1000 Solved Problems in Modern Physics

3.3 Solutions 227

3.93
u(θ,φ)=^1 / 4

√

15

π

sin 2θcos 2φ

=^1 / 4

√

15

π

sin^2 θ

[

(e^2 iφ+e−^2 iφ)

2

]

(1)

ButY 2 + 2 (θ,φ)=

√

15

32 π

sin^2 θe^2 iφ (2)

Y 2 − 2 (θ,φ)=

√

15

32 π

sin^2 θe−^2 iφ (3)

Adding (2) and (3)

Y 2 + 2 (θ,φ)+Y 2 − 2 (θ,φ)=

√

15

32 π

sin^2 θ(e^2 iφ+e−^2 iφ)(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 Ylm=^2 l(l+1)Ylm

SoL^2 u(θ,φ)=L

(^2) [Y 22 (θ,φ)+Y 2 − 2 (θ,φ)]
√
2
=2(2+1)^2 [Y^22 (θ,φ)+√Y 22 −^2 (θ,φ)]
= 6 ^2 u(θ,φ)
Thus the eigen value ofL^2 is 6^2
3.94 The wave function is identified asψ 322
LZψ=−i
∂ψ
∂φ
= 2 ψ
Thus the eigen value ofLZis 2.
3.95 (a) Using the values,Y 10 =

√

3

4 πcosθandY^1 ,±^1 =∓

√

3

8 πsinθexp(±iφ), we

can write

ψ=

√

1

3

(

−

√

2 Y 11 +Y 10

)

f(r)

Hence the possible values ofLZare+,0