1000 Solved Problems in Modern Physics

3.3 Solutions 167

force” is supplied by the potential energy, and hence adds to theV(r)

which appears in (13) for the radial motion. This will have exactly the

form indicated in (13) if we putL=

√

l(l+1)

3.16−

[

1

sinθ

∂

∂θ

(

sinθ∂θ∂

)

+sin^12 θ∂

2

∂φ^2

]

Y(θ,φ)=λY(θ,φ)

We solve the equation by the method of separation of variables.

LetY(θ,φ)=f(θ)g(φ) and multiply by sin^2 θ

−g(φ)sinθ

∂

∂θ

(

sinθ

∂f

∂θ

)

+f(θ)

∂^2 g

∂φ^2

=λsin^2 θ fg

Divide through out byf(θ)g(φ) and separate theθandφvariables.

1

f(θ)

sinθ

∂

∂θ

(

sinθ

∂f

∂θ

+λsin^2 θ

)

=−

1

g(φ)

∂^2 g

∂φ^2

=m^2 (1)

LHS is a function ofθonly and RHS function ofφonly. The only way the

above equation can be satisfied is to equate each side to a constant, say−m^2 ,

wherem^2 is positive.

1

g(φ)

d^2 g(φ)

dφ^2

=−m^2

Thereforeg(φ)=Aeimφ

We can now normalizeg(φ) by requiring

∫

g∗(φ)g(φ)dφ=^1

A^2

∫ 2 π

0

∣

∣eimφ

∣

∣^2 dφ= 2 πA^2 = 1

OrA=(2π)−^1 /^2

We shall now show thatmis an integer

g(φ+ 2 π)=g(φ)

g(φ+ 2 π)=(2π)−

(^12)
eim(φ+^2 π)=(2π)−
(^12)
eimφ.e^2 πmi
=g(φ)e^2 πmi
∴e^2 πmi=cos( 2 πm)+isin(2πm)
=cos(2πm)= 1
Thusmis any integer,m= 0 ,± 1 ,± 2 ...
3.17 (a) In Problem 3.16, going back to Eq. (1) and multiplying byf(θ) and divid-
ing by sin^2 θand puttingμ=cosθ.