notes-5 spin angular momentum
5-1. Communtation relations
For the orbital angular momentum we realize that it is an operator. In note-4 we
derived the commutation relation like
and
[ Lx , L y ]  iLz

[ L2 , L ]  0

Writing L in terms of differential operators, we obtained the eigenfunctions Ym . We
can consider Ym to be just basis functions, | m  .
In this basis set, the L operator is represented by a matrix. Thus Lz is a diagonal matrix
1 0 0 


L z   0 0 0 
 0 0  1


for L=1. Using matrix elements of L+ and L-, one can construct the matrix representations
of Lx and Ly. This will be left as an exercise.
5.2. Spin
One can define the spin operator following the commutation relation
[ S x , S y ]  iS z
where
1 0   
S z  
   z
 0  1 2 2
From the matrix representations of Sx and Sy, by defining
 1 
S  
2
one thus obtains these properties of Pauli matrices
0 1
0  i
1 0 
  y  
  z  

 x  
1 0
i 0 
 0  1
[ x ,  y ]  2i z
1 0
  1
 2x   2y   2z  
0 1
 x  y   y  x
 z  x   x  z
 y  z   z  y
Let    1  and    0  be the two eigenstates of  z, with eigenvalues +1 and -1,
 0
1
 
 
respectively. An arbitrary state in the spin ½ Hilbert space can be written as
  c1  c2  where the coefficients are complex numbers. Since | c1 |2  | c2 |2  1 and
the overall phase is not important, one can write a general spin ½ state as


  cos e i / 2  sin e i / 2 
(1)
2
2
The unit sphere with spherical angles (  ,  ) is called the Bloch sphere. Let the
Cartesian unit vectors for a vector u be given by uˆ  (u x , u y , u z ) , then an operator S in
the spin space can be expressed in the spinor space

Su  S  uˆ  S x sin  cos   S y sin  sin   S z cos 
  cos 
= 
2  sin e i
sin e i 

 cos  
For the state  , one can calculate


  | S x |   sin  cos  ,   | S y |   sin  sin  ,
2
2
  | S z |  

cos 
2
(2)
such that the expectation value of


 S  S x  xˆ   S y  yˆ   S z  zˆ = (sin  cos xˆ  sin  sin yˆ  cos zˆ ) ,
2
This shows that the state  has its spin pointing in the direction with spherical angles
( , ) .
That is, the state  in the 2D complex vector space (the Bloch sphere) is isomorphic to
the rotation in 3D space.
Note that for spin ½ particle, let   0 and  increases by 2  , the vector  changes
sign, i.e., it does not return to its initial state. This is true for any odd half-integer spins.
These objects do not have classical analogs. Since physical quantities involve matrix
elements of two  ' s , this is not a problem.
---------------------------------------------------------------------Homework #5
5-1. Calculate the matrix elements of Sx and Sy and Sz for S=1, 3/2 and 2 in the basis set
where Sz is diagonal.
5-2. Prove that eq. (2) given indeed is correct.
5-3. An electron with spin ½ in a magnetic field has the Hamiltonian given by
e  
 B
2mc
Let B be along the z-direction. What are the two eigenstates and their eigenvalues?
H
Suppose that at t=0 the spin is an eigenstate of Sx with eigenvalue +(  / 2) , calculate
the spin function at later time t, and the new "direction" of S, ie., calculate  S x ,
 S y ,  S z  at time t.
5-4.
Consider a spin ½ system represented by the normalized state vector
 cos  


 sin e i 


What is the probability that a measurement of Sy yields   / 2 ?
5-5.
Consider an agular momentum 1 system, represented by the state vector
 1 
1  
u
 4
26  
  3
What is the probability that a measurement of Lx yields the value 0?
5-6.
You are given the Hamiltonian
1 2
1 2
1 2
H
Lx 
Ly 
Lz .
2 I1
2I 2
2I 3
Find the eigenvalues of H : (a) when the angular momentum of the system is 1; (b)
when the angular momentum of the system is 2.
Note: The matrix representations of L x L y L z for angular momentum 2 are obtainable
from (you need to show this first)
0 
2 0 0 0


0 
0 1 0 0
Lz    0 0 0 0
0 


 0 0 0 1 0 
 0 0 0 0  2


0

0
L    0

0
0

2
0
0
0
0
0
6
0
0
0
0
0
6
0
0
0

0
0  L  ( L )  .

2
0 