1
ME 501 Homework #1 Due Friday, September 16, 2011
Prof. Lucht (E-mail address: [email protected])
Note: A useful table of integrals in posted on the class website.
1. Laurendeau, Problem 3.2, p. 147.
2
3
4
2. A particle of mass m is constrained to move along the x-axis in the region 0  x  L .
The normalized wavefunction for the particle is given by
 
( , ) ( ) ( ) sin exp
( , ) ,
x t x T t
L
n x
L
i
t
x  | L
  | x t x x L h n
m L
  F
H
IK

FH
IK
   

 

2
0
0 0
8
1 1
1
2
21
2

The uncertainty B in the determination of the value of a dynamical variable B is given by
the square root of the variance, B B B   2 2 . For a particle with quantum number
n1  3, determine the value of the product x px. Is this result consistent with Heisenberg's
uncertainty principle?
Solution:
   
           
2 2 2 2
2 * 2 *
0
* * *
2 2 1 1 1 1
2 2 2 1
2
sin
L
n x
x x dx

 

0
 |   | 
L  | L
   | L
 
 | L
  |  

L  | L
 , ,
, ,
2 2
sin exp exp sin
x x x
L
x x x p p p
x x t x x t dx
x t x t x T t x T t
n x t t n x
i i
 
   
     
  
  
       
           

From the table of integrals we have:
     
3 2
2 2
3 2
1
sin sin 2 cos 2
6 4 8 4
x x x
x ax dx ax ax
a a a
 
     
  
Evaluating the integral for a n1
 L
 we obtain:
5
3 2
2 1
3
2 1 2
sin
6 4 8
L a a L
   
4
a
   | 0
2
1
L
L
   

   
1 1
6 3 4 2
L n n
   
 
2 1
    | 2 2
3 18
    
 
3
2
L
 
    | 3 2
cos 2
L L
n
   | 2
L | 2
1
0.3277
L

x  | L L * * 2     x t x x t dx x dx , , sin 2 n x 1
 0  |  0 L  | L  
Again from the table of integrals we have
     
2
2
2
1
sin sin 2 cos 2
4 4 8
x x
x ax dx ax ax
a a
   
2
2 2 1
sin
x x n x
x | 1
4 4
L a  | L | 2 | 0x x n x
x
    
      
1
2
2
cos
L
x n x


  
   
    
        
  
   
 
   
cos 2 1 0.5000
       
2
2
4 8

 
 
  
   | 2 |  
   |   | 

8
2
 
 2 1
1
1 2
cos
L
n x
a L
L  | L
n | L n L



   
      x x x L L L 2 2 2 0.3277 0.5 0.279  2
The expectation value of the x-momentum is zero. The particle is just as likely to be moving in
the +x and –x direactions:
   
     
* 1 1
0 0
1 1 1
0
2
1 1 1
2
sin cos sin
L
n x n x n x L
dx
  
     
      | 0
0
L L
     | 1
2
n L

  
02
, , sin sin
2
sin cos
1
sin cos sin
2
L L
x
L
i n x n x
p x t i x t dx dx
x L L x L
i n n x n x
dx
L L L L
ax ax dx ax
a
 
  
         
             
         
   
     
   

 

 



0
L
px

 


The expectation value of px2 is nonzero
6
   
   
2 2 2
2 * 1 1
2
0 0
2 2
21 1 0
2 2
2 2 1 1
0
1 0
2
2
, , sin sin
2
sin
1
sin sin 2
2 4
2
sin sin
2 4 2
L L
x
L
L
L
x
n x n x
p x t i x t dx dx
x L L x L
n n x
dx
L L L
x
ax dx ax
a
n x n x L L L
dx
L n L
p
 
 
 

         
             
         
   
     
   
 
     
        
     
 

 



2 2 2
2 1 1
2
n n L
L L L
     
      
   
 
For
n1  3:
 
2 2
2
2
2 2
3
88.83
9.43
0.279 9.43 2.63 0.419 0.0796
4
 | x | L

 | 
 x
x x x
p
L L
p p p
L
h
x p L h h

 
   
 
   
 
        



