CHEM106: Assignment 6
Harmonic Oscillator
1. For the harmonic oscillator system, the solution of the Schrödinger equation leads to
1
the quantized energy EV  (V  h ) .
2
A. Define the zero-point energy of the system.
The zero-point energy (ZPE) for a quantum mechanical system is defined as the
lowest possible energy allowed. In the case of the harmonic oscillator, the energy
depends on the quantum number v, and v = 0, 1, 2, 3, ….
1
1
Thus,
ZPE  E0  (0  h )  h .
2
2
B. Determine the energy gap between state v + 1 and state v.
The energy gap between the state v + 1 and v can be calculated as
1
1
EV 1  EV  (V  1  )h  (V  )h  h .
2
2
It is worth noting that the energy gap is related to the frequency of molecular
vibration (see Unit 7, Section 3 later).
2. For the harmonic oscillator, the solution of the Schrödinger equation leads to the
2

ground state wave function 0 ( x )  ( )1 / 4 e x / 2 .

A. Show that the wave function n (x ) is normalized.

We need to evaluate the integral M   0 ( x )*0 ( x )dx . If M = 1, then the wave

function is normalized.
M=
ò
+¥
-¥
Y 0 (x)* Y 0 (x)dx
a
a
-¥
p
p
+¥ a 1/2 -a x
= ò ( ) e dx.
-¥
p
=
ò
+¥
( )1/4 e-a x /2 ( )1/4 e-a x /2 dx
2
2
2
Look up the integration table, and find that


0


( )1/ 2 e x dx 
2
1 
.
2 
 

2
2


1 

 1.
We have M   ( )1 / 2 e x dx  2( )1 / 2  e x dx  2
 
0


2 
Thus, 0 ( x ) is normalized.
B. Calculate the average value of the linear momentum.
According to the average value postulate of quantum mechanics, the expectation
value for any physical observable A is defined as
A   * ( x ) Aˆ( x )dx ,
where * ( x ) is the complex conjugate of  ( x ) , and the integration is done over
the entire space.
Thus, the average linear momentum can be evaluated by the following integral:
< Px >=
=
ò
+¥
-¥
ò
+¥
-¥
Y 0 (x)* P̂x Y 0 (x)dx
a
p
( )1/4 e-a x /2i
2
d a 1/4 -a x 2 /2
( ) e
dx
dx p
+¥ -a x /2
a
e
(a x)e-a x /2 dx
-¥
p
+¥
a
= i ( )1/2 a ò xe-a x dx.
-¥
p
= i ( )1/2 ò
2
2
2
Consider that xex is an odd function, and its integral over the whole range must
2
vanish, i.e.,



xex dx  0 .
2
Thus, we obtain  Px  0 .
3. Tunneling occurs in the quantum harmonic oscillator. For a classical harmonic
oscillator, the particle cannot go beyond the classical energy barriers (i.e., points
where the total energy equals the potential energy).
A. Identify these points for a quantum-mechanical harmonic oscillator in its ground
state.
1
For the ground state of the harmonic oscillator, the total energy is Etotal  h .
2
Panel (a) of the above figure plots the potential energy as a function of
displacement (x). At the classical barrier (position a and –a), the total energy of
1
1
the system is all in the form of potential energy. That is Etotal  h  ka 2 .
2
2
Solve for a, and we obtain
a
h
.
k
For the harmonic oscillator, the vibrational frequency is given by  
1
2
k

.
Plug into the above equation, and we obtain
h
h
a

.
k
2 k
B. Set up an integral defining the probability of tunneling (i.e., the probability that the
particle will go beyond the classical barriers). [You do not need to evaluate the
integral.]
2

The ground state wave function 0 ( x )  ( )1 / 4 e x / 2 for the harmonic oscillator is

plotted in panel (b) of the above figure. For the +x direction, the probability of
tunneling can be evaluated as

 
 
2
2
2

P direction   0 ( x )* 0 ( x )dx   ( )1 / 4 e x / 2 ( )1 / 4 e x / 2 dx   ( )1 / 2 e x dx .
a
a
a



Due to the symmetry, the overall probability of tunneling should be twice of that
of one (+) direction. Thus, the overall probability of tunneling is

a

h
where a 
.
2 k

Ptunneling  2  ( )1 / 2 e x dx ,
2
4. Vibration of a diatomic molecule can be modeled as a harmonic oscillator. Calculate
the ratio of the vibrational frequency between hydrogen chloride (H35Cl) and
deuterium chloride (D35Cl).
For the harmonic oscillator, the vibrational frequency is given by  
1
2
where k is the force constant, and  is the reduced mass defined by  
k

,
m1m2
.
m1  m2
The frequency for H35Cl and D35Cl are defined by  HCl 
 DCl 
1
2
k DCl
 DCl
1
2
k HCl
 HCl
and
, respectively.
For various isotopes of the same element, the force constant k stays the same. We
have k HCl  k DCl . Therefore, the frequency ratio is
 HCl
1
(
 DCl
2
  HCl 

 HCl
 DCl
k HCl
 HCl
) /(
1
2
k DCl
 DCl
)
 DCl
 HCl
35  1
 0.9722( amu) and
35  1
1.8919

 1.4 .
0.9722
 DCl 
35  2
 1.8919( amu)
35  2