VIBRATIONS OF DIATOMIC MOLECULES
CLASSICAL APPROACH
The classical one-dimensional harmonic oscillator is a particle which is bound to an
equilibrium position by a force which is proportional to the displacement from that
position and is expressed as
m
d 2x
dU ( x )
and
 Fe  ke x  
2
dt
dx
U ( x) 
1
ke x 2
2
(1)
mass of particle, m
displacement of particle, x
elastic restoring force, Fe
force constant (elastic spring constant), ke
potential energy of system, U
The solution to this classical problem is
x(t )  A sin(o t   )
(2)
amplitude, A
initial phase angle, 
ke
m
classical total vibrational energy of the oscillator,
classical oscillation frequency, o 
Evib 
1
1
k A2  m A2 o 2
2
2
(3)
(4)
The total vibrational energy of the particle can have any value and is proportional to the
square of the amplitude and the square of the oscillation frequency.
QUANTUM APPROACH
The motion of a vibrating system is governed by the Schrodinger equation [Atkins]. We
will consider the solutions of the Schrodinger equation for the vibrations of the centre of
mass of a diatomic molecule as shown in figure 1. Analytic and numerical solutions of
the Schrodinger equation will be compared.
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Fig. 1.
Vibrations of a diatomic molecule and
its centre of mass.
For small amplitude vibrations about the equilibrium separation distance Req of the two
atoms, the potential energy U(R) as a function of the separation distance R can be
approximated by a parabola
U ( R) 
1
ke ( R  Req ) 2
2
(5)
Assuming the two atoms with masses m1 and m2 are at locations x1 and x2 and are free to
move parallel to the X-axis, then the Schrodinger equation for the motion for the atoms
can be written as
 2   2   2   2 

 2   
 2   U   E
 2m1  x1   2m2  x2 
(6)
We can simplify equation (6) using a number of transformation of the variables
total mass of molecule, m = m1 + m2
reduced mass, 1/ = 1/m1 + 1/m2
separation of atoms, R = x1 – x2
location of centre of mass, x = (m1/m)x1 + (m2/m) x2

d  m1  d

 
x1 dR  m  dx

d  m2  d

 
x2
dR  m  dx
wavefunction,    tran s ( x )  vib ( R)
The Schrodinger equation can now be written as two separate equations:
(1) The free translational motion for a particle of mass m along the X-axis
 2  d 2 trans ( x) 


  Etrans  trans ( x )
2
2
m
dx



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2
(2)
The internal vibrational (harmonic) motion for a particle of mass 
 2  d 2 vib ( R)  1

 ke ( R  Req )2  Evib  vib ( R)


2
 2  dR
 2
(8)
We can ignore the free motion of the molecule and concentrate on the internal vibrational
molecular motion as described by equation (8). Equation (8) can be solved analytically
and its solutions can be expressed in terms of Hermite polynomials [McKelvey].
The particle executes harmonic motion due to the restoring force acting between the two
atoms. The motion of the atoms are restricted because as R increases the potential energy
U(R) rises to large values and the wavefunction vib(R) will tend towards zero. Since the
vibrations are restricted and for physically acceptable solutions of the wavefunction, the
total vibrational energy Evib = En is quantized and the discrete set of energy level
eigenvalues are
1

En   n   o
2

o 
n = 0, 1, 2, 3, …
(9)
ke
m
The equal spacing between adjacent energy levels
E  En1  En  o
(10)
is an unique property of the parabolic potential well.
The ground state (lowest energy state, n = 1) of the harmonic oscillator is characterized
1
o . The distance between the
by a minimum zero-point vibrational energy, E1 
2
particles is fluctuating incessantly around the equilibrium separation whereas from a
classical mechanics point of view, the particles would be perfectly stationary.
The wavefunction  vib for each state is related to a Hermite polynomials and the normalized
wavefunctions are
 vib,n  N n H n (b) e  b
2
/2
(11)
normalization constant, N n 
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1
a  2n n !
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a
mk
b
x

Gausian function, e b
2
/2
Hermite polynomial, Hn(b)
The Hermite polynomials satisfy the differential equation
d 2 H n ( b)
dH (b)
 2b n
 2 n H n ( b)  0
2
db
db
n  0,1,2,3,
1
2
(12)
and the first few polynomials are
n
0
1
2
3
4
5
6
Hn(b)
1
2b
4 b2 - 2
8 b3 – 12 b
16 b4 - 48 b2 +12
32 b5 -160 b3 + 120 b
64 b6 – 480 b4 + 720 b2 -120
However, the best way to calculate the Hermite polynomials is by the use of the recursion
relationship
H n1 (b)  2 b H n (b)  2 n H n1 (b)
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MATLAB EXPLORATIONS
(1) Classical vibrations:
shm.m
For a classical oscillator, executing simple harmonic motion (SHM) calculate the
following quantities for a macroscopic particle attached to a spring and set vibrating
(amplitude of oscillation, A = 1 m, mass of object, m = 1.00 kg and force constant,
ke= 25.0 N.m-1)

angular frequency (rad.s-1)

natural frequency (Hz)

period (s)

separation between energy levels (eV)
Animate the motion of the particle and plot the following graphs

x/ t, v / t, a / t and v / x

K / t, U / t and E / t

U / x and Fe / x
For our bound classical particle, the probability of locating it in an interval dx is
proportional to the time dt spent by the particle in that interval. The classical
probability density function DP is derived from equation (2)
x(t )  A cos(o t )
(14)
 1
 x 
t  abs  acos   
 A 
 o
(15)
DP  N
dt
dx
(16)
The probability of finding the particle is 1, therefore

