Download Lecture XX

Classical Model of Rigid Rotor
A particle rotating around a fixed point, as shown below, has angular momentum and rotational
kinetic energy (“rigid rotor”)
The classical kinetic energy is given by:
If the particle is rotating about a fixed point at radius r with a frequency ʋ (s−1 or Hz), the velocity o
the particle is given by:
where ω is the angular frequency (rad s−1 or rad Hz). The rotational kinetic energy can be now
expressed as:
Also
where
Consider a classical rigid rotor corresponding to a diatomic molecule. Here we consider only rotation
restricted to a 2-D plane where the two masses (i.e., the nuclei) rotate about their center of mass.
The rotational kinetic energy for diatomic molecule in terms of angular momentum
Note that there is no potential energy involved in free rotation.
Momentum Summary
Classical
Linear
Momentum
dr
p  mv  m
dt
2
Energy
Rotational Momentum
(Angular)
Energy
QM
p̂  i 
2
p
md r
K

2m 2 dt 2
L  r p
1
2
K  r  p
2I
Kˆ  
2
2
2
L̂  i r  
Kˆ  
2
2I
r  
2
Angular Momentum
L  rp
L  x
y
z    px
i
L x
px
j
y
py
py
k
z
pz
pz 
Angular Momentum
L  Lx i  Ly j  Lz k
Lx  ypz  zp y
Ly  zpx  xpz
Lz  xp y  ypx
Angular Momentum
ˆ
L  rˆ  pˆ  i rˆ  
L  x
y
d
z   i 
 dx
i
L  i x
d
dx
j
y
d
dy
k
z
d
dz
d
dy
d

dz 
Angular Momentum
L  Lx i  Ly j  Lz k
 d
d 
ˆ
Lx  i  y  z 
dy 
 dz
d
 d
ˆ
Ly  i  z  x 
dz 
 dx
 d
d
ˆ
Lz  i  x  y 
dx 
 dy
Two-Dimensional Rotational Motion
Polar Coordinates
y
x  r cos( )
r
y  r sin( )

x
d
d 
   i  j
 dx dy 
2
2


d
d
2
  2  2
dy 
 dx
Two-Dimensional Rotational Motion
d 1 d
1 d2
d2
d2
r
 2
 2 2
2
dr r dr
r d
dx
dy
2
d
1
d
1
d
2  r
 2
dr r dr r d 2
2


d
1
d
1
d
2
ˆ
H 
 
r
 2

2
2  dr r dr r d 2 
2
2
Two-Dimensional Rigid Rotor
Hˆ (r , )  
2
2
 2 (r , )  E (r , )
2


d
1
d
1
d
2
ˆ
H 
 
r
 2

2
2  dr r dr r d 2 
2
2
Assume r is rigid, ie. it is constant
2
2
2
1
d
Hˆ  
 2r  
2
2  r 2 d 2
Two-Dimensional Rigid Rotor
2
2
d
Hˆ ( )  
 ( )  E ( )
2
2 I d
 2 d2

  2 I d 2  E  ( )  0


d
2I 
 d 2  2 E  ( )  0


2
 d2
2
 2  m  ( )  0
 d

2I
m  2E

2
m2 2
E
2I
Solution of equation
Energy and Momentum
2
m 
E
2I
2
Z
2
2
L
m 

2I
2I
LZ  m
2
As the system is rotating
about the z-axis
Two-Dimensional Rigid Rotor
m2 2
Em 
2I
m
E
Lz m  m
 m E
 2
 
m I 
Lz m
6
 6
18.0
6
5
 5
12.5
5
4
 4
8.0
4
3
2
1
 3
 2
 1
4.5
3
2

2.0
0.5
Only 1 quantum number is require to determine the state of the system.
Spherical coordinates
Spherical polar coordinate
Hamiltonian in spherical polar
coordinate
1 d 2 d
1
d
d
1
d2
  2 r
 2
sin 
 2 2
r dr dr r sin  d
d r sin  d 2
2
Rigid Rotor in Quantum Mechanics
Transition from the above classical expression to quantum mechanics can be carried out by
replacing the total angular momentum by the corresponding operator:
Wave functions must contain both θ and Φ dependence:
are called spherical harmonics
Schrondinger equation
2I
Multiplyin g by sin  2