The result is consistent with the Heisenberg uncertainty principle.
7
3. A particle confined along a line between x = 0 and x = L is in a "mixed" quantum state,
i.e., a linear superposition of the eigenstates A and B with quantum numbers n1A  3 and n1B  5 ,
respectively. The normalized wavefunction ( , ) x t for the particle is given by
4 3
sin exp
x t
i  | 2 5
sin exp
x t
i | ( , )
x t c  | ( , )
x t c  | ( , )
x t
    |     |     |   3 3
A B
A A  | B B | L  | L
   | L  | L
   |   |    |   |  
for 0   x L . For x  0 or x  L, ( , ) x t  0. The energies  A and  B are given by
2 2  | 2 2                
1
8 2
A
h n
m L  | 1
8 2
B
h n
m L | A
   | B
 (a) Verify that the wavefunction is a solution of the Schrödinger wave equation.
(b) Determine the expectation value for the energy,  t, at a given time t.
(c) Plot the probability density function P x t x t x t a f a f a f , , ,  *  at t  0 and
t
  /  A.
Solution:
(a) SWE, Cartesian coordinates, V x t x L  , 0, 0    
       
     
, , ,
A A B B
x t c x t c x t
    
 
2
3 3
8 2
A
h
mL m L
 
   
 
  | 5 25
2 9
B A
m L
 
   
 

 
2
,
x t
   | 2
4 3 3
x
     | 2
2 5 5
A
x t
      | B
t
 
2
m x
  | x op  | op | t
2 2
2
, 2
2 2 2 2 2 2
2 | sin
   | exp
i
    | sin exp
i
   
  | x
  |    |  

 |   |   | 

3
L L
   | 3
L L
   | L
   | L
   |   |  , ,
, ,
x t x t
p x t x t i

 
  
     
  
      


 
   
2
2
3
,
L L
x t
   
  | 2
3
L
   | L
   | L
   |    |   | 2 2
2
4 3 3 2 5 2 2
sin exp sin exp
, ,
2
A A B B
A A A B B B
m t m t x x
i i
c x t c x t
m x
      
 
             
                  
    


   

 
 
     
2 2
, ,
, ,
x t x t
i c x t c x t
 
  
      

 | SWE is satisified
2
2
m x
 | A A A  | B B B | x

sin
  | exp
i
    | sin exp
i
     | i
    | i
     | t
  |   |   | 

 | 
3
L
 
  | 3
L L
    
   | L
   | L
  |   | 

   |  
     | 
   | 
   | 
   | 
  |  2
, 4 3 2 5 5
A A B B
x t x x t t

8
(b)
   
     
       
       
       
*
0
* * * *
0
* * * *
0 0
* * * *
0 0
*
, ,
, , ,
, , , ,
, , , ,
, , , ,
,
L
op
op A A A B B B
L
A A B B A A A B B B
L L
A A A A A A B A B A
L L
B A B A B B B B B B
A
x t x t dx
x t c x t c x t
c x t c x t c x t c x t dx
c c x t x t dx c c x t x t dx
c c x t x t dx c c x t x t dx
x
 
  
  
 
 
  
    
              
     
     

 
 
 
   
   
0  | 0 | A
L  | L | A  | A2
0
0
*
0 0
3 3
, sin exp sin exp
3 1 3
sin sin
2 12 2
5 5
, , sin exp sin
L
L
L  | L | B | B
B  | B | L
   | L |   |   |  x x t t
t x t dx i i dx
L L
x L x L
dx x
L L
x x t
x t x t dx i
   
 

  
           
               
           
     
         
     
      
         
 

 
 

   
 
2
0
0
*
0 0
exp
5 1 5
sin sin
2 20 2
3 5
, , sin exp sin exp
3
exp sin
L
x

   | x
dx

   | L
x | L |  L
L  | L
 
    | A
   
     | B
 
 
 
 | A  | B | 
  | 
  | 
 
  | 
 
  |   | 
L
    | L
    
 A B
t
i dx
L L
x x t t
x t x t dx i i dx
t x
i
L


   
  
    
      
     
         
     
   
    
  

 


   
       
0
0
*
0
* * * *
0
* *
4
L L L
c c c c
  
     
  
  |   | 
2
A A A
 
   | B B B A
    | 
B
     | A B
    
2
   | 3 2
L
 
   | 3 2 3
    | 35 1 2 1 7
sin sin sin 0
2 7
, , , 0
,
, , , ,
L
L
L
B A
L
A A B B A A A B B B
x x x
dx
L L L
Similarly x t x t dx
Therefore
c x t c x t c x t c x t dx
  
  
        
           
        
  
              



2 2 2 2 2 2 2 2
2 2 1
2 3 1 5 25 1
3 7
3 2 3 2 6 6
L
L
m L m L m L m L
   

     
               
