Quantum Mechanics II angular Momentum II : Addition of Angular Momentum Clebsch-Gordan Coefficients


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angularaddition

J = J1 ⊗ I ⊕ I ⊗ J2



abbreviated as


Jˆ = Jˆ1 + Jˆ2

[The hat or the vector signs (which is more commonly used) indicating the operator nature of angular momentum will be frequently omitted. The commutation relations satisfied by


J1 and J2 are

[J1i, J1j] = ikϵijkJ1k


[J2i, J2j] = ikϵijkJ2k
[J1i, J2j] = 0 (1)

Since the components of J1 commute with those of J2, they can have common eigenstates. We can choose, J2, J1z, J2 and J2z to be a set of operators which have common eigenstates.


1 2
Let us denote the common eigenstates by | j1, j2; m1, m2⟩ ≡| j1, m1⟩ | j2, m2⟩ [ Note : This
notation is not standard, several books will denote this as | j1, m1, j2, m2⟩. We follow Sakurai here.]. We then have

1

1
J2 | j1, j2; m1, m2⟩ = j1(j1 + 1)k2 | j1, j2; m1, m2Jz | j1, j2; m1, m2⟩ = m1k | j1, j2; m1, m2

2

2
J2 | j1, j2; m1, m2⟩ = j2(j2 + 1)k2 | j1, j2; m1, m2Jz | j1, j2; m1, m2⟩ = m2k | j1, j2; m1, m2
The dimension of the space to which J1 and J2 belong is (2j1 + 1)(2j2 + 1). The set of states | j1, j2; m1, m2⟩ form a complete and orthonormal set

Σ

Σ
+j1 +j2
| j1, j2; m1, m2⟩⟨j1, j2; m1, m2 |= 1 (2)

m1=−j1 m2=−j2
j1, j2; m1, m2 | j1j , j2j ; mj1, mj2 = δj1,j1 δj2,j2 δm1,m1 δm2,m2

(3)



k

k

2R
As mentioned for the case of the orbital and spin angular momenta, the rotation operator in this case is given by


1R
U (nˆ, θ) ⊗ U
(nˆ, θ) = exp −iJ1 · nˆθ ! exp −iJ2 · nˆθ !

As a consequence of (1), the components of the total angular momentum Jˆ satisfy


[Ji, Jj] = ikϵijkJk (4)
which shows that Jˆ is an angular momentum operator. It is easy to check that J2 com- mutes with Jz, J2 and J2. However, Jz does not commute with either J1z or J2z. As a re-
1 2

1

2
sult, it is possible to consider an alternate set of commuting operators, viz., {J2,z , J2, J2}


1

2
which would describe the same space as did our old set {J2, J2, J1z, J2z}. We denote the
simultaneous eigenstates represented by this new set by | j1, j2; j, m⟩. In terms of this
new set, we have


J2 | j1, j2; j, m⟩ = j(j + 1)k2 | j1, j2; j, m⟩ (5)
Jz | j1, j2; j, m⟩ = mk | j1, j2; j, m⟩ (6)
These kets are also eigenkets of J2 and J2, though not of J1z and J2z. They satisfy the
1 2
completeness relation
j

mΣ=−j
| j1, j2; j, m⟩⟨j1, j2; j, m |= 1 (7)

They can be taken to be orthonormalized

j1, j2; , j, m | j1j , j2j ; jj, mj = δm,m δj,j (8)


It is tacitly assumed that j1 and j2 in a given problem are given and fixed . We have not yet found out what values j can take. However, the sum over j in (7) must be over all values of j consistent with given values of j1 and j2. Using the completeness property(2) of the old set {| j1, j2; m1, m2⟩}, we may express a member of the new set {| j1, j2; j, m⟩} as



Σ

Σ
j1 j2
| j1, j2; j, m⟩ = | j1, j2; m1, m2⟩⟨j1, j2, m1, m2 | j1, j2; j, m⟩ (9)
m1=−j1 m2=−j2
The coefficients ⟨j1, j2, m1, m2 | j1, j2; j, m⟩ defined in (9) are called Clebsch-Gordan (C-G)Coefficients.. The basis defined here as represented by the kets on the left hand side of (9) will be also referred to as the Coupled Representation.


  1. Properties of C-G Coefficients


    1. The C-G coefficients ⟨j1, j2, m1, m2 | j1, j2; j, m⟩ are zero unless m = m1 + m2. To show this, note that Jz J1z J2z is a null operator,

j1, j2, m1, m2 | Jz J1z J2z | j1, j2; j, m⟩ = 0

because Jz acts to the ket on its right giving m times the ket, while J1z J2z acts to the bra on the left giving m1m2 times the bra,
(m m1m2)⟨j1, j2, m1, m2 | j1, j2; j, m⟩ = 0 which shows that of m =/ m1 + m2, the C-G coefficient would be zero.

    1. By convention, the C-G coefficients are taken to be real, so that

j1, j2, m1, m2 | j1, j2; j, m⟩ = ⟨j1, j2; jm | j1, j2; m1, m2



    1. Orthonormality of C-G Coefficients : In the expression (8) given above, if we insert a complete set given by (2) of old states, we get

j1, j2; , j, m | j1j , j2j ; jj, mj = δm,m δj,j

Σ

Σ
+j1 +j2
j1, j2; j, m | j1, j2; m1, m2⟩⟨j1, j2; m1, m2 | j1j , j2j ; jj, mj = δm,m δj,j

m1=−j1 m2=−j2


Substituting j = jj and m = mj in the above, we get

(10)

(11)


In a similar way may start with the orthonormality condition on the old basis set
(3) and insert the completeness condition (7) of the new basis set to obtain
(12)

    1. The C-G coefficients vanish unless



(13)

The inequality on the right is obvious because mmax = j1 and mmax = j2. Hence


1 2
mmax = j1 + j2. Since the maximum value of m is j, the right inequality follows

and jmax = j1 + j2.

Σ
The inequality on the left requires a bit of working out. We know that there are (2j1 + 1)(2j2 + 1) number of states. For each value of j, there are 2j + 1 states. Thus we must have
jmax
(2j + 1) = (2j1 = 1)(2j2 + 1)
j=jmin
Left hand side of above is an arithmetic series of finite number of terms. The sum works out as follows :

l.h.s. = jmax jmin + 1 (2j
2


max


+ 1 + 2j


min

+ 1)



= (jmax jmin + 1)(jmax + Jmin + 1)

min
= (jmax + 1)2j2

min
= (j1 + j2 + 1)2j2




min
Equating the above to (2j1 + 1)(2j2 + 1), we get j2
= (j1j2)2, which gives


Thus the values of j satisfies
jmin =| j1j2 |

| j1j2 |≤ j j1 + j2 (14)






    1. For a given j, the possible m values are −j m ≤ +j.

A notation, primarily used by the nuclear physicists is known as Wigner’s 3-j symbol and is related to C-G coefficients as follows. While the C-G coefficients are for adding two angular moments, the 3-j symbols are for addition of three angular momenta such that their sum gives zero angular momentum state.


1

2

1

1

2
j , j ; m , m
| j , j ; j, m⟩ = (−1)j1j2+m2j + 1 j1 j2 j (15)





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