Calculation of reduced matrix elements - IPNL

Calculation of reduced matrix elements
•〈 j‖1‖ j ′ 〉= ?
We have
〈 jm|1| j ′ m ′ 〉=〈 jm| j ′ m ′ 〉=δ j j ′δ mm ′. (1)
On the other hand, we can use the fact that 1 is a number≡ascalar≡atensor operator of
rank 0:
〈 jm|1| j ′ m ′ 〉=〈 jm|1 (0)
0 | j′ m ′ 〉
= (−1) 0〈j′ m ′ 00| jm〉
√
2 j+1
〈 j‖1‖ j ′ 〉
〈 jm 00| j ′ m ′ 〉 is zero unless| j−0| j ′ j+0 and m ′ = m+0 so
〈 jm|1| j ′ m ′ 〈 jm 00| jm〉
〉=δ j j ′δ mm ′ √ 〈 j‖1‖ j ′ 〉
2 j+1
Furthermore, we know that〈 jm 00| jm〉=1so finally
Using eq. (1) and eq. (2) we obtain
〈 jm|1| j ′ m ′ 〉= δ j j ′δ mm ′
√
2 j+1
〈 j‖1‖ j ′ 〉. (2)
〈 j‖1‖ j ′ 〉=δ j j ′√
2 j+1.
•〈 j‖j‖ j〉= ?
j=( j x , j y , j z ) is a vector i.e. a tensor of rank 1 with spherical components j (1)
0
= j z and
( )
j (1)
± = ∓ jx − i j y . We can write
√
1
2
The Wigner-Eckart theorem gives
〈 jm| j (1)
0 | jm′ 〉=m〈 jm| jm ′ 〉=mδ mm ′. (3)
〈 jm| j (1)
0 | jm′ 〉=(−1) 2〈jm′ 10| jm〉
√ 〈 j‖j‖ j〉,
2 j+1
Using〈 jm ′ 10| jm〉=(−1) j−1+m√ ( ) j 1 j
2 j+1
m ′ one has
0 -m
( )
〈 jm ′ | j (1)
j 1 j
0 | jm〉=(−1) j−1+m m ′ 〈 j‖j‖ j〉.
0 -m
1

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

•〈l a ‖Y (L) ‖l b 〉= ?
The spherical harmonics Y (L)
M
(ˆr) are the coordinate representation of the 2L+1 spherical
components of a tensor of rank L, the Wigner-Eckart theorem can be written (for example
for M= 0)
( )
〈l a m a |Y (L)
0 |l bm b 〉=(−1) l a−m a la L l b
〈l
-m a 0 m a ‖Y (L) ‖l b 〉. (6)
b
Furthermore using〈l b m b |ˆr〉=Y (l a)∗
m a
(ˆr) and〈ˆr|l a m a 〉=Y (l b)
m b
(ˆr) one has
∫
〈l a m a |Y (L)
0 |l bm b 〉= dˆr Y (l a)∗
m a
(ˆr)Y (L)
0 (ˆr)Y(l b)
m b
(ˆr)
where L andl b can be coupled tolusing
√
∑
Y (L)
(2L+1)(2lb + 1)(2l+1)
0 (ˆr)Y(l b)
m b
(ˆr)=
4π
so one has
∑
〈l a m a |Y (L)
0 |l bm b 〉=
lm
lm
√
(2L+1)(2lb + 1)(2l+1)
4π
The integral gives
∫ ∫
dˆr Y (l a)∗
m a
(ˆr)Y m
(l)∗ (ˆr)=(−1)m
so
〈l a m a |Y (L)
0 |l bm b 〉=(−1) −m a
Using the properties of the 3− j symbols
( )( )
L la l b L la l b
=
0 -m a m b 0 0 0
so
〈l a m a |Y (L)
0 |l bm b 〉=(−1) −m a
( )( )
L l lb L l lb
0 m m b 0 0 0
( )( )∫
L l lb L l lb
0 m m b 0 0 0
dˆr Y (l a)∗
m a
(ˆr)Y -m (l)∗ (ˆr)=(−1)m δ la lδ ma -m
√
(2L+1)(2la + 1)(2l b + 1)
4π
√
(2L+1)(2la + 1)(2l b + 1)
4π
Y m
(l)∗ (ˆr)
dˆr Y (l a)∗
m a
( )( )
L la l b L la l b
.
0 -m a m b 0 0 0
( )( )
la L l b la L l b
δ
-m a 0 m b 0 0 0
ma m b
Using eq. (6) and eq. (7) we obtain
√
〈l a ‖Y (L) ‖l b 〉=(−1) l (2L+1)(2la + 1)(2l
a
b + 1)
4π
(ˆr)Y (l)∗ (ˆr).
( )( )
la L l b la L l b
δ
-m a 0 m b 0 0 0
ma m b
.
(7)
( )
la L l b
.
0 0 0
m
3