Solutions to Problems in Sakurai's Quantum Mechanics - P. is B. Brinne

Solutions to Problems

in

Quantum Mechanics

P. Saltsidis, additions by B. Brinne

1995,1999

0

Most of the problems presented here are taken from the bookSakurai, J. J., Modern Quantum Mechanics, Reading, MA: Addison-Wesley,1985.

Contents

I Problems 3

1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . 52 Quantum Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 73 Theory of Angular Momentum . . . . . . . . . . . . . . . . . . 144 Symmetry in Quantum Mechanics . . . . . . . . . . . . . . . . 175 Approximation Methods . . . . . . . . . . . . . . . . . . . . . 19

II Solutions 23

1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . 252 Quantum Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 363 Theory of Angular Momentum . . . . . . . . . . . . . . . . . . 754 Symmetry in Quantum Mechanics . . . . . . . . . . . . . . . . 945 Approximation Methods . . . . . . . . . . . . . . . . . . . . . 103

1

2 CONTENTS

Part I

Problems

3

1. FUNDAMENTAL CONCEPTS 5

1 Fundamental Concepts

1.1 Consider a ket space spanned by the eigenkets fja0ig of a Her-mitian operator A. There is no degeneracy.(a) Prove that Y

a0(A� a0)

is a null operator.(b) What is the signi�cance of

Ya00 6=a0

(A� a00)a0 � a00

?

(c) Illustrate (a) and (b) using A set equal to Sz of a spin12 system.

1.2 A spin 12system is known to be in an eigenstate of ~S � n with

eigenvalue �h=2, where n is a unit vector lying in the xz-plane thatmakes an angle with the positive z-axis.(a) Suppose Sx is measured. What is the probability of getting+�h=2?(b) Evaluate the dispersion in Sx, that is,

h(Sx � hSxi)2i:

(For your own peace of mind check your answers for the specialcases = 0, �=2, and �.)

1.3 (a) The simplest way to derive the Schwarz inequality goes asfollows. First observe

(h�j + ��h�j) � (j�i + �j�i) � 0

for any complex number �; then choose � in such a way that thepreceding inequality reduces to the Schwarz inequility.

6

(b) Show that the equility sign in the generalized uncertainty re-lation holds if the state in question satis�es

�Aj�i = ��Bj�i

with � purely imaginary.

(c) Explicit calculations using the usual rules of wave mechanicsshow that the wave function for a Gaussian wave packet given by

hx0j�i = (2�d2)�1=4 exp

"ihpix0�h

� (x0 � hxi)24d2

#

satis�es the uncertainty relation

qh(�x)2i

qh(�p)2i = �h

2:

Prove that the requirement

hx0j�xj�i = (imaginary number)hx0j�pj�i

is indeed satis�ed for such a Gaussian wave packet, in agreementwith (b).

1.4 (a) Let x and px be the coordinate and linear momentum inone dimension. Evaluate the classical Poisson bracket

[x; F (px)]classical :

(b) Let x and px be the corresponding quantum-mechanical opera-tors this time. Evaluate the commutator�

x; exp�ipxa

�h

��:

(c) Using the result obtained in (b), prove that

exp�ipxa

�h

�jx0i; (xjx0i = x0jx0i)

2. QUANTUM DYNAMICS 7

is an eigenstate of the coordinate operator x. What is the corre-sponding eigenvalue?

1.5 (a) Prove the following:

(i) hp0jxj�i = i�h@

@p0hp0j�i;

(ii) h�jxj�i =Zdp0��(p

0)i�h@

@p0��(p

0);

where ��(p0) = hp0j�i and ��(p0) = hp0j�i are momentum-space wavefunctions.(b) What is the physical signi�cance of

exp�ix�

�h

�;

where x is the position operator and � is some number with thedimension of momentum? Justify your answer.

2 Quantum Dynamics

2.1 Consider the spin-procession problem discussed in section 2.1in Jackson. It can also be solved in the Heisenberg picture. Usingthe Hamiltonian

H = ��eB

mc

�Sz = !Sz;

write the Heisenberg equations of motion for the time-dependentoperators Sx(t), Sy(t), and Sz(t). Solve them to obtain Sx;y;z as func-tions of time.

2.2 Let x(t) be the coordinate operator for a free particle in onedimension in the Heisenberg picture. Evaluate

[x(t); x(0)] :

8

2.3 Consider a particle in three dimensions whose Hamiltonian isgiven by

H =~p2

2m+ V (~x):

By calculating [~x � ~p;H] obtain

d

dth~x � ~pi =

*p2

m

+� h~x � ~rV i:

To identify the preceding relation with the quantum-mechanicalanalogue of the virial theorem it is essential that the left-hand sidevanish. Under what condition would this happen?

2.4 (a) Write down the wave function (in coordinate space) for thestate

exp��ipa

�h

�j0i:

You may use

hx0j0i = ��1=4x�1=20 exp

24�1

2

x0

x0

!235 ;

0@x0 �

�h

m!

!1=21A :

(b) Obtain a simple expression that the probability that the stateis found in the ground state at t = 0. Does this probability changefor t > 0?

2.5 Consider a function, known as the correlation function, de�nedby

C(t) = hx(t)x(0)i;where x(t) is the position operator in the Heisenberg picture. Eval-uate the correlation function explicitly for the ground state of aone-dimensional simple harmonic oscillator.


2.6 Consider again a one-dimensional simple harmonic oscillator.Do the following algebraically, that is, without using wave func-tions.

(a) Construct a linear combination of j0i and j1i such that hxi is aslarge as possible.

(b) Suppose the oscillator is in the state constructed in (a) at t = 0.What is the state vector for t > 0 in the Schr�odinger picture?Evaluate the expectation value hxi as a function of time for t > 0using (i) the Schr�odinger picture and (ii) the Heisenberg picture.

(c) Evaluate h(�x)2i as a function of time using either picture.

2.7 A coherent state of a one-dimensional simple harmonic oscil-lator is de�ned to be an eigenstate of the (non-Hermitian) annihi-lation operator a:

aj�i = �j�i;where � is, in general, a complex number.

(a) Prove that

j�i = e�j�j2=2e�a

yj0iis a normalized coherent state.

(b) Prove the minimum uncertainty relation for such a state.

(c) Write j�i asj�i =

1Xn=0

f(n)jni:

Show that the distribution of jf(n)j2 with respect to n is of thePoisson form. Find the most probable value of n, hence of E.

(d) Show that a coherent state can also be obtained by applyingthe translation (�nite-displacement) operator e�ipl=�h (where p is themomentum operator, and l is the displacement distance) to theground state.

10

(e) Show that the coherent state j�i remains coherent under time-evolution and calculate the time-evolved state j�(t)i. (Hint: di-rectly apply the time-evolution operator.)

2.8 The quntum mechanical propagator, for a particle with massm, moving in a potential is given by:

K(x; y;E) =Z 1

0dteiEt=�hK(x; y; t; 0) = A

Xn

sin(nrx) sin(nry)

E � �h2r2

2mn2

where A is a constant.(a) What is the potential?

(b) Determine the constant A in terms of the parameters describingthe system (such as m, r etc. ).

2.9 Prove the relationd�(x)

dx= �(x)

where �(x) is the (unit) step function, and �(x) the Dirac deltafunction. (Hint: study the e�ect on testfunctions.)

2.10 Derive the following expression

Scl =m!

2 sin(!T )

h(x20 + 2x2T ) cos(!T )� x0xT

ifor the classical action for a harmonic oscillator moving from thepoint x0 at t = 0 to the point xT at t = T .

2.11 The Lagrangian of the single harmonic oscillator is

L =1

2m _x2 � 1

2m!2x2


(a) Show that

hxbtbjxatai = exp�iScl�h

�G(0; tb; 0; ta)

where Scl is the action along the classical path xcl from (xa; ta) to(xb; tb) and G is

G(0; tb; 0; ta) =

limN!1

Zdy1 : : : dyN

�m

2�i�h"

� (N+1)2

exp

8<: i�h

NXj=0

�m

2"(yj+1 � yj)

2 � 1

2"m!2y2j

�9=;

where " = tb�ta(N+1)

.

(Hint: Let y(t) = x(t) � xcl(t) be the new integration variable,xcl(t) being the solution of the Euler-Lagrange equation.)

(b) Show that G can be written as

G = limN!1

�m

2�i�h"

� (N+1)2

Zdy1 : : : dyNexp(�nT�n)

where n =

2664y1...yN

3775 and nT is its transpose. Write the symmetric

matrix �.

(c) Show that

Zdy1 : : : dyNexp(�nT�n) �

ZdNye�n

T �n =�N=2pdet�

[Hint: Diagonalize � by an orhogonal matrix.]

(d) Let�2i�h"m

�Ndet� � det�0N � pN . De�ne j � j matrices �0j that con-

sist of the �rst j rows and j columns of �0N and whose determinantsare pj . By expanding �0j+1 in minors show the following recursionformula for the pj :

pj+1 = (2 � "2!2)pj � pj�1 j = 1; : : : ; N (2.1)

12

(e) Let �(t) = "pj for t = ta + j" and show that (2.1) implies that inthe limit "! 0; �(t) satis�es the equation

d2�

dt2= �!2�(t)

with initial conditions �(t = ta) = 0; d�(t=ta)dt

= 1.

(f) Show that

hxbtbjxatai =s

m!

2�i�h sin(!T )exp

(im!

2�h sin(!T )[(x2b + x2a) cos(!T )� 2xaxb]

)

where T = tb � ta.

2.12 Show the composition propertyZdx1Kf (x2; t2;x1; t1)Kf (x1; t1;x0; t0) = Kf (x2; t2;x0; t0)

where Kf (x1; t1;x0; t0) is the free propagator (Sakurai 2.5.16), byexplicitly performing the integral (i.e. do not use completeness).

2.13 (a) Verify the relation

[�i;�j] =

i�he

c

!"ijkBk

where ~� � m ~xdt= ~p� e ~A

cand the relation

md2~x

dt2=d~�

dt= e

"~E +

1

2c

d~x

dt� ~B � ~B � d~x

dt

!#:

(b) Verify the continuity equation

@�

@t+ ~r0 �~j = 0


with ~j given by

~j =

�h

m

!=( �~r0 )�

�e

mc

�~Aj j2:

2.14 An electron moves in the presence of a uniform magnetic �eldin the z-direction ( ~B = Bz).

(a) Evaluate[�x;�y];

where

�x � px � eAx

c; �y � py � eAy

c:

(b) By comparing the Hamiltonian and the commutation relationobtained in (a) with those of the one-dimensional oscillator problemshow how we can immediately write the energy eigenvalues as

Ek;n =�h2k2

2m+

jeBj�hmc

!�n+

1

2

�;

where �hk is the continuous eigenvalue of the pz operator and n is anonnegative integer including zero.

2.15 Consider a particle of mass m and charge q in an impenetrablecylinder with radius R and height a. Along the axis of the cylin-der runs a thin, impenetrable solenoid carrying a magnetic ux �.Calculate the ground state energy and wavefunction.

2.16 A particle in one dimension (�1 < x < 1) is subjected to aconstant force derivable from

V = �x; (� > 0):

14

(a) Is the energy spectrum continuous or discrete? Write down anapproximate expression for the energy eigenfunction speci�ed byE.

(b) Discuss brie y what changes are needed if V is replaced be

V = �jxj:

3 Theory of Angular Momentum

3.1 Consider a sequence of Euler rotations represented by

D(1=2)(�; �; ) = exp��i�3�

2

�exp

�i�2�2

!exp

��i�3 2

�

=

e�i(�+ )=2 cos �

2�e�i(�� )=2 sin �

2

ei(�� )=2 sin �2

ei(�+ )=2 cos �2

!:

Because of the group properties of rotations, we expect that thissequence of operations is equivalent to a single rotation about someaxis by an angle �. Find �.

3.2 An angular-momentum eigenstate jj;m = mmax = ji is rotatedby an in�nitesimal angle " about the y-axis. Without using theexplicit form of the d

(j)m0m function, obtain an expression for the

probability for the new rotated state to be found in the originalstate up to terms of order "2.

3.3 The wave function of a patricle subjected to a sphericallysymmetrical potential V (r) is given by

(~x) = (x+ y + 3z)f(r):

3. THEORY OF ANGULAR MOMENTUM 15

(a) Is an eigenfunction of ~L? If so, what is the l-value? If

not, what are the possible values of l we may obtain when ~L2 ismeasured?

(b)What are the probabilities for the particle to be found in variousml states?

(c) Suppose it is known somehow that (~x) is an energy eigenfunc-tion with eigenvalue E. Indicate how we may �nd V (r).

3.4 Consider a particle with an intrinsic angular momentum (orspin) of one unit of �h. (One example of such a particle is the %-meson). Quantum-mechanically, such a particle is described by aketvector j%i or in ~x representation a wave function

%i(~x) = h~x; ij%iwhere j~x; ii correspond to a particle at ~x with spin in the i:th di-rection.

(a) Show explicitly that in�nitesimal rotations of %i(~x) are obtainedby acting with the operator

u~" = 1� i~"

�h� (~L+ ~S) (3.1)

where ~L = �hir � ~r. Determine ~S !

(b) Show that ~L and ~S commute.

(c) Show that ~S is a vector operator.

(d) Show that ~r� ~%(~x) = 1�h2( ~S � ~p)~% where ~p is the momentum oper-

ator.

3.5 We are to add angular momenta j1 = 1 and j2 = 1 to formj = 2; 1; and 0 states. Using the ladder operator method express all

16

(nine) j;m eigenkets in terms of jj1j2;m1m2i. Write your answer as

jj = 1;m = 1i = 1p2j+; 0i � 1p

2j0;+i; : : : ; (3.2)

where + and 0 stand for m1;2 = 1; 0; respectively.

3.6 (a) Construct a spherical tensor of rank 1 out of two di�erent

vectors ~U = (Ux; Uy; Uz) and ~V = (Vx; Vy; Vz). Explicitly write T(1)�1;0 in

terms of Ux;y;z and Vx;y;z.

(b) Construct a spherical tensor of rank 2 out of two di�erent

vectors ~U and ~V . Write down explicitly T(2)�2;�1;0 in terms of Ux;y;z

and Vx;y;z.

3.7 (a) EvaluatejX

m=�jjd(j)mm0(�)j2m

for any j (integer or half-integer); then check your answer for j = 12 .

(b) Prove, for any j,

jXm=�j

m2jd(j)m0m(�)j2 = 12j(j + 1) sin � +m02 + 1

2(3 cos2 � � 1):

[Hint: This can be proved in many ways. You may, for instance,examine the rotational properties of J2

z using the spherical (irre-ducible) tensor language.]

3.8 (a) Write xy, xz, and (x2 � y2) as components of a spherical(irreducible) tensor of rank 2.

4. SYMMETRY IN QUANTUM MECHANICS 17

(b) The expectation value

Q � eh�; j;m = jj(3z2 � r2)j�; j;m = ji

is known as the quadrupole moment. Evaluate

eh�; j;m0j(x2 � y2)j�; j;m = ji;

(where m0 = j; j�1; j�2; : : : )in terms of Q and appropriate Clebsch-Gordan coe�cients.

4 Symmetry in Quantum Mechanics

4.1 (a) Assuming that the Hamiltonian is invariant under timereversal, prove that the wave function for a spinless nondegeneratesystem at any given instant of time can always be chosen to bereal.

(b) The wave function for a plane-wave state at t = 0 is given bya complex function ei~p�~x=�h. Why does this not violate time-reversalinvariance?

4.2 Let �(~p0) be the momentum-space wave function for state j�i,that is, �(~p0) = h~p0j�i.Is the momentum-space wave function for thetime-reversed state �j�i given by �(~p0, �(�~p0), ��(~p0), or ��(�~p0)?Justify your answer.

4.3 Read section 4.3 in Sakurai to refresh your knowledge of thequantum mechanics of periodic potentials. You know that the en-ergybands in solids are described by the so called Bloch functions n;k full�lling,

n;k(x+ a) = eika n;k(x)

18

where a is the lattice constant, n labels the band, and the latticemomentum k is restricted to the Brillouin zone [��=a; �=a].

Prove that any Bloch function can be written as,

n;k(x) =XRi

�n(x�Ri)eikRi

where the sum is over all lattice vectors Ri. (In this simble one di-mensional problem Ri = ia, but the construction generalizes easilyto three dimensions.).

The functions �n are called Wannier functions, and are impor-tant in the tight-binding description of solids. Show that the Wan-nier functions are corresponding to di�erent sites and/or di�erentbands are orthogonal, i:e: proveZ

dx�?m(x�Ri)�n(x�Rj) � �ij�mn

Hint: Expand the �ns in Bloch functions and use their orthonor-mality properties.

4.4 Suppose a spinless particle is bound to a �xed center by apotential V (~x) so assymetrical that no energy level is degenerate.Using the time-reversal invariance prove

h~Li = 0

for any energy eigenstate. (This is known as quenching of orbitalangular momemtum.) If the wave function of such a nondegenerateeigenstate is expanded asX

l

Xm

Flm(r)Yml (�; �);

what kind of phase restrictions do we obtain on Flm(r)?

4.5 The Hamiltonian for a spin 1 system is given by

H = AS2z +B(S2

x � S2y):

5. APPROXIMATION METHODS 19

Solve this problem exactly to �nd the normalized energy eigen-states and eigenvalues. (A spin-dependent Hamiltonian of this kindactually appears in crystal physics.) Is this Hamiltonian invariantunder time reversal? How do the normalized eigenstates you ob-tained transform under time reversal?

5 Approximation Methods

5.1 Consider an isotropic harmonic oscillator in two dimensions.The Hamiltonian is given by

H0 =p2x2m

+p2y2m

+m!2

2(x2 + y2)

(a) What are the energies of the three lowest-lying states? Is thereany degeneracy?

(b) We now apply a perturbation

V = �m!2xy

where � is a dimensionless real number much smaller than unity.Find the zeroth-order energy eigenket and the corresponding en-ergy to �rst order [that is the unperturbed energy obtained in (a)plus the �rst-order energy shift] for each of the three lowest-lyingstates.

(c) Solve the H0+V problem exactly. Compare with the perturba-tion results obtained in (b).

[You may use hn0jxjni =q�h=2m!(

pn + 1�n0;n+1 +

pn�n0;n�1):]

5.2 A system that has three unperturbed states can be representedby the perturbed Hamiltonian matrix0

B@ E1 0 a0 E1 ba� b� E2

1CA

20

where E2 > E1. The quantities a and b are to be regarded as per-turbations that are of the same order and are small compared withE2 � E1. Use the second-order nondegenerate perturbation theoryto calculate the perturbed eigenvalues. (Is this procedure correct?)Then diagonalize the matrix to �nd the exact eigenvalues. Finally,use the second-order degenerate perturbation theory. Comparethe three results obtained.

5.3 A one-dimensional harmonic oscillator is in its ground statefor t < 0. For t � 0 it is subjected to a time-dependent but spatiallyuniform force (not potential!) in the x-direction,

F (t) = F0e�t=�

(a) Using time-dependent perturbation theory to �rst order, obtainthe probability of �nding the oscillator in its �rst excited state fort > 0). Show that the t ! 1 (� �nite) limit of your expression isindependent of time. Is this reasonable or surprising?

(b) Can we �nd higher excited states?


pn+ 1�n0;n+1 +

pn�n0;n�1):]

5.4 Consider a composite system made up of two spin 12 objects.

for t < 0, the Hamiltonian does not depend on spin and can betaken to be zero by suitably adjusting the energy scale. For t > 0,the Hamiltonian is given by

H =�4�

�h2

�~S1 � ~S2:

Suppose the system is in j + �i for t � 0. Find, as a function oftime, the probability for being found in each of the following statesj++i, j+�i, j �+i, j � �i:

(a) By solving the problem exactly.


(b) By solving the problem assuming the validity of �rst-ordertime-dependent perturbation theory with H as a perturbation switchedon at t = 0. Under what condition does (b) give the correct results?

5.5 The ground state of a hydrogen atom (n = 1,l = 0) is subjectedto a time-dependent potential as follows:

V (~x; t) = V0cos(kz � !t):

Using time-dependent perturbation theory, obtain an expressionfor the transition rate at which the electron is emitted with mo-mentum ~p. Show, in particular, how you may compute the angulardistribution of the ejected electron (in terms of � and � de�nedwith respect to the z-axis). Discuss brie y the similarities and thedi�erences between this problem and the (more realistic) photo-electric e�ect. (note: For the initial wave function use

n=1;l=0(~x) =1p�

�Z

a0

� 32

e�Zr=a0:

If you have a normalization problem, the �nal wave function maybe taken to be

f (~x) =�1

L32

�ei~p�~x=�h

with L very large, but you should be able to show that the observ-able e�ects are independent of L.)

Part II

Solutions

23


1 Fundamental Concepts

1.1 Consider a ket space spanned by the eigenkets fja0ig of a Her-mitian operator A. There is no degeneracy.

(a) Prove that Ya0

(A� a0)

is a null operator.