2
Two equations
Solution of second equation
Solution of First equation
   P
m
J
  J ( J  1)
Associated Legendre Polynomial
Associated Legendre Polynomial
m
 d

Pl (cos )  sin  
Pl (cos )

 d (cos ) 
m
m
l

1  d
2
l
Pl (cos )  l 
(cos


1)
2 l !  d (cos ) 
0
0
Y
0

1  d
2
0
Pl (cos )  0 
(cos


1)
1

2 0!  d (cos ) 
For l=0, m=0
0
 d

0
P0 (cos )  sin 0  
11

 d (cos ) 
Y 
0
0
1
4
First spherical harmonics
Spherical Harmonic, Y0,0
0 2
0
Y
 const  distance of surface from the origin
l= 1, m=0
0
 d

P1 (cos )  sin  
cos  cos

 d (cos ) 
0
0
1
3
3
i 0
Y1,0 ( , ) 
cos( )
 1 cos( ) e  
2
2
2 
l= 1, m=0
0 2
1
Y
 const  cos 2   distance of surface from the origin
θ
cos2θ
0
1
30
3/4
45
1/2
60
1/4
90
0
l=2, m=0
Y2,0 ( , ) 
5
4 
(3cos 2 ( )  1)
θ
0
cos2θ
1
3cos2θ-1
2
30
3/4
(9/4-1)=5/4
45
1/2
(3/2-1)=1/2
60
1/4
(3/4-1)=-1/4
90
0
-1
l = 1, m=±1
Y1,1 ( , ) 
3
 i
sin( ) e
2 2
Complex Value??
If Ф1 and Ф2 are degenerateeigenfunctions, their linear
combinations are also an eigenfunction with the same
eigenvalue.
l=1, m=±1
1
Y1,1 ( , )  Y1,1 ( , )  
2
3
sin( ) e  i  e  i  
4 2
3
sin( )cos( ) 
2 2
Along x-axis
1
Y1,1 ( , )  Y1,1 ( , )  
2i
3
sin( ) e  i  e  i  
4i 2
3
sin( )sin( ) 
2 2
E
Three-Dimensional Rigid Rotor States
El 
l
2
2I
l (l  1)
m
32
3
2
2
1
0
l
 m ,..,m
Y
Ll 
l (l  1)
 2
 
l I 
L Lz m
E
Lz  m
3
2
10
-1 -2
-3
3
6.0
Y3,2,1,0,
1, 2, 3
12

2
2
0
2
10
-1 -2
1 0
-1
0
2
2,1,0, 1, 2
Y
1
Y1,0,
1
0
Y0
3.0
6

2
1.0
0.5
2

0
0
0
Only 2 quantum numbers are required to determine the state of the system.
Rotational Spectroscopy
EJ 
2
2 r
2
o
J ( J  1)
EJ 
2
2I
J ( J  1)
J : Rotational quantum number
E  EJ 1  EJ

2
2 r
2
o
( J  1)( J  2)  J ( J  1)
2
J  1  2hcBJ  1
E 
I
2
B
2 Ihc
Rotational Constant
Rotational Spectroscopy
E  h 
hc

 hc
Wavenumber (cm-1)
h( J  1)

4 2 Ic
B
  2 B( J  1)
Rotational Constant
v  c  c  J 1  J 
h
8 2  ro2 c
Line spacing
 c  2 B( J  1  1)  2 B( J  1)   2cB
v
Frequency (v)
v
Bond length
• To a good approximation, the microwave
spectrum of H35Cl consists of a series of
equally spaced lines, separated by 6.26*1011
Hz. Calculate the bond length of H35Cl.