Notes on Lecture 2
Frank Merritt
(October 1, 2008)
In describing the state of a system of a single angular momentum J , it is usually
convenient to describe this in the j, m basis, which is the eigenbasis of the commuting
operators J 2 and J z . The quantum numbers j and m of these operators are defined by:
J 2 j, m 
2
 j ( j  1) j , m
(1.1)
J Z j, m   m j, m
We know (from Ph234) that j and m can only take on integer or half-integer values, and
that m   j ,  j  1,...., j  1, j . Note that J can represent either orbital angular
momentum (usually written as L ) or spin angular momentum (usually written as S ).
For orbital angular momentum, the eigenvalues of L2 and LZ must be integers; for spin,
they can be either integers (for bosons) or half-integers (for fermions).
Now consider a system of 2 angular momenta, j1 and j2 , which can be either
orbital angular momenta, or spin angular momenta, or one of each. Then the operators
J12 , J 22 , J1Z , J 2 Z can clearly operate on the state vector of this system. J12 and J1Z will
only operate on particle 1, and J 22 and J 2 Z will only operate on particle 2. Consequently
it is clear the 4 operators in this set all commute with each other, and therefore we can
define a common set of eigenvectors of all 4. The basis vectors of this Hilbert space can
be written as j1 , j2 , m1 , m2 .
But for such a system we can also define the total angular momentum J  J1  J 2 .
It is easy to show that each of the components of this operator are hermition, and that
 J i , J j   i  ijk J k
(1.2)
where the indices run over the three spatial axes. This means that J is indeed an angular
momentum operator (obvious from its definition), and so the eigenvalues of J 2 and J Z
must be integer or half-integer, with J  M  J . It is not difficult to show that the set
 J12 , J 22 , J 2 , J Z  also form a mutually commuting set of operators (just remember that any
component of J1 must commute with any component of J 2 ); consequently we can define
a set of basis vectors which are simultaneously eigenvectors of each of these 4 operators.
These basis vectors are written as j1 , j2 , J , M . [It may not be obvious that this set of
basis vectors is complete for a 2-particle system; in fact it is, and we will prove it below.]
This gives us two sets of basis vectors which span the same Hilbert space. Both
of these bases are important. We will encounter problems where one of these sets
commutes with the Hamiltonian H, and the other set does not (see Shankar, page X). In
that case, we clearly want to use the basis of the commuting set when finding energy
eigenstates.
In all of this, we will assume that j1 and j2 are fixed, and “given”; that is the
case in almost all problems that one encounters. Consequently, if there is no ambiguity I
will often write these bases as m1 , m2 and J , M ; but one has to remember that we are
always dealing with subspaces with well-defined j1 and j2 .
Clebsch-Gordon Coefficients:
Since we have two different bases, both spanning the same subspace, it’s clear
that we need to be able to convert between them. To do this, we need to find the
coefficients which give the basis vectors in one basis in terms of the basis vectors of the
other. For example, we must be able to expand the J , M basis vectors in term of the
m1 , m2 eigenvectors:
j1 , j2 ; J , M 
 C( j , j ; J , M , m , m )
1
2
1
2
j1 , j2 ; m1 , m2
(1.3)
m1 , m2
The constants C (each of which depends on the 6 eigenvalues indicated) are called
“Clebsch-Gordon” coefficients, and are a fundamental tool in atomic, nuclear, and
particle physics. Since m1 , m2 is an orthonormal basis (as is J , M ), multiplying both
sides by the bra m1 , m2 gives
(1.4)
C( j1 , j2 ; J , M , m1 , m2 )  j1 , j2 ; m1 , m2 j1 , j2 ; J , M
That is, the Clebsch-Gordon coefficient relating two eigenvectors in the two bases is just
the scalar product of the two eigenvectors. This makes perfect sense.
The CG coefficients can be chosen to be real (see below or Shankar p.412), and
so the expansion of m1 , m2 in terms of J , M can also be written in terms of CG
coefficients:
j1 , j2 ; m1 , m2 j1 , j2 ; J , M   C ( j1 , j2 ; J , M , m1 , m2 ) j1 , j2 ; J , M
(1.5)
J ,M
Formal derivation of CG coefficients for a two-particle system:
This may seem a bit abstract, so it is useful to consider a specific example and
work through it completely. The simplest non-trivial system is one consisting of two
spin-1/2 particles. The m1 , m2 basis (called the product basis) consists of 4
eigenvectors, which are the direct products of the 2-component eigenbasis of each of the
two particles. In general, we can write the product basis in the form
(1.6)
j1 , j2 ; m1 , m2  j1 , m1  j2 , m2  j1 , m1 j2 , m2
We can write most conveniently either as the first expression in equation (1.6), or for the
case of two spin-1/2 particles using the “   ” notation used in class and in Shankar. In
any case, the four basis vectors are
1
2
, 12 ;  12 ,  12   
1
2
, 12 ;  12 ,  12   
1
2
, 12 ;  12 ,  12   
1
2
, 12 ;  12 ,  12   
(1.7)
The problem of finding the CG coefficients which relate these to the J , M basis is
equivalent to finding the probabilities that measurement of total spin and total zcomponent of spiin, for a system that is initially in the eigenstate m1 , m2 , will yield the
values J and M. Because of that, this is often referred to as “the addition of angular
momenta”.
The operators which give the square of the total angular momentum and the zcomponent of the total angular momentum are clearly
J 2  ( J1  J 2 )2  J12  J 22  2 J1 J 2
(1.8)
J Z  J1Z  J 2 Z
First let’s find the eigenvalues of J Z . To do this, we need to first write J1Z and
J 2Z in the basis of equation (1.7). That is easy, since each of the eigenvectors is an
eigenstate of J1Z . Recall that for a 1-particle spin-1/2 system,
so that when J1Z
1 0 
J 1Z  

