α

Due to the Hermiticity the eigenvalues nk ,α of the operator Nk,α are real and the eigenvectors Nk,α |nk ,α i = nk ,α |nk ,α i form an orthonormal complete basis. With the help of the commutation rule [a† , Nk0 ,α0 ] = −δkk0 δαα0 a† k,α k,α we see that Nk,α a† |nk ,α i = k,α =

(a† Nk,α + a† )|nk ,α i k,α k,α (nk ,α + 1)a† |nk ,α i. k,α

Similarly we can show that Nk,α ak,α |nk ,α i = (nk ,α − 1)ak,α |nk ,α i.

= |c+ |2 hnk ,α + 1|nk ,α + 1i = hnk ,α |ak,α a† |nk ,α i k,α = hnk ,α |Nk,α + [ak,α , a† ]|nk ,α i k,α = nk ,α + 1,

= 0.

1

[ak,α , Nk0 ,α0 ]

|c+ |2

αα0

We define dimensionless combinations ak,α and a† of k,α the operators Pk,α and Qk,α as ak,α

Because the states |nk ,α i are normalized we can calculate the coefficients as

|c− |2

= hnk ,α |a† ak,α |nk ,α i = nk ,α . k,α

We choose the phase of the coefficients so that at the moment t = 0 we have q a† |nk ,α i = nk ,α + 1|nk ,α + 1i k,α p nk ,α |nk ,α − 1i. ak,α |nk ,α i = Because nk ,α = hnk ,α |Nk,α |nk ,α i = hnk ,α |a† ak,α |nk ,α i k,α and because the norm of a vectors is always non-negative we must have nk ,α ≥ 0. From this we can deduce that the only possible eigenvalues are nk ,α = 0, 1, 2, . . . . We interprete • the state |nk ,α i to contain exactly nk ,α photons, each of which is characterized by a wave vector k and a polarization ˆ(α) . • the operator a† to create a photon with the wave k,α vector k and the polarization ˆ(α) . • the operator ak,α to destroy a photon with the wave vector k and the polarizartion ˆ(α) . • the operator Nk,α to count the number of photons with the wave vector k and the polarization ˆ(α) in the state The state composed of various kind of photons is a direct product of individual vectors |nki ,αi i: |nk1 ,α1 , nk2 ,α2 , . . . , nki ,αi , . . .i = |nk1 ,α1 i ⊗ |nk2 ,α2 i ⊗ · · · ⊗ |nki ,αi i ⊗ · · · . The vector |0i stands for the state that has no kind of photons, i.e. |0i = |0k1 ,α1 i ⊗ |0k2 ,α2 i ⊗ · · · .

Application of any operator ak,α onto this results always zero. We say that |0i represents the vacuum. It is easy to see that a general normalized photon state can be constructed applying operations a† k,α consecutively:

where ω = |k|c. When we choose the energy scale so that H|0i = 0,

n

|nk1 ,α1 , nk2 ,α2 , . . .i =

† Y (ak ,α ) ki ,αi qi i |0i. n ! ki ,αi ki ,αi

†

and a† 0 0 commute the k,α k ,α order of operators does not matter. The many photon states are symmetric with respect to the exchange of photons. We say that the photons obey the Bose-Einstein statistics or that they are bosons. Since the numbers nk ,α tell us the number of photons of type (k, α) in the volume under consideration we call them the occupation numbers of the state. Correspondingly the space spanned by the state vectors is called the occupation number space. In the quantum theory the Fourier coefficients of a classical radiation field must be replaced by the corresponding non-commuting creation and annihilation operators. Substituting p ck,α 7→ c ¯h/2ω ak,α (t) p c∗k,α 7→ c ¯h/2ω a† (t) k,α Note Since the operators a

the Hamiltonian takes the form XX H= ¯hωNk,α . α k When it acts on a many photon state the result is H|nk1 ,α1 , nk2 ,α2 , . . .i X = nki ,αi ¯hωi |nk1 ,α1 , nk2 ,α2 , . . .i. i

