Z 1
À1
1 À x
2
À
Á l dx ¼
Z π
0
sin
2lþ1
θdθ:
ð3:231Þ
We have already estimate this integral in (3.132) to have
2
2lþ1 l!
ð Þ
2
2lþ1
ð
Þ! . Therefore,
f 0
ð Þ ¼
À1
ð Þ
2l 2l
ð Þ!
2
2l l!
ð Þ
2
2
2lþ1 l!
ð Þ
2
2l þ 1
ð
Þ!
¼
2
2l þ 1
:
ð3:232Þ
Thus, we get
f m
ð Þ ¼
l þ m
ð
Þ!
l À m
ð
Þ!
f 0
ð Þ ¼
l þ m
ð
Þ!
l À m
ð
Þ!
2
2l þ 1
:
ð3:233Þ
From (3.228) and (3.233), we have
Z 1
À1
P
m
l x
ð ÞP
m
l
0 x
ð Þdx ¼
l þ m
ð
Þ!
l À m
ð
Þ!
2
2l þ 1
δ ll
0 :
ð3:234Þ
Accordingly, putting
f
P
m
l x
ð Þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2l þ 1
ð
Þ l À m
ð
Þ!
2 l þ m
ð
Þ!
s
P
m
l x
ð Þ,
ð3:235Þ
we get
Z 1
À1
f
P
m
l x
ð Þ f
P
m
l
0 x
ð Þdx ¼ δ ll
0 :
ð3:236Þ
Normalized Legendre polynomials immediately follow. This is given by
e
P l x
ð Þ
ffiffiffiffiffiffiffiffiffiffiffiffi
2l þ 1
2
r
P l x
ð Þ ¼
À1
ð Þ
l
2
l l!
ffiffiffiffiffiffiffiffiffiffiffiffi
2l þ 1
2
r
d
l
dx l 1 À x
2
À
Á l
h
i
:
ð3:237Þ
Combining a normalized function (3.235) with
1 ffiffiffiffi
2π
p e
imϕ , we recover
Y
m
l θ, ϕ
ð
Þ ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2l þ 1
ð
Þ l À m
ð
Þ!
4π l þ m
ð
Þ!
s
P
m
l x
ð Þe
imϕ x ¼ cos θ; 0 θ π
ð
Þ :
ð3:238Þ
Notice in (3.238), however, we could not determine Condon–Shortley phase
(À1)
m ; see (3.212).
106
3 Hydrogen-Like Atoms