Solutions to Problems 2.1 and 2.2
1. [10] A free particle of mass m in one dimension takes the form at t = 0
  x, t  0     x    A   exp  iKx  12 Ax 2 
1/ 4
This is identical with chapter 1 problem 4. Find the wave at all subsequent
times.
The procedure, as discussed in class, is to first find the Fourier transform,   k  .
This was found in problem 1.4, part a:
  k    A 
1/4
  k  K 2 
exp  
.
2 A 

Then the answer to the question is simply
  x, t  




k 2 
dk
  k  exp  ikx  i
t
2m 
2

  A 
1/4



  A 
1/4
  A 
1/4
1/4
 A
 
 
1/4
 A
 
 
1/4
 A
 
 

k 2
dk
k 2  2kK  K 2 
exp  ikx  i
t

2m
2A
2


 K 2   dk
1  2 
  i t
K 

exp  
exp   

 k   ix   k 
A 

  2m 2 A 
 2 A   2
1
2
  ix  K A 2 
 K2 
exp  

 exp 
i t 2 m  1 2 A
 2A 
 4  it 2m  1 2 A  

 K 2  Ax 2  2iKx  K 2 A 
1

exp  

2
2 1  iAt m  
A
1  iAt m

  Ax 2  2iKx  K 2 A   K 2 A  1  iAt m  
1

exp 
2 1  iAt m 
1  iAt m


  Ax 2  2iKx  iK 2t m 
1
exp 
.
2 1  iAt m 
1  iAt m


It’s messy, but it’s finished, and there isn’t much you can do to simplify it.
2. [10] One solution of the 2D Harmonic oscillator Schrodinger equation takes the
form
  x, y, t    x  iy  e

 A x2  y 2

2  it
e
(a) [3] Find the probability density   x, y, t  at all times.
  x, y, t    *   x  iy  e

 A x2  y2
 2 eit x  iy e A x


2
 y2
 2 e  it  x 2  y 2 e A x


2
 y2
.
(b) [4] Find the probability current j  x, y, t  at all times.
j





Im   * 
m

 A x 2  y 2  2 it 
 A x 2  y 2  2  it  


 
e  xˆ  yˆ   x  iy  e
e
Im  x  iy  e
 
m 
y  
 x

2
2
 A x  y  2
 A x 2  y 2  2 

xˆ 1  Ax  x  iy    yˆ i  Ay  x  iy   e
Im  x  iy  e

m 
  A x 2  y 2 
e
Im xˆ  x  iy  Ax  x 2  y 2    yˆ ix  y  Ay  x 2  y 2  
m
 A x2  y2

  yxˆ  xyˆ  e   .
m




(c) [3] Check the local version of conservation of probability, i.e., show that your
solution satisfies  t    j  0
Since  is independent of time, the first term is zero.
 A x 2  y 2  
 A x 2  y 2   
j j



   j  x  y     ye
  xe
 y 
 
t
x y m  x 
 A x 2  y 2 
 A x 2  y 2  

  2 Axye
 2 Axye
 0.

m 