The quantum mechanical momentum operator is exactly of the same form as the classical function (the Poynting vector): Z 1 (E × B) d3 x P = c XX 1 = ¯hk(a† ak,α + ak,α a† ) k,α k,α 2 k α XX 1 = ¯hk(Nk,α + ). 2 k α

we get

Since here the summation goes over all wave vectors the term associated with the factor 1/2 will not appear in the h 1 X ¯ h ak,α (t)ˆ(α) eik·x final result the terms ¯hk and −¯hk cancelling each other. A(x, t) = √ c 2ω V For the momentum operator we get thus k,α i XX +a† (t)ˆ(α) e−ik·x . P = ¯hkNk,α . k,α α k Note Here A is an operator defined at every point of the space whereas A of the classical theory is a three For one photon states we have component field defined at every point. The variables x Ha† |0i = h ¯ ωa† |0i and t are both in classical and quantum mechanical cases k,α k,α variables parametrizing the fields. Fields like the operator † † P a |0i = h ¯ ka |0i, A are called field operators or quantized fields. k,α k,α Also in the quantum theory the Hamiltonian is of the so form Z 1 H= (B · B + E · E) d3 x. ¯hω = ¯h|k|c = photon energy 2 ¯hk = photon momentum. Substituting the field operator A into the equations r

E B

1 ∂ A c ∂t = ∇×A

= −

and noting that this time the Fourier coefficients do not commute we get H

1 XX hω(a† ak,α + ak,α a† ) ¯ k,α k,α 2 k α XX 1 = (Nk,α + )¯ hω, 2 α k

=

The photon mass will be (mass)2

1 2 (E − |p|2 c2 ) c4 1 = [(¯ hω)2 − (¯ h|k|c)2 ] c4 = 0. =

The photon state is also characterized by its polarization ˆ(α) . Since ˆ(α) transforms under rotations like a vector the photon is associated with one unit of angular momentum, i.e. the spin angular momentum of the

photon is one. We define the circularly polarized combinations 1 ˆ(±) = ∓ √ (ˆ(1) ± iˆ(2) ). 2 We rotate these vectors by an infinitesimal angle δφ around the progation direction k. Their changes are δφ = ∓ √ (ˆ(2) ∓ iˆ(1) ) 2 = ∓iδφ ˆ(±) .

δˆ(±)

We select the propagation direction k as the quantization axis and compare this expression with the transformation properties of angular momentum eigenstates   i |jmiR = 1 − Jz δφ |jmi = (1 − im δφ)|jmi. ¯h We see that • the spin components of the polarizations ˆ(±) are m = ±1. • the state m = 0 is missing due to the transversality condition. • our original linear polarization states are 50/50 mixtures of m = 1 and m = −1 states. Hence the photon spin is always either parallel or antiparallel to the direction of the propagation. The operators ak,α and a† are time dependent and so k,α they satisfy the Heisenberg equations of motion a˙ k,α

i [H, ak,α ] ¯ h i XX [¯ hω 0 Nk0 ,α0 , ak,α ] = h 0 0 ¯ k α = −iωak,α =

like also †

†

a˙ = iωa . k,α k,α These equations have solutions ak,α

= ak,α (0)e−iωt

a† k,α

= a† (0)eiωt . k,α

The final form of the field operator is then r 1 X h h ¯ A(x, t) = √ c ak,α (0)ˆ(α) eik·x−iωt 2ω V k,α i † + a (0)ˆ(α) e−ik·x+iωt . k,α We should note that • the operator A is Hermitean.

• x and t in the expression for the field operator A are not quantum mechanical variables but simply parameters which the operator A depends on. For example, it is not allowed to interprete the variables x and t as the space-time coordinates of a photon. • the quantized field A operates at every point (x, t) of the space where it with a certain probability creates and annihilates excitation states called photons. Thus photons can be interpreted as the quantum mechanical excitations of the radiation field. We consider photon emission and and absorption of non relativistic atomic electrons. The relevant interaction Hamiltonian is of the form X e A(xi , t) · pi Hint = − m ec i  e2 + 2 2 A(xi , t) · A(xi , t) , 2me c where the transversality condition is taken into account by replacing the operator pi · A with the operator A · pi . The summation goes over all electrons participating in the process. The symbols xi stand for their position coordinates. Note If we had to consider the interaction of spin and radiation we should also include the term (spin)

Hint

=−

X e¯h σ i · [∇ × A(x, t)]|x=xi . 2me c i

This time the Hamiltonian operator Hint operates not only on the atomic states but also on the photon states. In the quantum theory of radiation • the vector describing the initial state |ii is the direct product of an atomic state A and a (many) photon state characterized by the occupation numbers nk ,α : |ii = |Ai ⊗ |nk ,α i = |A; nk ,α i. • the vector describing the final state |f i is the direct product of an atomic state B and a (many) photon state characterized by the occupation numbers nk0 ,α0 : |f i = |Ai ⊗ |nk0 ,α0 i = |A; nk0 ,α0 i.

