Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 13
The Energy levels of a Rigid Rotator Are E=ħ2 Ɩ (Ɩ+1)/2I
- In this section we shall discuss a simple model for a
rotating diatomic molecule.
- The model consists of two point masses m1 and m2 at fixed
distance r1 and r2 from their center of mass (Fig.1)
- Because the distance between the two masses is fixed, this
model is referred to as the rigid rotator model.
Figure (1). Two masses m1 and m2 shown rotating about their
center of mass.
- We can treat a rigid rotator as having one mass fixed at
the origin with another mass, the reduced mass µ,
rotating about the origin at a fixed distance r.
- We discussed a rigid rotator classically before and
showed there that the energy of a rigid rotator is
1
K  I 2
(1)
2
where ω is the angular velocity and I is the moment of
inertia,
I = µ r2
(2)
[1]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 13
The angular momentum L is
L=Iω
(3)
And the kinetic energy can be written
L2
K 
2I
(4)
The Hamiltonian operator of a rigid rotator is just the
kinetic energy operator and using the correspondence
between linear and angular systems, we can replace m by
I and write it as:
2

Hˆ 
2
2I
(5)
Because one of the two masses of the rigid rotator is fixed
as the origin we shall express 𝛁2 is spherical coordinate
and so write Ĥ as :
 1  

Hˆ  
Sin


2I  Sin  

2
1 2 


2
2 
 Sin     (6)
There is no term in Ĥ here involving the partial
derivative with respect to r because r is fixed in the rigid
rotator model. By comparing equation (6) with the
classical expression equation (4) we see that:
L̂  
2
2
 1  
 
1 2 
 Sin   Sin    Sin 2  2  (7)




Note that the square of the angular momentum is a
naturally occurring operator in quantum mechanics
[2]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 13
- The rigid rotator wave function are customarily denoted
by Y(,) and so the Schrödinger equation for a rigid
rotator reads
ˆ ( ,  )  EY ( ,  )
HY
(8)
Or
 1  
 
1 2 
 
Y ( ,  )  EY ( ,  ) (9)
 Sin

2I  Sin  
  Sin 2  2 
2
If we multiply equation (9) by Sin2() and let

2IE
2
We find the partial differential equation
 
Y   2Y
2
Sin
Sin




Sin
 Y  0 (10)


2
 
  
β must obey the condition:
β=Ɩ (Ɩ +1) Ɩ = 0, 1, 2… (11)
Using definition of β, equation 11 is equivalent to
E1 
2
2I

 1
Ɩ =0, 1, 2… (12)
[3]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 13
The Rigid Rotator Is a Model for a Rotating Diatomic
Molecule
ΔƖ = ± 1
(13)
- Equation 13 is called a selection rule
- In the case of absorption of electromagnetic radiation, the
molecule goes from a state with a quantum number Ɩ to one
with Ɩ +1. The energy difference then is
ΔE = EƖ+1 – EƖ =
h2
4 2 I

 1
(14)
- The energy levels and absorption transitions are shown in
Figure 2 using Bohr frequency conditions E = h , the
frequencies at which the absorption transitions occur are

h
4 2 I

 1
=0,1,2….
(15)
It is common practice in microwave spectroscopy
equation 15as
  2B   1
where B 
to write
(16)
h
8 2 I
Hz is called the rotational constant of the
molecule.
If we use a relation   2B (  1)
[4]
=0,1,2,....
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 13
Where B is the rotational constant expressed in units of wave
numbers?
B 
h
8 2cI
Cm-1
(17)
Figure 2: The energy levels and absorption transitions of
a rigid rotator.
[5]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 13
Example
To a good approximation, the microwave spectrum of H 35Cl
consists of a series of equally spaced lines separated by
6.26×1011Hz. Calculate the bond length of H Cl.
Solution:
According to equation (16) the spacing of lines in microwave
spectrum is given by
2B 
h
4 2 I
h
4 2 I
 6.26  1011 Hz
Solving this equation for I
(6.626 1034 J .s )
47
2
I  2

2.68

10
Kg
.
m
4 (6.26 1011 s 1 )
The reduced mass of HCl is

(1.00)(35.00)
(1.66 1027 Kg )  1.611027 Kg
(36.00)
Using the fact that I=r2 we obtain
 2.68 1047 Kg .m 2 
r 

27
 1.6110 Kg 
1
2
 1.29 1010 m
[6]