A
A
DP dx  1
(17)
from which the normalizing constant, N can be found. The derivative dt/dx can be
found numerically using the Matlab function gradient and the value of N found by
numerically by performing the integration given in equation (17) by Simpson’s rule
with the function simpson1d.
Plot the probability density function DP and calculate the probability of locating the
particle within  A/10 of the origin and  A/10 of the extreme positions.
Comment on the significance of each result.
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Matlab output
shm.m
CLASSICAL OSCILLATOR - SIMPLE HARMONIC MOTION
Angular frequency (rad/s,) wo = 2.19e+001
Natural frequency (Hz), fo = 3.49e+000
Period (s), T = 2.87e-001
Total Energy (J), Etot = 2.40e+000
Separation of energy levels (eV), dE = 1.3910-15
stopped
0.1
0.08
0.06
displacement x (m)
0.04
0.02
0
-0.02
-0.04
-0.06
-0.08
-0.1
0
0.2
0.4
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0.6
0.8
time t (s)
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1.2
1.4
6
50
-
acceleration a (m.s2)
displaecment x (m)
0.1
0.05
0
-0.05
-0.1
0
0.5
1
0
-50
1.5
0
0.5
3
3
2
2
-
1
0
-1
-2
-3
kinetic energy K (J)
1.5
1
0
-1
-2
0
0.5
1
1.5
-3
-0.1
-0.05
0
0.05
displacement x (m)
time t (s)
potential energu U (J)
1
time t (s)
velocity v (m.s1)
-
velocity v (m.s1)
time t (s)
0.1
3
2
1
0
0
0.5
1
1.5
1
1.5
1
1.5
time t (s)
3
2
1
0
0
0.5
total energy Etot (J)
time t (s)
4
3
2
1
0
0
0.5
time t (s)
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kinetic energy K (J)
2.5
2
1.5
1
0.5
0
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
position x (m)
0.04
0.06
0.08
0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
position x (m)
0.04
0.06
0.08
0.1
force F (N)
50
0
-50
-0.1
Comments
At any time t, the particle’s trajectory, velocity, acceleration and energy can be predicted.
The energy separation E = o for any macroscopic system is negligible, for practical
purposes the total energy varies continuously and not in discrete jumps.
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Classical probability distribution function
Prob near origin = 6.3 %
Prob near extremes = 28.5 %
8
7
probabilty density
6
5
4
3
2
1
0
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
position x (m)
0.4
0.6
0.8
1
Comments
(2) Quantum vibration:
vibrations02.m
The force constants for typical diatomic molecules are in the range between 400 to
2000 N.m-1.
Molecule
HF
HCl
Force
constant, ke
970
480
-1
(N.m )
Atom
H
F
mass (amu)
1.008
19.00
1 amu = 1.6605610-27 kg
HBr
HI
CO
NO
410
320
1860
1530
Cl
35.45
Br
79.91
C
12.01
O
16.00
For the diatomic molecules listed above, calculate the following:

angular frequency (rad.s-1)

natural frequency (Hz)

period (s)

separation between energy levels (J and eV)
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
wavelength  of the electromagnetic radiation emitted in the transition
n + 1  n.
Comment on the significance of your calculations
Matlab output
Quantum oscillator - diatomic molecule
HCl
Angular frequency (rad/s,) wo = 5.45e+014
Natural frequency (Hz), fo = 8.68e+013
Period (s), T = 1.15e-014
Spacing between energy levels (J), dE = 5.75e-021
Spacing between energy levels (eV), dE = 3.59e-002
Wavelength (m) lambda = 3.46e-005
Wavelength (m) lambda = 3.46e+001
Comments
The radiation associated with the vibration of molecules is mainly in the infrared part of
the spectrum. Diatomic molecules will be strong absorbers and emitters in the infrared
part of the spectrum due to molecular vibrations.
For molecules such as HCl, the mass of chlorine atom is much greater than the mass of
the proton. The chlorine atom can be regarded as stationary and the proton vibrating
towards and away from the more massive atom.
(3) Hermite polynomials:
function Hn = hermite(n,Hx)
Hn is the value of the Hermite
hermitePlot.m
polynomial of order n evaluated at Hx ( )
Evaluate the Hermite polynomials Hn() using the recursion relationship given by
equation (13) and plot them for the range -5    +5
Matlab output
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Hermite Polynomial, H(B): order = 3
1000
800
Hermite Polynomial, H
600
400
200
0
-200
-400
-600
-800
-1000
-5
0
B
5
(3) Hermite polynomials:
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