Absorption

Now |ii = |A; nk ,α i |f i = |B; nk ,α − 1i. In the first order perturbation theory the amplitude of the process |ii −→ |f i is the matrix element of the interaction operator HI between the states |ii and |f i. Up to this order

• only ak,α leads to a nonzero matrix element, eventhough the field operator A is a linear superposition of creation and annihilation operators a† and ak,α , respectively. k,α • the term A · A is out of the question in this process because it either changes the number of photons by two or does not change it at all. The first order transition matrix element is now hB; nk ,α − 1|Hint |A; nk ,α i e =− hB; nk ,α − 1| me c r X ¯h c ak,α (0)eik·xi −iωt pi · ˆ(α) |A; nk ,α i 2ωV i s nk ,α ¯h X e hB|eik·xi pi · ˆ(α) |Aie−iωt . =− me 2ωV i Comparing this with the matrix element of the semiclassical perturbation potential  eA0  i(ω/c)n † ˆ ·x ˆ · p Vni =− e me c ni we see that they both give exactly the same result provided we use in the semiclassical theory the equivalent radiation field (abs) ik·x−iωt

A(abs) = A0

e

,

(abs)

A0

=c

nk ,α ¯ h 2ωV

hB; nk ,α + 1|Hint |A; nk ,α i s (nk ,α + 1)¯ hX e =− hB|e−ik·xi pi · ˆ(α) |Aieiωt . me 2ωV i If nk ,α is very large then q p nk ,α + 1 ≈ nk ,α , and the semiclassical and quantum mechanical treatment coincide. If nk ,α is small the semiclassical method fails completely. In particular, the semicalssical treatment of the spontaneous emission, nk ,α = 0, is impossible. The semiclassical method can be applied if we insert the atom into the fictitious radiation field (emis) −ik·x+iωt

A(emis) = A0 where

s (emis)

A0

=c

e

(nk ,α + 1)¯ h 2ωV

,

ˆ(α) .

The field A(emis) is not • directly proportional to the number of photons nk ,α , • the complex conjugate of the field A(abs) . Example Spontaneous emission from the state A to the state B. In the first order the transition rate is

where the amplitude is s

is a† which adds one photon to the final state. The k,α relevant matrix element is now

ˆ(α) .

Because the transition probability is • according to the semiclassical theory directly proportional to the intensity of the radiation,

wA→B 2π |hB; 1k,α |Hint |A; 0i|2 δ(EB − EA + ¯ hω) ¯h 2 2π e2 ¯h X (α) = hB|e−ik·xi ˆ · pi |Ai 2 ¯h 2me ωV =

i

|A0 |2 ∝ nk ,α , • according to the quantum theory directly proportional to the occupation number nk ,α , both the semiclassical and quantum mechanical results give equivalent results also at low intensities, i.e. when nk ,α is small. Emission

Now |ii = |A; nk ,α i |f i = |B; nk ,α + 1i and in the first order the only potential term of the field r h h ¯ 1 X A(x, t) = √ c ak,α (0)ˆ(α) eik·x−iωt 2ω V k,α i † + a (0)ˆ(α) e−ik·x+iωt k,α

×δ(EB − EA + h ¯ ω). Like in the photoelectric efect we can deduce that the number of the allowed photon states ρ(E, dΩ) in the energy interval (¯hω, ¯hω + d(¯ hω)) and in the solid angle dΩ is V ω2 ρ(E, dΩ) = n2 dn dΩ = d(¯ hω) dΩ. (2π)3 ¯hc3 The transition rate of photons emitting into a certain solid angle is thus 2 V ω 2 dΩ 2π e2 ¯h X −ik·xi (α) ˆ hB|e  · p |Ai wdΩ = , i (2π)3 ¯hc3 ¯h 2m2e ωV i where h ¯ ω = E A − EB . We consider only hydrogen like atoms so that only one electron participates in the process and we restrict to the dipole approximation. Then wdΩ =

e2 ω |hB|p|Ai · ˆ(α) |2 dΩ. 8π 2 m2e ¯hc3

Earlier we saw that ime (EB − EA ) hB|x|Ai ¯h = −ime ωxBA .

hB|p|Ai =

We let the symbol Θ(α) stand for the angle between the vector xBA and the polarization direction ˆ(α) , i.e. cos Θ(1) cos Θ(2)

= =

sin θ cos φ sin θ sin φ,

• in the term A · A creation and annihilation operators appear as quadratic, only the quadratic term A · A contributes in the first order perturbation theory. Only two of the terms of the form aa† , a† a, aa, a† a† in the operator A · A have non zero matrix elements provided that 0

when θ and φ are the direction angles of the vector x. Then e2 ω 3 |xBA |2 cos2 Θ(α) dΩ. wdΩ = 8π 2 ¯ hc3 The total transition rate is obtained by integrating over all propagation directions k/|k| and summing over both polarizations: e2 ω 3 w= |xBA |2 . 3π¯hc3 The life time of a state was obtained from the formula X 1 = wA→Bi , τA i where we have to sum also over the magnetic quantum numbers m. For example the life time of the hydrogen 2p state is τ (2p −→ 1s) = 1.6 × 10−9 s. Electron photon scattering

We consider the process |1k,α i −→ |1k0 ,α0 i, i.e.

