3.3 Solutions
167
force” is supplied by the potential energy, and hence adds to the V (r )
which appears in (13) for the radial motion. This will have exactly the
form indicated in (13) if we put L =
√
l(l + 1)
3.16 −
1
sin θ
∂
∂θ
sin θ
∂
∂θ
+
1
sin
2 θ
∂
2
∂ϕ 2
Y (θ, ϕ) = λY (θ, ϕ)
We solve the equation by the method of separation of variables.
Let Y (θ, ϕ) = f (θ )g(ϕ) and multiply by sin
2
θ
−g(ϕ) sin θ
∂
∂θ
sin θ
∂ f
∂θ
+ f (θ )
∂
2 g
∂ϕ 2 = λ sin
2
θ fg
Divide through out by f (θ )g(ϕ) and separate the θ and ϕ variables.
1
f (θ )
sin θ
∂
∂θ
sin θ
∂ f
∂θ
+ λ sin
2
θ
= −
1
g(ϕ)
∂
2 g
∂ϕ 2 = m
2
(1)
LHS is a function of θ only and RHS function of ϕ only. The only way the
above equation can be satisfied is to equate each side to a constant, say −m
2 ,
where m
2 is positive.
1
g(ϕ)
d
2 g(ϕ)
dϕ 2 = −m
2
Therefore g(ϕ) = A e
imϕ
We can now normalize g(ϕ) by requiring
g
∗ (ϕ)g(ϕ)dϕ = 1
A
2
2π
0
e
imϕ
2 dϕ = 2π A
2
= 1
Or A = (2π )
−1/2
We shall now show that m is an integer
g(ϕ + 2π ) = g(ϕ)
g(ϕ + 2π ) = (2π)
−
1
2 e
im(ϕ+2π)
= (2π)
−
1
2 e
imϕ
.e
2π mi
= g(ϕ)e
2π mi
∴ e
2π mi
= cos (2πm) + i sin(2πm)
= cos(2π m) = 1
Thus m is any integer, m = 0, ±1, ±2 . . .
3.17 (a) In Problem 3.16, going back to Eq. (1) and multiplying by f (θ ) and dividing by sin
2
θ and putting μ = cos θ.