130_notes.dvi

Its easy to find functions that give the eigenvalue ofLz.

Yℓm(θ,φ) = Θ(θ)Φ(φ) = Θ(θ)eimφ

LzYℓm(θ,φ) =

̄h

i

∂

∂φ

Θ(θ)eimφ=

̄h

i

imΘ(θ)eimφ=m ̄hYℓm(θ,φ)

To find theθdependence, we will use the fact that there are limits onm. The state with maximum
mmust give zero when raised.

L+Yℓℓ= ̄heiφ

(

∂

∂θ

+icotθ

∂

∂φ

)

Θℓ(θ)eiℓφ= 0

This gives us a differential equation for that state.

dΘ(θ)

dθ

+iΘ(θ) cotθ(iℓ) = 0

dΘ(θ)

dθ

=ℓcotθΘ(θ)

The solution is
Θ(θ) =Csinℓθ.

Check the solution.
dΘ
dθ

=Cℓcosθsinℓ−^1 θ=ℓcotθΘ

Its correct.

Here we should note that only the integer value ofℓwork for these solutions. If we were to use
half-integers, the wave functions would not be single valued, for example atφ= 0 andφ= 2π. Even
though the probability may be single valued, discontinuities in the amplitude would lead to infinities
in the Schr ̈odinger equation. We will find later that thehalf-integer angular momentum states
are usedfor internal angular momentum (spin), for which noθorφcoordinates exist.

Therefore,the eigenstateYℓℓis.

Yℓℓ=Csinℓ(θ)eiℓφ

We can compute the next state down by operating withL−.

Yℓ(ℓ−1)=CL−Yℓℓ

We can continue to lowermto get all of the eigenfunctions.

We call these eigenstates theSpherical Harmonics. The spherical harmonics arenormalized.

∫^1

− 1

d(cosθ)

∫^2 π

0

dφ Yℓm∗Yℓm= 1

∫

dΩYℓm∗Yℓm= 1

Since they are eigenfunctions of Hermitian operators, they areorthogonal.