• a† creates a photon of the type (k0 , ˆ(α ) ), • a annihilates a photon of the type (k, ˆ(α ), and then h1k0 ,α0 |ak,α a† 0 0 |1k,α i = 1. k ,α Now hB; 1k0 ,α0 |Hint |A; 1k,α i 2 e = hB; 1k0 ,α0 2 A(x, t) · A(x, t) A; 1k,α i 2me c 2 e = hB; 1k0 ,α0 (a a† 0 + a† 0 0 ak,α ) k ,α 2me c2 k,α k ,α0 0 0 0 c2 ¯h ˆ(α) · ˆ(α ) ei(k−k )·x−i(ω−ω )t A; 1k,α i × √ 2V ωω 0 0 0 e2 c2 ¯h √ = 2ˆ(α) · ˆ(α ) e−i(ω−ω )t hB|Ai, 2 2me c 2V ωω 0 where again the exponential functions e±k·x are replaced by the constant 1 (the long wave length approximation). In the first order we have thus Z i t iωf i t0 e Vf i (t0 ) dt0 c(1) (t) = − ¯h t0 =

• before the scattering the atom is in the state A, and k and ˆ(α) are the wave vector and polarization of the incoming photon. • after the scattering the atom is in the state B, k0 is 0 the wave vector and ˆ(α ) the polarization vector of the outgoing photon. The Hamiltonian of the interaction is Hint = −

e2 e A(x, t) · p + A(x, t) · A(x, t). me c 2m2e c2

Because • the number of photons does not change in the scattering, • in order to be non zero the matrix element of the interaction must contain products of photon creation and annihilation operators, • in the term A · p creation and annihilation operators appear as linear,

0 1 e2 c2 ¯h √ 2δAB ˆ(α) · ˆ(α ) 2 0 i¯h 2me c 2V ωω Z t 0 0 × ei(¯hω +EB −¯hω−EA )t /¯h dt0 ,

0

where ω = |k|c and ω 0 = |k0 |c. Now • in the transition amplitude c(1) (t) the interaction is in fact of second order: A · A. • in the second order correction c(2) (t) the term A · p is also of second order. To collect all contributions up to the second order in the interaction we have to consider also the correction c(2) (t), into which we take all double actions of the operator A · p. Now (2)

c

 (t)

=

i − ¯h

2 X Z m

t

dt

0

Z

t0

t0

0

dt00 eiωf m t Vf m (t0 )

t0

00

×eiωmi t Vmi (t00 ). Thus there are two possibilities: the interaction A · p can

• at the moment t1 annihilate the incoming photon (k, ˆ(α) ) and at some later time t2 create the 0 outgoing photon (k0 , ˆ(α ) ) or

photons c/V , so the differential cross section is  0 0 ω dσ 2 δAB ˆ(α) · ˆ(α ) = r0 dΩ ω 0  • at the moment t1 create the outgoing photon 1 X (p · ˆ(α ) )BI (p · ˆ(α) )IA 0 (α0 ) − (k , ˆ ) and at some later time t2 annihilate the me EI − EA − ¯hω I incoming photon (k, ˆ(α) ). 0  2 (p · ˆ(α) )BI (p · ˆ(α ) )IA + Between the moments t1 and t2 the atom is in the state I, , EI − EA + ¯hω 0 which normally is neither of the states A and B. In the intermediate state I there are thus two where r0 ≈ 2.82 × 10−13 cm is the classical radius of possibilities: either there are no photons present or both electron. This expression is known as the incoming and outgoing photons are present. We get thus Kramers-Heisenberg formula. (in the dipole approximation) Example Elastic scattering. Now A = B ja h ¯ω = h ¯ ω 0 . Using the commutation  2 Z t Z t2 2 c ¯ h e 1 relations of the operators x and p, the completeness of √ − dt2 dt1 c(2) (t) = 2 0 m c the intermediate states and the relation (i¯h) 2V ωω e 0 0 X 0 0 pAB = ime ωAB xAB × hB|p · ˆ(α ) |Iiei(EB −EI +¯hω )t2 /¯h I

we can write

×hI|p · ˆ(α) |Aiei(EI −EA −¯hω)t1 /¯h +

X

(α)

hB|p · ˆ

0

ˆ(α) · ˆ(α )

i(EB −EI +¯ hω)t2 /¯ h

|Iie

0 1 Xh (x · ˆ(α) )AI (p · ˆ(α ) )IA i¯h I i 0 −(p · ˆ(α) )AI (x · ˆ(α ) )IA 0 1 X 2 (p · ˆ(α) )AI (p · ˆ(α ) )IA , me ¯h ωIA

