1
Angular Momentum
Classical Angular Momentum
Recall the classical definition for angular momentum. L  r  p  mr  v
r
v
L
Quantum mechanically, angular momentum operator can be expressed as similarly in terms
Lˆ  rˆ  pˆ
of the position and velocity operators.
For a particle constrained to the x-y plane
L  rp
2
p2  L  1
L2
TE
  

2m  r  2m 2mr 2
Recall the definition for the moment of inertia: I  mr 2
L2
Therefore, the energy of particle rotating in a plane can be written as E 
2I
Quantization of Angular Momentum
Recall the deBroglie relationship p 
h

rh

Thus, as the particle is moving on the ring it travels as a wave.
A particle rotating in a plane has a wavelength. L  r p 
- unless momentum has correct value, the wave will destructively interfere with itself
- constructive interference occurs only at discrete values of 
- circumference of ring is 2r.
2r

where m is an integer.
m
2
Rearrangement yields
r
m
2
Substitute into the deBroglie relationship.
r h h m
L
 
m
  2
**Thus, we realize that a particle with angular momentum and wave-like properties must
have its angular momentum quantized.**
Particle-on-a-Ring
R

Note that for this model, the particle has a single degree of freedom which is the angle, .
Schrödinger Equation
The potential energy function for a particle-on-a-ring is
V  r   {0 ififr0RrR,rR
Note: V    0
Therefore the Schrödinger equation is
 2 1  2
 2  2
 2 2IE

E



E


0

 2 0
2m r 2 2
2I 2
2
2IE
Let 2  m 2
 2  2IE
 2 0 
2

D
2
 2
 m 2   0  D 2  m 2  0
2

 m 2    0    im
Particle-on-a-Ring Wavefunction
The normalized solution for a particle-on-a-ring is
2IE
1 im
where m  
m   
e
2
2
Particle-on-a-Ring Energies
The energy level for a particle-on-a-ring can be written as. E 
- Note again that L  m
m 2 2 L2

2I
2I
3
Quantization of a Particle-on-a-Ring
To make the wavefunction physically meaningful, we can apply a boundary condition to the
solution, that is m    2  m   .
1 im  2 
1 im
e

e
2
2

1 im im2 
1 im
e e

e
2
2

eim2   1
The above boundary condition can only be satisfied when m  0, 1, 2, 3,...
Thus, quantization is imposed by requiring the requirements of the wavefunction.
*Therefore, two simple arguments can be found for the quantization of angular momentum.*
Particle-on-a-sphere (Rigid Rotor)
z
(R,,)

R
R sin 
y

x
Now the particle has two degrees of freedom, , .
Schrödinger Equation
The potential energy function for a particle-on-a-sphere is
V  r   {0 ififr0RrR,rR
Note: V  ,   0
Therefore the Schrödinger equation is
 2  1  2
1  
  

sin

 2

   E  0
2I  sin  2 sin   
  
4
To begin to solve the equation, we will employ the separation of variables technique.
First we will assume a solution of the form   ,        
 2
2I
 1  2        
         
1  

sin

 2

   E         0
2
sin   

 
 sin 
Allow the differential operators to operate.
2
     
 2 
1    
1  


   
   2
 sin 
   E         0
2
2I 
sin  
sin   
  
 2      
Now divide each side of the equation by
2I
sin 2 
2
     2IE 2
1      sin   

sin


  2 sin   0
    2
     
 
Separate terms with  from terms with .
2
     2IE 2
sin   
1    
sin


sin





2
     
 
    2
Let each side of the equation be equal to a constant.
2
     2IE 2
sin   
1    
 m 2
 sin 
  2 sin   
2
     
 
    
Note: It will be convenient to make the perspicacious choice -m2 as a constant rather
than m to find the solution to the differential equation.
Now the two-variable differential equation above can be split into two separate singlevariable differential equations.
     2IE 2
sin   
2
 sin 
  2 sin   m  0
     
 
2
1    
 m2  0
2
    
The  equation can be rewritten as
 2   
 m2     0
2

The solutions are the same particle-on-a-ring solutions with the same restrictions on the m
quantum number.
m   
1 im
e
2
5
We make another perspicacious choice for a constant: Let
2IE
2
 l  l  1
By dividing by sin2, and multiply by (), the  equation can be rewritten as
    
1  
m2 
sin


l
l

1



  
     0
sin   
  
sin 2  

Finally, allow the
operator to operate

 2    cos     
m2 

  l  l  1  2       0
2
sin  
sin  

Associated Legendre Equation
In the late 18th century, French mathematician Adrien-Marie Legendre found the solution the
for the below equation.
2
d2P  x 
dP x 

1  x  dx 2  2x dx    l  l  1  1 mx 2  P  x   0


2
The solutions, P(x), are finite only if P(x) remains between 1 and –1 and the values of l = 0,
1, 2, 3, … and m = 0, 1, 2, …,  l.
Given the above condition, one can make the substitution for x, x = cos  and manipulate the
associated Legendre equation and make it identical to the  equation being discussed above.
In other words, the  equation is the associated Legendre equation.
6
The solutions to the associated Legendre equation are the associated Legendre polynomials,
Plm  x  .
  