(b) What is the signi�cance of

Ya00 6=a0

(A� a00)a0 � a00

?

(c) Illustrate (a) and (b) using A set equal to Sz of a spin12system.

(a) Assume that j�i is an arbitrary state ket. Then

Ya0(A� a0)j�i =

Ya0(A� a0)

Xa00ja00i ha00j�i| {z }

ca00

=Xa00ca00

Ya0(A� a0)ja00i

=Xa00ca00

Ya0(a00 � a0)ja00i a

002fa0g= 0: (1.1)

(b) Again for an arbitrary state j�i we will have24 Ya00 6=a0

(A� a00)a0 � a00

35 j�i =

24 Ya00 6=a0

(A� a00)a0 � a00

35

1z }| {Xa000ja000iha000 j�i

=Xa000ha000j�i Y

a00 6=a0

(a000 � a00)a0 � a00

ja000i =

=Xa000

ha000j�i�a000a0ja000i = ha0j�ija0i )24 Ya00 6=a0

(A� a00)a0 � a00

35 = ja0iha0j � �a0: (1.2)

So it projects to the eigenket ja0i.

26

(c) It is Sz � �h=2(j+ih+j� j�ih�j). This operator has eigenkets j+i and j�iwith eigenvalues �h=2 and -�h=2 respectively. SoY

a0(Sz � a0) =

Ya0(Sz � a01)

=

"�h

2(j+ih+j � j�ih�j)� �h

2(j+ih+j+ j�ih�j)

#

�"�h

2(j+ih+j � j�ih�j) + �h


#

= [��hj�ih�j][�hj+ih+j] = ��h2j�i0z }| {

h�j+ih+j = 0; (1.3)

where we have used that j+ih+j+ j�ih�j = 1.For a0 = �h=2 we have

Ya00 6=a0

(Sz � a00)a0 � a00

=Y

a00 6=�h=2

(Sz � a001)�h=2 � a00

=Sz +

�h21

�h=2 + �h=2

=1

�h

"�h

2(j+ih+j � j�ih�j) + �h


#

=1

�h�hj+ih+j = j+ih+j: (1.4)

Similarly for a0 = ��h=2 we have

Ya00 6=a0

(Sz � a00)a0 � a00

=Y

a00 6=��h=2

(Sz � a001)��h=2 � a00

=Sz � �h

21

��h=2� �h=2

= �1

�h

"�h

2(j+ih+j � j�ih�j)� �h


#

= �1

�h(��hj�ih�j) = j�ih�j: (1.5)

1.2 A spin 12 system is known to be in an eigenstate of ~S � n with

eigenvalue �h=2, where n is a unit vector lying in the xz-plane thatmakes an angle with the positive z-axis.(a) Suppose Sx is measured. What is the probability of getting+�h=2?


(b) Evaluate the dispersion in Sx, that is,

h(Sx � hSxi)2i:(For your own peace of mind check your answers for the specialcases = 0, �=2, and �.)

Since the unit vector n makes an angle with the positive z-axis and islying in the xz-plane, it can be written in the following way

n = ez cos + ex sin (1.6)

So

~S � n = Sz cos + Sx sin = [(S-1.3.36),(S-1.4.18)]

=

"�h

2(j+ih+j � j�ih�j)

#cos +

"�h

2(j+ih�j+ j�ih+j)

#sin :(1.7)

Since the system is in an eigenstate of ~S � n with eigenvalue �h=2 it has tosatisfay the following equation

~S � nj~S � n; +i = �h=2j~S � n; +i: (1.8)

From (1.7) we have that

~S � n �=

�h

2

cos sin sin � cos

!: (1.9)

The eigenvalues and eigenfuncions of this operator can be found if one solvesthe secular equation

det(~S � n� �I) = 0) det

�h=2 cos � � �h=2 sin �h=2 sin ��h=2 cos � �

!= 0)

��h2

4cos2 + �2 � �h2

4sin2 = 0) �2 � �h2

4= 0) � = ��h

2: (1.10)

Since the system is in the eigenstate j~S � n; +i � ab

!we will have that

�h

2

cos sin sin � cos

! ab

!=

�h

2

ab

!)(a cos + b sin = aa sin � b cos = b

))

b = a1� cos

sin = a

2 sin2 2

2 sin 2 cos

2

= a tan

2: (1.11)

28

But we want also the eigenstate j~S � n; +i to be normalized, that is

a2 + b2 = 1 ) a2 + a2 tan2

2= 1) a2 cos2

2a2 sin2

2= cos2

2

) a2 = cos2

2) a = �

rcos2

2= cos

2; (1.12)

where the real positive convention has been used in the last step. This meansthat the state in which the system is in, is given in terms of the eigenstatesof the Sz operator by

j~S � n; +i = cos

2j+i+ sin

2j�i: (1.13)

(a) From (S-1.4.17) we know that

jSx; +i = 1p2j+i+ 1p

2j�i: (1.14)

So the propability of getting +�h=2 when Sx is measured is given by

��hSx; +j~S � n; +i��2 =

��

1p2h+j+ 1p

2h�j

!�cos

2j+i+ sin

2j�i

��2

=

�� 1p2 cos

2+

1p2sin

2

��2

=1

2cos2

2+1

2sin2

2+ cos

2sin

2= 1

2 +12 sin =

12(1 + sin ): (1.15)

For = 0 which means that the system is in the jSz; +i eigenstate we have

jhSx; +jSz; +ij2 = 12(1) =

12 : (1.16)

For = �=2 which means that the system is in the jSx; +i eigenstate wehave

jhSx; +jSx; +ij2 = 1: (1.17)

For = � which means that the system is in the jSz;�i eigenstate we have

jhSx; +jSz;�ij2 = 12(1) =

12 : (1.18)


(b) We have that

h(Sx � hSxi)2i = hS2xi � (hSxi)2: (1.19)

As we know

Sx =�h

2(j+ih�j+ j�ih+j))

S2x =

�h2

4(j+ih�j + j�ih+j) (j+ih�j+ j�ih+j))

S2x =

�h2

4(j+ih+j + j�ih�j)| {z }

1

=�h2

4: (1.20)

So

hSxi =�cos

2h+j + sin

2h�j

��h

2(j+ih�j+ j�ih+j)

�cos

2j+i+ sin

2j�i

�

=�h

2cos

2sin

2+�h

2sin

2cos

2=

�h

2sin )

(hSxi)2 =�h2

4sin2 and

hS2xi =

�cos

2h+j + sin

2h�j

��h2

4

�cos

2j+i+ sin

2j�i

�

=�h2

4[cos2

2+ sin2

2] =

�h2

4: (1.21)

So substituting in (1.19) we will have

h(Sx � hSxi)2i = �h2

4(1 � sin2 ) =

�h2

4cos2 : (1.22)

and �nally

h(�Sx)2i =0;jSz;+i =�h2

4; (1.23)

h(�Sx)2i =�=2;jSx ;+i = 0; (1.24)

h(�Sx)2i =0;jSz;�i =�h2

4: (1.25)

30

1.3 (a) The simplest way to derive the Schwarz inequality goes asfollows. First observe

(h�j + ��h�j) � (j�i + �j�i) � 0

for any complex number �; then choose � in such a way that thepreceding inequality reduces to the Schwarz inequility.

(b) Show that the equility sign in the generalized uncertainty re-lation holds if the state in question satis�es

�Aj�i = ��Bj�iwith � purely imaginary.

(c) Explicit calculations using the usual rules of wave mechanicsshow that the wave function for a Gaussian wave packet given by

hx0j�i = (2�d2)�1=4 exp

"ihpix0�h

� (x0 � hxi)24d2

#

satis�es the uncertainty relation

qh(�x)2i

qh(�p)2i = �h

2:

Prove that the requirement

hx0j�xj�i = (imaginary number)hx0j�pj�iis indeed satis�ed for such a Gaussian wave packet, in agreementwith (b).

(a) We know that for an arbitrary state jci the following relation holds

hcjci � 0: (1.26)

This means that if we choose jci = j�i+ �j�i where � is a complex number,we will have

(h�j + ��h�j) � (j�i + �j�i) � 0) (1.27)

h�j�i + �h�j�i + ��h�j�i+ j�j2h�j�i � 0: (1.28)


If we now choose � = �h�j�i=h�j�i the previous relation will be

h�j�i � h�j�ih�j�ih�j�i � h�j�ih�j�i

h�j�i + jh�j�ij2h�j�i � 0 )

h�j�ih�j�i � jh�j�ij2: (1.29)

Notice that the equality sign in the last relation holds when

jci = j�i + �j�i = 0) j�i = ��j�i (1.30)

that is if j�i and j�i are colinear.(b) The uncertainty relation is

h(�A)2ih(�B)2i � 1

4jh[A;B]ij2 : (1.31)

To prove this relation we use the Schwarz inequality (1.29) for the vectorsj�i = �Ajai and j�i = �Bjai which gives

h(�A)2ih(�B)2i � jh�A�Bij2: (1.32)

The equality sign in this relation holds according to (1.30) when

�Ajai = ��Bjai: (1.33)

On the other hand the right-hand side of (1.32) is

jh�A�Bij2 = 1

4jh[A;B]ij2 + 1

4jhf�A;�Bgij2 (1.34)

which means that the equality sign in the uncertainty relation (1.31) holds if

1

4jhf�A;�Bgij2 = 0) hf�A;�Bgi = 0

) haj�A�B +�B�Ajai = 0(1:33)) ��haj(�B)2jai+ �haj(�B)2jai = 0

) (� + ��)haj(�B)2jai = 0: (1.35)

Thus the equality sign in the uncertainty relation holds when

�Ajai = ��Bjai (1.36)

with � purely imaginary.

32

(c) We have

hx0j�xj�i � hx0j(x� hxi)j�i = x0hx0j�i � hxihx0j�i= (x0 � hxi)hx0j�i: (1.37)

On the other hand

hx0j�pj�i � hx0j(p� hpi)j�i= �i�h @

@x0hx0j�i � hpihx0j�i (1.38)

But

@

@x0hx0j�i = hx0j�i @

@x0

"ihpix0�h

� (x0 � hxi)24d2

#

= hx0j�i"ihpi�h� 1

2d2(x0 � hxi)

#(1.39)

So substituting in (1.38) we have

hx0j�pj�i = hpihx0j�i + i�h

2d2(x0 � hxi) hx0j�i � hpihx0j�i

=i�h

2d2(x0 � hxi) hx0j�i = i�h

2d2hx0j�xj�i )

hx0j�xj�i =�i2d2�h

hx0j�pj�i: (1.40)

1.4 (a) Let x and px be the coordinate and linear momentum inone dimension. Evaluate the classical Poisson bracket

[x; F (px)]classical :

(b) Let x and px be the corresponding quantum-mechanical opera-tors this time. Evaluate the commutator�

x; exp�ipxa

�h

��:

(c) Using the result obtained in (b), prove that

exp�ipxa

�h

�jx0i; (xjx0i = x0jx0i)


is an eigenstate of the coordinate operator x. What is the corre-sponding eigenvalue?

(a) We have

[x; F (px)]classical � @x

@x

@F (px)

@px� @x

@px

@F (px)

@x

=@F (px)

@px: (1.41)

(b) When x and px are treated as quantum-mechanical operators we have

�x; exp

�ipxa

�h

��=

"x;

1Xn=0

(ia)n

�hnpnxn!

#=

1Xn=0

1

n!

(ia)n

�hn[x; pnx]

=1Xn=0

1

n!

(ia)n

�hn

n�1Xk=0

pkx [x; px] pn�k�1x

=1Xn=1

1

n!

(ia)n

�hn(i�h)

n�1Xk=0

pkxpn�k�1x =

1Xn=1

n

n!

(ia)n�1

�hn�1pn�1x (�a)

= �a1Xn=1

1

(n� 1)!

�ia

�hpx

�n�1= �a exp

�ipxa

�h

�: (1.42)

(c) We have now

x�exp

�ipxa

�h

��jx0i (b)

= exp�ipxa

�h

�xjx0i � a exp

�ipxa

�h

�jx0i

= x0 exp�ipxa

�h

�jx0i � a exp

�ipxa

�h

�jx0i

= (x0 � a) exp�ipxa

�h

�jx0i: (1.43)

So exp�ipxa�h

�jx0i is an eigenstate of the operator x with eigenvalue x0 � a.

So we can write

jx0 � ai = C exp�ipxa

�h

�jx0i; (1.44)

where C is a constant which due to normalization can be taken to be 1.

34

1.5 (a) Prove the following:

(i) hp0jxj�i = i�h@

@p0hp0j�i;

(ii) h�jxj�i =Zdp0��(p

0)i�h@

@p0��(p

0);

where ��(p0) = hp0j�i and ��(p0) = hp0j�i are momentum-space wavefunctions.

(b) What is the physical signi�cance of

exp�ix�

�h

�;

where x is the position operator and � is some number with thedimension of momentum? Justify your answer.

(a) We have(i)

hp0jxj�i = hp0jx1z }| {Z

dx0jx0ihx0j�i =Zdx0hp0jxjx0ihx0j�i

=Zdx0x0hp0jx0ihx0j�i (S�1:7:32)

=Zdx0x0Ae�

ip0x0

�h hx0j�i

= AZdx0

@

@p0

�e�

ip0x0

�h

�(i�h)hx0j�i = i�h

@

@p0

�Zdx0Ae�

ip0x0

�h hx0j�i�

= i�h@

@p0

�Zdx0hp0jx0ihx0j�i

�= i�h

@

@p0hp0j�i )

hp0jxj�i = i�h@

@p0hp0j�i: (1.45)

(ii)

h�jxj�i =Zdp0h�jp0ihp0jxj�i =

Zdp0��(p

0)i�h@

@p0��(p

0); (1.46)

where we have used (1.45) and that h�jp0i = ��(p0) and hp0j�i = ��(p0).


(b) The operator exp�ix��h

�gives translation in momentum space. This can

be justi�ed by calculating the following operator

�p; exp

�ix�

�h

��=

"p;

1Xn=0

1

n!

�ix�

�h

�n#=

1Xn=0

1

n!

�i�

�h

�n[p; xn]

=1Xn=1

1

n!

�i�

�h

�n nXk=1

xn�k[p; x]xk�1

=1Xn=1

1

n!

�i�

�h

�n nXk=1

(�i�h)xn�1 =1Xn=1

1

n!

�i�

�h

�nn(�i�h)xn�1

=1Xn=1

1

(n� 1)!

�i�

�h

�n�1xn�1(�i�h)

�i�

�h

�= �

1Xn=0

1

n!

�ix�

�h

�n

= �exp�ix�

�h

�: (1.47)

So when this commutator acts on an eigenstate jp0i of the momentum oper-ator we will have�

p; exp�ix�

�h

��jp0i = p

�exp

�ix�

�h

�jp0i

��exp

�ix�

�h

��p0jp0i )

�exp�ix�

�h

�= p

�exp

�ix�

�h

�jp0i

�� p0

�exp

�ix�

�h

��jp0i )

p�exp

�ix�

�h

�jp0i

�= (p0 + �)

�exp

�ix�

�h

�jp0i

�: (1.48)

Thus we have that

exp�ix�

�h

�jp0i � Ajp0 + �i; (1.49)

where A is a constant which due to normalization can be taken to be 1.

36

2 Quantum Dynamics

2.1 Consider the spin-procession problem discussed in section 2.1in Jackson. It can also be solved in the Heisenberg picture. Usingthe Hamiltonian

H = ��eB

mc

�Sz = !Sz;

write the Heisenberg equations of motion for the time-dependentoperators Sx(t), Sy(t), and Sz(t). Solve them to obtain Sx;y;z as func-tions of time.

Let us �rst prove the following

[AS; BS] = CS ) [AH ; BH] = CH : (2.1)

Indeed we have

[AH; BH] =hUyASU ;UyBSU

i= UyASBSU � UyBSASU

= Uy [AS; BS]U = UyCSU = CH : (2.2)

The Heisenberg equation of motion gives

dSxdt

=1

i�h[Sx;H] =

1

i�h[Sx; !Sz]

(S�1:4:20)=

!

i�h(�i�hSy) = �!Sy; (2.3)

dSydt

=1

i�h[Sy;H] =

1

i�h[Sy; !Sz]

(S�1:4:20)=

!

i�h(i�hSx) = !Sx; (2.4)

dSzdt

=1

i�h[Sz;H] =

1

i�h[Sz; !Sz]

(S�1:4:20)= 0) Sz = constant: (2.5)

Di�erentiating once more eqs. (2.3) and (2.4) we get

d2Sxdt2

= �!dSydt

(2:4)= �!2Sx ) Sx(t) = A cos!t+B sin!t) Sx(0) = A

d2Sydt2

= !dSxdt

(2:3)= �!2Sy ) Sy(t) = C cos!t+D sin!t) Sy(0) = C:

But on the other hand

dSxdt

= �!Sy )�A! sin!t+B! cos !t = �C! cos!t�D! sin!t)

A = D C = �B: (2.6)


So, �nally

Sx(t) = Sx(0) cos!t� Sy(0) sin!t (2.7)

Sy(t) = Sy(0) cos!t+ Sx(0) sin!t (2.8)

Sz(t) = Sz(0): (2.9)

2.2 Let x(t) be the coordinate operator for a free particle in onedimension in the Heisenberg picture. Evaluate

[x(t); x(0)] :

The Hamiltonian for a free particle in one dimension is given by

H =p2

2m: (2.10)

This means that the Heisenberg equations of motion for the operators x andp will be

@p(t)

@t=

1

i�h[p(t);H(t)] =

1

i�h

"p(t);

p2(t)

2m

#= 0)

p(t) = p(0) (2.11)

@x(t)

@t=

1

i�h[x;H] =

1

i�h

"x(t);

p2(t)

2m

#=

1

2mi�h2p(t)i�h =

p(t)

m

(2:11)=

p(0)

m)

x(t) =t

mp(0) + x(0): (2.12)

Thus �nally

[x(t); x(0)] =�t

mp(0) + x(0); x(0)

�=

t

m[p(0); x(0)] = �i�ht

m: (2.13)

2.3 Consider a particle in three dimensions whose Hamiltonian isgiven by

H =~p2

2m+ V (~x):

38

By calculating [~x � ~p;H] obtain

d

dth~x � ~pi =

*p2

m

+� h~x � ~rV i:

To identify the preceding relation with the quantum-mechanicalanalogue of the virial theorem it is essential that the left-hand sidevanish. Under what condition would this happen?

Let us �rst calculate the commutator [~x � ~p;H]

[~x � ~p;H] =

"~x � ~p; ~p

2

2m+ V (~x)

#=

24 3Xi=1

xipi;3X

j=1

p2j2m

+ V (~x)

35

=Xij

"xi;

p2j2m

#pi +

Xi

xi [pi; V (~x)] : (2.14)

The �rst commutator in (2.14) will give"xi;

p2j2m

#=

1

2m[xi; p

2j ] =

1

2m(pj [xi; pj ] + [xi; pj]pj) =

1

2m(pji�h�ij + i�h�ijpj)

=1

2m2i�h�ijpj =

i�h

m�ijpj : (2.15)

The second commutator can be calculated if we Taylor expand the functionV (~x) in terms of xi which means that we take V (~x) =

Pn anx

ni with an

independent of xi. So

[pi; V (~x)] =

"pi;

1Xn=0

anxni

#=Xn

an [pi; xni ] =

Xn

ann�1Xk=0

xki [pi; xi]xn�k�1i

=Xn

ann�1Xk=0

(�i�h)xn�1i = �i�hXn

annxn�1i = �i�h @

@xi

Xn

anxni

= �i�h @

@xiV (~x): (2.16)

The right-hand side of (2.14) now becomes

[~x � ~p;H] =Xij

i�h

m�ijpjpi +

Xi

(�i�h)xi @@xi

V (~x)

=i�h

m~p2 � i�h~x � ~rV (~x): (2.17)


The Heisenberg equation of motion gives

d

dt~x � ~p =

1

i�h[~x � ~p;H]

(2:17)=

~p2

m� ~x � ~rV (~x))

d

dth~x � ~pi =

*p2

m

+� h~x � ~rV i; (2.18)

where in the last step we used the fact that the state kets in the Heisenbergpicture are independent of time.

The left-hand side of the last equation vanishes for a stationary state.Indeed we have

d

dthnj~x � ~pjni = 1

i�hhnj [~x � ~p;H] jni = 1

i�h(Enhnj~x � ~pjni � Enhnj~x � ~pjni) = 0:

So to have the quantum-mechanical analogue of the virial theorem we cantake the expectation values with respect to a stationaru state.

2.4 (a) Write down the wave function (in coordinate space) for thestate

exp��ipa

�h

�j0i:

You may use

hx0j0i = ��1=4x�1=20 exp

24�1

2

x0

x0

!235 ;

0@x0 �

�h

m!

