CHEM106: Assessment 4
Wave Functions and Probability
Answer Key
1. Define the complex conjugate of the following functions.
The complex conjugate for a complex number Z = a + ib is defined as Z* = a  ib. In
general, to find the complex conjugate of any function, all we need to do is to
reverse the sign of the imaginary component.
A. Z *  3  ib
B. f * ( x )  eix
C. f * ( x )  sin( 2 x ) , noticing that for a real function, f * ( x )  f ( x )
2. What are the requirements for an acceptable wavefunction?
Since  2   *  represents the probability density for finding the particle, the
wavefunction  itself must meet the following requirements: (1) finite (or
quadratically integrable); (2) single valued; and (3) continuous.
3. Which of the following is an acceptable wave function over the indicated interval?
A.
1
with 0  x  
x
1
is not an acceptable wavefunction since it approaches infinity as x = 0.
x
B. 2cos(x) with -¥ £ x £ 0
2cos(x) is an acceptable wavefunction since it meets all three requirements listed
in question 2 in the specified domain.
C. e  x with    x  
e  x is not an acceptable wavefunction since it approaches infinity as
x approaches   .
4. Which of the following functions are normalizable over the indicated intervals?
Normalize those functions that can be normalized. (Hint: In normalizing
wavefunctions, the integration is over all space in which the wave function is
defined.)
A. f ( )  e i 0    2
The function is finite everywhere, and thus normalizable.
2
Given M 

2
2
0
0
0
1
.
M
the normalization constant is
Thus, the normalized function is f N ( ) 
B. f ( x)  e2 x
2
f * ( ) f ( )d   e i e i d   d   0  2 ,
1 i
e .
2
with    x  
The function f ( x)  e2 x is infinite at x   , and thus not normalizable.
2
x ) with 0  x  L
L
The function is finite everywhere, and thus normalizable.
C. f ( x )  sin(
L
Given M  
0
2
2
2
f ( x ) f ( x )d x   sin(
x ) sin(
x )d x   sin 2 ( x )d x ,
L
L
L
0
0
L
L
*
look up the integration table, and find that  sin 2 (x )d x 
Thus, M 
1
L
4
x
sin(
x)
2
8
L
The normalization constant is
L
0
1
1
x
sin( 2x ) .
2
4
L L
4
L

[sin(
L)  sin( 0)]  .
2 8
L
2
1
.
M

Thus, the normalized function is f N ( x ) 
2
2
sin(
x) .
L
L
5. Consider the following normalized wavefunction  ( x ) 
1
3
cos( x ) , for a oneL
2L
dimensional system confined in the region  L  x  L .
A. Calculate the probability that the particle will be found in the region between x = 0
and x = L/2.
L/2
L/2
1
3
P   * ( x ) ( x )d x   cos2 ( x )d x
L 0
2L
0
Look up the integration table, and find that
1 1
2L
6
P  [ x
sin(
x )]
L 2
12
2L
0
L/2
Thus,
1 1
 
( 1)  0.197
4 6

1
1
 cos (x)d x  2 x  4 sin( 2x) .
2
1 L 2L
6 L
{ 
[sin(
)  0]}
L 4 12
2L 2
.
B. Determine the probability that the particle will not be located in the region
between x = 0 and x = L/2.
The probability in case B and that in case A above should add up to 100%.
Thus, P = 1  0.197 = 0.803.