Bringing together the above results, we have.
∂
∂x
= sinθcosφ∂
∂r+
1
r
cosφcosθ∂
∂θ−
1
rsinφ
sinθ∂
∂φ
∂
∂y= sinθsinφ∂
∂r+
1
rsinφcosθ∂
∂θ+
1
rcosφ
sinθ∂
∂φ
∂
∂z= cosθ∂
∂r−
1
rsinθ∂
∂θNow simply plug these into the angular momentum formulae.
Lz =̄h
i(
x∂
∂y−y∂
∂x)
=
̄h
i∂
∂φL± = Lx±iLy=
̄h
i(
y∂
∂z
−z∂
∂y
±i(
z∂
∂x
−x∂
∂z))
=
̄h
i(
(y∓ix)∂
∂z
−z(
∂
∂y
∓i∂
∂x))
= ̄h(
∓(x±iy)∂
∂z
±z(
∂
∂x
±i∂
∂y))
= ̄hr(
∓sinθe±iφ∂
∂z±cosθ(
∂
∂x±i∂
∂y))
= ̄h(
∓sinθe±iφ(
rcosθ∂
∂r−sinθ∂
∂θ)
±cosθ(
rsinθe±iφ∂
∂r+ cosθe±iφ∂
∂θ±
ie±iφ
sinθ∂
∂φ))
= ̄he±iφ(
±sin^2 θ∂
∂θ±
(
cos^2 θ∂
∂θ±
icosθ
sinθ∂
∂φ))
= ̄he±iφ(
±
∂
∂θ+icotθ∂
∂φ)
We will use these results to find the actual eigenfunctions of angular momentum.
Lz =̄h
i∂
∂φL± = ̄he±iφ(
±
∂
∂θ+icotθ∂
∂φ)
14.4.5 The OperatorsL±
The next step is to figure out how theL±operators change the eigenstateYℓm. What eigenstates
ofL^2 are generated when we operate withL+orL−?
L^2 (L±Yℓm) =L±L^2 Yℓm=ℓ(ℓ+ 1) ̄h^2 (L±Yℓm)BecauseL^2 commutes withL±, we see that we have the sameℓ(ℓ+ 1) after operation. This is also
true for operations withLz.
L^2 (LzYℓm) =LzL^2 Yℓm=ℓ(ℓ+ 1) ̄h^2 (LzYℓm)