QuantumPhysics.dvi

ApplyingJ 3 on either side gives (1).

• To prove (2), we shall assume without loss of generality thatj′−j≥2. Now consider the

matrix element of the vanishing operator [J+,V+] = 0,

〈j′,j+ 2,α′|[J+,V+]|j,j,α〉= 0 (9.65)

In the commutator, the term withJ+on the right ofV+vanishes, so we are left with

√

j′(j′+ 1)−(j+ 1)(j+ 2)〈j′,j+ 1,α′|V+|j,j,α〉= 0 (9.66)

Sincej′(j′+ 1)−(j+ 1)(j+ 2) 6 = 0 forj′−j≥2, it follows that

〈j′,j+ 1,α′|V+|j,j,α〉= 0 (9.67)

Next, using [J+,V 0 ] =− ̄h

√

2 V+, and then [J+,V−] =

√

2 ̄hV 0 , we also find that

〈j′,j,α′|V 0 |j,j,α〉=〈j′,j− 1 ,α′|V−|j,j,α〉= 0 (9.68)

As a result, we then have

〈j′,m,α′|Vq|j,j,α〉= 0 (9.69)

for allqand allm. Next, we evaluate

√

2 j〈j′,m,α′|Vq|j,j− 1 ,α〉 = 〈j′,m,α′|VqJ−|j,j,α〉

= 〈j′,m,α′|J−Vq|j,j,α〉

−

∑

q′

D(1)(J−)qq′〈j′,m,α′|Vq′|j,j,α〉 (9.70)

The second term on the right hand side vanishes in view of (9.69). Thefirst term may be

re-expressed in terms of〈j′,m+ 1,α′|Vq|j,j,α〉which vanishes in view of (9.69). Thus, we

also have〈j′,m,α′|Vq|j,j− 1 ,α〉= 0 for allm,q. It is now clear that this argument may be

recursively repeated and leads to

〈j′,m′,α′|Vq|j,m,α〉= 0 (9.71)

for allm,m′,qas long asj′−j≥2, which proves assertion (2).

• To prove (3), we first show that the matrix elements ofV±are related to those ofV 0 , by

using the commutator relations [J±,V 0 ] =∓

√

2 ̄hV±, and we find,

√

2 〈j′,m′,α′|V±|j,m,α〉 = ±Nj,m± 〈j′,m′,α′|V 0 |j,m± 1 ,α〉

∓Nj∓′,m′〈j′,m′∓ 1 ,α′|V 0 |j,m,α〉 (9.72)