130_notes.dvi

This leads to a great simplification of the 3D problem.

It is possible toseparate the Schr ̈odinger equationsincerandL^2 appear separately. Write the
solution as a product

uE(~r) =REℓ(r)Yℓm(θ,φ)

whereℓlabels the eigenvalue of theL^2 operator andmlabels the eigenvalue of theLzoperator.
SinceLzdoes not appear in the Schr ̈odinger equation, we only label the radial solutions with the
energy and the eigenvalues ofℓ.

We get the three equations.

L^2 Yℓm(θ,φ) =ℓ(ℓ+ 1) ̄h^2 Yℓm(θ,φ)

LzYℓm(θ,φ) =m ̄hYℓm(θ,φ)

− ̄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)

By assuming the eigenvalues ofL^2 have the formℓ(ℓ+ 1) ̄h^2 , we haveanticipated the solution
but not constrained it, since the units of angular momentum are those of ̄hand since we expect
L^2 to have positive eigenvalues.

〈Yℓm|L^2 |Yℓm〉=〈LxYℓm|LxYℓm〉+〈LyYℓm|LyYℓm〉+〈LzYℓm|LzYℓm〉≥ 0

The assumption that the eigenvalues ofLz are some (dimensionless) number times ̄hdoes not
constrain our solutions at all.

We will use the algebra of the angular momentum operators to help ussolve the angular part of
the problem in general.

For any given problem with rotational symmetry, we will need tosolve a particular differential
equation in one variabler. This radial equation can be simplified a bit.

− ̄h^2

2 μ

[

∂^2

∂r^2

+

2

r

∂

∂r

]

REℓ(r) +

(

V(r) +

ℓ(ℓ+ 1) ̄h^2

2 μr^2

)

REℓ(r) =EREℓ(r)

We have grouped the term due to angular momentum with the potential. It is often called a pseudo-
potential. Forℓ 6 = 0, it is like a repulsive potential.

14.2 Angular Momentum Algebra: Raising and Lowering Operators

We have already derived thecommutators of the angular momentum operators

[Lx,Ly] = i ̄hLz