3.3 Solutions
223
Thus
ψ
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)
Similarly
ψ
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 = r sin θ cos ϕ ; y = r sin θ sin ϕ ; z = r cos θ
(1)
So that xy + yz + zx = r
2 sin
2
θ sin ϕ cos ϕ + r
2 sin θ cos θ sin ϕ
+ r
2 sin θ cos θ cos ϕ
(2)
The spherical harmonics are
Y 00 =
1
4π
1/2
; Y 10 =
3
4π
1/2
cos θ ; Y 1±1 = ∓
3
8π
1/2
sin θ e
±ϕ
Y 20 =
5
16π
1
2
(3 cos
2
θ − 1); Y 2±1 = ∓
15
8π
1
2
sin θ cos θ e
±iϕ ;
Y 2±2 =
15
32π
1/2
sin
2
θ e
±2iϕ
(3)
Expressing (2) in terms of (3),
sin
2
θ sin ϕ cos ϕ =
1
/ 2 sin
2
θ sin 2ϕ =
Y 22 − Y 2−2
4i
32π
15
1/2
Similarly sin θ cos θ sin ϕ =
8π
15
1/2 (Y 21 −Y2−1)
2i
sin θ cos θ cos ϕ =
8π
15
1
2
(Y 21 − Y 2−1 )/2
Hence, xy + yz + zx = r
2
8π
15
1
2 [(Y 22 − Y 2−1 )/i + (Y 21 + Y 2−1 )/2i
+(Y 21 − Y 2−1 )/2]
The above expression does not contain Y 00 corresponding to l = 0, nor
Y 10 and Y 1±1 corresponding to l = 1. All the terms belong to l = 2, and the
probability for finding l = 2 and therefore L
2
= l(l + 1) = 6
2 is unity.
3.88 L z = −i
∂
∂ϕ
x = r sin θ cos ϕ ; y = r sin θ sin ϕ ; z = r cos θ
ψ 1 = (x + iy) f (r ) = r sin θ(cos ϕ + i sin ϕ) f (r )
= (cos ϕ + i sin ϕ) sin θ F(r )