1.19 Angular Momentum
For the common problem of central potentials (See section 14.1)V(r), we use the obviousrotational
symmetryto find that theangular momentum,L~=~x×~p, operators commute withH,
[H,Lz] = [H,Lx] = [H,Ly] = 0but they do not commute with each other.
[Lx,Ly] 6 = 0We want to findtwo mutually commuting operatorswhich commute withH, so we turn to
L^2 =L^2 x+L^2 y+L^2 zwhich does commute with each component ofL.
[L^2 ,Lz] = 0We chose our two operators to beL^2 andLz.
Some computation reveals that we can write
p^2 =1
r^2(
L^2 + (~r·~p)^2 −i ̄h~r·~p)
.
With this the kinetic energy part of our equation will only have derivatives inrassuming that we
have eigenstates ofL^2.
− ̄h^2
2 μ[
1
r^2(
r∂
∂r) 2
+
1
r∂
∂r−
L^2
̄h^2 r^2]
uE(~r) +V(r)uE(~r) =EuE(~r)TheSchr ̈odinger equation thus separatesinto an angular part (theL^2 term) and a radial part
(the rest). With this separation we expect (anticipating the angular solution a bit)
uE(~r) =REℓ(r)Yℓm(θ,φ)will be a solution. TheYℓm(θ,φ) will be eigenfunctions ofL^2
L^2 Yℓm(θ,φ) =ℓ(ℓ+ 1) ̄h^2 Yℓm(θ,φ)so the radial equation becomes
− ̄h^2
2 μ[
1
r^2(
r∂
∂r) 2
+
1
r∂
∂r−
ℓ(ℓ+ 1)
r^2]
REℓ(r) +V(r)REℓ(r) =EREℓ(r)We must come back to this equation for eachV(r) which we want to solve.
Wesolve the angular part of the problem in generalusing angular momentum operators. We
find thatangular momentum is quantized.
LzYℓm(θ,φ) =m ̄hYℓm(θ,φ)
L^2 Yℓm(θ,φ) =ℓ(ℓ+ 1) ̄h^2 Yℓm(θ,φ)withℓandmintegers satisfying the condition−ℓ≤m≤ℓ. The operators thatraise and lower
thezcomponentof angular momentum are
L±=Lx±iLyL±Yℓm= ̄h√
ℓ(ℓ+ 1)−m(m±1)Yℓ(m±1)We derive the functional form of theSpherical HarmonicsYℓm(θ,φ) using the differential form
of the angular momentum operators.