3.3 Solutions
227
3.93
u(θ, ϕ) =
1
/ 4
15
π
sin 2θ cos 2ϕ
=
1
/ 4
15
π
sin
2
θ
(e
2iϕ
+ e
−2iϕ )
2
(1)
But Y 2+2 (θ, ϕ) =
15
32π
sin
2
θe
2iϕ
(2)
Y 2−2 (θ, ϕ) =
15
32π
sin
2
θe
−2iϕ
(3)
Adding (2) and (3)
Y 2+2 (θ, ϕ) + Y 2−2 (θ, ϕ) =
15
32π
sin
2
θ(e
2iϕ
+ e
−2iϕ )
( 4 )
Dividing (1) by (4) and simplifying we get
u(θ, ϕ) =
1
√
2
(Y 2+2 (θ, ϕ) + Y 2−2 (θ, ϕ))
We know that
L
2 Y lm =
2 l(l + 1)Y lm
So L
2 u(θ, ϕ) =
L
2 [Y22(θ,ϕ)+Y2−2(θ,ϕ)]
√
2
= 2(2 + 1)
2 [Y 22 (θ,ϕ)+Y2−2(θ,ϕ)]
√
2
= 6
2 u(θ, ϕ)
Thus the eigen value of L
2 is 6
2
3.94 The wave function is identified as ψ 322
L Z ψ = −i
∂ψ
∂ϕ
= 2ψ
Thus the eigen value of L Z is 2.
3.95 (a) Using the values, Y 10 =
3
4π
cos θ and Y 1,±1 = ∓
3
8π
sin θ exp(±iϕ), we
can write
ψ =
1
3
−
√
2Y 11 + Y 10
f (r )
Hence the possible values of L Z are +, 0