=

I 0

0

×hI|p · ˆ(α ) |Aiei(EI −EA −¯hω )t1 /¯h

 =

2  c2 ¯h e √ = − i¯h2V ωω 0 me c X (p · ˆ(α0 ) )BI (p · ˆ(α) )IA × EI − E A − ¯ hω i ! 0 (p · ˆ(α) )BI (p · ˆ(α ) )IA + EI − EA + ¯ hω 0 Z t 0 × dt2 ei(EB −EA +¯hω −¯hω)t2 /¯h .

I

where ωIA = (EI − EA )/¯h. We see that 0

δAA ˆ(α) · ˆ(α ) " 0 1 X (p · ˆ(α ) )AI (p · ˆ(α) )IA − me ¯h ωIA − ω I # 0 (p · ˆ(α) )AI (p · ˆ(α ) )IA + ωIA + ω " 0 1 X ω(p · ˆ(α ) )AI (p · ˆ(α) )IA =− me ¯h ωIA (ωIA − ω) I # 0 ω(p · ˆ(α) )AI (p · ˆ(α ) )IA − . ωIA (ωIA + ω)

0

For the transition rate we get combining the terms c(1) (t) and c(2) (t) and taking into account the relation Z t 2 ixt0 0 lim e dt = 2πtδ(x) t→∞ 0

the expression Z wdΩ = (|c(1) + c(2) |2 /t)ρ(E, dΩ) dE =

2π ¯h



c2 ¯ h √ 2V ωω 0

2 

e2 me c2

2

If ω is small then 1 1 ± (ω/ωIA ) ≈ . ωIA ∓ ω ωIA 2

V ω0 dΩ (2π)3 ¯hc3

0 × δAB ˆ(α) · ˆ(α ) 0  1 X (p · ˆ(α ) )BI (p · ˆ(α) )IA − me EI − EA − ¯ hω I 0  2 (p · ˆ(α) )BI (p · ˆ(α ) )IA + . EI − EA + h ¯ ω0

Then X 1 h 0 (p · ˆ(α ) )AI (p · ˆ(α) )IA 2 ωIA I i 0 −(p · ˆ(α) )AI (p · ˆ(α ) )IA Xh 0 = m2e (x · ˆ(α ) )AI (x · ˆ(α) )IA I 0

−(x · ˆ(α) )AI (x · ˆ(α ) )IA 0

Because in the initial state there was exactly one photon in the volume V and the flux density of the incoming

= m2e ([x · ˆ(α ) , x · ˆ(α) ])AA = 0.

i

The differential cross section is now  2 X  3 1 dσ r0 ω 4 = dΩ me ¯h ωIA I

(α0 )

×[(p · ˆ

)AI (p · ˆ(α) )IA

(α)

(α0 )

+ (p · ˆ )AI (p · ˆ  r m 2 X 1 0 e = ω 4 ¯h ωIA

2 )IA ]

I

0

×[(x · ˆ(α ) )AI (x · ˆ(α) )IA 2 0 +(x · ˆ(α) )AI (x · ˆ(α ) )IA ] . At long wave lengths the differential cross section obeys the Rayleigh law or dσ 1 ∝ 4. dΩ λ Now • for ordinary colourless gases ωIA corresponds to wave lengths in the ultraviolet, • for the visible light we have then ω  ωIA , so our approximations are valid in the atmossphere. The theory explains why the sky is blue and the sunset red.

Dirac’s equation We construct relativistically covariant equation that takes into account also the spin. The kinetic energy operator is H (KE) =

p2 . 2m

Previously we derived for Pauli spin matrices the relation (σ · a)2 = |a|2 , so we can also write H (KE) =

(σ · p)(σ · p) . 2m

Here φ is a two component wave function (spinor). We define new two component wave functions   ∂ 1 (R) i¯h − i¯hσ · ∇ φ φ = mc ∂x0 φ(L)

It is easy to see that these satisfy the set of simultaneous equations   ∂ φ(L) = −mcφ(R) i¯hσ · ∇ − i¯h ∂x0   ∂ −i¯hσ · ∇ − i¯h φ(R) = −mcφ(L) . ∂x0 We define yet new two component wave functions

However, when the particle moves under the influence of a vector potential these expressions differ. Substituting

ψA ψB

p 7→ p − eA/c the latter takes the form     eA eA 1 σ· p− σ· p− 2m c c  2 1 eA = p− 2m c     eA eA i σ· p− × p− + 2m c c  2 1 eA e¯ h = p− − σ · B, 2m c 2mc

= φ.