1
cos 
sin 
Plm  x 
1
x
l
m
0
1
1
0
0
1
2
0
2
1
3x 1  x 2 
2
2
3 1  x 2 
3sin 2 
3
0
3
1
3
2
1
5x 3  3x 

2
3
5x 2  1 1  x 2 

2
15x 1  x 2 
1
5cos3   3cos  

2
3
5cos 2   1 sin 

2
15cos  sin 2 
3
3
15 1  x 2 
15sin 3 
4
0
4
1
4
2
1
35x 4  30x 2  3 

8
5
 7x 3  3x  1  x 2 
2
15
 7x 2  11  x 2 
2
4
3
4
4
1  x 
1
 3x  1
2
2
1
3cos 2   1

2
3cos  sin 
2
3
1  x 
105 1  x 
2 3
105x
2 2
1
35cos 4   30 cos 2   3 

8
5
 7 cos3   3cos   sin 
2
15
 7 cos2   1 sin 2 
2
105cos  sin 3 
105sin 4 
Emphasis should be made that the lm   functions depend on l and m which are numbers
that can only be discrete values. Thus the angular momentum for a particle-on-a-sphere
remains quantized.
7
Particle-on-a-Sphere Wavefunction
The normalized solution for a particle-on-a-sphere are called the spherical harmonics.
The spherical harmonics are simply the normalized product of the    functions and the
    functions.
  ,   Ylm  ,   Nlmlm   m   where Nlm is a normalization constant
One can think of the spherical harmonics as the set of possible standing waves on a sphere.
Y00  ,   
1
4
1
1
 3 2
Y10  ,      cos 
 4 
1
 3 2
Y11  ,      sin  ei
 8 
 3 2
Y11  ,      sin  ei
 8 
1
 5 2
2
Y  ,    
  3cos   1
 16 
0
2
1
 15  2
Y21  ,      sin  cos ei
 8 
1
1
 15  2
Y21  ,      sin  cos ei
 8 
1
 15  2 2 2i
Y  ,    
 sin e
 32 
 15  2 2 2i
Y  ,    
 sin e
 32 
2
2
2
2
- We shall soon see that the spherical harmonics dictate the shape of orbitals in a
hydrogen atom.
Particle-on-a-Sphere Energies
The energy level for a particle-on-a-sphere can be written as. E 
- Note that L2  l  l  1
2
l  l  1
2I
2

L2
2I
8
Eigenvalues of Angular Momentum Operators
L̂2
2I
Since spherical harmonics are eigenfunctions of T̂ , they are eigenfunctions of L̂2 .
- Note Ylm  ,  are not eigenfunctions of L̂ .
L̂2 is proportional to p̂ 2 which to proportional to T̂ .
L̂2  
2
T̂ 
 1 2
1  
 
2 2 2

 2
 sin      r 
2
  
 sin   sin   
L̂2 Ylm  ,   
2
l  l  1 Ylm  ,  
Spherical harmonics are also eigenfunctions of L̂ z
 
 

L̂z  i  x  y   i
x 

 y
L̂z Ylm  ,    m Ylm  ,  
It can be shown that  Lˆ 2 , Lˆ z   0 , L2 and Lz can be known at the same time.
However,  Lˆ x , Lˆ y   i Lˆ z  0 . Also  Lˆ y , Lˆ z   i Lˆ x  0 and  Lˆ z , Lˆ x   i Lˆ y  0
Thus only one component of the angular momentum can be known as a time.
- If Lz is known, then Lx and Ly are completely undetermined.
Vector Model of Angular Momentum
z
Lz
L  L2 
Lz  m
L
y
x
l  l  1
Since Lx and Ly are completely unknown, we can think of the
angular momentum vector as precessing about the z-axis.
- Caution! One should not take the precession too literally.
The angular momentum vector, L , sweeps out a cone in space
with the axis of the cone being the L z vector.
Later, we will consider how angular momenta add together and
vector model will be very helpful.
9
Spin
Many particles (including e-) have intrinsic angular momentum.
For lack of better word, we call it spin angular momentum.
The eigenvalues of spin are similar to orbital angular momentum.
However, the wavefunctions and the operators that describe spin are much different.
The wavefunctions that describe spin are called spin functions or spinors.
- More about spinors when we discuss the hydrogen atom.
Electron Spin
Electron spin has one l value that is relabeled as s. s 
1
2
Electron spin has two m values that are relabeled as ms. ms  
Ŝ2  
2
11 
3
  1  
22 
4
1
Ŝz   ms    
2
s  s  1  
2
1
2
2

Nuclear Spin
1
2
Because nucleus is composite particle(made of quarks and gluons), predicting spin is very
difficult (impossible).
Both proton and neutron have s 
Experimentally measured spins found for naturally occurring nuclei.
1 3 5 7 9
s  0, ,1, , ,3, , 4, ,5, 6, 7
2 2 2 2 2
Possible values for ms.
ms  s,s 1,s  2,...,  s 1 , s
Eigenvalues of nuclear spin same as electron spin!!
Ŝ2 
2
s  s  1 
Ŝz   ms 
We will come back to nuclear spins during discussion of nuclear magnetic spectroscopy.