Physics 741 – Graduate Quantum Mechanics 1
Solutions to Chapter 12
3. [15] Joe isn’t getting any smarter. He is attempting to find the ground state energy of
an infinite square well with allowed region  a  x  a using the trial wave function (in
the allowed region)   x   1  x 2 a 2  B 1  x 4 a 4  , where B is a variational parameter.
Estimate the ground state energy, and compare to the exact value.
Since there is no potential, we need to calculate only the normalization and kinetic terms,
which are
2
 x2
 x4 
    1  2  B 1  4   dx
a
 a 
a 
a
 2x2 x4
 x2 x4 x6 
 2 x 4 x8  
 2 1  2  4  2 B  1  2  4  6   B 2  1  4  8   dx
a
a
a 
a
a 
 a a

0 
2
256
64
 2a 1  23  15  2 B 1  13  15  17   B 2 1  52  19    a  16
15  105 B  45 B  ,
a
 P   P
2
2
a


a
2
2
a
 x4 
 2 x 4 Bx 3 
d  x2
i 1  2  B  1  4   dx  2  i  2  4  dx
dx  a
a 
 a 
a
0
 4 x 2 16 Bx 4 16 B 2 x 6 
8 2 1 4
2
4
 2   4 


 dx 
3  5 B 7 B .
6
8
a
a
a
a

0
a
2
The expectation value of the energy, as a function of B, is therefore
2
1
4
4
2  P 2 
 H
8 2
3 2 60 B 2  84 B  35
3  5 B 7 B
E  B 





.
2
256
64
2m  
2ma 2 16
4ma 2 28 B 2  48 B  21

15  105 B  45 B
To minimize this, we set the derivative equal to zero, which yields
28B 2  48B  21 120 B  84    60 B 2  84 B  35   56 B  48 

0
,
2
 28B 2  48B  21
0  528B 2  560 B  84  16  33B 2  35 B  214  ,
35  352  4  33  214
35  532
 0.1808 or  0.8798.
2  33
66
We now substitute each of these into the expression for E  B  , to yield
B

E  0.1808  
3 2 60  0.1808   84  0.1808   35 1.233719 2
,


4ma 2 28  0.1808 2  48  0.1808   21
ma 2
2
3 2 60  0.8798   84  0.8798   35 12.76628 2
.
E  0.8798  


4ma 2 28  0.8798 2  48  0.8798   21
ma 2
2
We are trying to minimize the energy, which clearly corresponds to the first case, not the second
(which is a maximum).
Since the well has width 2a, the exact energy is
E
 2 2
2 m  2a 
2

 22
8ma 2

1.233701 2
,
ma 2
or a difference of about 15 parts per million. Not bad, for a simple polynomial estimate!
4. [5] A particle lies in one dimension with Hamiltonian H  P 2 2m  F X . Using the
WKB method, our goal is to find the eigenenergies of this Hamiltonian.
(a) [1] For energy E, find the classical turning points a and b.
We first find the classical turning points, which are solutions to E  F x . The solutions
are x  E F , x   E F , so a   E F and b  E F .
(b)[2] Perform the integral required by the WKB method.
The WKB formula for the energy is
   n  12   
2m  E  V  x   dx  
b
E F
E F
a
 2
E F
0
2m  E  F x dx
3
2 2
2m  E  Fx dx  
  2m  E  Fx   2
2mF 3
E F
0
2  2mE 

3mF
(c) [2] Solve the resulting equation for En
Solving for E, we have
 2mE 
3/2
 32    n  12  mF ,
 32    n  12  mF 
En 
2m
2/3
 3   n  12  F 


 4 2m

2/3
.
3/2
.