2  0 1
acts on each of the two eigenstates we get:
(1.9)
(1.10)
J1Z " "   2 " "
and
J1Z " "   2 " "
When it acts on one of the basis vectors of equation (1.7), its action is determined entirely
by the first particle. Consequently,
J 1Z     2  
and
J 1Z     2  
(1.11)
J 1Z     2  
and
J 1Z     2  
To find where these coefficients go in the 4  4 matrix of J1Z , everything we need is in
equations (1.11). Each of these equations, find the column corresponding to the initial
state and the row corresponding to the final state. Place the coefficient of the right-hand
side of the equation in that location. This gives
 
J 1Z
Similarly,
1

0
 
2 0

0
 
 
 
0 0 0  

1 0 0  
0 1 0   

0 0 1  
(1.12)
 
J 2Z
 
1 0

0 1
 
2 0 0

0 0
 
 
0 0  

0 0  
1 0 

0 1  
(1.13)
Since J Z is the sum of these,
 
 
 
 
2 0 0 0  
1 0 0 0 




(1.14)
0 0 0 0  
0 0 0 0


JZ 

0 0 0 0 
2 0 0 0 0  




 0 0 0 2   
 0 0 0 1
We can follow the same procedure for finding J X . Here, the operator for a single
particle is
0 1
(1.15)
J1 X  

2 1 0
When this acts on either of the two basis spinors, we get
(1.16)
J1X " "   2 " "
and
J1X " "   2 " "
Consequently
J1 X     2  
and
J1 X     2  
(1.17)
J1 X     2  
and
J1 X     2  
The coefficients of the right-hand sides (all = +1 in this case) go at the intersection of the
column corresponding to the initial state and the row corresponding to the final state.
This gives
 
 
 
J 1Z
0

0
 
2 1

0
0
0
0
1
J2X
0

1
 
2 0

0
1 0 0  

0 0 0  
0 0 1 

0 1 0  
1
0
0
0
 
0  

1 
0  

0  
(1.18)
Similarly
and so the sum of these is
(1.19)
0

1
JX  
2 1

0
In a similar manner we find
 0 0 i

0 0 0
J1Y  
2i 0 0

0 i 0
As a check, recall that J   J X  iJY
1 1 0 

0 0 1 
0 0 1 

1 1 0  
(1.20)
0
 0 i 0 0 



i 
i 0 0 0

(1.21)
and J 2Y 
0
2  0 0 0 i 



0
0 0 i 0 
is the spin raising/lowering operator. If we compute
J1 from the equations above, we get
0 0 1 0


0 0 0 1

J1 
(1.22)
0 0 0 0


0 0 0 0
Check that this gives the expected result when acting on each of the 4 basis vectors.
Make similar checks for J1 and J 2 .
From the results above, we can calculate the J 2 operator for the total angular
momentum of the system. The easiest way of doing this is to derive the formula
suggested by Shankar (problem 15.1.12). Whatever method we use, the result is
2 0 0 0


1 1 0
2
2 0
J 
(1.23)
0 1 1 0


0 0 0 2
Using the usual techniques, we can find both the eigenvalues and the eigenvectors of this
operator. And we find, as expected, that these are also eigenvalues of J Z . The common
eigenvectors (which are of course the basis vectors of the J , M basis, are:
1,1 
1
2
, 12
 12 , 

1
1, 1   2 ,  12
1, 0 
1
2
1
2
  12 , 12 

(1.24)
 12 ,  1   12 , 12 
2


1
1
All of the CG coefficients for 2  2 can be read directly from these equations.
0, 0 
1
2
Note that the J , M states consist of all of the J=1 multiplet plus the J=0 multiplet. We
can write (using Shankar’s notation)
1
1
(1.25)
2  2  1 0
where the integers are the spins of the multiplets involved. A better notation, I think, is
the one normally used in HEP, where the multiplets are identified by their multiplicity
(2J+1) rather than their spin. In this notation, equation (1.25) becomes
(1.26)
2  2  3 1
Also note that the J=1 states are all symmetrical under exchange of particles 1 and
2, while the J=0 state is antisymmetric under this exchange.