                                        
9
(c)
 
,
P x t  |    
* 2
4 3
, , sin
x
x t x t
  |  
8 3 5
sin sin exp
A B
x x t
i
   
 
2
8 3 5 2 5
sin sin exp sin
tx x x
    
        
2 2
4 3 2 5 8 3
sin sin sin s
x x x
  
     
    |  
t
  | x

3
L  |    |   |   |    |     |   |   |    | 

3
L L L L
        | 
3
L L L
      | 3
L
 | L
   | 
B A
i
     |   |   |    |   | 
3
L  | 3
L L
   | 3
L L
   | L
  | L | 
  |   |   |   |   |  | B  | A        
  2 2  
2 2
5
in
  | 2cos |    |    |   4 3 2 5 2 8 3 5
, sin sin sin sin cos
3 3 3
4 3 2 5 4 2 3 5 16
sin sin sin sin cos
3 3 3 9
B A
A
x x x x t
LP x t
L L L L
x x x x t
L L L L
     
    
   
    
          
            
         
        
          
        



  
0
0.5
1
1.5
2
2.5
3
3.5
0 0.2 0.4 0.6 0.8 1
t=0 t = h/(2
A
)
x/L
L P(x,t)
10
4. Incropera, Problem 4.5 (revised). A particle's motion in the x-direction is described by the
amplitude function
 
x | 5 3 2
4 8
x i
x i



(a) Determine the normalization constant.
(b) Plot the probability density as a function of x .
(c) Where is the particle most likely to be found?
(d) Determine the expectation value x .
(e) Discuss your results from (c) and (d).
Solution:
(a)
   
   
2
*
2 2 4
2
25 3
x
C

  | 2
25 3
2
x

 | dx  | dx even function
4
16 8
x


4
0
16 8
x
 |   | 
2  | 2
  |   |   | 
4 4 4 4
0 0 0 0
50 2 3 25 2 3 1
8 2 1 8 2 1 2 1 4 2 1 4 2
x x
C dx dx dx dx
   
   1/4 1/4
5 3 5 3 25 3
4 8 4 8 16 8
2 2
x i x i x
x x
x i x i x
x x x x
d
Let u x du dx dx
 
     
             
   
  
 
 
   
2
1/4 1/4 4 4
0 0
u
C du du
u u
  
     | 2 2  | 4  | 4 | 4
0
du
u
 
2
4
0
u
du
u

 
From the table of definite integrals:1/4
2 2
2
25 3 1 1
3.716 0.6307
4 2 2 4 2 1 1 1 1
u
u x u x
   
 
 
1
0
2
0
1
sin
3.716 0.6307
3.716 0.6307 4.828
3 2 2
4sin 4sin 4 4
4 2 2 2
2.197
1  | 5 3

norm | 2.197 4 8
m
n
u
du for m n
u
  | m
n
 
  | n
C
C
x
x  | i
ix


   
 



  
 
    
       
               



11
(b)      
2
*
4
1 25 3
norm norm 4.828 16 8
x
P x x x
x
 

 

0
0.05
0.1
0.15
0.2
0.25
0.3
-8 -6 -4 -2 0 2 4 6 8
P(x)
x
(c) Find maxima of P x   .
    
1
2 4
1
dP x d
 |  
2 3
1 50 25 3 64
x x x
  
 
4.828 16 8 16 8
x
x
      | 4.828
dx dx
 
4
   | 2
4
 | 25 3 16 8
x x
      |     
 
 
4 2
5 5 3
2 2 2
4 4 4
1 800 400 1600 192 1 800 192 400
4.828 4.828 16 8 16 8 16 8
x x x x x x x
x x x
 
               
      
   
The probability is a minimum or maximum when dP x   0
dx
 . A minimum obviously occurs at
x  0 .
12
 
    
 
4 2 2 2
2
0 800 192 400 800 192 400 0,
192 192 4 800 400
0.1200 0.7172
2 800
dP x
also when x x w w where w x
dx
w
  |    | 
  |   |   |        
Need to choose the positive root since x is real. The particle is most likely to be found when
w  0.5972  | x   0.7728
(d)    
*
x x x x dx
 

  |    
*
x x x dx
 

  | 2
1 25 3
0
x
x dx

4
4.828 16 8
x

norm  | norm  | norm
  | norm  |   |   | 
   
The expectation value is zero because x  norm norm * x x    is an odd function.