!1=21A :

(b) Obtain a simple expression that the probability that the stateis found in the ground state at t = 0. Does this probability changefor t > 0?

(a) We have

j�; t = 0i = exp��ipa

�h

�j0i )

hx0j�; t = 0i = hx0 exp��ipa

�h

�j0i (Pr:1:4:c)

= hx0 � aj0i

= ��1=4x�1=20 exp

24�1

2

x0 � a

x0

!235 : (2.19)

40

(b) This probability is given by the expression

jh0j�; t = 0ij2 = jhexp��ipa

�h

�j0ij2: (2.20)

It is

hexp��ipa

�h

�j0i =

Zdx0h0jx0ihx0j exp

��ipa�h

�j0i

=Zdx0��1=4x�1=20 exp

24�1

2

x0

x0

!235��1=4x�1=20

� exp

24�1

2

x0 � a

x0

!235

=Zdx0��1=2x�10 exp

"� 1

2x20

�x02 + x02 + a2 � 2ax0

�#

=1p�x0

Zdx0 exp

"� 2

2x20

x02 � 2x0

a

2+a2

4+a2

4

!#

= exp

� a2

4x20

!1p�x0

p�x0 = exp

� a2

4x20

!: (2.21)

So

jh0j�; t = 0ij2 = exp

� a2

2x20

!: (2.22)

For t > 0

jh0j�; tij2 = jh0jU(t)j�; t = 0ij2 = jh0j exp��iHt

�h

�j�; t = 0ij2

=��e�iE0t=�hh0j�; t = 0i

��2 = jh0j�; t = 0ij2: (2.23)

2.5 Consider a function, known as the correlation function, de�nedby

C(t) = hx(t)x(0)i; (2.24)

where x(t) is the position operator in the Heisenberg picture. Eval-uate the correlation function explicitly for the ground state of aone-dimensional simple harmonic oscillator.


The Hamiltonian for a one-dimensional harmonic oscillator is given by

H =p2(t)

2m+ 1

2m!2x2(t): (2.25)

So the Heisenberg equations of motion will give

dx(t)

dt=

1

i�h[x(t);H] =

1

i�h

"x(t);

p2(t)

2m+ 1

2m!2x2(t)

#

=1

2mi�h

hx(t); p2(t)

i+ 1

2m!2 1

i�h

hx(t); x2(t)

i=

2i�h

2i�hmp(t) =

p(t)

m(2.26)

dp(t)

dt=

1

i�h[p(t);H] =

1

i�h

"p(t);

p2(t)

2m+ 1

2m!2x2(t)

#

=m!2

2i�h

hp(t); x2(t)

i=m!2

2i�h[�2i�hx(t)] = �m!2x(t): (2.27)

Di�erentiating once more the equations (2.26) and (2.27) we get

d2x(t)

dt2=

1

m

dp(t)

dt

(2:27)= �!2x(t)) x(t) = A cos!t+B sin!t) x(0) = A

d2p(t)

dt2=

1

m

dx(t)

dt

(2:26)= �!2p(t)) p(t) = C cos!t+D sin!t) p(0) = C:

But on the other hand from (2.26) we have

dx(t)

dt=

p(t)

m)

�!x(0) sin !t+B! cos!t =p(0)

mcos!t+

D

msin!t)

B =p(0)

m!D = �m!x(0): (2.28)

So

x(t) = x(0) cos!t+p(0)

m!sin!t (2.29)

and the correlation function will be

C(t) = hx(t)x(0)i (2:29)= hx2(0)i cos !t+ hp(0)x(0)i 1

m!sin!t: (2.30)

42

Since we are interested in the ground state the expectation values appearingin the last relation will be

hx2(0)i = h0j �h

2m!(a+ ay)(a+ ay)j0i = �h

2m!h0jaayj0i = �h

2m!(2.31)

hp(0)x(0)i = i

sm�h!

2

s�h

2m!h0j(ay � a)(a+ ay)j0i

= �i�h2h0jaayj0i = �i�h

2: (2.32)

Thus

C(t) =�h

2m!cos!t� i

�h

2m!sin!t =

�h

2m!e�i!t: (2.33)

2.6 Consider a one-dimensional simple harmonic oscillator. Do thefollowing algebraically, that is, without using wave functions.

(a) Construct a linear combination of j0i and j1i such that hxi is aslarge as possible.

(b) Suppose the oscillator is in the state constructed in (a) at t = 0.What is the state vector for t > 0 in the Schr�odinger picture?Evaluate the expectation value hxi as a function of time for t > 0using (i) the Schr�odinger picture and (ii) the Heisenberg picture.

(c) Evaluate h(�x)2i as a function of time using either picture.

(a) We want to �nd a state j�i = c0j0i + c1j1i such that hxi is as large aspossible. The state j�i should be normalized. This means

jc0j2 + jc1j2 = 1) jc1j =q1 � jc0j2: (2.34)

We can write the constands c0 and c1 in the following form

c0 = jc0jei�0c1 = jc1jei�1 (2:34)

= ei�1q1� jc0j2: (2.35)


The average hxi in a one-dimensional simple harmonic oscillator is givenby

hxi = h�jxj�i = (c�0h0j + c�1h1j) x (c0j0i + c1j1i)= jc0j2h0jxj0i + c�0c1h0jxj1i + c�1c0h1jxj0i + jc1j2h1jxj1i

= jc0j2s

�h

2m!h0ja+ ayj0i+ c�0c1

s�h

2m!h0ja + ayj1i

+c�1c0

s�h

2m!h1ja + ayj0i + jc1j2

s�h

2m!h1ja + ayj1i

=

s�h

2m!(c�0c1 + c�1c0) = 2

s�h

2m!<(c�0c1)

= 2

s�h

2m!cos(�1 � �0)jc0j

q1 � jc0j2; (2.36)

where we have used that x =q

�h2m!

(a+ ay).What we need is to �nd the values of jc0j and �1 � �0 that make the

average hxi as large as possible.@hxi@jc0j = 0 )

q1 � jc0j2 � jc0j2q

1 � jc0j2jc0j6=1) 1 � jc0j2 � jc0j2 = 0

) jc0j = 1p2

(2.37)

@hxi@�1

= 0 ) � sin(�1 � �0) = 0) �1 = �0 + n�; n 2 Z: (2.38)

But for hxi maximum we want also

@2hxi@�21

��1=�1max

< 0) n = 2k; k 2 Z: (2.39)

So we can write that

j�i = ei�01p2j0i + ei(�0+2k�)

1p2j1i = ei�0

1p2(j0i + j1i): (2.40)

We can always take �0 = 0. Thus

j�i = 1p2(j0i + j1i): (2.41)

44

(b) We have j�; t0i = j�i. So

j�; t0; ti = U(t; t0 = 0)j�; t0i = e�iHt=�hj�i = 1p2e�iE0t=�hj0i + 1p

2e�iE1t=�hj1i

=1p2

�e�i!t=2j0i+ e�i!3t=2j1i

�=

1p2e�i!t=2

�j0i+ e�i!tj1i

�:(2.42)

(i) In the Schr�odinger picture

hxiS = h�; t0; tjxSj�; t0; tiS=

"1p2

�ei!t=2h0j+ ei!3t=2h1j

�#x

"1p2

�e�i!t=2j0i + e�i!3t=2j1i

�#

= 12e

i(!t=2�!3t=2)h0jxj1i + 12e

i(!3t=2�!t=2)h1jxj0i

= 12e

�i!ts

�h

2m!+ 1

2ei!t

s�h

2m!=

s�h

2m!cos !t: (2.43)

(ii) In the Heisenberg picture we have from (2.29) that

xH(t) = x(0) cos!t+p(0)

m!sin!t:

So

hxiH = h�jxH j�i=

"1p2h0j + 1p

2h1j#

x(0) cos!t+p(0)

m!sin!t

!"1p2j0i+ 1p

2j1i#

= 12cos !th0jxj1i+ 1

2cos !th1jxj0i + 1

2

1

m!sin !th0jpj1i

+12

1

m!sin!th1jpj0i

= 12

s�h

2m!cos!t+ 1

2

s�h

2m!cos!t+

1

2m!sin!t(�i)

sm�h!

2

+1

2m!sin!ti

sm�h!

2

=

s�h

2m!cos!t: (2.44)

(c) It is known that

h(�x)2i = hx2i � hxi2 (2.45)


In the Sch�odinger picture we have

x2 =

24s

�h

2m!(a+ ay)

352 = �h

2m!(a2 + ay

2+ aay+ aya); (2.46)

which means that

hxi2S = h�; t0; tjx2j�; t0; tiS=

"1p2

�ei!t=2h0j + ei!3t=2h1j

�#x2"1p2

�e�i!t=2j0i+ e�i!3t=2j1i

�#

=h12ei(!t=2�!t=2)h0jaayj0i + 1

2ei(!3t=2�!3t=2)h1jaayj1i + 1

2h1jayaj1i

i �h

2m!

=h12+ 21

2+ 1

2

i �h

2m!=

�h

2m!: (2.47)

So

h(�x)2iS (2:43)=

�h

2m!� �h

2m!cos2 !t =

�h

2m!sin2 !t: (2.48)

In the Heisenberg picture

x2H(t) =

"x(0) cos!t+

p(0)

m!sin!t

#2

= x2(0) cos2 !t+p2(0)

m2!2sin2 !t

+x(0)p(0)

m!cos!t sin!t+

p(0)x(0)

m!cos !t sin!t

=�h

2m!(a2 + ay

2+ aay+ aya) cos2 !t

� m�h!

2m2!2(a2 + ay

2 � aay� aya) sin2 !t

+i

m!

s�hm�h!

4m!(a+ ay)(ay� a)

sin 2!t

2

+i

m!

s�hm�h!

4m!(ay� a)(a+ ay)

sin 2!t

2

=�h

2m!(a2 + ay

2+ aay+ aya) cos2 !t

46

� �h

2m!(a2 + ay

2 � aay� aya) sin2 !t+i�h

2m!(ay

2 � a2) sin 2!t

=�h

2m!(aay+ aya) +

�h

2m!a2 cos 2!t+

�h

2m!ay

2cos 2!t

+i�h

2m!(ay

2 � a2) sin 2!t; (2.49)

which means that

hx2HiH = h�jx2H j�iH=

�h

2m!

"1p2h0j+ 1p

2h1j#

�haay+ aya+ a2 cos 2!t+ ay

2cos 2!t+ i(ay

2 � a2) sin 2!ti

�"1p2j0i + 1p

2j1i#

=�h

4m!

hh0jaayj0i+ h1jaayj1i + h1jayaj1i

i=

�h

4m![1 + 2 + 1] =

�h

m!: (2.50)

So

h(�x)2iH (2:44)=

�h

2m!� �h

2m!cos2 !t =

�h

2m!sin2 !t: (2.51)

2.7 A coherent state of a one-dimensional simple harmonic oscil-lator is de�ned to be an eigenstate of the (non-Hermitian) annihi-lation operator a:

aj�i = �j�i;where � is, in general, a complex number.

(a) Prove that

j�i = e�j�j2=2e�a

yj0iis a normalized coherent state.

(b) Prove the minimum uncertainty relation for such a state.


(c) Write j�i asj�i =

1Xn=0

f(n)jni:

Show that the distribution of jf(n)j2 with respect to n is of thePoisson form. Find the most probable value of n, hence of E.

(d) Show that a coherent state can also be obtained by applyingthe translation (�nite-displacement) operator e�ipl=�h (where p is themomentum operator, and l is the displacement distance) to theground state.

(e) Show that the coherent state j�i remains coherent under time-evolution and calculate the time-evolved state j�(t)i. (Hint: di-rectly apply the time-evolution operator.)

(a) We have

aj�i = e�j�j2=2ae�a

yj0i = e�j�j2=2ha; e�a

yij0i; (2.52)

since aj0i = 0. The commutator is

ha; e�a

yi

=

"a;

1Xn=0

1

n!(�ay)n

#=

1Xn=0

1

n!�nha; (ay)n

i

=1Xn=1

1

n!�n

nXk=1

(ay)k�1ha; ay

i(ay)n�k =

1Xn=1

1

n!�n

nXk=1

(ay)n�1

=1Xn=1

1

(n� 1)!�n(ay)n�1 = �

1Xn=0

1

n!(�ay)n = �e�a

y

: (2.53)

So from (2.52)

aj�i = e�j�j2=2�e�a

yj0i = �j�i; (2.54)

which means that j�i is a coherent state. If it is normalized, it should satisfyalso h�j�i = 1. Indeed

h�j�i = h0je��ae�j�j2e�ayj0i = e�j�j2h0je��ae�ayj0i

48

= e�j�j2 Xn;m

1

n!m!(��)n�mh0janj(ay)mj0i [(ay)mj0i =

pm!jmi]

= e�j�j2 Xn;m

pn!

n!

pm!

m!(��)n�mhnjmi = e�j�j

2Xn

1

n!(j�j2)n

= e�j�j2ej�j

2= 1: (2.55)

(b) According to problem (1.3) the state should satisfy the following relation

�xj�i = c�pj�i; (2.56)

where �x � x � h�jxj�i, �p � p � h�jpj�i and c is a purely imaginarynumber.

Since j�i is a coherent state we have

aj�i = �j�i ) h�jay = h�j��: (2.57)

Using this relation we can write

xj�i =s

�h

2m!(a+ ay)j�i =

s�h

2m!(�+ ay)j�i (2.58)

and

hxi = h�jxj�i =s

�h

2m!h�j(a+ ay)j�i =

s�h

2m!(h�jaj�i + h�jayj�i)

=

s�h

2m!(� + ��) (2.59)

and so

�xj�i = (x� hxi)j�i =s

�h

2m!(ay� ��)j�i: (2.60)

Similarly for the momentum p = iq

m�h!2 (ay� a) we have

pj�i =pi

sm�h!

2(ay � a)j�i = i

sm�h!

2(ay� �)j�i (2.61)


and

hpi = h�jpj�i = i

sm�h!

2h�j(ay � a)j�i = i

sm�h!

2(h�jayj�i � h�jaj�i)

= i

sm�h!

2(�� ) (2.62)

and so

�pj�i = (p� hpi)j�i = i

sm�h!

2(ay� ��)j�i )

(ay� ��)j�i = �is

2

m�h!�pj�i: (2.63)

So using the last relation in (2.60)

�xj�i =s

�h

2m!(�i)

s2

m�h!�pj�i = � i

m!| {z }purely imaginary

�pj�i (2.64)

and thus the minimum uncertainty condition is satis�ed.

(c) The coherent state can be expressed as a superposition of energy eigen-states

j�i =1Xn=0

jnihnj�i =1Xn=0

f(n)jni: (2.65)

for the expansion coe�cients f(n) we have

f(n) = hnj�i = hnje�j�j2=2e�ayj0i = e�j�j2=2hnje�ayj0i

= e�j�j2=2hnj

1Xm=0

1

m!(�ay)mj0i = e�j�j

2=21X

m=0

1

m!�mhnj(ay)mj0i

= e�j�j2=2

1Xm=0

1

m!�mpm!hnjmi = e�j�j

2=2 1pn!�n ) (2.66)

jf(n)j2 =(j�j2)nn!

exp(�j�j2) (2.67)

whichmeans that the distribution of jf(n)j2 with respect to n is of the Poissontype about some mean value n = j�j2.

50

The most probable value of n is given by the maximumof the distributionjf(n)j2 which can be found in the following way

jf(n + 1)j2jf(n)j2 =

(j�j2)n+1

(n+1)!exp(�j�j2)

(j�j2)nn!

exp(�j�j2) =j�j2n+ 1

� 1 (2.68)

which means that the most probable value of n is j�j2.(d) We should check if the state exp (�ipl=�h) j0i is an eigenstate of the an-nihilation operator a. We have

a exp (�ipl=�h) j0i =ha; e(�ipl=�h)

ij0i (2.69)

since aj0i = 0. For the commutator in the last relation we have

ha; e(�ipl=�h)

i=

1Xn=0

1

n!

�il�h

!n

[a; pn] =1Xn=1

1

n!

�il�h

!n nXk=1

pk�1[a; p]pn�k

=1Xn=1

1

n!

�il�h

!n nXk=1

pn�1i

sm�h!

2

= i

sm�h!

2

�il�h

! 1Xn=1

1

(n � 1)!

�ilp�h

!n�1

= l

rm!

2�he(�ipl=�h); (2.70)

where we have used that

[a; p] = i

sm�h!

2[a; ay� a] = i

sm�h!

2: (2.71)

So substituting (2.70) in (2.69) we get

a [exp (�ipl=�h) j0i] = l

rm!

2�h[exp (�ipl=�h) j0i] (2.72)

which means that the state exp (�ipl=�h) j0i is a coherent state with eigen-

value lq

m!2�h .

(e) Using the hint we have

j�(t)i = U(t)j�i = e�iHt=�hj�i (2:66)= e�iHt=�h

1Xn=0

e�j�j2=2 1p

n!�njni


=1Xn=0

e�iEnt=�he�j�j2=2 1p

n!�njni (2:3:9)

=1Xn=0

e�it�h �h!(n+

12)e�j�j

2=2 1pn!�njni

=1Xn=0

�e�i!t

�ne�i!t=2e�j�j

2=2 1pn!�njni

= e�i!t=21Xn=0

e�j�e�i!tj2=2 (�e

�i!t)npn!

jni (2:66)= e�i!t=2j�e�i!ti (2.73)

Thus

aj�(t)i = e�i!t=2aj�e�i!ti = �e�i!te�i!t=2j�e�i!ti= �e�i!tj�(t)i: (2.74)

2.8 The quntum mechanical propagator, for a particle with massm, moving in a potential is given by:

K(x; y;E) =Z 1

0dteiEt=�hK(x; y; t; 0) = A

Xn

sin(nrx) sin(nry)

E � �h2r2

2mn2

where A is a constant.(a) What is the potential?

(b) Determine the constant A in terms of the parameters describingthe system (such as m, r etc. ).

We have

K(x; y;E) �Z 1

0dteiEt=�hK(x; y; t; 0) �

Z 1

0dteiEt=�hhx; tjy; 0i

=Z 1

0dteiEt=�hhxje�iHt=�hjyi

=Z 1

0dteiEt=�h

Xn

hxje�iHt=�hjnihnjyi

=Z 1

0dteiEt=�h

Xn

e�iEnt=�hhxjnihnjyi

=Xn

�n(x)��n(y)

Z 1

0ei(E�En)t=�hdt

52

=Xn

�n(x)��n(y) lim"!0

" �i�hE � En + i"

ei(E�En+i")t=�h

#10

=Xn

�n(x)��n(y)

i�h

E � En: (2.75)

So

Xn

�n(x)��n(y)

i�h

E �En= A

Xn

sin(nrx) sin(nry)

E � �h2r2

2mn2

)

�n(x) =

sA

i�hsin(nrx); En =

�h2r2

2mn2: (2.76)

For a one dimensional in�nite square well potential with size L the energyeigenvalue En and eigenfunctions �n(x) are given by

�n(x) =

s2

Lsin

�n�x

L

�; En =

�h2

2m

��

L

�2n2: (2.77)

Comparing with (2.76) we get �L= r) L = �

rand

V =

(0 for 0 < x < �

r

1 otherwise(2.78)

while

A

i�h=

2r

�) A = i

2�hr

�: (2.79)

2.9 Prove the relationd�(x)

dx= �(x)

where �(x) is the (unit) step function, and �(x) the Dirac deltafunction. (Hint: study the e�ect on testfunctions.)

For an arbitrary test function f(x) we have

Z +1

�1d�(x)

dxf(x)dx =

Z +1

�1d

dx[�(x)f(x)]dx �

Z +1

�1�(x)

df(x)

dxdx


= �(x)f(x)

��+1�1 �Z +1

0

df(x)

dxdx

= limx!+1 f(x)� f(x)

��+10

= f(0)

=Z +1

�1�(x)f(x)dx)

d�(x)

dx= �(x): (2.80)

2.10 Derive the following expression

Scl =m!

2 sin(!T )

h(x20 + 2x2T ) cos(!T )� x0xT

i

for the classical action for a harmonic oscillator moving from thepoint x0 at t = 0 to the point xT at t = T .

The Lagrangian for the one dimensional harmonic oscillator is given by

L(x; _x) = 12m _x2 � 1

2m!2x2: (2.81)

From the Lagrange equation we have

@L@x

� d

dt

@L@ _x

= 0(2:81)) �m!2x� d

dt(m _x) = 0)

�x+ !2x = 0: (2.82)

which is the equation of motion for the system. This can be solved to give

x(t) = A cos!t+B sin!t (2.83)

with boundary conditions

x(t = 0) = x0 = A (2.84)

x(t = T ) = xT = x0 cos!T +B sin!T ) B sin!T = xT � x0 cos!T )B =

xT � x0 cos!T

sin!T: (2.85)

54

So

x(t) = x0 cos!t+xT � x0 cos!T

sin!Tsin!t

=x0 cos!t sin!T + xT sin!t� x0 cos!T sin!t

sin!T

=xT sin!t+ x0 sin!(T � t)

sin!T) (2.86)

_x(t) =xT! cos!t� x0! cos!(T � t)

sin!T: (2.87)

With these at hand we have

S =Z T

0dtL(x; _x) =

Z T

0dt�12m _x2 � 1

2m!2x2

�

=Z T

0dt

"12m

d

dt(x _x)� 1

2mx�x� 12m!

