Calculation of reduced matrix elements - IPNL

The 3− j symbol has a simple value
( ) j 1 j
m ′ =−(−1) j+m m
√ δ
0 -m
mm ′
j(2 j+1)( j+1)
we end up with ( j+m is integer so 2( j+m) is even)
〈 jm| j (1)
0 | jm′ 〉=
Using eq. (3) and eq. (4) we obtain
m
√
j(2 j+1)( j+1)
δ mm ′〈 j‖j‖ j〉. (4)
〈 j‖j‖ j〉= √ j(2 j+1)( j+1). (5)
•〈 1‖σ‖ 1〉= ?
2 2
The spin (vector) operator is s= σ so 2
〈 1 ‖σ‖ 1 〉= 2 2 2
〈 1 ‖s‖ 1 〉 2 2
and〈 1 ‖s‖ 1 〉 is a special case of eq. (5) with j=s and j= 1 so
2 2 2
〈 1 2 ‖σ‖ 1 2 〉= 2 √
1
2 (2× 1 2 + 1)( 1 2 + 1)=√ 6.
2