Physics 742 – Graduate Quantum Mechanics 2
Midterm Exam, Spring 2014
1. [15 points] A spinless particle in three dimensions has potential
V  A  X 2  Y 2  Z 2   BXYZ . Consider the unitary rotation operator R which
2
cyclically permutes the position operators x, y, and z, so that in the position basis,
we would have R x, y, z  y, z , x , or equivalently:
R † XR  Y , R †YR  Z , R † ZR  X ,
and similar equations apply to the momentum as well.
(a) [6] Convince yourself (and me) that R commutes with the Hamiltonian.
The momentum term is invariant under all rotations. The distance squared is also
invariant, as is easily demonstrated, for example
R †  X 2  Y 2  Z 2  R  R † X 2 R  R †Y 2 R  R † Z 2 R  R † XRR † XR  R †YRR †YR  R † ZRR † ZR
 YY  ZZ  XX  X 2  Y 2  Z 2 .
We therefore have
R †VR  AR †  X 2  Y 2  Z 2  R  BR † XYZR
2
 AR †  X 2  Y 2  Z 2  RR †  X 2  Y 2  Z 2  R  BR † XRR †YRR † ZR
 A  X 2  Y 2  Z 2  X 2  Y 2  Z 2   BYZX  V .
So R †VR  V , or multiplying on the left by R, we have VR  RV , and hence the whole
Hamiltonian commutes with R.
(b) [6] Show that R3 = 1. Find all possible eigenvalues of R .
It is pretty easy to see that
R 3 x, y , z  R 2 y , z , x  R z , x , y  x , y , z ,
which, since x, y, z forms a complete basis set, shows that R 3  1 . If we have any
eigenstate  such that R     , then it is easy to see that   R 3    3  ,
which implies that  3  1 . This can be rewritten as 0   3  1     1   2    1 , and
we can then solve this equation to find the three possible eigenvalues namely
  1 or    12  i
3
2
.
(c) [3] Certain of the eigenvalues you found in part (b) will result in twofold
degeneracies of the Hamiltonian. Which eigenvalues is this true for?
Let  be an eigenstate of the Hamiltonian, so that H   E  . If we think in
a coordinate basis, then H is real, as is E, and hence H  *  E  * , where  * is the
state vector whose wave function is the complex conjugate of  .
We also know that R     , but thinking in a coordinate basis again, R is
purely real, so that if we take the complex conjugate of this relation, we have
R  *  *  *
If    * (which means   1 ), then this doesn’t tell us anything; indeed, it is likely that
 and  * are the same thing, but if    * , then this means that  and  * are
orthogonal, and hence must be different states. It follows that for    12  i
3
2
, we will
get pairs of states,  and  * , with identical energies.
2. [20 points] The first two eigenstates of the Harmonic oscillator have wave
functions of the form
 0  x   x 0   1/ 4 A1/4 e
1 Ax 2
2
and  1  x   x 1   1/4 2 A3/ 4 xe
 12 Ax 2
.
(a) [3] For two non-identical non-interacting particles, write the normalized
wave function x1 , x2  if the first particle is in state n1 = 0 and the second in
n2 = 1.
Normalized states of two particles just look like tensor products of the individual
states, and hence are of the form 0  1  0,1 , or in this case,  1 , 2 , and the
resulting wave functions are just direct products, so
x1 , x2   x1 , x2 0,1   0  x1  1  x2  
2

Ax2 e

 12 A x12  x22

.
(b) [5] What is the probability that both particles have xi  0 simultaneously?
Taking advantage of the given integrals, this is given by
P  x1  0, x2  0   

0
 A
2



0
  x1 , x2  dx1dx2  2 A2  e Ax dx1  x22 e Ax dx2
2


2
0
2
A
1/2

4
A
3/2
 .
1
4
2
1
0
2
2
(c) [5] Repeat part (a), but this time assume the two particles are (identical spin½) electrons in the   spin state. I want x1 , x2 ;    properly normalized.
This time we need to make the wave function anti-symmetric, because electrons
are fermions. This is done simply by making
 
1
2
 0,1  1, 0    

 0,1;  
1
2
 1, 0;  

Writing out the properly normalized wave function, we then have
x1 , x2 ;    
1
2

1


x1 , x2 0,1  x1 , x2 1, 0  
Ax2 e

 12 A x12  x22

1

Ax1e
1
2

 12 A x12  x22
 0  x1  1  x2   1  x1  0  x2  

1

A  x2  x1  e

 12 A x12  x22
.
(d) [7] What is the probability that both particles have xi  0 simultaneously?
We simply perform the same computation as before, except the wave function is
more complicated, so we get more terms.
P  x1  0, x2  0   

