CHEM 342. Spring 2002. Problem Set #2. Answers.
Orthogonality
 x 
 2x 
1. Prove that the functions  1  Asin   and  2  A sin 
 are orthogonal
 a 
 a 
0  x  a.
sin n  mx sin n  mx
Hint:  sin nx sin mxdx 

C
2n  m
2n  m

a
 1*  2 dx 
0

a
0
 x 
 2x 
2
A sin   A sin 
dx  A
a
a
 



a
0
 x   2x 
sin   sin 
dx 
 a   a 
a
    2 
 
   2  
 3  
 sin  a  x sin  a  x 
 sin  a a  sin  a a 



 


  
 A2 

 A2 


    2 
   2 
   2  
 6  
2
 
  
 2 a 
  a 

 a  0

 a  
 
 


 sin   sin 3
0
 A2 

   2   6  
 a   a  
   

2. Give a mathematical definition for the Kronnecker delta  nm  . What is the numerical
value of the Kronnecker delta when the two eigenfunctions are orthogonal? What is the
numerical value of the Kronnecker delta when n and m are the same eigenfunction (i.e. n = m)?
In addition to these two values, can the Kronnecker delta be equal to any other numerical values?
 nm    *n  m d
The eigenfunctions  n and  m are orthogonal when  nm  0 . The other possible value
of the Kronnecker delta is 1, and it occurs when n = m. The Kronnecker delta can only equal zero
or one; no other values are possible.
Operators


d 
d 
3. Find the result of operating with A  y    and B  y    on the function
 dy 
 dy 
f y  e  y
2
/2
. Is f(y) an eigenfunction of  or of B̂ ?
2
d  y2 / 2
A f  y   ye  y / 2 
e
dy

A f  y   ye  y




2
/2
A f  y   2 ye  y
2

 ey
2
/2
 y 
/2
1
CHEM 342. Spring 2002. Problem Set #2. Answers.
This function is not an eigenfunction of  .


2
/2


2
/2
 ey

2
/2
 ye  y
B f  y   ye  y
B f  y   ye  y
B f  y   ye  y
d  y2 / 2
e
dy


B f  y    y  y e  y
2
/2
2
/2
2

 y 
/2

 0  e  y
2
/2

This function is an eigenfunction of B̂ with eigenvalue of zero.
4. Find the following commutators for any function f(x).
(a)
 2 
dˆ , x 


(b)
 2 
d , xˆ 




d
d2
Hint: dˆ 
, d 2  2 , xˆ  x , and x 2  x 2 .
dx
dx
(a)


 2 
2
2
ˆ
ˆ
d , x  f x   d x f x   x dˆf x 


 2 
d 2
 df x   2
x f x   
x
dˆ , x  f x  
dx
 dx 


 2 
df x   df x   2
d x2
ˆ




d
,
x
f
x

f
x
 x2

x


dx
dx
 dx 


 2 
dˆ , x  f x   2 x  f  x 




   
 
Therefore dˆ , x 2   2 x


(b)
2
CHEM 342. Spring 2002. Problem Set #2. Answers.


 2 
2
2
ˆ
ˆ
ˆ




d
,
x
f
x

d
x
f
x

x
d
f x 




 2 
d 2 f x 
d2
d , xˆ  f x   2 xf  x   x
dx
dx 2


 2 
df  x 
d 2 f x 
d 
dx


f
x

x

x
d , xˆ  f x  
dx 
dx
dx 
dx 2


 2 
d , xˆ  f x  


 2 
d , xˆ  f x  


df  x  
d 2 f x 
d 


f
x

x

x
dx 
dx 
dx 2
df  x  dx df x 
d 2 f x 
d 2 f x 

x
x
dx
dx dx
dx 2
dx 2
 2 
df  x  df  x 

d , xˆ  f x  
dx
dx


 2 
 df  x  

d , xˆ  f x   2
 dx 


 2 
d 
Therefore d , xˆ   2 
 dx 



5. Find the result of operating with the operator O 
1  d  2 d  2
 r    on the function
r 2  dr   dr  r

  Ae br . What values must the constants have for  to be an eigenfunction of O ?


















O Ae br 

1  d   2  d
Ae br
  r 
2 
r  dr    dr
 
O Ae br
  

 



 




2 Abe br 2
 Ae br
r
r
2b 2 

 Ae br b 2 
 
r
r

O Ae br  Ab 2 e br 
O Ae br
br
1  d  2
2
br 
r

bAe

Ae br




2
r  dr 
 r
1
2
 2  bA r 2  be br  2re br  Ae br
r
r

br

2e  2
br
 bA  be br 
  Ae
r  r

O Ae br 
O Ae br
   2r Ae 



3
CHEM 342. Spring 2002. Problem Set #2. Answers.

For  to be an eigenfunction of O , the constant b must equal 1, while A can be any real
number. In this case, the eigenvalue is 1.


