3.3 Solutions 223
Thusψ(
1
2
,
1
2
)
=
√
2
3
φ(1,1)φ(
1
2
,−
1
2
)
−
√
1
3
φ(1,0)φ(
1
2
,
1
2
)
(11)
Similarlyψ(
1
2
,−
1
2
)
=
√
1
3
φ(1,0)φ(
1
2
,−
1
2
)
−
√
2
3
φ(1,−1)φ(
1
2
,
1
2
)
(12)
The coefficients appearing in (1), (2), (5), (6), (11) and (12) are known as
Clebsch – Gordon coefficients. These are displayed in Table 3.3.3.87 In spherical coordinates
x=rsinθ cosφ;y=rsinθ sinφ;z=rcosθ (1)
So thatxy+yz+zx=r^2 sin^2 θsinφcosφ+r^2 sinθcosθsinφ
+r^2 sinθcosθcosφ (2)
The spherical harmonics areY 00 =
(
1
4 π) 1 / 2
;Y 10 =
(
3
4 π) 1 / 2
cosθ;Y 1 ± 1 =∓(
3
8 π) 1 / 2
sinθe±φY 20 =
(
5
16 π)^12
(3 cos^2 θ−1);Y 2 ± 1 =∓(
15
8 π)^12
sinθ cosθe±iφ;Y 2 ± 2 =
(
15
32 π) 1 / 2
sin^2 θe±^2 iφ (3)Expressing (2) in terms of (3),sin^2 θsinφcosφ=^1 / 2 sin^2 θsin 2φ=Y 22 −Y 2 − 2
4 i(
32 π
15) 1 / 2
Similarly sinθcosθsinφ=( 8 π
15) 1 / (^2) (Y 21 −Y 2 − 1 )
2 i
sinθ cosθ cosφ=
(
8 π
15) (^12)
(Y 21 −Y 2 − 1 )/ 2
Hence, xy+yz+zx = r^2
( 8 π
15
)^12
[(Y 22 −Y 2 − 1 )/i+(Y 21 +Y 2 − 1 )/ 2 i
+(Y 21 −Y 2 − 1 )/2]
The above expression does not containY 00 corresponding tol =0, nor
Y 10 andY 1 ± 1 corresponding tol=1. All the terms belong tol=2, and the
probability for findingl=2 and thereforeL^2 =l(l+1)= 6 ^2 is unity.3.88 Lz=−i
∂
∂φ
x=rsinθcosφ;y=rsinθ sinφ;z=rcosθ
ψ 1 =(x+iy)f(r)=rsinθ(cosφ+isinφ)f(r)
=(cosφ+isinφ)sinθF(r)