Physics 742 – Graduate Quantum Mechanics 2
Solutions to Final Exam, Spring 2014
Due Monday, May 5 by 5 PM
You may use (1) class notes, (2) former homeworks and solutions (available online), or
(3) any math references, such as integral tables, Maple, etc, including any routines I
provided for you. If you cannot do an integral or sum, or if you cannot solve an equation,
try to go on, as if you knew the answer. Some useful integral appear on the last page.
The approximate points (out of 200) are marked for each problem and part. Feel free to
contact me with questions.
Work: 758-4994
Home: 724-2008
Cell:
407-6528
1. [15] A particle of mass m and initial momentum k moving in the +z direction
scatters from a weak potential V  r     z  exp   12   x 2  y 2   . Calculate the
differential cross-section d d  and total cross-section  in the first Born
approximation. Note that the potential is not spherically symmetric, and hence
you cannot choose the momentum change K to be along the z-axis. I recommend
working in Cartesian coordinates. For  , you may leave one integral undone.
The first step is we need the Fourier transform of V, which is given by
3
iK r
 d rV  r  e 



iK x  x 2
iK z
   z  e dz  e e dx  e
2
x
z


iK y y  y 2 2
e
dy

  iK  
  iK x   
 K  Ky

y
  2 exp   x

exp  
exp  


 4  2  
 2

2
 4  2    2

2
2

2

2



We now plug this into the formula for the first Born approximation, namely
d
m2

d  4 2  4
 d re
3
V r 
iK r
2
2
 K x2  K y2 
m 2  2 
 2 4

 exp  
4    



We have K  k F  k I , but kI has no component in the x- or y-direction, so only kF
contributes, for which k F  k  sin  cos  xˆ  sin  sin  yˆ  cos  zˆ  , so we have
 k2
 m2 2
 k 2 sin 2  
d m 2  2
 4 2 exp    sin 2  cos 2   sin 2  sin 2     4 2 exp  
.

d  
 
 


We now integrate this over all angles to yield
d
m2 2
 
d  4 2
d

2
 k 2 sin 2  
2 m 2  2
d

exp
d
cos



0 1   
 4 2
1
 k2
2 
exp
1   1  w  dw
1
The final integral is some sort of error function, but this doesn’t really help much.
2. [20] A particle of mass m lies in the spherical infinite square well, with potential
 0 if r  R ,
V r   
 if r  R .
The radius R of the confining sphere will actually be a function of time, so that it
changes.
(a) [10] The ground state is given in the allowed region r < R by
 r  
N
r 
sin  
r
 R 
Show that this is, in fact, an eigenstate of the Hamiltonian, and determine the
energy and the correct normalization factor N.
We simply substitute it into Schrodinger’s equation. It is obvious that the wave
function is continuous at the boundary r = R, and within this boundary there is no
potential, so we have
1
 2 


 2 2
2 
 r sin   
2
2
2
2 2
2 1  2 
  r  N   
  r  N 
r   
sin
cos
sin




,
N
 
 
 
2
2
2m r r 2 
 R   2mr R r
 R  2mr R
 R  2mR
 22
.
E
2mR 2
E  r   
2 2
2  1  2
1
 

 
r   2
 sin 

2 
2m
2m  r r

r sin   
To get the normalization, we merely demand that the normalization integral yield 1, so
we have
1    r 
2
1
r 
d r  N  r dr  d  2 sin 2    4 N
r
 R 
0
R
3
2
1
,
2 R
1
r 
sin   .
 r  
r 2 R
 R 
N 
2
R
2
 sin
0
2
2
r 
  dr  2 N R ,
 R 
(b) [10] The radius R of the allowed region is slowly decreased from R to 12 R .
What is the probability that it will remain in the ground state? The radius is
then suddenly increased from 12 R to R. What is the probability now that it
remains in the ground state?
Because the change is occurring slowly, we use the adiabatic approximation. This
implies that the ground state will continuously evolve into the ground state. Hence for
this process, we have P  g  g    1 . The wave function at the end will be the same
expression as before, except R  12 R , so it is now
 r  
 2 r 
sin 
.
r R
 R 
1
This is only in the allowed region r  12 R .
The radius now suddenly increases, which means that we can use the sudden
approximation, P  g   g   g g  . So we need to find
2
g g   
*
 r    r  d
3
r
1R
2
 r dr  d  r
2
0
1
r  1
 2 r 
sin  
sin 