2x2#

= �12mZ T

0dtx[�x+ !2x] +

m

2x _x��T0

(2:82)=

m

2[x(T ) _x(T )� x(0) _x(0)]

=m

2

�xT!

sin!T(xT cos!T � x0)� x0!

sin!T(xT � x0 cos!T )

�=

m!

2 sin!T

hx2T cos!T � x0xT � x0xT + x20 cos!T

i=

m!

2 sin!T

h(x2T + x20) cos!T � 2x0xT

i: (2.88)

2.11 The Lagrangian of the single harmonic oscillator is

L =1

2m _x2 � 1

2m!2x2

(a) Show that


�G(0; tb; 0; ta)

where Scl is the action along the classical path xcl from (xa; ta) to(xb; tb) and G is

G(0; tb; 0; ta) =


limN!1

Zdy1 : : : dyN

�m

2�i�h"

� (N+1)2

exp

8<: i�h

NXj=0

�m

2"(yj+1 � yj)

2 � 1

2"m!2y2j

�9=;

where " = tb�ta(N+1)

.

[Hint: Let y(t) = x(t) � xcl(t) be the new integration variable,xcl(t) being the solution of the Euler-Lagrange equation.]

(b) Show that G can be written as

G = limN!1

�m

2�i�h"

� (N+1)2

Zdy1 : : : dyNexp(�nT�n)

where n =

2664y1...yN

3775 and nT is its transpose. Write the symmetric

matrix �.

(c) Show that

Zdy1 : : : dyNexp(�nT�n) �

ZdNne�n

T �n =�N=2pdet�

[Hint: Diagonalize � by an orthogonal matrix.]

(d) Let�2i�h"m

�Ndet� � det�0N � pN . De�ne j � j matrices �0j that con-

sist of the �rst j rows and j columns of �0N and whose determinantsare pj . By expanding �0j+1 in minors show the following recursionformula for the pj :

pj+1 = (2 � "2!2)pj � pj�1 j = 1; : : : ; N (2.89)

(e) Let �(t) = "pj for t = ta+ j" and show that (2.89) implies that inthe limit "! 0; �(t) satis�es the equation

d2�

dt2= �!2�(t)

with initial conditions �(t = ta) = 0; d�(t=ta)dt

= 1.

56

(f) Show that

hxbtbjxatai =s

m!


(im!

2�h sin(!T )[(x2b + x2a) cos(!T )� 2xaxb]

)

where T = tb � ta.

(a) Because at any given point the position kets in the Heisenberg pictureform a complete set, it is legitimate to insert the identity operator writtenas Z

dxjxtihxtj = 1 (2.90)

So

hxbtbjxatai = limN!1

Zdx1dx2 : : : dxNhxbtbjxNtNihxN tN jxN�1tN�1i : : :�

hxi+1ti+1jxitii : : : hx1t1jxatai: (2.91)

It is

hxi+1ti+1jxitii = hxi+1je�iH(ti+1�ti)=�hjxii = hxi+1je�iH"=�hjxii= hxi+1je�i

"�h(

12mp2+

12m!2x2)jxii (since " is very small)

= hxi+1je�i "�hp2

2m e�i"�h12m!2x2jxii

= e�i"�h

12m!2x2

i hxi+1je�i "�hp2

2m jxii: (2.92)

For the second term in this last equation we have

hxi+1je�i "�hp2

2m jxii =Zdpihxi+1je�i "�h

p2

2m jpiihpijxii

=Zdpie

�i "�hp2i

2m hxi+1jpiihpijxii

=1

2��h

Zdpie

�i "�hpi2

2m eipi(xi+1�xi)=�h

=1

2��h

Zdpie

�i "2m�h

hp2i�2pi m" (xi+1�xi)+m2

"2(xi+1�xi)2�m2

"2(xi+1�xi)2

i

=1

2��he

i"2m�h

m2

"2(xi+1�xi)2

Zdpie

�i "2m�h[pi�pi m" (xi+1�xi)]2


=1

2��he

i"2m�h

m2

"2(xi+1�xi)2

s�2�hm

i"

=

rm

2��hi"eim2�h" (xi+1�xi)2: (2.93)

Substituting this in (2.92) we get

hxi+1ti+1jxitii =�

m

2�i�h"

�12ei�h[

m2" (xi+1�xi)2� 1

2 "m!2xi] (2.94)

and this into (2.91):

hxbtbjxatai �ZDx exp

�i

�hS[x]

�=

limN!1

Zdx1 : : : dxN

�m

2�i�h"

� (N+1)2

exp

8<: i�h

NXj=0

�m

2"(xj+1 � xj)

2 � 1

2"m!2x2j

�9=; :

Let y(t) = x(t)� xcl(t)) x(t) = y(t)+ xcl(t)) _x(t) = _y(t) + _xcl(t) withboundary conditions y(ta) = y(tb) = 0: For this new integration variable wehave Dx = Dy and

S[x] = S[y + xcl] =Z tb

taL(y + xcl; _y + _xcl)dt

=Z tb

ta

24L(xcl; _xcl) + @L

@x

��xcl

y +@L@ _x

��xcl

_y + 12

@2L@2x

��xcl

y2 + 12

@2L@ _x2

��xcl

_y2

35

= Scl +@L@ _x

y

��tb

ta

+Z tb

ta

"@L@x

� d

dt

@L@ _x

!#��xcl

y +Z tb

ta

h12m _y2 � 1

2m!2y2

idt:

So

hxbtbjxatai =ZDy exp

�i

�hScl +

i

�h

Z tb

ta

h12m _y2 � 1

2m!2y2

idt�

= exp�iScl�h

�G(0; tb; 0; ta) (2.95)

with

G(0; tb; 0; ta) =

limN!1

Zdy1 : : : dyN

�m

2�i�h"

� (N+1)2

exp

8<: i�h

NXj=0

�m

2"(yj+1 � yj)

2 � 1

2"m!2y2j

�9=; :

58

(b) For the argument of the exponential in the last relation we have

i

�h

NXj=0

�m

2"(yj+1 � yj)

2 � 1

2"m!2y2j

�(y0=0)=

i

�h

NXj=0

m

2"(y2j+1 + y2j � yj+1yj � yjyj+1)� i

�h

NXi;j=1

1

2"m!2yi�ijyj

(yN+1=0)=

� m

2"i�h

NXi;j=1

(2yi�ijyj � yi�i;j+1yj � yi�i+1;jyj)� i"m!2

2�h

NXi;j=1

yi�ijyj :(2.96)

where the last step is written in such a form so that the matrix � will besymmetric. Thus we have

G = limN!1

�m

2�i�h"

� (N+1)2

Zdy1 : : : dyNexp(�nT�n) (2.97)

with

� =m

2"i�h

26666666664

2 �1 0 : : : 0 0�1 2 �1 : : : 0 00 �1 2 : : : 0 0...

......

......

0 0 0 : : : 2 �10 0 0 : : : �1 2

37777777775+i"m!2

2�h

26666666664

1 0 0 : : : 0 00 1 0 : : : 0 00 0 1 : : : 0 0...

......

......

0 0 0 : : : 1 00 0 0 : : : 0 1

37777777775:(2.98)

(c) We can diagonalize � by a unitary matrix U . Since � is symmetric thefollowing will hold

� = Uy�DU ) �T = UT�D(Uy)T = UT�DU

� = �) U = U�: (2.99)

So we can diagonalize � by an orthogonal matrix R. So

� = RT�DR and detR = 1 (2.100)

which means thatZdNne�n

T �n =ZdNne�n

TRT�Rn Rn=�=

ZdN�e��

T ��

=�Z

d�1e��21a1

� �Zd�2e

��22a2�: : :�Z

d�N e��2

NaN

�

=

s�

a1

s�

a2: : :

s�

aN=

�N=2qQNi=1 ai

=�N=2pdet�D

=�N=2pdet�

(2.101)


where ai are the diagonal elements of the matrix �D.

(d) From (2.98) we have

2i�h"

m

!N

det� =

det

8>>>>>>>>><>>>>>>>>>:

26666666664

2 �1 0 : : : 0 0�1 2 �1 : : : 0 00 �1 2 : : : 0 0...

......

......

0 0 0 : : : 2 �10 0 0 : : : �1 2

37777777775� "2!2

26666666664

1 0 0 : : : 0 00 1 0 : : : 0 00 0 1 : : : 0 0...

......

......

0 0 0 : : : 1 00 0 0 : : : 0 1

37777777775

9>>>>>>>>>=>>>>>>>>>;=

det�0N � pN : (2.102)

We de�ne j � j matrices �0j that consist of the �rst j rows and j columns of�0N . So

det�0j+1 = det

2666666666664

2� "2!2 �1 : : : 0 0 0�1 2� "2!2 : : : 0 0 00 �1 : : : 0 0 0...

......

......

0 0 : : : 2� "2!2 �1 00 0 : : : �1 2 � "2!2 �10 0 : : : 0 �1 2 � "2!2

3777777777775:

From the above it is obvious that

det�0j+1 = (2� "2!2) det�0j � det�0j�1 )pj+1 = (2� "2!2)pj � pj�1 for j = 2; 3; : : : ; N (2.103)

with p0 = 1 and p1 = 2� "2!2.

(e) We have

�(t) � �(ta + j") � "pj

) �(ta + (j + 1)") = "pj+1 = (2 � "2!2)"pj � "pj�1= 2�(ta + j")� "2!2�(ta + j")� �(ta + (j � 1)")

) �(t+ ") = 2�(t)� "2!2�(t)� �(t� "): (2.104)

60

So

�(t+ ")� �(t) = �(t)� �(t� ")� "2!2�(t))�(t+")��(t)

"� �(t)��(t�")

"

"= �!2�(t))

lim"!0

�0(t)� �0(t� ")

"= �!2�(t)) d2�

dt2= �!2�(t): (2.105)

From (c) we have also that

�(ta) = "p0 ! 0 (2.106)

and

d�

dt(ta) =

�(ta + ")� �(ta)

"="(p1 � p0)

"= p1 � p0

= 2 � "2!2 � 1! 1: (2.107)

The general solution to (2.105) is

�(t) = A sin(!t+ �) (2.108)

and from the boundary conditions (2.106) and (2.107) we have

�(ta) = 0) A sin(!ta + �) = 0) � = �!ta + n� n 2 Z (2.109)

which gives that �(t) = A sin!(t� ta), while

d�

dt= A! cos(t� ta)) �0(ta) = A!

(2:107))

A! = 1) A =1

!(2.110)

Thus

�(t) =sin!(t� ta)

!: (2.111)

(f) Gathering all the previous results together we get

G = limN!1

"�m

2�i�h"

�(N+1) �Npdet�

#1=2


=�

m

2�i�h

�1=2 24 limN!1

"

2i�h"

m

!N

det�

35�1=2

(d)=

�m

2�i�h

�1=2 �limN!1

"pN

��1=2(e)=�m

2�i�h

�1=2[�(tb)]

�1=2

(2:111)=

sm!

2�i�h sin(!T ): (2.112)

So from (a)


�G(0; tb; 0; ta)

(2:88)=

sm!


�im!

2�h sin !T

h(x2b + x2a) cos!T � 2xbxa

i�:

2.12 Show the composition propertyZdx1Kf (x2; t2;x1; t1)Kf (x1; t1;x0; t0) = Kf (x2; t2;x0; t0)

where Kf (x1; t1;x0; t0) is the free propagator (Sakurai 2.5.16), byexplicitly performing the integral (i.e. do not use completeness).

We haveZdx1Kf (x2; t2;x1; t1)Kf (x1; t1;x0; t0)

=Zdx1

sm

2�i�h(t2 � t1)exp

"im(x2 � x1)2

2�h(t2 � t1)

#�

sm


"im(x1 � x0)2

2�h(t1 � t0)

#

=m

2�i�h

s1

(t2 � t1)(t1 � t0)exp

imx222�h(t2 � t1)

expimx20

2�h(t2 � t1)�

Zdx1 exp

"im

2�h(t2 � t1)x21 +

im

2�h(t2 � t1)x21 �

im

2�h(t2 � t1)2x1x2 � im

2�h(t2 � t1)2x1x0

#

62

=m

2�i�h

s1

(t2 � t1)(t1 � t0)exp

(im

2�h

"x22

(t2 � t1)+

x20(t1 � t0)

#)�

Zdx1 exp

(im

2�h

"1

(t2 � t1)+

1

(t1 � t0)

#x21 �

im

�hx1

"x2

(t2 � t1)+

x0(t1 � t0)

#)

=m

2�i�h

s1

(t2 � t1)(t1 � t0)exp

(im

2�h

"x22

(t2 � t1)+

x20(t1 � t0)

#)�

Zdx1 exp

(�m2i�h

"t2 � t0

(t2 � t1)(t1 � t0)

#"x21 �

2�h

im

t2 � t0(t2 � t1)(t1 � t0)

im

�hx1

"x2(t1 � t0) + x0(t2 � t1)

(t2 � t1)(t1 � t0)

##)

=m

2�i�h

s1

(t2 � t1)(t1 � t0)exp

(im

2�h

"x22(t1 � t0) + x20(t2 � t1)

(t2 � t1)(t1 � t0)

#)�2

4Z dx1 exp

8<:�m2i�h

"t2 � t0

(t2 � t1)(t1 � t0)

# "x1 � x2(t1 � t0) + x0(t2 � t1)

(t2 � t0)

#29=;35�

exp

(�im2�h

1

(t2 � t1)(t1 � t0)

[x2(t1 � t0) + x0(t2 � t1)]2

(t2 � t0)

)

=m

2�i�h

s1

(t2 � t1)(t1 � t0)

vuut�2i�h(t2 � t1)(t1 � t0)

m(t2 � t0)exp

(im

2�h

1

(t2 � t1)(t1 � t0)�

"x22(t1 � t0)(t2 � t0) + x20(t2 � t1)(t2 � t0)

(t2 � t0)�

x22(t1 � t0)2 � x20(t2 � t1)2 � 2x2x02(t1 � t0)(t2 � t1)

(t2 � t0)

#)

=

sm

2�i�h(t2 � t1)�

exp

(im

2�h

"x22(t1 � t0)(t2 � t0 � t1 + t0) + x20(t2 � t1)(t2 � t0 � t2 + t1)

(t2 � t0)(t2 � t1)(t1 � t0)�

2x2x02(t1 � t0)(t2 � t1)

(t2 � t0)(t2 � t1)(t1 � t0)

#)

=

sm


"im(x2� x0)2

2�h(t2 � t0)

#

= Kf (x2; t2;x0; t0): (2.113)


2.13 (a) Verify the relation

[�i;�j] =

i�he

c

!"ijkBk

where ~� � m ~xdt= ~p � e ~A

cand the relation

md2~x

dt2=d~�

dt= e

"~E +

1

2c

d~x

dt� ~B � ~B � d~x

dt

!#:

(b) Verify the continuity equation

@�

@t+ ~r0 �~j = 0

with ~j given by

~j =

�h

m

!=( �~r0 )�

�e

mc

�~Aj j2:

(a) We have

[�i;�j] =�pi � eAi

c; pj � eAj

c

�= �e

c[pi; Aj]� e

c[Ai; pj]

=i�he

c

@Aj

@xi� i�he

c

@Ai

@xj=i�he

c

@Aj

@xi� @Ai

@xj

!

=

i�he

c

!"ijkBk: (2.114)

We have also that

dxidt

=1

i�h[xi;H] =

1

i�h

24xi; ~�2

2m+ e�

35 = 1

i�h

24xi; ~�2

2m

35

=1

i�h2mf[xi;�j] �j +�j [xi;�j]g = 1

i�h2mf[xi; pj ]�j +�j [xi; pj]g

=2i�h

2i�hm�j�ij =

�i

m)

64

d2xidt2

=1

i�h

"dxidt;H

#=

1

i�hm

24�i;

~�2

2m+ e�

35

=1

i�h2m2f[�i;�j] �j +�j [�i;�j]g+ e

i�hm[�i; �]

(2:114)=

1

2m2i�h

"i�he

c"ijkBk�j +

i�he

c"ijk�jBk

#+

e

i�hm

�pi � eAi

c; ��

=e

2m2c(�"ikjBk�j + "ijk�jBk) +

e

i�hm[pi; �]

=e

2m2cm�"ijk

xjdtBk � "ikjBk

xjdt

�� e

m

@�

@xi)

md2xidt2

= eEi +e

2c

" ~x

dt� ~B

!i

� ~B � ~x

dt

!i

#)

md2~x

dt2= e

"~E +

1

2c

~x

dt� ~B � ~B � ~x

dt

!#: (2.115)

(b) The time-dependent Schr�odinger equation is

i�h@

@thx0j�; t0; ti = hx0jHj�; t0; ti = hx0j 1

2m

0@~p� e ~A

c

1A2

+ e�j�; t0; ti

=1

2m

24�i�h~r0 � e ~A(~x0)

c

35 �24�i�h~r0 � e ~A(~x0)

c

35 hx0j�; t0; ti+ e�(~x0)hx0j�; t0; ti

=1

2m

"��h2~r0 � ~r0 +

e

ci�h~r0 � ~A(~x0) + i�h

e

c~A(~x0) � ~r0 +

e2

c2A2(~x0)

# (~x0; t)

+e�(~x0) (~x0; t)

=1

2m

��h2r02 (~x0; t)0 +

e

ci�h�~r0 � ~A

� (~x0; t) +

e

ci�h ~A(~x0) � ~r0 (~x0; t)

+ i�he

c~A(~x0) � ~r0 (~x0; t) +

e2

c2A2(~x0) (~x0; t)

#+ e�(~x0) (~x0; t)

=1

2m

"��h2r02 +

e

ci�h�~r0 � ~A

� + 2i�h

e

c~A � ~r0 +

e2

c2A2

#+ e� : (2.116)

Multiplying the last equation by � we get

i�h � @@t =


1

2m

"��h2 �r02 +

e

ci�h�~r0 � ~A

�j j2 + 2i�h

e

c~A � �~r0 +

e2

c2A2j j2

#+ e�j j2:

The complex conjugate of this eqution is

�i�h @@t � =

1

2m

"��h2 r02 � � e

ci�h�~r0 � ~A

�j j2 � 2i�h

e

c~A � ~r0 � +

e2

c2A2j j2

#+ e�j j2:

Thus subtracting the last two equations we get

� �h2

2m

h �r02 � r02 �i

+�e

mc

�i�h�~r0 � ~A

�j j2 +

�e

mc

�i�h ~A � ( �~r0 + ~r0 �)

= i�h

� @@t +

@

@t �!)

� �h2

2m~r0 �

h �~r0 � ~r0 �i+ �

e

mc

�i�h�~r0 � ~A

�j j2 +

�e

mc

�i�h ~A � (~r0j j2)

= i�h@

@tj j2 )

@

@tj j2 = � �h

m~r0 �

h=( �~r0 )

i+�e

mc

�~r0 �

h~Aj j2

i)

@

@tj j2 + ~r0 �

"�h

m=( �~r0 )�

�e

mc

�~Aj j2

#= 0)

@�

@t+ ~r0 �~j = 0 (2.117)

with ~j =��hm

�=( �~r0 )�

�emc

�~Aj j2: and � = j j2

2.14 An electron moves in the presence of a uniform magnetic �eldin the z-direction ( ~B = Bz).

(a) Evaluate[�x;�y];

where

�x � px � eAx

c; �y � py � eAy

c:

66

(b) By comparing the Hamiltonian and the commutation relationobtained in (a) with those of the one-dimensional oscillator problemshow how we can immediately write the energy eigenvalues as

Ek;n =�h2k2

2m+

jeBj�hmc

!�n+

1

2

�;

where �hk is the continuous eigenvalue of the pz operator and n is anonnegative integer including zero.