6. Find the result of operating with the operator  2 on the function x 2  y 2  z 2 . Is it
an eigenfunction?
2
2
2
2
  2 2 2
x
y
z


x
x

 2  2  2
 6
2 x2  y2  z2
2
2
x

2 x2  y2  z2 
2
 y2  z2
2
 y2  z2
2
2
 y2  z2

y, z

2
x
y



 2 x   2 y   2 z 
x
y
z
2
2
x
2
 y2  z2

x, z

2
z
2
x
2
 y2  z2

x, y
This is not an eigenfunction of  2 .
7. The function   Ax1  x is a well-behaved wave function in the interval 0  x  1 .
Calculate the normalization constant (A), and the average value of a series of measurements of x
(i.e find the expectation value:  x  ).
1
0 
*
dx  1
A 2  x 2 1  x 2 dx  1
1
0

1

 x dx  1
A 2  x 2 1  2 x  x 2 dx  1
0
1

A2  x 2  2x 3
0
4
13 214 15 
A  
  1
4
5 
 3
 1 
A2    1
 30 
2
A  30
4
CHEM 342. Spring 2002. Problem Set #2. Answers.
 xˆ    *  xˆ dx
1
0
1
 xˆ   x * dx
0
 xˆ   x30 x 2 1  x  dx
1
2
0
1


 xˆ  30  x 3  2 x 4  x 5 dx
0
14 215 16 
 xˆ  30  
 
5
6 
 4
1
 xˆ 
2
Expectation Values
8. For the wave function  and the operator ̂ , give an expression that could be used to
calculate the average value obtained from repeated measurements (i.e. show an expression for
 ˆ  ).
 * ˆ  d

 ˆ 
*
  d
or
 ˆ    * ˆ  d , where  is normalized.
Particle In a Box
9. Calculate the value of A so that  n  A sin
Hint:

sin 2 bx dx 
nx
is normalized in the region 0  x  a .
a
x  1 
   sin 2bx   C
2  4b 
5
CHEM 342. Spring 2002. Problem Set #2. Answers.
   dx  1
a
0
*
n
n
2
nx 

A sin
dx  1

a 
0 
a
nx
A 2 sin 2
dx  1
a
0

a

a
2x
a
 2nx 
A  
sin 
  1
 2 4n  a  0
a
a
sin 2n  sin 0  1
A2  
 2 4n

a
A2    1
2
2
Therefore, A 
a
Note: sin 2n  0 for n = integer.
10. For a particle in a one-dimensional box 0  x  a  , we used eigenfunctions of the
form   Asin kx . Explain why we could not use
(a)   Ae kx
(b)   Acoskx
(a) The wave function must be zero for x = a. If k is a real number, then e kx cannot fit
this boundary condition.
(b) The boundary conditions also require that   0 when x = 0. This cannot be true for
the cosine function because cos 0 = 1. Therefore, the allowed eigenfunction must be
of the form   Asin kx .
11. The ground-state wave function for a particle confined to a one-dimensional box of
1/ 2
 x 
sin   . The box is 10.0 nm long. Calculate the probability that the
 L
x  1 
particle is between 4.95 nm and 5.05 nm. Hint: sin 2 bx dx     sin 2bx   C
2  4b 
2
length L is    
L

6
CHEM 342. Spring 2002. Problem Set #2. Answers.

P

5.05
P   * dx
4.95
 2  2  x 
  sin  L dx
4.95  L 
 
5.05
2
P 
L
 2  x 
sin  L dx
4.95 
 

5.05
5.05
 2  x L
 2x 
P    
sin 

 L   2 4  L  4.95
5.05
x 1
 2x 
P 
sin 

 L 2  L  4.95
24.95 
5.05  4.95 1  25.05

sin
 sin

10.0
2 
10.0
10.0 
P  0.020
P
12. What is the ground state energy (i.e. n = 1) for an electron that is confined to a box
which is 0.2 nm wide. [Hint: Planck's constant, h,is 6.626 10 34 J s; the mass of an electron, me,
is 9.109 10 31 kg]
h2n2
E
8ma 2
E

6.626  10
34

Js
 1
2
2
8 9.109  10 31 kg 0.2  10 9 m

2
E  1.506  10 18 J
Uncertainty
13. The speed of a certain proton is 4.5  105 m/s along the x-axis. If the uncertainty in its
momentum along the x-axis is 0.010 %, what is the maximum uncertainty in its location along
the x-axis (i.e. x )?
p x  0.010% p 0
p x  0.00010mv
7
CHEM 342. Spring 2002. Problem Set #2. Answers.
xp x 
h
4
x 
h
4p x
x 
h
40.00010 mv
x 

6.626 10
40.00010  1.673 10
34
 27
Js


kg 4.5 10 5 m s 1

x  7.0110 10 m
Tunneling
14. The wave function inside an infinitely long barrier of height V is   Ae  kx .
Calculate (a) the probability that the particle is inside the barrier; and (b) the average penetration
depth of the particle into the barrier (i.e. the expectation value  x  ). Because the barrier is
n!
infinitely long, this wave function is valid for 0  x   . Hint:  x n e ax dx  n1
a


P   dx 
*
0

A e
2  2 kx
0


 x  x * dx 
0


0
2 e


1  A2
2
dx  A 

A
0



  2k

2
k

2
k



0
xA2 e  2 kx dx 
 2 kx
A2
4k 2
8