2 R
 R r R
 R 
1R
2
4
4 2R 4 2
  r   2 r 
sin   sin 
,


 dr 

3
R 2 3
R 2 0
 R   R 
32
P  g   g   2  36.03%.
9

Maple was happy to do that last integral for us.
> integrate(sin(Pi*r/R)*sin(2*Pi*r/R),r=0..R/2);
3. [15] A particle is in a one-dimensional infinite square well, with allowed region
0  x  a . It is initially in the ground, or n  1 state. At t = 0, a perturbation of
the form W  APe t , where A is a small positive number and P is the
momentum operator, is turned on. Calculate to leading order (that’s order A2),
the probability that the particle ends up in the state n at t   for all n > 1.
The eigenstates and energies of the one-dimensional infinite square well can be
found in equations (2.38), so we know that in the allowed region
 n  x 
2
 22n2
  nx 
sin 
,
E
.

n

a
2ma 2
 a 
The matrix element Wn1  t  is then given by
 d
2    t
  nx 
x 
Wn1  t   n W  t  1    x 
 1  x  dx  
Ae  sin 
 cos 
 dx ,
i dx
a ia
 a 
 a 
0
a
*
n
 4 n A   t
e
 2
Wn1  t    i  n  1 a

0

if n even,
if n odd.
We now simply use equation (15.14) to get the amplitude, keeping only the first
non-trivial term. Keeping only the even cases, we have
S n1 
1  in1t
1  A 4n   t in1t
4 An
1
e Wn1  t  dt 
e e dt  2
.
2


i 0
i i a n  1 0
n  1 a    in1 
We then have
P 1  n even   Sn1 
16 A2 n 2
2
a 2  n 2  1   2  n21 
2
, P 1  n odd   0 .
The frequency difference is given by
2
2
En  E1    n  1

n1 
.
2ma 2

This allows us, if we wish, to rewrite our final answer as
P 1  n even  
64m 2 a 2 A2 n 2
2
2
 n2  1 4 2 m2 a 4   2 2  n2  1 
, P 1  n odd   0 .
4. [15] A particle of mass m in one dimension is in the bound state of the Dirac
delta function. This state has wave function and energy given by
 g  x    e  x , Eg   ,
The values of  and  depend on the mass and the strength of the delta
function. It is then affected by a time-dependent perturbation of the form
W  t    P cos t  , where P is the momentum operator and  is very small.
The frequency is high enough such that    .
(a) [2] Write W  t  in the form W  t   We  it  W †eit . What is W?
Using cos   12 ei  12 e  i , it is clear that W  12  P .
(b)[5] For sufficiently high energies, the potential is negligible, and the
eigesntates are approximately momentum eigenstates k . Work out the
matrix element k W  g for these high energy final states. Hint – let any
momentum operators act to the left.
Momentum eigenstates, properly normalized, have wave function given by eq.
(3.14) in three dimensions, or k  x   eikx 2 in one. The matrix elements will be
k W  g  12  k P  g  12  k k  g  12  k

2
2
 k 

2
2

 k
 e

0
 x  ikx
e

2

0
dx   e x e  ikx dx 


e
 ikx  x
dx

 k
2

2


0
e x  e ikx  eikx  dx
 k 3/2
1 
 1
.




2  2  k 2 
   ik   ik 

(c) [8] Calculate the rate   g  k
 for a given final state wave number k.
Integrate it over k to get the rate for the transition.
The rate is just given by equation (15.21):

 g  k

2
  2k 2
2

2
2   k 3/2

 
Wkg   Ek  Eg    

    
2
2

  2   k    2m




 2 3k 2   2 k 2

     .

2
2 2

  k   2 m
We then integrate this over k to get the total rate:

 2 3 k 2
 2k 2

 2 3k 2  2 k
 

     dk  
2
2 2
2
2 2
m
k   k 
 2m

   k 
1

2m 2 3 k
  2  k 2 
2
.
The expression in absolute values comes from the general rule about integrating delta
functions. The factor of 2 comes because there are two values of k where the delta
function vanishes. The value of k in this expression is the place where the delta function
argument vanishes, explicitly
k
1
2m      .