The magentic �eld ~B = Bz can be derived from a vector petential ~A(~x)of the form

Ax = �By2; Ay =

Bx

2; Az = 0: (2.118)

Thus we have

[�x;�y] =�px � eAx

c; py � eAy

c

�(2:118)=

�px +

eBy

2c; py � eBx

2c

�

= �eB2c

[px; x] +eB

2c[y; py] =

i�heB

2c+i�heB

2c

= i�heB

c: (2.119)

(b) The Hamiltonian for this system is given by

H =1

2m

0@~p � e ~A

c

1A2

=1

2m�2x +

1

2m�2y +

1

2mp2z = H1 +H2 (2.120)

where H1 � 12m�

2x +

12m�

2y and H2 � 1

2mp2z. Since

[H1;H2] =1

4m2

"�px +

eBy

2c

�2+�; py � eBx

2c

�2; p2z

#= 0 (2.121)

there exists a set of simultaneous eigenstates jk; ni of the operators H1 andH2. So if �hk is the continious eigegenvalue of the operator pz and jk; ni itseigenstate we will have

H2jk; ni = p2z2m

jk; ni = �h2k2

2m: (2.122)


On the other hand H1 is similar to the Hamiltonian of the one-dimensionaloscillator problem which is given by

H =1

2mp2 + 1

2m!2x2 (2.123)

with [x; p] = i�h. In order to use the eigenvalues of the harmonic oscillator

En = �h!�n+ 1

2

�we should have the same commutator between the squared

operators in the Hamiltonian. From (a) we have

[�x;�y] = i�heB

c) [

��xc

eB

�;�y] = i�h: (2.124)

So H1 can be written in the following form

H1 � 1

2m�2x +

1

2m�2y =

1

2m�2y +

1

2m

��xc

eB

�2 jeBj2c2

=1

2m�2y +

12m

jeBjmc

!2 ��xc

eB

�2: (2.125)

In this form it is obvious that we can replace ! with jeBjmc

to have

Hjk; ni = H1jk; ni+H2jk; ni = �h2k2

2mjk; ni+

jeBj�hmc

!�n+

1

2

�jk; ni

=

"�h2k2

2m+

jeBj�hmc

!�n+

1

2

�#jk; ni: (2.126)

2.15 Consider a particle of mass m and charge q in an impenetrablecylinder with radius R and height a. Along the axis of the cylin-der runs a thin, impenetrable solenoid carrying a magnetic ux �.Calculate the ground state energy and wavefunction.

In the case where ~B = 0 the Schr�odinger equation of motion in thecylindrical coordinates is

��h2

m[r2 ] = 2E )

��h2

m

h@2

@�2+ 1

�@@�+ 1

�2@2

@�2+ @2

@z2

i(~x) = 2E(~x) (2.127)

68

If we write (�; �; z) = �(�)R(�)Z(z) and k2 = 2mE�h2

we will have

�(�)Z(z)d2R

d�2+ �(�)Z(z)

1

�

dR

d�+R(�)Z(z)

�2d2�

d�2

+R(�)�(�)d2Z

dz2+ k2R(�)�(�)Z(z) = 0)

1

R(�)

d2R

d�2+

1

R(�)�

dR

d�+

1

�2�(�)

d2�

d�2+

1

Z(z)

d2Z

dz2+ k2 = 0(2.128)

with initial conditions (�a; �; z) = (R;�; z) = (�; �; 0) = (�; �; a) = 0.So

1

Z(z)

d2Z

dz2= �l2 ) d2Z

dz2+ l2Z(z) = 0) Z(z) = A1e

ilz +B1e�ilz (2.129)

with

Z(0) = 0) A1 + B1 = 0) Z(z) = A1

�eilz � e�ilz

�= C sin lz

Z(a) = 0) C sin la = 0) la = n�) l = ln = n�

an = �1;�2; : : :

So

Z(z) = C sin lnz (2.130)

Now we will have

1

R(�)

d2R

d�2+

1

R(�)�

dR

d�+

1

�2�(�)

d2�

d�2+ k2 � l2 = 0

) �2

R(�)

d2R

d�2+

�

R(�)

dR

d�+

1

�(�)

d2�

d�2+ �2(k2 � l2) = 0

) 1

�(�)

d2�

d�2= �m2 ) �(�) = e�im�: (2.131)

with

�(�+ 2�) = �(�)) m 2 Z: (2.132)

So the Schr�odinger equation is reduced to

�2

R(�)

d2R

d�2+

�

R(�)

dR

d��m2 + �2(k2 � l2) = 0


) d2R

d�2+1

�

dR

d�+

"(k2 � l2)� m2

�2

#R(�) = 0

) d2R

d(pk2 � l2�)2

+1p

k2 � l2�

dR

d(pk2 � l2�)

+

"1� m2

(k2 � l2)�2

#R(�) = 0

) R(�) = A3Jm(pk2 � l2�) +B3Nm(

pk2 � l2�) (2.133)

In the case at hand in which �a ! 0 we should take B3 = 0 since Nm ! 1when �! 0. From the other boundary condition we get

R(R) = 0) A3Jm(Rpk2 � l2) = 0) R

pk2 � l2 = �m� (2.134)

where �m� is the �-th zero of the m-th order Bessel function Jm. This meansthat the energy eigenstates are given by the equation

�m� = Rpk2 � l2 ) k2 � l2 =

�2m�

R2) 2mE

�h2��n�

a

�2=�2m�

R2

) E =�h2

2m

"�2m�

R2+�n�

a

�2#(2.135)

while the corresponding eigenfunctions are given by

nm�(~x) = AcJm(�m�

R�)eim� sin

�n�z

a

�(2.136)

with n = �1;�2; : : : and m 2 Z.Now suppose that ~B = Bz. We can then write

~A =

B�2a2�

!� =

�

2��

!�: (2.137)

The Schr�odinger equation in the presence of the magnetic �eld ~B can bewritten as follows

1

2m

24�i�h~r� e ~A(~x)

c

35 �24�i�h~r� e ~A(~x)

c

35 (~x) = E (~x)

) � �h2

2m

"�@

@�+ z

@

@z+ �

1

�

@

@�� ie

�hc

�

2�

!#�"

�@

@�+ z

@

@z+ �

1

�

@

@�� ie

�hc

�

2�

!# (~x) = E (~x): (2.138)

70

Making now the transformation D� � @@�� ie

�hc�2�

we get

� �h2

2m

"�@

@�+ z

@

@z+ �

1

�D�

#�"�@

@�+ z

@

@z+ �

1

�D�

# (~x) = E (~x)

) � �h2

2m

"@2

@�2+

1

�

@

@�+

1

�2D2

� +@2

@z2

#(~x) = E(~x); (2.139)

where D2� =

�@@�� ie

�hc�2�

�2. Leting A = e

�hc�2�

we get

D2� =

@2

@�2� 2ie

�hc

�

2�

@

@��A2

!=

@2

@�2� 2iA

@

@��A2

!: (2.140)

Following the same procedure we used before (i.e. (�; �; z) = R(�)�(�)Z(z))we will get the same equations with the exception of"@2

@�2� 2iA

@

@��A2

#� = �m2�) d2�

d�2� 2iA

d�

d�+ (m2 �A2)� = 0:

The solution to this equation is of the form el�. So

l2el� � 2iAlel� + (m2 �A2)el� = 0) l2 � 2iAl+ (m2 �A2)

) l =2iA�

q�4A2 � 4(m2 �A2)

2=

2iA� 2im

2= i(A�m)

which means that

�(�) = C2ei(A�m)�: (2.141)

But

�(�+ 2�) = �(�)) A�m = m0 m0 2 Z) m = �(m0 �A) m0 2 Z: (2.142)

This means that the energy eigenfunctions will be

nm�(~x) = AcJm(�m�

R�)eim

0� sin�n�z

a

�(2.143)

but now m is not an integer. As a result the energy of the ground state willbe

E =�h2

2m

"�2m�

R2+�n�

a

�2#(2.144)


where now m = m0 �A is not zero in general but it corresponds to m0 2 Zsuch that 0 � m0�A < 1. Notice also that if we require the ground state tobe unchanged in the presence of B, we obtain ux quantization

m0 �A = 0) e

�hc

�

2�= m0 ) � =

2�m0�hce

m0 2 Z: (2.145)

2.16 A particle in one dimension (�1 < x < 1) is subjected to aconstant force derivable from

V = �x; (� > 0):

(a) Is the energy spectrum continuous or discrete? Write down anapproximate expression for the energy eigenfunction speci�ed byE.

(b) Discuss brie y what changes are needed if V is replaced be

V = �jxj:

(a) In the case under construction there is only a continuous spectrum andthe eigenfunctions are non degenerate.

From the discussion on WKB approximation we had that for E > V (x)

I(x) =A

[E � V (x)]1=4exp

�i

�h

Z q2m[E � V (x)]dx

�

+B

[E � V (x)]1=4exp

�� i

�h

Z q2m[E � V (x)]dx

�

=c

[E � V (x)]1=4sin

1

�h

Z x0

x

q2m[E � V (x)]dx � �

4

!

=c

[E � V (x)]1=4sin

p2m�

�h

Z x0=E=�

x

�E

�� x

�1=2dx� �

4

!

=c

[E � V (x)]1=4sin

24�2

3

�E

�� x

�3=2s2m�

�h� �

4

35

=c1

[q]1=4sin

�2

3q3=2 +

�

4

�(2.146)

72

where q = �hE�� x

iand � =

�2m��h2

�1=3.

On the other hand when E < V (x)

II(x) =c2

[�x� E]1=4exp

�1

�h

Z x

x0=E=�

q2m(�x � E)dx

!

=c2

[�x� E]1=4exp

"� 1

�h2m�

Z x

x0=E=�

q2m(�x �E)d(2m�x)

#

=c3

[�q]1=4 exp��2

3(�q)2=3

�: (2.147)

We can �nd an exact solution for this problem so we can compare withthe approximate solutions we got with the WKB method. We have

Hj�i = Ej�i ) hpjHj�i = hpjEj�i) hpj p

2

2m+ �xj�i = Ehpj�i

) p2

2m�(p) + i�h�

d

dp�(p) = E�(p)

) d

dp�(p) =

�i�h�

E � p2

2m

!�(p)

) d�(p)

�(p)=�i�h�

E � p2

2m

!dp

) ln�(p) =�i�h�

Ep � p3

6m

!+ c1

) �E(p) = c exp

"i

�h�

p3

6m� Ep

!#: (2.148)

We also have

�(E � E0) = hEjE 0i =ZdphEjpihpjE 0i =

Z��E(p)�E0(p)dp

(2:148)= jcj2

Zdp exp

�i

�h�(E � E0)p

�jcj22��h��(E � E0))

c =1p2��h�

: (2.149)

So

�E(p) =1p2��h�

exp

"i

�h�

p3

6m�Ep

!#: (2.150)


These are the Hamiltonian eigenstates in momentum space. For the eigen-functions in coordinate space we have

(x) =ZdphxjpihpjEi (2:150)

=1

2��hp�

Zdpe

ipx�h e

i�h�

�p3

6m�Ep�

=1

2��hp�

Zdp exp

"ip3

�h�6m� i

�h

�E

�� x

�p

#: (2.151)

Using now the substitution

u =p

(�h2m�)1=3) p3

�h�6m=u3

3(2.152)

we have

(x) =(�h2m�)1=3

2��hp�

Z +1

�1du exp

"iu3

3� i

�h

�E

�� x

�u(�h2m�)1=3

#

=�

2�p�

Z +1

�1du exp

"iu3

3� iuq

#; (2.153)

where � =�2m��h2

�1=3and q = �

hE�� x

i. So

(x) =�

2�p�

Z +1

�1du cos

u3

3� uq

!=

�

�p�

Z +1

0cos

u3

3� uq

!du

sinceR+1�1 sin

�u3

3 � uq�du = 0. In terms of the Airy functions

Ai(q) =1p�

Z +1

0cos

u3

3� uq

!du (2.154)

we will have

(x) =�p��Ai(�q): (2.155)

For large jqj, leading terms in the asymptotic series are as follows

Ai(q) � 1

2p�q1=4

exp��2

3q3=2

�; q > 0 (2.156)

Ai(q) � 1p�(�q)1=4 sin

�2

3(�q)3=2+ �

4

�; q < 0 (2.157)

74

Using these approximations in (2.155) we get

(q) � �

�p�

1

q1=4sin

�2

3q3=2 +

�

4

�; for E > V (x)

(q) � �

2�p�

1

(�q)1=4 exp��2

3(�q)3=2

�; for E < V (x) (2.158)

as expected from the WKB approximation.

(b) When V = �jxj we have bound states and therefore the energy spec-trum is discrete. So in this case the energy eigenstates heve to satisfy theconsistency relationZ x1

x2dxq2m[E � �jxj] =

�n+ 1

2

��h; n = 0; 1; 2; : : : (2.159)

The turning points are x1 = �E�and x2 =

E�. So

�n + 1

2

��h =

Z E=�

�E=�dxq2m[E � �jxj] = 2

Z E=�

0

q2m[E � �x]dx

= �2p2m�

Z E=�

0

�E

�� x

�1=2d(�x)

= �2p2m�

2

3

�E

�� x

�3=2��E=�

0

= 2p2m�

2

3

�E

�

�3=2)

�E

�

�3=2=

3�n+ 1

2

��h

4p2m�

)�E

�

�=

[3�n+ 1

2

��h]2=3

42=3(2m�)1=3)

En =

243�n + 1

2

��h�

4p2m

352=3

: (2.160)


3 Theory of Angular Momentum

3.1 Consider a sequence of Euler rotations represented by

D(1=2)(�; �; ) = exp��i�3�

2

�exp

�i�2�2

!exp

��i�3 2

�

=

e�i(�+ )=2 cos �

2�e�i(�� )=2 sin �

2

ei(�� )=2 sin �2

ei(�+ )=2 cos �2

!:

Because of the group properties of rotations, we expect that thissequence of operations is equivalent to a single rotation about someaxis by an angle �. Find �.

In the case of Euler angles we have

D(1=2)(�; �; ) =

e�i(�+ )=2 cos �

2�e�i(�� )=2 sin �

2

ei(�� )=2 sin �2 ei(�+ )=2 cos �

2

!(3.1)

while the same rotation will be represented by

D(1=2)(�; n)(S�3:2:45)

=

0@ cos

��2

�� inz sin

��2

�(�inx � ny) sin

��2

�(�inx + ny) sin

��2

�cos

��2

�+ inz sin

��2

�1A : (3.2)

Since these two operators must have the same e�ect, each matrix elementshould be the same. That is

e�i(�+ )=2 cos�

2= cos

�

2

!� inz sin

�

2

!

) cos

�

2

!= cos

(� + )

2cos

�

2

) cos � = 2 cos2�

2cos2

(�+ )

2� 1

) � = arccos

"2 cos2

�

2cos2

(� + )

2� 1

#: (3.3)

3.2 An angular-momentum eigenstate jj;m = mmax = ji is rotatedby an in�nitesimal angle " about the y-axis. Without using the

76

explicit form of the d(j)m0m function, obtain an expression for the

probability for the new rotated state to be found in the originalstate up to terms of order "2.

The rotated state is given by

jj; jiR = R("; y)jj; ji = d(j)(")jj; ji =�exp

��iJy"

�h

��jj; ji

=

"1 � iJy"

�h+(�i)2"22�h2

J2y

#jj; ji (3.4)

up to terms of order "2. We can write Jy in terms of the ladder operators

J+ = Jx + iJyJ� = Jx � iJy

)) Jy =

J+ � J�2i

: (3.5)

Subtitution of this in (3.4), gives

jj; jiR =

"1 � "

2�h(J+ � J�) +

"2

8�h2(J+ � J�)2

#jj; ji (3.6)

We know that for the ladder operators the following relations hold

J+jj;mi = �hq(j �m)(j +m+ 1)jj;m+ 1i (3.7)

J�jj;mi = �hq(j +m)(j �m+ 1)jj;m� 1i (3.8)

So

(J+ � J�)jj; ji = �J�jj; ji = ��hq2jjj; j � 1i (3.9)

(J+ � J�)2jj; ji = ��hq2j(J+ � J�)jj; j � 1i

= ��hq2j (J+jj; j � 1i � J�jj; j � 1i)

= ��hq2j�q

2jjj; ji �q2(2j � 1)jj; j � 2i

�and from (3.6)

jj; jiR = jj; ji+ "

2

q2jjj; j � 1i � "2

82jjj; ji+ "2

82qj(2j � 1)jj; j � 2i

=

1� "2

4j

!jj; ji+ "

2

q2jjj; j � 1i + "2

4

qj(2j � 1)jj; j � 2i:


Thus the probability for the rotated state to be found in the original statewill be

jhj; jjj; jiRj2 =�� 1� "2

4j

!��2

= 1� "2

2j +O("4): (3.10)

3.3 The wave function of a particle subjected to a sphericallysymmetrical potential V (r) is given by

(~x) = (x+ y + 3z)f(r):

(a) Is an eigenfunction of ~L? If so, what is the l-value? If

not, what are the possible values of l we may obtain when ~L2 ismeasured?

(b)What are the probabilities for the particle to be found in variousml states?

(c) Suppose it is known somehow that (~x) is an energy eigenfunc-tion with eigenvalue E. Indicate how we may �nd V (r).

(a) We have

(~x) � h~xj i = (x+ y + 3z)f(r): (3.11)

So

h~xj~L2j i (S�3:6:15)= ��h2

"1

sin2 �

@2

@�2+

1

sin �

@

@�

sin �

@

@�

!# (~x): (3.12)

If we write (~x) in terms of spherical coordinates (x = r sin � cos�; y =r sin � sin�; z = r cos �) we will have

(~x) = rf(r) (sin � cos �+ sin � sin�+ 3 cos �) : (3.13)

Then

1

sin2 �

@2

@�2 (~x) =

rf(r) sin �

sin2 �

@

@�(cos �� sin�) = �rf(r)

sin �(cos �+ sin �)(3.14)

78

and

1

sin �

@

@�

sin �

@

@�

! (~x) =

rf(r)

sin �

@

@�

h�3 sin2 � + (cos�+ sin�) sin � cos �

i=

rf(r)

sin �

h�6 sin � cos � + (cos�+ sin �)(cos2 � � sin2 �)

i:(3.15)

Substitution of (3.14) and (3.15) in (3.11) gives

h~xj~L2j i = ��h2rf(r)�� 1

sin �(cos �+ sin�)(1� cos2 � + sin2 �) � 6 cos �

�

= �h2rf(r)�

1

sin �2 sin2 �(cos �+ sin�) + 6 cos �

�= 2�h2rf(r) [sin � cos �+ sin � sin�+ 3 cos �] = 2�h2 (~x))

L2 (~x) = 2�h2 (~x) = 1(1 + 1)�h2 (~x) = l(l+ l)�h2 (~x) (3.16)

which means that (~x) is en eigenfunction of ~L2 with eigenvalue l = 1.

(b) Since we already know that l = 1 we can try to write (~x) in terms ofthe spherical harmonics Y m

1 (�; �). We know that

Y 01 =

s3

4�cos � =

s3

4�

z

r) z = r

s4�

3Y 01

Y +11 = �

q38�

(x+iy)r

Y �11 =

q38�

(x�iy)r

9=;)

8<: x = r

q2�3

�Y �11 � Y +1

1

�y = ir

q2�3

�Y �11 + Y +1

1

�So we can write

(~x) = r

s2�

3f(r)

h3p2Y 0

1 + Y �11 � Y +1

1 + iY +11 + iY �1

1

i

=

s2�

3rf(r)

h3p2Y 0

1 + (1 + i)Y �11 + (i� 1)Y +1

1

i: (3.17)

But this means that the part of the state that depends on the values of mcan be written in the following way

j im = Nh3p2jl = 1;m = 0i + (1 + i)jl = 1;m = �1i + (1� i)jl = 1;m = 1i

iand if we want it normalized we will have

jN j2(18 + 2 + 2) = 1) N =1p22: (3.18)


So

P (m = 0) = jhl = 1;m = 0j ij2 = 9 � 222

=9

11; (3.19)

P (m = +1) = jhl = 1;m = +1j ij2 = 2

22=

1

11; (3.20)

P (m = �1) = jhl = 1;m == 1j ij2 = 2

22=

1

11: (3.21)

(c) If E(~x) is an energy eigenfunction then it solves the Schr�odinger equation

��h22m

"@2

@r2 E(~x) +

2

r

@

@r E(~x)� L2

�h2r2 E(~x)

#+ V (r) E(~x) =

E E(~x)

) ��h22m

Y ml

"d2

dr2[rf(r)] +

2

r

d

dr[rf(r)]� 2

r2[rf(r)]

#+ V (r)rf(r)Y m

l =

Erf(r)Y ml )

V (r) = E +1

rf(r)

�h2

2m

"d

dr[f(r) + rf 0(r)] +

2

r[f(r) + rf 0(r)]� 2

rf(r)

#)

V (r) = E +1

rf(r)

�h2

2m[f 0(r) + f 0(r) + rf 00(r) + 2f 0(r)]])

V (r) = E +�h2

2m

rf 00(r) + 4f 0(r)rf(r)

: (3.22)

3.4 Consider a particle with an intrinsic angular momentum (orspin) of one unit of �h. (One example of such a particle is the %-meson). Quantum-mechanically, such a particle is described by aketvector j%i or in ~x representation a wave function

%i(~x) = h~x; ij%iwhere j~x; ii correspond to a particle at ~x with spin in the i:th di-rection.