0


0
  x1 , x2  dx1dx2  1 A2 
2

0

 x
0
2
 x1  e Ax1  Ax2 dx1dx2
2
2
2
2





2
2
2
2
2
A    Ax12
e dx1  x22 e  Ax2 dx2   x12 e  Ax1 dx1  e Ax2 dx2  2  x1e  Ax1 dx1  x2 e  Ax2 dx2 

0
0
0
0
0

  0
 1 A2 

2
A1/2

4
A3/ 2 

4
A3/2

2
A1/2  2  12 A1 12 A1   18  18  21  14  21 .
3. [20 points] An electron lies in a region with magnetic and electric fields
A  r   a  xyˆ  yxˆ  , U  r   u  x 2  y 2  2 z 2 
where a and u are positive constants.
(a) [5] What are the electric and magnetic fields from these potentials?
We calculate these using the given formulas:
 A A 
 A A   A A 
B    A  xˆ  z  y   yˆ  x  z   zˆ  y  x   zˆ  a  a   2azˆ ,
z 
x   x
y 
 z
 y
A
U
U
U
 xˆ
 yˆ
 zˆ
 2uxxˆ  2uyyˆ  4uzzˆ
E  U 
t
x
y
z
(b) [5] Write the Hamiltonian for an electron in these fields.
We simply substitute directly into the formula for the Hamiltonian:
2
1
ge
 P  eA  R, t    eU  R, t  
B  R, t   S
2m
2m
2
1 
gea
2
2
2
2
2









P
eaY
P
eaX
P
eu
X
Y
2
Z
Sz






x
y
z

2m 
m
1
 Px2  Py2  Pz2  2eaYPx  2eaXPy  e2 a 2 X 2  e 2 a 2Y 2 

2m
gea
 eu  X 2  Y 2  2Z 2  
Sz .
m
H
(c) [10] Show that the Hamiltonian you found in part (b) into something of the
form
H
1 2
P  c  L  12 m x2 X 2   y2Y 2   z2 Z 2   d  S ,
2m
and determine the vectors c and d and the three frequencies i2 . Determine
the range of value of u for which all three frequencies i2 are positive.
It is a moment’s work to rewrite the expression above at
H
 e2 a 2

1 2 ea
gea
P  Lz  
 eu   X 2  Y 2   2euZ 2 
Sz .
2m
m
m
 2m

We therefore have
c
ea
gea
zˆ , d 
zˆ ,
m
m
1
2
mx2  12 m y2 
e2 a 2
 eu ,
2m
1
2
m z2  2eu .
These last two equations can be solved, if we want to:
x2   y2 
e2 a 2 2eu
4eu

,  z2 
.
2
m
m
m
These frequencies will all be positive provided 0  u 
ea 2
.
2m
(d) [3] Find a spin and an angular momentum operator that commute with H.
You don’t have to prove they commute.
The spin operator Sz and Lz both commute with H. This is obvious for Sz. For Lz,
we simply note that the quantities X 2  Y 2 and Z2 are both unchanged by rotations
around the z-axis.
4. [20 points] Consider the ground state of the Harmonic oscillator   0 .
(a) [5] Find the expectation value of the position operator X, the position
squared X2, and the uncertainty X for a single particle in the ground state
of the Harmonic oscillator. If we were to make a measurement of a single
particle, can we predict what value x will be obtained? If so, what value is it,
or if not, predict what value it would average to if many measurements were
made.
We can easily see that
X 





a  a†  0 
0 1 1 
1 .

2m
2m
2m
We therefore have

1 0,
2m
X   X  0
X 2   X 2     X   X   1
X 
X2  X
2




1 
,
2m 2m
2m

.
2m
Because the uncertainty is non-zero, this implies that this is not an eigenstate of X.
However, over a series of measurements, the value x should average to X  0 .
(b) [15] Now consider a large collection of N particles, all in the ground state of
the Harmonic oscillator,   000 0 . Define the average position
operator
X ave 
1
N
N
X
i 1
i
2
Find the expectation value of X ave , X ave
, and the uncertainty X ave for this
quantum state. In the limit N   , if we measured X ave , can we predict
what value xave will be obtained, and if so, what value would we obtain?
We similarly first find
X ave  
1
N
 N
  ai  ai†  000 0
2m i 1