These in turn satisfy the matrix equation       −i¯h ∂ −i¯hσ · ∇ ψA ∂x0  ψA  = −mc . ψB ψB i¯hσ · ∇ i¯h ∂ ∂x0 We now define the four component wave function    (R)  ψA φ + φ(L) ψ= = ψB φ(R) − φ(L) and the 4 × 4-matrices 

γk where we have used the identities γ4

(σ · a)(σ · b) = a · b + iσ · (a × b) and p × A = −i¯ h(∇ × A) − A × p. Let us suppose that for the relativistically invariant expression (E 2 /c2 ) − p2 = (mc)2 the operator analogy 1 (op) 2 E − p2 = (mc)2 c2

−iσk 0   1 0 = . 0 −1

=

0 iσk



We end up with the Dirac’s equation   mc ∂ ψ+ ψ=0 γ · ∇ + γ4 ∂(ix0 ) ¯h for free spin- 21 particles. Employing the four vector notation the equation takes the form   mc ∂ γµ + ψ = 0. ∂xµ ¯h Note The Dirac equation is in fact a set of four coupled

holds. Here E (op)

∂ ∂ = i¯ h = i¯ hc ∂t ∂x0

and p = −i¯ h∇. We write the operator equation into the form ! ! E (op) E (op) −σ·p + σ · p = (mc)2 c c or 

= φ(R) + φ(L) = φ(R) − φ(L) .

∂ i¯h + σ · i¯h∇ ∂x0

  ∂ i¯ h − σ · i¯ h∇ φ = (mc)2 φ. ∂x0

linear differential equations. The wave function ψ is the four component vector   ψ1  ψ2   ψ=  ψ3  . ψ4 This kind of a four component object is called bispinor or Dirac’s spinor. Explicitely written down the Dirac equation is  4 X 4   mc  X ∂ (γµ )αβ + δαβ ψβ = 0. ∂xµ ¯h µ=1 β=1

Note The fact that the Dirac spinor happens to have

four components has nothing to do with our four dimensional space-time; ψβ does not transform like a four vector under Lorentz transformations. It is easy to verify that the gamma-matrices (Dirac matrices) γµ satisfy the anticommutation rule {γµ , γν } = γµ γν + γν γµ = 2δµν .

Forming the Hermitean conjugate of the Dirac equation   mc ∂ + ψ=0 γµ ∂xµ ¯h we get mc † ∂ ∂ † ψ γk + ∗ ψ † γ4 + ψ = 0. ∂xk ∂x4 ¯h We multiply this from right by the matrix γ4 and end up with the adjungated equation

Furthermore, every 㵠is Hermitian, 㵆 = 㵠,

−

and traceless, i.e. Tr γµ = 0.

mc ¯ ∂ ¯ ψγµ + ψ = 0. ∂xµ ¯h

Here we have used the relations

Let’s multiply the equation   mc ∂ ψ+ ψ=0 γ · ∇ + γ4 ∂(ix0 ) h ¯

∂ ∂x∗4 γk γ4

∂ ∂ =− ∂(ict)∗ ∂x4 = −γ4 γk .

=

on both sides by the matrix γ4 and we get   ∂ c¯hγ4 γ · ∇ − i¯ h ψ + γ4 mc2 ψ = 0. ∂t

Let’s multiply the original Dirac equation   mc ∂ γµ ψ=0 + ∂xµ ¯h

Denote

from left with the adjungated spinor ψ¯ and the adjungated equation



β αk



1 0 0 −1   0 σk = iγ4 γk = , σk 0 = γ4 =

−

∂ ¯ mc ¯ ψγµ + ψ=0 ∂xµ ¯h

from right with the spinor ψ and subtract the resulting equations. We then get

which satisfy the relations {αk , β} = 0 β2 = 1 {αk , αl } = 2δkl . When we now write