(a) Show explicitly that in�nitesimal rotations of %i(~x) are obtainedby acting with the operator

u~" = 1� i~"

�h� (~L+ ~S) (3.23)

80

where ~L = �hir � ~r. Determine ~S !

(b) Show that ~L and ~S commute.

(c) Show that ~S is a vector operator.

(d) Show that ~r� ~%(~x) = 1�h2( ~S � ~p)~% where ~p is the momentum oper-

ator.

(a) We have

j%i =3X

i=1

Zj~x; iih~x; ij%i =

3Xi=1

Zj~x; ii%i(~x)d3x: (3.24)

Under a rotation R we will have

j%0i = U(R)j%i =3X

i=1

ZU(R) [j~xi jii] %i(~x)d3x

=3X

i=1

ZjR~xi jiiD(1)

il (R)%l(~x)d3x

detR=1=

3Xi=1

Zj~x; iiD(1)

il (R)%l(R�1~x)d3x

=3X

i=1

Zj~x; ii%i0~x)d3x)

%i0(~x) = D(1)il (R)%

l(R�1~x)) ~%0(~x) = R~%(R�1~x): (3.25)

Under an in�nitesimal rotation we will have

R(��; n)~r = ~r + �~r = ~r + ��(n� ~r) = ~r + ~"� ~r: (3.26)

So

~%0(~x) = R(��)~%(R�1~x) = R(��)~%(~x� ~"� ~x)

= ~%(~x� ~"� ~x) + ~"� ~%(~x� ~"� ~x): (3.27)

On the other hand

~%(~x� ~"� ~x) = ~%(~x)� (~"� ~x) � ~r~%(~x) = ~%(~x) � ~" � (~x� ~r)~%(~x)= ~%(~x)� i

�h~" � ~L~%(~x) (3.28)


where ~r~%(~x) �h~r%i(~x)

ijii. Using this in (3.27) we get

~%0(~x) = ~%(~x)� i

�h~" � ~L~%(~x) + ~"�

�~%(~x)� i

�h~" � ~L~%(~x)

�

= ~%(~x)� i

�h~" � ~L~%(~x) + ~"� ~%(~x): (3.29)

But

~"� ~% =�"y%

3 � "z%2�ex +

�"z%

1 � "x%1�ey +

�"x%

2 � "y%1�ez (3.30)

or in matrix form0B@ %10

%20

%30

1CA =

0B@ 0 �"z "y

"z 0 �"x�"y "x 0

1CA0B@ %1

%2

%3

1CA =

264"x

0B@ 0 0 0

0 0 �10 1 0

1CA+ "y

0B@ 0 0 1

0 0 0�1 0 0

1CA+ "z

0B@ 0 �1 0

1 0 00 0 0

1CA3750B@ %1

%2

%3

1CA =

� i

�h

264"x

0B@ 0 0 0

0 0 �i�h0 i�h 0

1CA + "y

0B@ 0 0 i�h

0 0 0�i�h 0 0

1CA+ "z

0B@ 0 �i�h 0i�h 0 00 0 0

1CA3750B@ %1

%2

%3

1CA

which means that

~"� ~% = � i

�h~" � ~S~%(~x)

with (S�)kl = �i�h��kl.Thus we will have that

~%0(~x) = U~"~%(~x) =�1 � i

�h~" � (~L+ ~S)

�~%(~x)) U~" = 1 � i

�h~" � (~L+ ~S): (3.31)

(b) From their de�nition it is obvious that ~L and ~S commute since ~L acts

only on the j~xi basis and ~S only on jii.(c) ~S is a vector operator since

[Si; Sj]km = [SiSj � SjSi]km =X

[(�i�h)�ikl(�i�h)�jlm � (�i�h)�jkl(�i�h)�ilm]=

Xh�h2�ikl�jml � �h2�jkl�iml

i= �h2

X(�ij�km � �im�jk � �ij�km + �jm�ki)

= �h2X

(�jm�ki � �im�jk)

= �h2X

�ijl�kml =X

i�h�ijl(�i�h�kml) =X

i�h�ijl(Sl)km: (3.32)

82

(d) It is

~r� ~%(~x) =i

�h~p� ~%(~x) =

i

�h�i�lp�%

l(~x)jii = 1

�h2(S�)

lmp�%

mjii

=1

�h2(~S � ~p)~%: (3.33)

3.5 We are to add angular momenta j1 = 1 and j2 = 1 to formj = 2; 1; and 0 states. Using the ladder operator method express all(nine) j;m eigenkets in terms of jj1j2;m1m2i. Write your answer as

jj = 1;m = 1i = 1p2j+; 0i � 1p

2j0;+i; : : : ; (3.34)

where + and 0 stand for m1;2 = 1; 0; respectively.

We want to add the angular momenta j1 = 1 and j2 = 1 to form j =jj1 � j2j; : : : ; j1 + j2 = 0; 1; 2 states. Let us take �rst the state j = 2, m = 2.This state is related to jj1m1; j2m2i through the following equation

jj;mi = Xm=m1+m2

hj1j2;m1m2jj1j2; jmijj1j2;m1m2i (3.35)

So setting j = 2, m = 2 in (3.35) we get

jj = 2;m = 2i = hj1j2; + + jj1j2; jmij++i norm.= j++i (3.36)

If we apply the J� operator on this statet we will get

J�jj = 2;m = 2i = (J1� + J2�)j++i) �h

q(j +m)(j �m+ 1)jj = 2;m = 1i =

�hq(j1 +m1)(j1 �m1 + 1)j0+i+ �h

q(j2 +m2)(j2 �m2 + 1)j+ 0i

)p4jj = 2;m = 1i =

p2j0+i+

p2j+ 0i

) jj = 2;m = 1i = 1p2j0+i+ 1p

2j+ 0i: (3.37)


In the same way we have

J�jj = 2;m = 1i =1p2(J1� + J2�)j0+i + 1p

2(J1� + J2�)j+ 0i )

p6jj = 2;m = 0i =

1p2

hp2j �+i +

p2j00i

i+

1p2

hp2j00i +

p2j+�i

i)

p6jj = 2;m = 0i = 2j00i + j �+i + j+�i )

jj = 2;m = 0i =

s2

3j00i + 1p

6j+�i+ 1p

6j �+i (3.38)

J�jj = 2;m = 0i =

s2

3(J1� + J2�)j00i

+1p6(J1� + J2�)j+�i+ 1p

6(J1� + J2�)j �+i )

p6jj = 2;m = �1i =

s2

3

hp2j � 0i +

p2j0�i

i+

1p6

p2j0�i+ 1p

6

p2j � 0i )

jj = 2;m = �1i =2

6

p2j0�i + 2

6

p2j � 0i + 1

6

p2j0�i+ 1

6

p2j � 0i )

jj = 2;m = �1i =1p2j0�i+ 1p

2j � 0i (3.39)

J�jj = 2;m = �1i =1p2(J1� + J2�)j0�i+ 1p

2(J1� + J2�)j � 0i )

p4jj = 2;m = �2i =

1p2

p2j � �i + 1p

2

p2j � �i )

jj = 2;m = �2i = j � �i: (3.40)

Now let us return to equation (3.35). If j = 1, m = 1 we will have

jj = 1;m = 1i = aj+ 0i + bj0+i (3.41)

This state should be orthogonal to all jj;mi states and in particular to jj =2;m = 1i. So

hj = 2;m = 1jj = 1;m = 1i = 0) 1p2a+ 1p

2b = 0)

a+ b = 0) a = �b : (3.42)

84

In addition the state jj = 1;m = 1i should be normalized so

hj = 1;m = 1jj = 1;m = 1i = 1) jaj2 + jbj2 = 1(3:42)) 2jaj2 = 1) jaj = 1p

2:

By convention we take a to be real and positive so a = 1p2and b = � 1p

2.

That is

jj = 1;m = 1i = 1p2j+ 0i � 1p

2j0+i: (3.43)

Using the same procedure we used before

J�jj = 1;m = 1i =1p2(J1� + J2�)j+ 0i � 1p

2(J1� + J2�)j0+i )

p2jj = 1;m = 0i =

1p2

hp2j00i +

p2j+�i

i� 1p

2

hp2j �+i +

p2j00i

i)

jj = 1;m = 0i =1p2j+�i � 1p

2j �+i (3.44)

J�jj = 1;m = 0i =1p2(J1� + J2�)j+�i � 1p

2(J1� + J2�)j �+i )

p2jj = 1;m = �1i =

1p2

p2j0�i � 1p

2

p2j � 0i )

jj = 1;m = �1i =1p2j0�i � 1p

2j � 0i: (3.45)

Returning back to (3.35) we see that the state jj = 0;m = 0i can be writtenas

jj = 0;m = 0i = c1j00i + c2j+�i+ c3j �+i: (3.46)

This state should be orthogonal to all states jj;mi and in particulat to jj =2;m = 0i and to j = 1;m = 0i. So

hj = 2;m = 0jj = 0;m = 0i = 0)s2

3c1 +

1p6c2 +

1p6c3

) 2c1 + c2 + c3 = 0 (3.47)

hj = 1;m = 0jj = 0;m = 0i = 0) 1p2c2 � 1p

2c3

) c2 = c3: (3.48)


Using the last relation in (3.47), we get

2c1 + 2c2 = 0) c1 + c2 = 0) c1 = �c2: (3.49)

The state jj = 0;m = 0i should be normalized so

hj = 0;m = 0jj = 0;m = 0i = 1) jc1j2 + jc2j2 + jc3j2 = 1) 3jc2j2 = 1

) jc2j = 1p3: (3.50)

By convention we take c2 to be real and positive so c2 = c3 = 1p3and

c1 = � 1p3. Thus

jj = 0;m = 0i = 1p3j+�i + 1p

3j �+i � 1p

3j00i: (3.51)

So gathering all the previous results together

jj = 2;m = 2i = j++ijj = 2;m = 1i = 1p

2j0+i+ 1p

2j+ 0i

jj = 2;m = 0i =q

23 j00i + 1p

6j+�i+ 1p

6j �+i

jj = 2;m = �1i = 1p2j0�i+ 1p

2j � 0i

jj = 2;m = �2i = j � �ijj = 1;m = 1i = 1p

2j+ 0i � 1p

2j0+i

jj = 1;m = 0i = 1p2j+�i � 1p

2j �+i

jj = 1;m = �1i = 1p2j0�i � 1p

2j � 0i

jj = 0;m = 0i = 1p3j+�i+ 1p

3j �+i � 1p

3j00i:

(3.52)

3.6 (a) Construct a spherical tensor of rank 1 out of two di�erent

vectors ~U = (Ux; Uy; Uz) and ~V = (Vx; Vy; Vz). Explicitly write T(1)�1;0 in

terms of Ux;y;z and Vx;y;z .

(b) Construct a spherical tensor of rank 2 out of two di�erent

vectors ~U and ~V . Write down explicitly T(2)�2;�1;0 in terms of Ux;y;z

and Vx;y;z.

86

(a) Since ~U and ~V are vector operators they will satisfy the following com-mutation relations

[Ui; Jj] = i�h"ijkUk [Vi; Jj] = i�h"ijkVk: (3.53)

From the components of a vector operator we can construct a spherical tensorof rank 1 in the following way. The de�ning properties of a spherical tensorof rank 1 are the following

[Jz; U(1)q ] = �hqU (1)

q ; [J�; U (1)q ] = �h

q(1 � q)(2� q)U

(1)q�1: (3.54)

It is

[Jz; Uz](3:53)= 0�hUz

(3:54))Uz = U0 (3.55)

[J+; U0](3:54)=

p2�hU+1 = [J+; Uz] = [Jx + iJy; Uz]

(3:53)= �i�hUy + i(i�h)Ux = ��h(Ux + iUy))

U+1 = � 1p2(Ux + iUy) (3.56)

[J�; U0](3:54)=

p2�hU�1 = [J�; Uz] = [Jx � iJy; Uz]

(3:53)= �i�hUy � i(i�h)Ux = �h(Ux � iUy))

U�1 =1p2(Ux � iUy) (3.57)

So from the vector operators ~U and ~V we can construct spherical tensorswith components

U0 = Uz V0 = VzU+1 = � 1p

2(Ux + iUy) V+1 = � 1p

2(Vx + IVy)

U�1 = 1p2(Ux � iUy) V�1 = 1p

2(Vx � iVy)

(3.58)

It is known (S-3.10.27) that ifX(k1)q1

and Z(k2)q2

are irreducible spherical tensorsof rank k1 and k2 respectively then we can construct a spherical tensor ofrank k

T (k)q =

Xq1q2

hk1k2; q1q2jk1k2; kqiX(k1)q1

Z(k2)q2

(3.59)


In this case we have

T(1)+1 = h11;+10j11; 11iU+1V0 + h11; 0 + 1j11; 11iU0V+1

(3:52)=

1p2U+1V0 � 1p

2U0V+1

= �12(Ux + iUy)Vz +

12Uz(Vx + iVy) (3.60)

T(1)0 = h11; 00j11; 10iU0V0 + h11;�1 + 1j11; 10iU�1V+1

+h11;+1 � 1j11; 10iU+1V�1(3:52)= � 1p

2U�1V+1 +

1p2U+1V�1

=1p212(Ux � iUy)(Vx + iVy)� 1p

212(Ux + iUy)(Vx � iVy)

=1

2p2[UxVx + iUxVy � iUyVx + UyVy � UxVx + iUxVy � iUyVx � UyVy]

=ip2(UxVy � UyVx) (3.61)

T(1)�1 = h11;�10j11; 11iU�1V0 + h11; 0 � 1j11; 11iU0V�1

(3:52)= � 1p

2U�1V0 +

1p2U0V�1

= �12(Ux � iUy)Vz +

12Uz(Vx � iVy): (3.62)

(b) In the same manner we will have

T(2)+2 = h11;+1 + 1j11; 2 + 2iU+1V+1

(3:52)= U+1V+1 =

12(Ux + iUy)(Vx + iVy)

= �12(UxVx � UyVy + iUxVy + iUyVx) (3.63)

T(2)+1 = h11; 0 + 1j11; 2 + 1iU0V+1 + h11;+10j11; 2 + 1iU+1V0

(3:52)=

1p2U0V+1 +

1p2U+1V0

= �12(UzVx + UxVz + iUzVy + iUyVz) (3.64)

T(2)0 = h11; 00j11; 20iU0V0 + h11;�1 + 1j11; 20iU�1V+1

+h11;+1 � 1j11; 20iU+1V�1

88

(3:52)=

s2

3U0V0 +

s1

6U�1V+1 +

s1

6U+1V�1

=

s2

3UzVz �

s1

612(Ux � iUy)(Vx + iVy)�

s1

6

1p212(Ux + iUy)(Vx � iVy)

=

s1

6

�2UzVz � 1

2UxVx � i

2UxVy +

i

2UyVx

�12UyVy � 1

2UxVx +

i

2UxVy � i

2UyVx � 1

2UyVy

�

=

s1

6(2UzVz � UxVx � UyVy) (3.65)

T(2)�1 = h11; 0 � 1j11; 2 � 1iU0V�1 + h11;�10j11; 2 + 1iU�1V0

(3:52)=

1p2U0V�1 +

1p2U�1V0

= 12(UzVx + UxVz � iUzVy � iUyVz) (3.66)

T(2)�2 = h11;�1 � 1j11; 2 � 2iU�1V�1 (3:52)

= U�1V�1 = 12(Ux � iUy)(Vx � iVy)

= 12(UxVx � UyVy � iUxVy � iUyVx): (3.67)

3.7 (a) EvaluatejX

m=�jjd(j)mm0(�)j2m

for any j (integer or half-integer); then check your answer for j = 12 .

(b) Prove, for any j,

jXm=�j

m2jd(j)m0m(�)j2 = 12j(j + 1) sin � +m02 + 1

2(3 cos2 � � 1):

[Hint: This can be proved in many ways. You may, for instance,examine the rotational properties of J2

z using the spherical (irre-ducible) tensor language.]


(a) We have

jXm=�j

jd(j)mm0(�)j2m

=jX

m=�jmjhjmje�iJy�=�hjjm0ij2

=jX

m=�jmhjmje�iJy�=�hjjm0i

�hjmje�iJy�=�hjjm0i

��

=jX

m=�jmhjmje�iJy�=�hjjm0ihjm0jeiJy�=�hjjmi

=jX

m=�jhjm0jeiJy�=�hmjjmihjmje�iJy�=�hjjm0i

=1

�hhjm0jeiJy�=�hJz

24 jXm=�j

jjmihjmj35 e�iJy�=�hjjm0i

=1

�hhjm0jeiJy�=�hJze�iJy�=�hjjm0i

=1

�hhjm0jD�(�; ey)JzD(�; ey)jjm0i: (3.68)

But the momentum ~J is a vector operator so from (S-3.10.3) we will havethat

D�(�; ey)JzD(�; ey) =Xj

Rzj(�; ey)Jj: (3.69)

On the other hand we know (S-3.1.5b) that

R(�; ey) =

0B@ cos � 0 sin�

0 1 0� sin � 0 cos �

1CA : (3.70)

SojX

m=�jjd(j)mm0(�)j2m =

1

�h[� sin�hjm0jJxjjm0i+ cos �hjm0jJzjjm0i]

=1

�h

�� sin�hjm0jJ+ + J�

2jjm0i + �hm0 cos�

�= m0 cos�: (3.71)

90

For j = 1=2 we know from (S-3.2.44) that

d(1=2)mm0 (�) =

cos �

2� sin �

2

sin �2

cos �2

!: (3.72)

So for m0 = 1=2

1=2Xm=�1=2

jd(j)m1=2(�)j2m = �12sin2

�

2+ 1

2cos2

�

2

= 12 cos� = m0 cos� (3.73)

while for m0 = �1=21=2X

m=�1=2jd(j)m1=2(�)j2m = �1

2 cos2 �

2+ 1

2 sin2 �

2

= �12cos � = m0 cos �: (3.74)

(b) We have

jXm=�j

m2jd(j)m0m(�)j2

=jX

m=�jm2jhjm0je�iJy�=�hjjmij2

=jX

m=�jm2hjm0je�iJy�=�hjjmi

�hjm0je�iJy�=�hjjmi

��

=jX

m=�jm2hjm0je�iJy�=�hjjmihjmjeiJy�=�hjjm0i

=jX

m=�jhjm0je�iJy�=�hm2jjmihjmjeiJy�=�hjjm0i

=1

�h2hjm0je�iJy�=�hJ2

z

24 jXm=�j

jjmihjmj35 eiJy�=�hjjm0i

=1

�h2hjm0je�iJy�=�hJ2

z eiJy�=�hjjm0i

=1

�hhjm0jD(�; ey)J2

zDy(�; ey)jjm0i: (3.75)


From (3.65) we know that

T(2)0 =

s1

6(3J2

z � J2) (3.76)

where T (2)0 is the 0-component of a second rank tensor. So

J2z =

p6

3T(2)0 +

1

3J2 (3.77)

and since D(R)J2Dy(R) = J2D(R)Dy(R) = J2 we will have

Pjm=�j m2jd(j)mm0(�)j2 =

1

�h31

3hjm0jJ2jjm0i+

s2

3

1

�h2hjm0jD(�; ey)J2

zDy(�; ey)jjm0i:(3.78)

We know that for a spherical tensor (S-3.10.22b)

D(R)T (k)q Dy(R) =

kXq0=�k

D(k)q0q (R)T

(k)q0 (3.79)

which means in our case that

hjm0jD(�; ey)J2zDy(�; ey)jjm0i = hjm0j

2Xq0=�2

T(2)q0 D(2)

q00(�; ey)jjm0i

=2X

q0=�2D(2)q00(�; ey)hjm0jT (2)

q0 jjm0i: (3.80)

But we know from the Wigner-Eckart theorem that hjm0jT (2)q0 6=0jjm0i = 0. So

jXm=�j

m2jd(j)mm0(�)j2

=1

3�h2�h2j(j + 1) +

1

�h2

s2

3D(2)

00 (�; ey)hjm0jT (2)0 jjm0i

=1

3j(j + 1) +

1

2d(2)00 (�)hjm0jJ2

z �1

3J2jjm0i

=1

3j(j + 1) +

1

2d(2)00 (�)

�m02 � 1

3j(j + 1)

�

92

=1

3j(j + 1) +

1

2(3 cos2 � � 1)

�m02 � 1

3j(j + 1)

�

= �1

2j(j + 1) cos2 � +

1

6j(j + 1) +

1

3j(j + 1) +

m02

2(3 cos2 � � 1)

= 12j(j + 1) sin2 � +m02 1

2(3 cos2 � � 1) (3.81)

where we have used d(2)00 (�) = P2(cos �) =12(3 cos2 � � 1).

3.8 (a) Write xy, xz, and (x2 � y2) as components of a spherical(irreducible) tensor of rank 2.