5. [20] A system consists of a superposition of a zero photon state and two one photon
states, so that
 t  0  N

2 0  q,1  i q, 2

where N is a normalization constant, q  qzˆ , and the two polarizations are
ε  q,1  xˆ and ε  q, 2   yˆ .
(a) [3] What is the normalization constant N?
We simply demand that the inner product of this state with itself be one, so
1    N
2

2 0  q,1  i q, 2


2 0  q,1  i q, 2  N
2
 2  1  1
4 N ,
2
N  12 .
(b) [5] What is the state vector   t  at an arbitrary time? Work in the
convention where the energy of the vacuum is 0.
Each of these states is an eigenstate of the Hamiltonian, and hence they simply
pick up a phase given by e iEt  . The energy of the vacuum is zero, and the energy of the
one-photon states is E    qc . Hence the last two terms each pick up a factor of
e iqct . Including our normalization, the state vector at arbitrary time is
 t  
1
2

2 0  q,1 e iqct  i q, 2 e iqct

(c) [12] what is the expectation value of the electric field E  r  at arbitrary
time and arbitrary position for the state you found in part (b)?
We simply sandwich the operator between the initial and final state. The electric
field operator has creation and annihilation operators for all photons, but since the state
has only photons with momentum q, any term that creates or annihilates the wrong type
of photon will automatically vanish. We are also dealing with real polarizations. It
follows that
k
E r     t 
i  ε k eik r ak  ε*k e ik r ak†    t 
2 0V
k ,
i

cq
2 0V
 t  ε  e


q
i cq
4 2 0V


iqr
aq  e iqr aq†    t 

2 0  q,1 eiqct  i q, 2 eiqct  eiqz xˆ aq1  eiqz yˆ aq 2  e  iqz xˆ aq†1  e  iqz yˆ aq†2 

2 0  q,1 e iqct  i q, 2 e  iqct .
As messy as this expression is, the creation and annihilation operators are non-vanishing
only if they are creating or annihilating the exact right photon. Hence if you multiply it
out, the only non-vanishing terms are
0 aq1 q,1  0 aq 2 q, 2  q,1 aq†1 0  q, 2 aq† 2 0  1.
Hence when you multiply everything out, you get
E r  

i cq
 2xˆ eiqz iqct  2iyˆ eiqz iqct  2xˆ eiqct iqz  2iyˆ eiqct iqz
4 2 0V

1 cq
 xˆ  ieiqz iqct  ieiqct iqz   yˆ  eiqz iqct  eiqct iqz  

4  0V 

1 cq
 xˆ sin  qz  qct   yˆ cos  qz  qct   .
2  0V 
The expectation value turned out to be real, as it had to be.

6. [20] An electron is trapped in a 3D infinite square well, with allowed region
0  x  a , 0  y  a , 0  z  a . It is in the quantum state nx , ny , nz  1,1, 4 .
(a) [12] Calculate every non-vanishing matrix element of the form
nx , ny , nz R 1,1, 4 where the final state is lower in energy than the initial
state (there will be very few of these).
The eigenstates of this Hamiltonian take the form
 n ,n
x

y , nz
8
  nx x    n y y    nz z 
sin
 sin 
.

 sin 
a3
 a   a   a 
The energy is given by
Enx ,ny , nz 
 2 2
2ma
2
n
2
x
 n y2  nz2  .
To lower the energy, we must therefore lower some of the three labels nx, ny¸and nz.
However, we can’t lower the first two, so we must lower the last one. We then need to
calculate the messy integrals nx , ny , nz R 1,1, 4 . To understand what is going on,
consider a specific one, such as the X operator, for which we would have
nx , n y , nz X 1,1, 4 
8
a3
a
  ny y    y 
  nx x    x 
sin
sin
sin
x
dx