∂ ¯ (ψγµ ψ) = 0. ∂xµ The quantity ¯ µ ψ = (cψ † αψ, icψ † ψ) sµ = icψγ

H = −ic¯ hα · ∇ + βmc2 , the Dirac equation takes the familiar form Hψ = i¯ h

∂ψ . ∂t

We define the adjungated spinor ψ¯ like: ψ¯ = ψ † γ4 . Explicitely, if ψ is a column vector   ψ1  ψ2   ψ=  ψ3  , ψ4 then ψ † and ψ¯ are row vectors ψ † = (ψ1∗ , ψ2∗ , ψ3∗ , ψ4∗ ) ψ¯ = (ψ1∗ , ψ2∗ , −ψ3∗ , −ψ4∗ ).

thus satisfies a continuity equation. According to Green’s theorem we have Z Z 3 ¯ ψγ4 ψ d x = ψ † ψ d3 x = constant, where the constant can be taken to be 1 with a suitable ¯ 4 ψ = ψ † ψ is positively normalization of ψ. Because ψγ definite it can be interpreted as a probability density. Then ¯ k ψ = cψ † αk ψ sk = icψγ can be identified as the density of the probability current. Note It can be shown that sµ transforms like a four vector, so the continuity equation is relativistically covariant. It can be proved that any sets of four matrices γµ and γµ0 satisfying the anticommutation relations {γµ , γν } = 2δµν {γµ0 , γν0 } = 2δµν ,

are related to eachother through a similarity transformation with a non-singular 4 × 4-matrix S:

Suppose now that E ≈ mc2 ,

Sγµ S −1 = γµ0 .

|eA0 |  mc2

and measure the energy starting from the rest energy: E (NR) = E − mc2 .

With the help of the matrices γµ0 the original Dirac equation can be written as   mc ∂ + S −1 Sψ = 0. S −1 γµ0 S ∂xµ ¯h

We expand c2 E − eA0 + mc2

Multiplying this from left with the matrix S we get   ∂ mc γµ0 + ψ 0 = 0, ∂xµ h ¯

= =

  1 2mc2 2m 2mc2 + E (NR) − eA0 " # E (NR) − eA0 1 1− + ··· . 2m 2mc2

This can be taken to be the power series in (v/c)2 since E (NR) − eA0 ≈ [p − (eA/c)]2 /2m ≈ mv 2 /2.

where ψ 0 = Sψ. Thus Dirac’s equation is independent on the explicit form of the matrices γµ ; only the anticommutation of the matrices is relevant. If the matrices γµ0 are Hermitean the transformation matrix S can be taken to be unitary. It is easy to show that then the probability density and current, for example, are independent on the representation: ¯ µ ψ. ψ¯0 γµ0 ψ 0 = ψγ Vector potential

When the system is subjected to a vector potential Aµ = (A, iA0 ), we make the ordinary substitutions

Taking into account only the leading term we get     1 eA eA σ· p− σ· p− ψA = (E (NR) − eA0 )ψA . 2m c c This can be written as # "  2 eA e¯h 1 σ · B + eA0 ψA = E (NR) ψA . p− − 2m c 2mc Up to the zeroth order of (v/c)2 the component ψA is thus the two component Schr¨odinger-Pauli wave function 2 (multiplied with the factor e−imc t ) familiar from the non-relativistic quantum mechanics. The equation    eA 1 − σ· p− ψA = − (E − eA0 + mc2 )ψB c c tells us that the component ψB is roughly by the factor |p − (eA/c)|/2mc ≈ v/2c

−i¯h(∂/∂xµ ) 7→ −i¯ h(∂/∂xµ ) − eAµ /c. The Dirac equation takes now the form   ∂ ie mc ψ = 0. − Aµ γµ ψ + ∂xµ ¯hc h ¯ Assuming that Aµ does not depend on time the time dependence of the spinor ψ can be written as ψ = ψ(x, t)|t=0 e−iEt/¯h . Let us write now the Dirac equation for the components ψA and ψB :    eA 1 σ· p− ψB = (E − eA0 − mc2 )ψA c c    eA 1 − σ· p− ψA = − (E − eA0 + mc2 )ψB . c c With the help of the latter equation we eliminate ψB from the upper equation and get        c2 eA eA σ· p− σ · p − ψA c c E − eA0 + mc2

”less” than ψA . Due to this, provided that E ≈ mc2 , ψA is known as the big and ψB as the small component of the Dirac wave function ψ. We obtain relativistic corrections only when we consider the second or higher order terms of the expansion   1 2mc2 c2 = 2m 2mc2 + E (NR) − eA0 E − eA0 + mc2 " # E (NR) − eA0 1 1− = + ··· . 2m 2mc2 Let us suppose now that A = 0. The wave equation is then HA ψA = E (NR) ψA , where 1 HA = (σ · p) 2m

E (NR) − eA0 1− 2mc2

! (σ · p) + eA0 .