(b) The expectation value

Q � eh�; j;m = jj(3z2 � r2)j�; j;m = ji

is known as the quadrupole moment. Evaluate

eh�; j;m0j(x2 � y2)j�; j;m = ji;

(where m0 = j; j�1; j�2; : : : )in terms of Q and appropriate Clebsch-Gordan coe�cients.

(a) Using the relations (3.63-3.67) we can �nd that in the case where ~U =~V = ~x the components of a spherical tensor of rank 2 will be

T(2)+2 = 1

2 (x2 � y2) + ixy T

(2)�2 = 1

2 (x2 � y2)� ixy

T(2)+1 = �(xz + izy) T

(2)�1 = xz � izy

T(2)0 =

q16(2z

2 � x2 � y2) =q

16(3z

2 � r2)

(3.82)

So from the above we have

�x2 � y2

�= T

(2)+2 + T

(2)�2 ; xy =

T(2)+2 � T

(2)�2

2i; xz =

T(2)�1 � T

(2)+1

2: (3.83)

(b) We have

Q = eh�; j;m = jj(3z2 � r2)j�; j;m = ji


(3:82)=

p6eh�; j;m = jjT (2)

0 j�; j;m = ji (W:�E:)= hj2; j0jj2; jjih�jkT

(2)k�jip2j + 1

p6e

) h�jkT (2)k�ji = Qp6e

p2j + 1

hj2; j0jj2; jji : (3.84)

So

e h�; j;m0j(x2 � y2)j�; j;m = ji(3:83)= eh�; j;m0jT (2)

+2 j�; j;m = ji+ eh�; j;m0jT (2)�2 j�; j;m = ji

= e

0z }| {hj2; j2jj2; jm0i h�jkT

(2)k�jip2j + 1

+ e�m0;j�2hj2; j � 2jj2; jj � 2ih�jkT(2)k�jip

2j + 1

(3:84)=

Qp6

hj2; j;�2jj2; j; j � 2ihj2; j; 0jj2; j; ji �m0;j�2: (3.85)

94

4 Symmetry in Quantum Mechanics

4.1 (a) Assuming that the Hamiltonian is invariant under timereversal, prove that the wave function for a spinless nondegeneratesystem at any given instant of time can always be chosen to bereal.

(b) The wave function for a plane-wave state at t = 0 is given bya complex function ei~p�~x=�h. Why does this not violate time-reversalinvariance?

(a) Suppose that jni in a nondegenerate energy eigenstate. Then

H�jni = �Hjni = Enjni ) �jni = ei�jni) �jn; t0 = 0; ti = �e�itH=�hjni = �e�itEn=�hjni =

eitEn=�h�jni = ei(Ent�h +�)jni = ei(

2Ent�h +�)jn; t0 = 0; ti

) ��Z

d3xj~xih~xj�jn; t0 = 0; ti = ei(

2Ent�h +�)

�Zd3xj~xih~xj

�jn; t0 = 0; ti

)Zd3xh~xjn; t0 = 0; ti�j~xi =

Zd3xei(

2Ent�h +�)h~xjn; t0 = 0; tij~xi

) ��n(~x; t) = ei(2Ent�h +�)�n(~x; t): (4.1)

So if we choose at any instant of time � = �2Ent�h the wave function will be

real.

(b) In the case of a free particle the Schr�odinger equation is

p2

2mjni = Ejni ) � �h2

2m~r�n(x) = E�n(x)

) �n(x) = Aei~p�~x=�h +Be�i~p�~x=�h (4.2)

The wave functions �n(x) = e�i~p�~x=�h and �0n(x) = ei~p�~x=�h correspond to the

same eigenvalue E = p2

2m and so there is degeneracy since these correspondto di�erent state kets j~pi and j � ~pi. So we cannot apply the previous result.

4.2 Let �(~p0) be the momentum-space wave function for state j�i,that is, �(~p0) = h~p0j�i.Is the momentum-space wave function for the


time-reversed state �j�i given by �(~p0), �(�~p0), ��(~p0), or ��(�~p0)?Justify your answer.

In the momentum space we have

j�i =Zd3p0h~p0j�ij~p0i ) j�i =

Zd3p0�(~p0)j~p0i

) �j�i =Zd3p0� [h~p0j�ij~p0i] =

Zd3p0h~p0j�i��j~p0i: (4.3)

For the momentum it is natural to require

h�j~pj�i = �h~�j~pj~�i )h~�j�~p��1j~�i ) �~p��1 = �~p (4.4)

So

�~pj~p0i (4:4)= �~p�j~p0i ) �j~p0i = j � ~p0i (4.5)

up to a phase factor. So �nally

�j�i =Zd3p0h~p0j�i�j � ~p0i =

Zd3p0h�~p0j�i�j~p0i

) h~p0j�j�i = ~�(~p0) = h�~p0j�i� = ��(�~p0): (4.6)

4.3 Read section 4.3 in Sakurai to refresh your knowledge of thequantum mechanics of periodic potentials. You know that the en-ergybands in solids are described by the so called Bloch functions n;k full�lling,

n;k(x+ a) = eika n;k(x)

where a is the lattice constant, n labels the band, and the latticemomentum k is restricted to the Brillouin zone [��=a; �=a].

Prove that any Bloch function can be written as,

n;k(x) =XRi

�n(x�Ri)eikRi

96

where the sum is over all lattice vectors Ri. (In this simble one di-mensional problem Ri = ia, but the construction generalizes easilyto three dimensions.).

The functions �n are called Wannier functions, and are impor-tant in the tight-binding description of solids. Show that the Wan-nier functions are corresponding to di�erent sites and/or di�erentbands are orthogonal, i:e: prove

Zdx�?m(x�Ri)�n(x�Rj) � �ij�mn

Hint: Expand the �ns in Bloch functions and use their orthonor-mality properties.

The de�ning property of a Bloch function n;k(x) is

n;k(x+ a) = eika n;k(x): (4.7)

We can show that the functionsP

Ri�n(x�Ri)eikRi satisfy the same relation

XRi

�n(x+ a�Ri)eikRi =

XRi

�n[x� (Ri � a)]eik(Ri�a)eika

Ri�a=Rj= eika

XRj

�n(x�Rj)eikRj (4.8)

which means that it is a Bloch function

n;k(x) =XRi

�n(x�Ri)eikRi: (4.9)

The last relation gives the Bloch functions in terms of Wannier functions.To �nd the expansion of a Wannier function in terms of Bloch functions wemultiply this relation by e�ikRj and integrate over k.

n;k(x) =XRi

�n(x�Ri)eikRi

)Z �=a

��=adke�ikRj n;k(x) =

XRi

�n(x�Ri)Z �=a

��=aeik(Ri�Rj)dk (4.10)


But

Z �=a

��=aeik(Ri�Rj)dk =

eik(Ri�Rj)

i(Ri �Rj)

��=a

��=a=

2 sin [�=a(Ri �Rj)]

Ri �Rj

= �ij2�

a(4.11)

where in the last step we used that Ri �Rj = na, with n 2 Z. SoZ �=a

��=adke�ikRj n;k(x) =

XRi

�n(x�Ri)�ij2�

a

) �n(x�Ri) =a

2�

Z �=a

��=ae�ikRi n;k(x)dk (4.12)

So using the orthonormality properties of the Bloch functionsZdx��m(x�Ri)�n(x�Rj)

=Z Z Z

a2

(2�)2eikRi �

m;k(x)e�ik0Rj n;k0(x)dkdk

0dx

=Z Z

a2

(2�)2eikRi�ik0Rj

Z �m;k(x) n;k0(x)dxdkdk

0

=Z Z a2

(2�)2eikRi�ik0Rj�mn�(k � k0)dkdk0

=a2

(2�)2�mn

Z �=a

��=aeik(Ri�Rj)dk =

a

2��mn�ij: (4.13)

4.4 Suppose a spinless particle is bound to a �xed center by apotential V (~x) so assymetrical that no energy level is degenerate.Using the time-reversal invariance prove

h~Li = 0

for any energy eigenstate. (This is known as quenching of orbitalangular momemtum.) If the wave function of such a nondegenerateeigenstate is expanded asX

l

Xm

Flm(r)Yml (�; �);

98

what kind of phase restrictions do we obtain on Flm(r)?

Since the Hamiltonian is invariant under time reversal

H� = �H: (4.14)

So if jni is an energy eigenstate with eigenvalue En we will have

H�jni = �Hjni = En�jni: (4.15)

If there is no degeneracy jni and �jni can di�er at most by a phase factor.Hence

j~ni � �jni = ei�jni: (4.16)

For the angular-momentum operator we have from (S-4.4.53)

hnj~Ljni = �h~nj~Lj~ni (4:16)= �hnj~Ljni )

hnj~Ljni = 0 : (4.17)

We have

�jni = �Zd3xj~xih~xjni =

Zd3xh~xjni��j~xi

=Zd3xh~xjni�j~xi (4:16)

= ei�jni )h~x0j�jni = h~x0jni� = ei�h~x0jni: (4.18)

So if we use h~xjni = Pl

Pm Flm(r)Y

ml (�; �)X

ml

F �lm(r)Y

m�l (�; �) = ei�

Xml

Flm(r)Yml (�; �)

(S�4:4:57)) Xml

F �lm(r)(�1)mY �m

l (�; �) = ei�Xml

Flm(r)Yml (�; �)

)ZY m0�l0

Xml

F �lm(r)(�1)mY �m

l (�; �)d = ei�ZY m0�l0

Xml

Flm(r)Yml (�; �)d

) Xml

F �lm(r)(�1)m�m0;�m�l0l = ei�

Xml

Flm(r)�m0;m�l0l

) F �l0;�m0(r)(�1)�m0

= ei�Fl0m0(r)) F �l0;�m0(r) = (�1)m0

Fl0m0(r)ei�: (4.19)


4.5 The Hamiltonian for a spin 1 system is given by

H = AS2z +B(S2

x � S2y ):

Solve this problem exactly to �nd the normalized energy eigen-states and eigenvalues. (A spin-dependent Hamiltonian of this kindactually appears in crystal physics.) Is this Hamiltonian invariantunder time reversal? How do the normalized eigenstates you ob-tained transform under time reversal?

For a spin 1 system l = 1 and m = �1; 0;+1. For the operator Sz wehave

Szjl;mi = �hmjl;mi ) hlnjSzjl;mi = �hmhnjmi ) (Sz)nm = �hm�nm (4.20)

So

Sz�= �h

0B@ 1 0 0

0 0 00 0 �1

1CA) S2

z�= �h2

0B@ 1 0 0

0 0 00 0 1

1CA

For the operator Sx we have

Sxjl;mi =S+ + S�

2j1;mi = 1

2S+j1;mi + 1

2S�j1;mi !

h1; njSxj1;mi = 12h1; njS+j1;mi+ 1

2h1; njS�j1;mi

(S�3:5:39)= 1

2�hq(1 �m)(2 +m)�n;m+1 +

12�hq(1 +m)(2�m)�n;m�1:

So

Sx�=

�h

2

0B@ 0

p2 0

0 0p2

0 0 0

1CA +

�h

2

0B@ 0 0 0p

2 0 0

0p2 0

1CA

=�h

2

0B@ 0

p2 0p

2 0p2

0p2 0

1CA)

S2x =

�h2

4

0B@ 2 0 2

0 4 02 0 2

1CA = �h2

0B@

12

0 12

0 1 012 0 1

2

1CA : (4.21)

100

In the same manner for the operator Sy =S+�S�

2iwe �nd

Sx�=

�h

2i

0B@ 0

p2 0

�p2 0p2

0 �p2 0

1CA)

S2x

�= ��h2

4

0B@ �2 0 2

0 �4 02 0 �2

1CA = �h2

0B@

12 0 �1

2

0 1 0�1

20 1

2

1CA : (4.22)

Thus the Hamiltonian can be represented by the matrix

H�= �h2

0B@ A 0 B

0 0 0B 0 A

1CA : (4.23)

To �nd the energy eigenvalues we have to solve the secular equation

det(H � �I) = 0) det

0B@ A�h2 � � 0 B�h2

0 �� 0B�h2 0 A�h2 � �

1CA = 0

) (A�h2 � �)2(��) + (B�h2)2� = 0) �h(A�h2 � �)2 � (B�h2)2

i= 0

) �(A�h2 � � �B�h2)(A�h2 � �+B�h2) = 0

) �1 = 0; �2 = �h2(A+B); �3 = �h2(A�B): (4.24)

To �nd the eigenstate jn�ci that corresponds to the eigenvalue �c we have tosolve the following equation

�h2

0B@ A 0 B

0 0 0B 0 A

1CA0B@ abc

1CA = �c

0B@ abc

1CA : (4.25)

For �1 = 0

�h2

0B@ A 0 B

0 0 0B 0 A

1CA0B@ abc

1CA = 0)

(aA+ cB = 0aB + cA = 0

)(

a = �cBA

�cB2

A+ cA = 0

)(a = 0c = 0

(4.26)


So

jn0i �=

0B@ 0b0

1CA norm:

=

0B@ 0

10

1CA)

jn0i = j10i: (4.27)

In the same way for � = �h2(A+B)

0B@ A 0 B

0 0 0B 0 A

1CA0B@ abc

1CA = (A+B)

0B@ abc

1CA)

8><>:aA+ cB = a(A+B)

0 = b(A+B)aB + cA = c(A+B)

)(a = cb = 0

(4.28)

So

jnA+Bi �=

0B@ c

0c

1CA norm:

=1p2

0B@ 1

01

1CA)

jnA+Bi =1p2j1;+1i + 1p

2j1;�1i: (4.29)

For � = �h2(A�B) we have

0B@ A 0 B

0 0 0B 0 A

1CA0B@ abc

1CA = (A�B)

0B@ abc

1CA)

8><>:aA+ cB = a(A�B)

0 = b(A�B)aB + cA = c(A�B)

)(a = �cb = 0

(4.30)

So

jnA+Bi �=

0B@ c

0�c

1CA norm:

=1p2

0B@ 1

0�1

1CA)

jnA�Bi =1p2j1;+1i � 1p

2j1;�1i: (4.31)

102

Now we are going to check if the Hamiltonian is invariant under time reversal

�H��1 = A�S2z�

�1 +B(�S2x�

�1 ��S2y�

�1)

= A�Sz��1�Sz��1 +B(�Sx�

�1�Sx��1 ��Sy��1�Sy��1)

= AS2z +B(S2

x � S2y ) = H: (4.32)

To �nd the transformation of the eigenstates under time reversal we use therelation (S-4.4.58)

�jl;mi = (�1)mjl;�mi: (4.33)

So

�jn0i = �j10i (4:33)= j10i

= jn0i (4.34)

(4.35)

�jnA+Bi =1p2�j1;+1i + 1p

2�j1;�1i

(4:33)= � 1p

2j1;�1i � 1p

2j1;+1i

= �jnA+Bi (4.36)

(4.37)

�jnA�Bi =1p2�j1;+1i � 1p

2�j1;�1i

(4:33)= � 1p

2j1;�1i+ 1p

2j1;+1i

= jnA�Bi: (4.38)


5 Approximation Methods

5.1 Consider an isotropic harmonic oscillator in two dimensions.The Hamiltonian is given by

H0 =p2x2m

+p2y2m

+m!2

2(x2 + y2)

(a) What are the energies of the three lowest-lying states? Is thereany degeneracy?

(b) We now apply a perturbation

V = �m!2xy

where � is a dimensionless real number much smaller than unity.Find the zeroth-order energy eigenket and the corresponding en-ergy to �rst order [that is the unperturbed energy obtained in (a)plus the �rst-order energy shift] for each of the three lowest-lyingstates.

(c) Solve the H0+V problem exactly. Compare with the perturba-tion results obtained in (b).


pn + 1�n0;n+1 +

pn�n0;n�1):]

De�ne step operators:

ax �rm!

2�h(x+

ipxm!

);

ayx �rm!

2�h(x� ipx

m!);

ay �rm!

2�h(y +

ipym!

);

ayy �rm!

2�h(y � ipy

m!): (5.1)

From the fundamental commutation relations we can see that

[ax; ayx] = [ay; a

yy] = 1:

104

De�ning the number operators

Nx � ayxax; Ny � ayyay

we �nd

N � Nx +Ny =H0

�h!� 1)

H0 = �h!(N + 1): (5.2)

I.e. energy eigenkets are also eigenkets of N :

Nx jm;n i = m jm;n i;Ny jm;n i = n jm;n i )N jm;n i = (m+ n) jm;n i (5.3)

so that

H0 jm;n i = Em;n jm;n i = �h!(m+ n+ 1) jm;n i:

(a) The lowest lying states are

state degeneracyE0;0 = �h! 1E1;0 = E0;1 = 2�h! 2E2;0 = E0;2 = E1;1 = 3�h! 3

(b) Apply the perturbation V = �m!2xy.

Full problem: (H0 + V ) j l i = E j l iUnperturbed problem: H0 j l0 i = E0 j l0 i

Expand the energy levels and the eigenkets as

E = E0 +�1 +�2 + : : :

j l i = j l0 i+ j l1 i + : : : (5.4)

so that the full problem becomes

(E0 �H0)hj l0 i + j l1 i+ : : :

i= (V ��1 ��2 : : :)

hj l0 i + j l1 i+ : : :

i:


To 1'st order:

(E0 �H0) j l1 i = (V ��1) j l0 i: (5.5)

Multiply with h l0 j to �nd

h l0 jE0 �H0 j l1 i = 0 = h l0 jV ��1 j l0 i )�1h l0 j l0 i = �1 = h l0 jV j l0 i (5.6)

In the degenerate case this does not work since we're not using the right basiskets. Wind back to (5.5) and multiply it with another degenerate basis ket

hm0 jE0 �H0 j l1 i = 0 = hm0 jV ��1 j l0 i )�1hm0 j l0 i = hm0 jV j l0 i: (5.7)

Now, hm0 j l0 i is not necessarily �kl since only states corresponding to di�er-ent eigenvalues have to be orthogonal!

Insert a 1: Xk2D

hm0 jV j k0 ih k0 j l0 i = �1hm0 j l0 i:

This is the eigenvalue equation which gives the correct zeroth order eigen-vectors!

Let us use all this:

1. The ground state is non-degenerate )

�100 = h 0; 0 jV j 0; 0 i = �m!2h 0; 0 jxy j 0; 0 i � h 0; 0 j (ax+ayx)(ay+ayy) j 0; 0 i = 0

2. First excited state is degenerate j 1; 0 i, j 0; 1 i. We need the matrixelements h 1; 0 jV j 1; 0 i, h 1; 0 jV j 0; 1 i, h 0; 1 jV j 1; 0 i, h 0; 1 jV j 0; 1 i.

V = �m!2xy = �m!2 �h

2m!(ax+a

yx)(ay+a

yy) =

��h!

2(axay+a

yxay+axa

yy+a

yxa

yy)

and

ax jm;n i =pm jm� 1; n i ayx jm;n i =

pm+ 1 jm+ 1; n i etc:

106

Together this gives

V10;10 = V01;01 = 0;

V10;01 =��h!

2h 1; 0 j axayy j 0; 1 i =

��h!

2;

V01;10 =��h!

2h 1; 0 j ayxay j 0; 1 i =

��h!

2:

(5.8)

The V -matrix becomes��h!

2

0 11 0

!

and so the eigenvalues (= �1) are

�1 = ��h!2:

To get the eigenvectors we solve 0 11 0

! xy

!= �

xy

!

and get

j � � � i+ =1p2( j 0; 1 i + j 1; 0 i); E+ = �h!(2 +

�

2);

j � � � i� =1p2( j 0; 1 i � j 1; 0 i); E� = �h!(2 � �

2): (5.9)

3. The second excited state is also degenerate j 2; 0 i, j 1; 1 i, j 0; 2 i, sowe need the corresponding 9 matrix elements. However the only non-vanishing ones are:

V11;20 = V20;11 = V11;02 = V02;11 =��h!p2

(5.10)

(where thep2 came from going from level 1 to 2 in either of the oscil-

lators) and thus to get the eigenvalues we evaluate

0 = det

0B@ �� 1 0

1 �� 10 1 ��

1CA = ��(�2 � 1) + � = �(2 � �2)


which means that the eigenvalues are f0;��h!g. By the same methodas above we get the eigenvectors

j � � � i+ = 12( j 2; 0 i +

p2 j 1; 1 i + j 0; 2 i); E+ = �h!(3 + �);

j � � � i0 =1p2(� j 2; 0 i + j 0; 2 i); E0 = 3�h!;

j � � � i� = 12( j 2; 0 i �

p2 j 1; 1 i + j 0; 2 i); E� = �h!(3 � �):

(c) To solve the problem exactly we will make a variable change. The poten-tial is

m!2h12(x2 + y2) + �xy

i=

= m!2

"1

4((x+ y)2 + (x� y)2) +

�

4(x+ y)2 � (x� y)2)

#: (5.11)

Now it is natural to introduce

x0 � 1p2(x+ y); p0x �

1p2(p0x + p0y);

y0 � 1p2(x� y); p0y �

1p2(p0x � p0y): (5.12)

Note: [x0; p0x] = [y0; p0y] = i�h, so that (x0, p0x) and (y0, p0y) are canonicallyconjugate.

