1000 Solved Problems in Modern Physics

3.3 Solutions 217

x=rsinθcosφ

y=rsinθsinφ (2)

z=rcosθ

r^2 =x^2 +y^2 +z^2 (3)

tan^2 θ=

x^2 +y^2

z^2

(4)

tanφ=

y

x

(5)

Differentiating (3), (4) and (5) partially with respect tox

∂r

∂x

=sinθcosφ;

∂r

∂y

=sinθsinφ;

∂r

∂z

=cosθ (6)

∂θ

∂x

=

(

1

r

)

cosθcosφ;

∂θ

∂y

=

(

1

r

)

cosθsinφ;

∂θ

∂z

=−

sinθ

r

(7)

∂φ

∂x

=−

(

1

r

)

cosecθsinφ;

∂φ

∂y

=

cosφ

rsinθ

;

∂φ

∂z

=0(8)

Lzψ(r,θφ)=−i

(

x∂ψ

∂y

−

y∂ψ

∂x

)

=−i

[

x

(

∂ψ

∂r

)

·

(

∂r

∂y

)

+

(

∂ψ

∂θ

)

·

(

∂θ

∂y

)

+

(

∂ψ

∂φ

)

·

(

∂φ

∂y

)

−y

(

∂ψ

∂r

)

·

(

∂r

∂x

)

+

(

∂ψ

∂θ

)

·

(

∂θ

∂x

)

+

(

∂ψ

∂φ

)

·

(

∂φ

∂x

)]

Lzψ(r,θ,φ)

=−i

[

∂ψ

∂r

(

x

∂r

∂y

−y

∂r

∂x

)

+

∂ψ

∂θ

(

x

∂θ

∂y

−y

∂θ

∂x

)

+

∂ψ

∂φ

(

x

∂φ

∂y

−y

∂φ

∂x

)]

(9)

Substituting (2), (6), (7) and (8) in (9) and simplifying, the first two terms drop

off and the third one reduces to∂ψ/∂φ, yielding

Lz=−i

∂

∂φ

(10)

In Problem 3.15 it was shown that the Schrodinger equation was separated

into radial (r) and angular parts (θandφ). The angular part was shown to be

separated intoθandφcomponents. The solution toφwas shown to be

g(φ)=

1

√

2 π

eimφ

wheremis an integer.