1
N

 100 0  010 0  001 0    0001 
2m 
We therefore have
X ave   X ave 

1
000 0
N

 100 0  010 0  001 0    0001   0 ,
2m
2
  X 2     X ave  X ave 
X ave

1   100 0  010 0  001 0    0001  


N 2 2m   100 0  010 0  001 0    0001  


1 

N 1   N 2  N   0 
 2
2 Nm
N 2m

2
2
.
 X ave
 X ave 
2 Nm


X ave

In the limit, N   , the uncertainty drops to zero, which implies we are in an eigenstate
of Xave, and the value we will get will be xave  X ave  0 .
5. [10 points] A particle of mass m in one dimension lies in potential V  x    A x ,
where A is a positive constant. The bound states will have negative
energy, En   . Estimate the energy of the n’th bound state by using the WKB
approximation.
The first step is to locate the classical turning points; those points where
V  x    A x  E . This will be the points where x   A E  A  , or x   A  . We
therefore have
   n  12   
A
A 
2m  E  V  x   dx  
A
A 
A
2m  A x    dx  2 2m 
0
A
  dx
x
x A 
 A

x
 2 2m 
sin 1
 Ax   x 2 
A
 
 x 0
 A

A
A2
A
 A
1
sin
sin 1 1
 2 2m 
 A   2  0  0   2 2m
A



 

 2 A 2m  12    A 2m  ,
 
A 2m
,
  n  12 
En   n 
2mA2
 2  n  12 
2
.
6. [15 points] A particle of mass m scatters from a weak potential V  V0 e   r .
Particles are incident on it with momentum k . Find the differential crosssection d d  as a function of the angle of deflection  , and the total cross
section  .
We first need to calculate the Fourier transform of V. Because the potential is
spherically symmetric, it doesn’t matter which direction K is in, so we might as well put
it along the z-axis. So we have
1
2

V  K    e iK rV  r  d 3r  V0  e iK r e  i r d 3r  V0  d  cos    d  e  iKr cos e   r r 2 dr
1
 2 V0  2    iK cos   d  cos   
3
1
1


2 V0
iK
2
0
cos 1
2 V0
2
   iK cos  
iK
cos 1

 2 V    iK 2     iK 2
2 V0 4 iK
1
1
0




2
2
2
2
2
2
iK
iK   2  K 2 
  K 
    iK     iK  
8 V0 

0
 K2
2
.
We then substitute this into the formula for the differential cross-section.
2
d
m
 2 4
d  4 
2
 d rV  r  e
3
 iK r
2


16m 2V02  2
m  8 V0  
.
 2 4

4
4
2
2
2
4     2  K 2 2 
k
k
2
2
cos









2
We then simply integrate this over angles to yield
1
4
16m 2V02  2 2
d
2
2
2
d 
d
k
k
d cos 
2
2
cos
 







0 1
d
4
3 cos 1
32 m 2V02  2 
1  2
2
2
k
k
2
2
cos
,









2 
4
cos 1
 3  2k 


16 m 2V02  2  1
1
.


  6   2  4 k 2 3 
3 4 k 2


Helpful Formulas:
EM Fields
B   A

E   A  U
t
EM Hamiltonian
π2
ge
H
 eU 
B S
2m
2m
Rotation Operators:
R    r  r
Gauge
Transformation

U U 
t
A   A  
   e
Symmetrization
1

P
N! P
1

 P P
N! P
R†    R     1
WKB Quantization

2m  E  V  x   dx   n  12   
b
a
Kinematic
Momentum
π  P  eA
Ly  ZPx  xPz
Lz  XPy  YPx
Harmonic Oscillator

X
a  a† 

2m
R†    R R      R
ie 
Angular
Momentum
Lx  YPz  ZPy
Pi
m †
a  a
2
a n  n n 1
First Born Approximation
d

 2 4  d 3rV  r  e  iK r
d  4 
K 2  2k 2 1  cos  
2
a† n  n  1 n  1
2
Helpful Integrals:
Gaussian:



0

0
Exponential:
Power Law:
e  ax dx 
2

2



a 1/2 ,
0
x 3e  ax dx  12 a 2 ,
2


0
e  ax dx  a 1 ,
  ax  b 
n
dx 

0


0
xe  ax dx  12 a 1 ,
2
x 4 e ax dx 
2
3 
8


0
a 5/2 ,
x 2 e  ax dx 
2


0

4
a 3/2 ,
x5e  ax dx  a 3 .
2
x n e  ax dx  n !a  n 1
1 1
n 1
 ax  b   C ,
n 1 a
Radical:

ax  b dx 

2
3/2
 ax  b   C ,
3a
2
dx

ax  b  C
ax  b a


a x  b dx 
a
bx
sin 1
 ax  bx 2  C
a
b
dx
a
bx 1
sin 1


ax  bx 2  C
a b
a xb b b