Rotational Motion:
a)
Rotation about a circle. Very simple but not many applications.
b)
Rotation about a sphere. More complex mathematically, essential result for
molecular rotational motion (and for the hydrogen atom).
Find quantized energies, quantized angular momentum, degenerate (same energy)
wavefunctions.
Rotation about a circle:
Particle mass m, moving in x, y plane a constant distance r from the origin.
y
r
V(x, y) = 0
m
with condition:
x2+y2=r2
x
Schrödinger Equation:
H ( x, y) ( x, y)  E ( x, y)
φ
where:
 
H ( x, y )  
 2m
2




 x
2


y
2
2
2



  V ( x, y )



this is not a 2 dimensional problem; x and y are not independent.
Change co-ordinates:
(x, y)  (r, φ) where r is constant
Can show that:
 
  1  d 

  


r
d


x

y




2
2
2
2
2
2
2
Therefore the Hamiltonian becomes:
 

 d

 2mr 
 d
H ( )  
and
2
2
2
2



  V ( )



V ( )  0 for all 
in the Schrödinger Equation
H ( ) ( )  E ( )
  d   

2mr d
2
2
2
E  
2
d     2mr

d

2
2
2
E
2
  
    A exp  ik ,
k
 2mr
 
 
2
E
2
1/ 2


Boundary Condition:
The wavefunction must obey  ( )
  (2   )
otherwise it will interfere destructively with itself!
     2   
A exp  ik   A exp  ik 2   
 A exp  ik  exp  ik 2 
1  exp  ik 2 
 k  0,1,2,3...
Again, quantization arises from the Boundary Conditions.
 2mr E 

k  
 

k 
E 
, k
2mr
2
1/ 2
2
2
k
2
2
the wavefunctions are:
 0,1,2,3...
    A exp  ik , k  integer

2
2
  d
 
2
0
*
   d
2
 A  e e d
ik
2
0
 ik
0
A  d  A 2   1
2
2
2
0
A  2 
    2 
1 / 2
1 / 2
k
exp  ik , k  integer
Notes:
    2  , E  0
the particle is equally probable at all values of  .
1) k = 0 is allowed,
2)
1 / 2
0
0
   and    have the same energy E
k
k
. They are degenerate.
k 
l
E 

2mr
2I
2
3) What is different?
k
k
2
2
2
l – Angular momentum, I – moment of inertia


l  l , l  k and is different (and quantized).
2
2
l = +k(h/2π)
Different rotation
directions.
l = -k(h/2π)
How does this connect to (say) a diatomic molecule with fixed bond length, rotating in
the x, y plane?
Rotation on a sphere.
Particle mass m
V(x, y, z) = 0 if
x2 + y2 + z2 = r2
2 dimensional problem.
Change coordinates
(x, y, z) (θ, φ)

   
    
H 


  , 


y
z   m r 
 m  x
is called the Legrengian and the solution of the
2
2
2
2
2
2
2
  ,  
2
2
2
2
 
H ( ,  )  
  ,   ( ,  )  E ( ,  )
m
r


2
2
2
is mathematically fairly complex.
Results: E  
2
l (l  1), l  0,1,2
2mr
 ( , )  f ( ) exp( im ),  l  m  l
l
l ,m
2
l ,m
Each energy level is (2l + 1) degenerate. 2 quantum numbers l and m. l defines the
angular momentum of the particle, m defines the direction of the angular momentum.

 l

E 
l (l  1) 
2mr
2mr

angular momentum, l  l (l  1)  
l  0,1,2
2
2
2
l
2
2
1/ 2
Angular momentum is quantized.
Relationship to a diatomic molecule, fixed bond length R, moving in 3 dimensional
space.
2 particles, m1 at (x1, y1, z1) and m2 at
(x2, y2, z2),
(x1 – x2)2 + (y1 – y2)2 + (z1 – z2)2 = R2
V(x1, y1, z1, x2, y2, z2) = 0
Change coordinates
(x1, y1, z1, x2, y2, z2) (x, y, z, θ, φ)
(x, y, z) – centre of mass


 m
 

      

 






  

y
z   m  x
y
z 
 x
 
   

 
 


  , 

  
x
y
z 
 R 
m







m  total mass
  reduced mass
translation
rotation
2
2
2
2
1
2
1
2
2
2
2
1
2
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
Can separate translation from rotation.
2
2
2
2