0  a   a  0  a  sin  a  dy
a
a
  n z   4 z 
 sin  z  sin 
 dz .
0
 a   a 
But the last integral vanishes unless nz  4 , so this term clearly vanishes. Similarly, we
will get zero if we used Y instead. So we must use Z, and furthermore, to get the other
terms to not vanish, we must choose nx  n y  1 . Hence we need only calculate
a
a
x 
 y 
  n z   4 z 
sin 2 
dx  sin 2 
dy  sin  z  sin 


 zdz
0
0
0
 a 
 a 
 a   a 
8a 2 n
16an 
8 a a
n
1   1n  
1   1  .
 3  
2
2
  2 n 2  16 

a 2 2  2 n 2  16 
1,1, n Z 1,1, 4 
8
a3

a




The last factor is 2 for odd n and zero for even n. Since we are only interested in values
smaller than n = 4, this narrows it down to
1,1,1 R 1,1, 4 
32a
zˆ ,
225 2
1,1,3 R 1,1, 4 
96a
zˆ .
49 2

(b) [8] Calculate the decay rate  1,1, 4  nx , n y , nz
 for this decay in the
dipole approximation for every possible final state. What is the ratio of the
two rates?
In the dipole approximation, these rates are
4IF3 rFI

3c 2
2
The frequencies are given by
IF 
EI  EF
 2

16  n 2 
2 

2ma
Assuming the final state is odd, then we have
2
3


32an 
4   2  
512 2 3
n2
2 3
  1,1, 4  1,1, n   2 



16
n
  2 2 2  3c 2m3a 4 16  n2 .
 
3c  2ma 2 
   n  16  
This formula is correct only for n = 1 and n = 3, for which this becomes
512 2 3 1
512 2 3 9
  1,1, 4  1,1,1  
 ,   1,1, 4  1,1,3  
 ,
3c 2 m3 a 4 15
3c 2 m3 a 4 7
  1,1, 4  1,1,3  135

 19.29 .
7
  1,1, 4  1,1,1 
7. [30] A hydrogen atom is at rest in a strong magnetic field oriented in the zdirection. This field adds an additional perturbation to the hydrogen atom as
given in Eq. (9.23):
W
eB
 S z  12 Lz 
m
The field is sufficiently strong that we can neglect any hyperfine effects. The
atom is initially in the state n, l , m, ms  1, 0, 0, 12 , but it will transition to
1, 0, 0,  12 .
(a) [2] Explain why this transition cannot occur in the electric dipole
approximation.
In the electric dipole approximation, the rate for such a transition is governed by
the electric dipole matrix element, F R I , but R only connects states whose l values
differ by one, so clearly this is not going to work. Also, R doesn’t change the spin state
anyway.
(b) [8] Find all non-vanishing matrix elements coming from the magnetic dipole
transition F  S  12 L  I .
Because our initial state (and final state) has l = 0, it has no angular dependence,
so L I  0 . Hence we need to look only at S. The operator S doesn’t change the space
wave function at all, so we focus exclusively on the spin part. Keeping in mind that
S  12 σ , and we can figure out the effects of the Pauli matrices on the spin states with
the help of eq. (7.17), so we have
  0 1  1    0  
Sx   
       ,
2  1 0  0  2  1  2
  0 i   1  i  1  i
 ,
Sy   
     
2  i 0 0 2 0 2
 1 0 1  1 
Sz   
       .
2  0 1   0  2  0  2
It follows that
F S x I  12  ,
F S y I  12 i ,
F  S  12 L  I  12   xˆ  iyˆ  .
F Sz I  0 ,
(c) [20] Find a formula for the differential decay rate d  d  for this decay into
a particular polarization ε . You do not need to sum it or integrate it over
polarizations.
We start with an equation from the middle of 316, supplemented by the last
equation on page 317, keeping only the relevant spin part. We have, therefore,
WFI1 

e

F e ik R ε*  P  i  k  ε*   S  I
m 2 0V 
ie
ie


F  k  ε*    S  12 L  I 
k  ε*    xˆ  iyˆ 

2m 2 0V 
m 2 0V 
We then substitute this into (18.1), yielding
I  F  
2
2
2 e 2 3
2
k  ε*    xˆ  iyˆ     F     I  .
WFI   EF  EI  