In these new variables the problem takes the form

H =1

2m(p02x + p02y ) +

m!2

2[(1 + �)x02+ (1 � �)y02]:

So we get one oscillator with !0x = !p1 + � and another with !0y = !

p1 � �.

The energy levels are:

E0;0 = �h!;

E1;0 = �h! + �h!0x = �h!(1 +p1 + �) =

= �h!(1 + 1 + 12� + : : :) = �h!(2 + 1

2�) +O(�2);

108

E0;1 = �h! + �h!0y = : : : = �h!(2� 12�) +O(�2);

E2;0 = �h! + 2�h!0x = : : : = �h!(3 + �) +O(�2);

E1;1 = �h! + �h!0x + �h!0y = : : : = 3�h! +O(�2);

E0;2 = �h! + 2�h!0y = : : : = �h!(3 � �) +O(�2):

(5.13)

So �rst order perturbation theory worked!

5.2 A system that has three unperturbed states can be representedby the perturbed Hamiltonian matrix0

B@ E1 0 a0 E1 ba� b� E2

1CA

where E2 > E1. The quantities a and b are to be regarded as per-turbations that are of the same order and are small compared withE2 � E1. Use the second-order nondegenerate perturbation theoryto calculate the perturbed eigenvalues. (Is this procedure correct?)Then diagonalize the matrix to �nd the exact eigenvalues. Finally,use the second-order degenerate perturbation theory. Comparethe three results obtained.

(a) First, �nd the exact result by diagonalizing the Hamiltonian:

0 =

��E1 � E 0 a

0 E1 � E ba� b� E2 � E

�� == (E1 � E)

h(E1 �E)(E2 � E)� jbj2

i+ a [0 � a�(E1 � E)] =

= (E1 � E)2(E2 � E)� (E1 � E)(jbj2 + jaj2): (5.14)

So, E = E1 or (E1 � E)(E2 � E)� (jbj2 + jaj2) = 0 i.e.

E2 � (E1 + E2)E + E1E2 � (jaj2 + jbj2) = 0)

E =E1 + E2

2�sE1 + E2

2

2

� E1E2 + jaj2 + jbj2 =


=E1 + E2

2�sE1 � E2

2

2

+ jaj2 + jbj2: (5.15)

Since jaj2 + jbj2 is small we can expand the square root and write the threeenergy levels as:

E = E1;

E =E1 + E2

2+E1 �E2

2

�1 + 1

2(jaj2 + jbj2)(2

E1 � E2)2 + : : :

�=

= E1 +jaj2 + jbj2E1 � E2

;

E =E1 + E2

2� E1 � E2

2(: : :) = E2 � jaj2 + jbj2

E1 � E2:

(5.16)

(b) Non degenerate perturbation theory to 2'nd order. The basis we use is

j 1 i =0B@ 1

00

1CA ; j 2 i =

0B@ 0

10

1CA ; j 3 i =

0B@ 0

01

1CA :

The matrix elements of the perturbation V =

0B@ 0 0 a

0 0 ba� b� 0

1CA are

h 1 jV j 3 i = a; h 2 jV j 3 i = b; h 1 jV j 2 i = h k jV j k i = 0:

Since �(1)k = h k jV j k i = 0 1'st order gives nothing. But the 2'nd order

shifts are

�(2)1 =

Xk 6=1

jVk1j2E01 � E0

k

=jh 3 jV j 1 ij2E1 � E2

=jaj2

E1 � E2;

�(2)2 =

Xk 6=2

jVk2j2E02 � E0

k

=jh 3 jV j 2 ij2E1 � E2

=jbj2

E1 � E2;

�(2)3 =

Xk 6=3

jVk3j2E03 � E0

k

=jaj2

E2 �E1+

jbj2E2 � E1

= �jaj2 + jbj2

E1 � E2:

(5.17)

110

The unperturbed problem has two (degenerate) states j 1 i and j 2 i withenergy E1, and one (non-degenerate) state j 3 i with energy E2. Using non-

degenerate perturbation theory we expect only the correction to E2 (i.e. �(2)3 )

to give the correct result, and indeed this turns out to be the case.

(c) To �nd the correct energy shifts for the two degenerate states we haveto use degenerate perturbation theory. The V -matrix for the degenerate

subspace is

0 00 0

!, so 1'st order pert.thy. will again give nothing. We have

to go to 2'nd order. The problem we want to solve is (H0 + V ) j l i = E j l iusing the expansion

j l i = j l0 i+ j l1 i+ : : : E = E0 +�(1) +�(2) + : : : (5.18)

where H0 j l0 i = E0 j l0 i. Note that the superscript index in a bra or ket de-notes which order it has in the perturbation expansion. Di�erent solutions tothe full problem are denoted by di�erent l's. Since the (sub-) problem we arenow solving is 2-dimensional we expect to �nd two solutions correspondingto l = 1; 2. Inserting the expansions in (5.18) leaves us with

(E0 �H0)hj l0 i+ j l1 i + : : :

i=

(V ��(1) ��(2) : : :)hj l0 i+ j l1 i+ : : :

i: (5.19)

At �rst order in the perturbation this says:

(E0 �H0) j l1 i = (V ��(1)) j l0 i;where of course �(1) = 0 as noted above. Multiply this from the left with abra h k0 j from outside the deg. subspace

h k0 jE0 �H0 j l1 i = h k0 jV j l0 i) j l1 i = X

k 6=D

j k0 ih k0 jV j l0 iE0 � Ek

: (5.20)

This expression for j l1 i we will use in the 2'nd order equation from (5.19)

(E0 �H0) j l2 i = V j l1 i ��(2) j l0 i:To get rid of the left hand side, multiply with a degenerate bra hm0 j(H0 jm0 i = E0 jm0 i)

hm0 jE0 �H0 j l2 i = 0 = hm0 jV j l1 i ��(2)hm0 j l0 i:


Inserting the expression (5.20) for j l1 i we getXk 6=D

hm0 jV j k0 ih k0 jV j l0 iE0 �Ek

= �(2)hm0 j l0 i:

To make this look like an eigenvalue equation we have to insert a 1:

Xn2D

Xk 6=D

hm0 jV j k0 ih k0 jV jn0 iE0 � Ek

hn0 j l0 i = �(2)hm0 j l0 i:

Maybe it looks more familiar in matrix formXn2D

Mmnxn = �(2)xm

where

Mmn =Xk 6=D

hm0 jV j k0 ih k0 jV jn0 iE0 � Ek

;

xm = 0hm j l0 i

are expressed in the basis de�ned by j l0 i. Evaluate M in the degeneratesubspace basis D = f j 1 i; j 2 ig

M11 =V13V31E1 �E0

3

=jaj2

E1 �E2; M12 =

V13V32E1 � E0

3

=ab�

E1 � E2;

M21 =V23V31E1 � E0

3

=a�b

E1 � E2; M22 =

jV23j2E1 � E0

3

=jbj2

E1 � E2:

With this explicit expression for M , solve the eigenvalue equation (de�ne� = �(2)(E1 � E2), and take out a common factor 1

E1�E2)

0 = det

jaj2 � � ab�

a�b jbj2 � �

!=

= (jaj2 � �)(jbj2 � �) � jaj2jbj2 == �2 � (jaj2 + jbj2)�) � = 0; jaj2 + jbj2

) �(2)1 = 0 �(2)

2 =jaj2 + jbj2E1 � E2

: (5.21)

112

From before we knew the non-degenerate energy shift, and now we see thatdegenerate perturbation theory leads to the correct shifts for the other twolevels. Everything is as we would have expected.

5.3 A one-dimensional harmonic oscillator is in its ground statefor t < 0. For t � 0 it is subjected to a time-dependent but spatiallyuniform force (not potential!) in the x-direction,

F (t) = F0e�t=�

(a) Using time-dependent perturbation theory to �rst order, obtainthe probability of �nding the oscillator in its �rst excited state fort > 0. Show that the t ! 1 (� �nite) limit of your expression isindependent of time. Is this reasonable or surprising?

(b) Can we �nd higher excited states?


pn+ 1�n0;n+1 +

pn�n0;n�1):]

(a) The problem is de�ned by

H0 =p2

2m+m!2x2

2V (t) = �F0xe

�t=� (F = �@V@x

)

At t = 0 the system is in its ground state j�; 0 i = j 0 i. We want to calculate

j�; t i =Xn

cn(t)e�Ent=�h jn i

E0n = �h!(n+ 1

2)

where we get cn(t) from its di�. eqn. (S. 5.5.15):

i�h@

@tcn(t) =

Xm

Vnmei!nmtcm(t)

Vnm = hn jV jm i!nm =

En � Em

�h= !(n�m) (5.22)

We need the matrix elements Vnm

Vnm = hn j � F0xe�t=� jm i = �F0e

�t=�hn jx jm i =

= �F0e�t=�

s�h

2m!(pm�n;m�1 +

pm+ 1�n;m+1):


Put it back into (5.22)

i�h@

@tcn(t) = �F0e

�t=�s

�h

2m!

�pn+ 1e�i!tcn+1(t) +

pnei!tcn�1(t)

�:

Perturbation theory means expanding cn(t) = c(0)n + c(1)n + : : :, and to zerothorder this is

@

@tc(0)n (t) = 0 ) c(0)n = �n0

To �rst order we get

c(1)n (t) =1

i�h

Z t

0dt0Xm

Vnm(t0)ei!nmt0c(0)m =

= �F0

i�h

s�h

2m!

Z t

0dt0e�t=�

�pn+ 1e�i!t

0

c(0)n+1(t) +

pnei!t

0

c(0)n�1(t)

�We get one non-vanishing term for n = 1, i.e. at �rst order in perturbationtheory with the H.O. in the ground state at t = 0 there is just one non-zeroexpansion coe�cient

c(1)1 (t) = �F0

i�h

s�h

2m!

Z t

0dt0ei!t

0�t0=�p1�1�1;0 =

= �F0

i�h

s�h

2m!

"1

i! � 1�

e(i!�1�)t0#t0

=F0

i�h

s�h

2m!

1

i! � 1�

�1� e(i!�

1�)t�

andj�; t i =X

n

c(1)n (t)e�iEnt

�h jn i = c(1)1 (t)e

�iE1t�h j 1 i:

The probability of �nding the H.O. in j 1 i is

jh 1 j�; t ij2 = jc(1)1 (t)j2:As t!1

c(1)1 ! F0

i�h

s�h

2m!

1

i! � 1�

= const:

This is of course reasonable since applying a static force means that thesystem asymptotically �nds a new equilibrium.

114

(b) As remarked earlier there are no other non-vanishing cn's at �rst order,so no higher excited states can be found. However, going to higher order inperturbation theory such states will be excited.

5.4 Consider a composite system made up of two spin 12objects.

for t < 0, the Hamiltonian does not depend on spin and can betaken to be zero by suitably adjusting the energy scale. For t > 0,the Hamiltonian is given by

H =�4�

�h2

�~S1 � ~S2:

Suppose the system is in j + �i for t � 0. Find, as a function oftime, the probability for being found in each of the following statesj++i, j+�i, j �+i, j � �i:

(a) By solving the problem exactly.

(b) By solving the problem assuming the validity of �rst-ordertime-dependent perturbation theory withH as a perturbation switchedon at t = 0. Under what condition does (b) give the correct results?

(a) The basis we are using is of course jS1z; S2z i. Expand the interactionpotential in this basis:

~S1 � ~S2 = S1xS2x + S1yS2y + S1zS2z = fin this basisg

=�h2

4

"( j+ ih� j + j � ih+ j )1( j+ ih� j + j � ih+ j )2+

+ i2(� j+ ih� j + j � ih+ j )1(� j+ ih� j + j � ih+ j )2 ++ ( j+ ih+ j � j� ih� j )1( j+ ih+ j � j� ih� j )2

#=

=�h2

4

"j ++ ih� � j + j +�ih� + j++ j �+ ih+ � j + j � � ih+ + j +

+ i2( j ++ ih� � j � j +�ih� + j +� j �+ ih+ � j + j � � ih+ + j ) +


+ j ++ ih+ + j � j +�ih+ � j +� j �+ ih� + j + j � � ih� � j

#=

In matrix form this is (using j 1 i = j ++ i j 2 i = j +�ij 3 i = j �+ i j 4 i = j � � i)

H = �

0BBB@

1 0 0 00 �1 2 00 2 �1 00 0 0 1

1CCCA : (5.23)

This basis is nice to use, since even though the problem is 4-dimensional weget a 2-dimensional matrix to diagonalize. Lucky us! (Of course this luck isdue to the rotational invariance of the problem.)

Now diagonalize the 2� 2 matrix to �nd the eigenvalues and eigenkets

0 = det

�1� � 2

2 �1� �

!= (�1� �)2 � 4 = �2 + 2� � 3

) � = 1;�3� = 1 :

�1 22 �1

! xy

!=

xy

!

)�x+ 2y = x) x = y =1p2

� = �3 : �1 22 �1

! xy

!= �3

xy

!

) �x+ 2y = �3x) x = �y = 1p2

So, the complete spectrum is:8>><>>:j ++ i; j � � i; 1p

2( j +�i+ j �+ i with energy�

1p2( j +�i � j �+ i with energy � 3�

116

This was a cumbersome but straightforward way to calculate the spectrum.A smarter way would have been to use ~S = ~S1 + ~S2 to �nd

~S2 = S2 = ~S21 + ~S2

2 + 2~S1 � ~S2 ) ~S1 � ~S2 = 12

�~S2 � ~S2

1 � ~S22

�

We know that ~S21 = ~S2

2 = �h2 12

�12+ 1

�= 3�h2

4so

~S1 � ~S2 = 12

S2 � 3�h2

2

!

Also, we know that two spin12 systems add up to one triplet (spin 1) and one

singlet (spin 0), i.e.

S = 1 (3 states)) ~S1 � ~S2 = 12(�h

21(1 + 1) � 3�h2

2 ) = 14�h

2

S = 0 (1 state)) ~S1 � ~S2 = 12(�3�h2

2) = �3

4�h2

: (5.24)

Since H = 4��h2~S1 � ~S2 we get

E(spin=1) =4�

�h21�h2

4= �;

E(spin=0) =4�

�h2�3�h24

= �3�:(5.25)

From Clebsch-Gordan decomposition we know thatnj ++ i; j � � i;

1p2( j +�i+ j �+ i)

oare spin 1, and 1p

2( j + �i � j �+ i) is spin 0!

Let's get back on track and �nd the dynamics. In the new basis H is diagonaland time-independent, so we can use the simple form of tthe time-evolutionoperator:

U(t; t0) = exp�� i

�hH(t� t0)

�:

The initial state was j +�i. In the new basis

nj 1 i = j ++ i; j 2 i = j � � i; j 3 i = 1p

2( j +�i + j �+ i);

j 4 i = 1p2( j +�i � j �+ i)

o


the initial state is

j +�i = 1p2( j 3 i + j 4 i):

Acting with U(t; 0) on that we get

j�; t i =1p2exp

�� i

�hHt�( j 3 i + j 4 i) =

=1p2

�exp

�� i

�h�t�j 3 i + exp

�3i

�h�t�j 4 i

�=

=

"exp

��i�t�h

�1p2( j +�i + j �+ i)+

+exp�3i�t

�h

�1p2( j +�i � j �+ i)

#=

= 12

h(e�i!t + e3i!t) j +�i+ (e�i!t + e3i!t) j �+ i

i

where

! � �

�h: (5.26)

The probability to �nd the system in the state j� i is as usual jh� j�; t ij28>>>>>><>>>>>>:

h+ + j�; t i = h� � j�; t i = 0

jh+ � j�; t ij2 = 14(2 + e4i!t + e�4i!t) = 1

2(1 + cos4!t) ' 1� 4(!t)2 : : :

jh� + j�; t ij2 = 14 (2 � e4i!t � e�4i!t) = 1

2 (1� cos4!t) ' 4(!t)2 : : :

(b) First order perturbation theory (use S. 5.6.17):

c(0)n = �ni;

c(1)n (t) =�i�h

Z t

t0dt0ei!nit

0

Vni(t0): (5.27)

Here we have (using the original basis) H0 = 0, V given by (5.23)

j i i = j +�i;

118

j f i = j �+ i;!ni =

En � Ei

�h= fEn = 0g = 0;

Vfi = 2�;

Vni = 0; n 6= f:

Inserting this into (5.27) yields

c(0)i = c

(0)j+�i = 1;

c(1)f = c

(1)j �+ i = � i

�h

Z t

0dt2� = �2i!t: (5.28)

as the only non-vanishing coe�cients up to �rst order. The probability of�nding the system in j � � i or j + + i is thus obviously zero, whereas forthe other two states

P ( j +�i) = 1

P ( j �+ i) = jc(1)f (t) + c(2)f (t) + : : : j2 = j2i!tj2 = 4(!t)2

to �rst order, in correspondence with the exact result.The approximation breaks down when !t� 1 is no longer valid, so for a

given t:

!t� 1) �� h

t:

5.5 The ground state of a hydrogen atom (n = 1,l = 0) is subjectedto a time-dependent potential as follows:

V (~x; t) = V0 cos(kz � !t):

Using time-dependent perturbation theory, obtain an expressionfor the transition rate at which the electron is emitted with mo-mentum ~p. Show, in particular, how you may compute the angulardistribution of the ejected electron (in terms of � and � de�nedwith respect to the z-axis). Discuss brie y the similarities and thedi�erences between this problem and the (more realistic) photo-electric e�ect. (note: For the initial wave function use

n=1;l=0(~x) =1p�

�Z

a0

� 32

e�Zr=a0:


If you have a normalization problem, the �nal wave function maybe taken to be

f (~x) =�1

L32

�ei~p�~x=�h

with L very large, but you should be able to show that the observ-able e�ects are independent of L.)

To begin with the atom is in the n = 1; l = 0 state. At t = 0 the perturbation

V = V0 cos(kz � !t)

is turned on. We want to �nd the transition rate at which the electron isemitted with momentum ~pf . The initial wave-function is

i(~x) =1p�

�1

a0

�3=2e�r=a0

and the �nal wave-function is

f(~x) =�

1

L3=2

�ei~p�~x=�h:

The perturbation is

V = V0hei(kz�!t) + e�i(kz�!t)

i= Vei!t + Vye�i!t: (5.29)

Time-dependent perturbation theory (S.5.6.44) gives us the transition rate

wi!n =2�

�h

��Vyni

��2 �(En � (Ei + �h!))

because the atom absorbs a photon �h!. The matrix element is��Vyni

��2 = V 20

4

��eikz�ni

��2and�eikz

�ni

= h~kf j eikz jn = 1; l = 0 i =Zd3xh~kf j eikz jx ihx jn = 1; l = 0 i =

=Zd3x

e�i~kf �~x

L3=2eikx3

1p�

�1

a0

�3=2e�r=a0 =

=1

L3=2p�a

3=20

Zd3xe�i(

~kf �~x�kx3)�r=a0: (5.30)

120

So�eikz

�niis the 3D Fourier transform of the initial wave-function (and some

constant) with ~q = ~kf � k~ez. That can be extracted from (Sakurai problem5.39) �

eikz�ni=

64�2

L3a50

1h1a20+ (~kf � k~ez)2

i4The transition rate is understood to be integrated over the density of states.We need to get that as a function of ~pf = �h~kf . As in (S.5.7.31), the volumeelement is

n2dnd = n2ddn

dpfdpf :

Using

k2f =p2f�h2

=n2(2�)2

L2

we getdn

dpf=

1

2n

2L2pf(2��h)2

=2��h

Lpf

L2pf(2��h)2

=L

2��h

which leaves

n2dnd =L3k2f(2�)3�h

ddpf =L3p2f(2��h)3

ddpf

and this is the sought density.Finally,

wi!~pf =2�

�h

V 20

4

64�2

L3a50

1h1a20+ (~kf � k~ez)2

i4 L3p2f(2��h)3

ddpf :

Note that the L's cancel. The angular dependence is in the denominator:

�~kf � k~ez

�2= [(jkf jcos� � k)~ez + jkf jsin� (cos'~ex + sin'~ey)]

2 =

= jkf j2cos2� + k2 � 2kjkf jcos� + jkf j2sin2� == k2f + k2 � 2kjkf jcos�: (5.31)

In a comparison between this problem and the photoelectric e�ect as dis-cussed in (S. 5.7) we note that since there is no polarization vector involved,w has no dependence on the azimuthal angle �. On the other hand we didnot make any dipole approximation but performed the x-integral exactly.

Solutions to Problems in Sakurai's Quantum Mechanics - P. is B. Brinne

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