This wave equation looks like a time independent Schr¨odinger equation for the wave function ψA . = (E However, − eA0 − mc2 )ψA .

• evaluating corrections up to the order (v/c)2 the wave function ψA is not normalized because the probability interpretation of Dirac’s theory requires that Z † † (ψA ψA + ψB ψB ) d3 x = 1, where ψB already of the order v/c. • explicitely writing down the expression for the operator HA we see that it contains the non-Hermitian term i¯hE · p. • the equation is not an eigenvalue equation since HA itself contains the term E (NR) . Up to the order (v/c)2 the normalization condition can now be written as   Z p2 † ψA d3 x ≈ 1, ψA 1+ 4m2 c2 because according to the equation    eA 1 − σ· p− ψA = − (E − eA0 + mc2 )ψB c c we have

σ·p ψA . 2mc It is worthwhile to define the new two component wave function Ψ: Ψ = ΩψA , ψB ≈

where

p . 8m2 c2

Now Ψ is up to the order (v/c)2 normalized correctly because   Z Z p2 † † 3 ψA d3 x. Ψ Ψ d x ≈ ψA 1 + 4m2 c2 We multiply the equation HA ψA = E (NR) ψA , on both sides with the operator Ω−1 = 1 − (p2 /8m2 c2 ), and get Ω−1 HA Ω−1 Ψ = E (NR) Ω−2 Ψ. Explicitely, up to the order (v/c)2 this can be written as 

p2 + eA0 − 2m

 p2 + eA0 2m !  (NR) (σ · p) E − eA0 − (σ · p) Ψ 2m 2mc2   p2 (NR) =E 1− Ψ. 4m2 c2 

p2 , 8m2 c2



p2 p4 + eA0 − 2m 8m3 c2

1  2 {p , (E (NR) − eA0 )} 8m2 c2  (NR) −2(σ · p)(E − eA0 )(σ · p) Ψ +

= E (NR) Ψ. Because for arbitrary operators A and B {A2 , B} − 2ABA = [A, [A, B]] holds we can, by setting

E

(NR)

σ·p = A − eA0 = B,

reduce the equation into the form  2 p p4 + eA0 − 2m 8m3 c2  e¯h2 e¯hσ · (E × p) − ∇·E Ψ − 4m2 c2 8m2 c2 = E (NR) Ψ. In the derivation of the equation we have employed the relations [σ · p, (E (NR) − eA0 )] = −ie¯hσ · E [σ · p, −ie¯hσ · E] = −e¯h2 ∇ · E −2e¯hσ · (E × p),

2

Ω=1+

Writing E (NR) p2 in the form 12 {E (NR) , p2 } we get



the validity of which can be verified by noting that ∇A0 ∇×E

= −E = 0.

The resulting equation is a proper Schr¨odinger equation for a two component wave function. Physical interpretation We look at the meaning of the terms in the equation  2 p4 p + eA0 − 2m 8m3 c2  e¯hσ · (E × p) e¯h2 − − ∇ · E Ψ 4m2 c2 8m2 c2 = E (NR) Ψ. p2 1. The term 2m + eA0 gives the non-realtivistic Schr¨odinger equation. p4 2. The term − 3 2 is a relativistic correction to the 8m c kinetic energy as can be seen from the expansion p |p|2 |p|4 (mc2 )2 + |p|2 c2 − mc2 = − + ···. 2m 8m3 c2

e¯hσ · (E × p) 3. The term − describes the interaction 4m2 c2 between the spin of a moving electron and electric field. Intuitively this, so called Thomas term, is due to the fact that the moving electron experiences an apparent magnetic field E × (v/c). If the electric field is a central field, eA0 = V (r), it can be written in the form −

e¯ h σ · (E × p) 4m2 c2

  ¯ h 1 dV − σ · (x × p) r dr 4m2 c2 1 1 dV S · L, 2m2 c2 r dr

= − =

where we have substituted S=h ¯ σ/2. So we actually have a spin orbit interaction. 2 4. The term − e¯h2 2 ∇ · E is known as the Darwin term. 8m c Its meaning can be deduced when we note that ∇ · E is the charge density. For example, in the hydrogen atom where ∇ · E = −eδ(x) it causes the energy shift Z e2 ¯h2 e2 ¯ h2 (Schr¨ o) 2 3 (Schr¨ o) 2 , δ(x)|ψ | d x = |ψ | 2 2 2 2 8m c 8m c x=0 which differs from zero only in the s-state.