2

 8m  0V 
We will need the energies of the initial and final states, which are easy to determine
because our states are eigenstates of the perturbation, with eigenvalues
I 
eB 
m 2
and  F  
eB 
eB
, I F 
.
m 2
m
The probability of going to a particular final state in an infinite universe is small,
which is why there is a factor of V in the denominator. To get over this problem, we sum
over final state momenta and/or polarizations. For now, let’s just sum over momenta.
We use the usual trick for converting a sum over states into an integral, and then also use
the identity   e 2  4 0 c  , and write at least some of the wave numbers using   kc .
We’ll also write k  kkˆ to separate the magnitude and direction of k.
I  F  
2
2 e 2 3
1
*
ˆ
ˆ
k

ε

x

y
   F     I 
i





 8m 2 0V  k 

2
 e2  2 k 2 d 3k 1 ˆ *
eB 

ˆ
ˆ
k

ε

x

y
   
i



3
2

4m  0V  2  
m 


2
 3c  k 4 dk 
eB 
ˆ  ε*   xˆ  iyˆ 
k





d


8 m 2 0 
m 


2
  2 3
d  kˆ  ε*   xˆ  iyˆ 
2 4 
8 m c






We now delay doing the angular integral and find
2
d    2 3 ˆ *
ˆ
ˆ
i
k
ε
x
y






d  8 m 2 c 4


The value of  comes from the delta function, namely   eB m .
8. [15] The Dirac equation in the presence of electromagnetic fields is given by
i

  cα   P  eA  R, t    eU  R, t   mc 2  
t


where P  i . Prove or disprove: The Dirac equation is invariant under
gauge transformations.
A gauge transformation is, according to eqs. (9.11),
U   r, t   U  r, t     r, t  t
A   r , t   A  r , t     r , t 
   r, t     r, t  exp   ie  r, t   
The question is whether, if the Dirac equation is satisfied by the unprimed functions, will
they be satisfied by the primes functions? To find out, plug it in on both sides, then
simplify until they are identical


i    r, t  t   cα   i  eA  r, t    eU   r, t   mc 2    r, t 
i
cα   i  eA  r, t   e  r, t   
 
 ie   r ,t  
 ie   r ,t  
  


r
,
t
e


    r, t  e



2
t





eU
t
e
t
t
mc
r
,

r
,







Now, let the derivatives act on each side. The resulting equation will have a common
factor of e  ie  r ,t   , which we cancel

ie  r, t 


 eA  r, t   e  r, t   
   r, t  ie   r, t 
 cα   i  i


  r, t    
i 

    r, t  ,
t

 t
 

2
eU  r, t   e    r, t  t   mc 

  r, t 
  r, t 
  r, t 


e
  r, t   cα   i  eA  r, t    eU  r, t   e
 mc 2     r, t  ,
i
t
t
t


i
  r, t 
 cα   i  eA  r, t    eU  r, t   mc 2    r, t  .
t


This is simply the Dirac equation back again, so we have shown they are equivalent.
Definite Integrals:
For each of the following formulas, m and n and p are assumed to be positive integers.

e
 Ax
x n dx 
0

  n  1
if Re  A   0,
An 1
n
 Ax
 e x dx 
2
0
  n21 
n 1
2A 2
if Re  A   0,
  n  1  n !,   12    ,   32  
a
1
2
 ,   52  
3
4
.
 2a n if n is odd,
  nx 
 dx  0
if n is even,
a 

 sin 
0
a
  nx 
 dx  0
a 
 cos 
0
  nx 
  mx 
  nx    mx 
0 cos  a  cos  a  dx  0 sin  a  sin  a  dx  12 a nm
a
a
2na   n 2  m 2  if n  m is odd
  nx 
  mx 
dx

sin
cos
0  a   a  
0
if n  m is even

a
 4a 2 mn

  nx    mx 
0 sin  a  sin  a  x dx  

a
 2  n 2  m 2 2  if n  m is odd


1 2
if n  m is even
4 a  nm
a 2 n  1
  nx 
  mx 
0 sin  a  cos  a  x dx    m2  n2 
a
m n
  2a 2 m 2  n 2  2 n 2  m 2 2  if n  m is odd

  
 

  nx 
  mx 

x
dx
cos
cos





0  a   a 
1 2

if n  m is even
4 a  nm
a