Proceeding exactly as before we have the dimensionless form

Since the total Hamiltonian can be written as
the sum of three Hamiltonians (one just in *x'*, the second
just in *y'*, and the third just in *z'*;
each Hamiltonian has the same form as a
1-d oscillator), the solution is just the product of three
1-d oscillator wavefunctions (one just in *x'* the second
just in *y'*, the third just in *z'*):

with *E'*=
(2*n _{x}*+1 + 2

Since the problem is rotationally symmetric rather
than square-symmetric, it makes more sense to use a coordinate
system that reflects the symmetry in the problem. Hence,
I'd like to solve this problem is spherical coordinates
(*r'*--).

Changing coordinates changes the Schrödinger's equation to:

The stuff in the square brackets (with the overall minus sign) is the (dimensionless) angular momentum
squared operator: *L' ^{2}*. The eigenfunctions of this
operator are degenerate, so there is a choice to make in defining the
basis states. In physics folks conventionally use the

*L' ^{2}*

*L' _{z}*

Chemists not uncommonly use a different basis which has the advantage that the functions are real, but the disadvantage that they do not display spherical symmetry.

The probability density *Y _{lm}^{*}*

Here is the corresponding plot just as a function of .

On the other hand, if *m*=0, the angular momentum vector must be
in the *x-y* plane and the particle must be moving perpendicular
to the angular momentum vector and hence in part in the
*z* direction. For example, if ** L**~

Here is the corresponding plot just as a function of .

(Here is a bit more information about the *Y _{lm}*
and other choices.)

Seeking a solution where the dependence on *r'* factors
with all the ,
dependence in *Y _{lm}* ("separation of variables"),
i.e., =

As usual we have to work a bit to solve for the *r'* wavefunction.
Start by seeing how the equation must work for large *r'*.
The only term that has a chance of matching the ever growing
*r'*^{2} is the two-derivative term. Just as
in the 1-d oscillator, an approximate cancellation requires
exp(-½*r' ^{2}*) behavior.

For small *r'* only the derivative terms have
a chance to match the ever growing
*l*(*l*+1)/*r'*^{2} term.
If we try a power-law solution *r'*^{n}

we find *n*=*l* or *n*=-(*l*+1); the former
is not singular as *r'* nears 0

Factoring out all the required behavior for large and
small *r'*, we hope to find a simple function
(polynomial with luck) *F(r')* that contains the
behavior for intermediate *r'*.

=*r' ^{l}*
exp(-½

So the next step is to
rewrite Schrödinger's equation for *F(r')*.
In what follows I've decided to call *r'*
just *r* for notational ease.

So:

If we now try to write *F* as a polynomial:

we can plug the polynomial form into the differential equation:

The result is a __two__ term recursion relation (i.e., the result
has just two *a*s so, for example, given *a*_{0} we can
calculate *a*_{2}, from which we can calculate *a*_{4},
etc. until we're done.)

Note that the the recursion relation connects
even *k* to even *k*. Since *a*_{0}
may not be zero (as if it were we'd factor out *r*, thus increasing
the *r ^{l}* term to

It turns out that the polynomial must end if the
wavefunction is to be normalizable. One can show that the non-terminating
sum gets at least as big as exp(+*r ^{2}*), so
instead of going to zero for large

Thus if *E'*=4*n _{r}*+2

Note that if we had tried for a polynomial solution for
itself (rather than factoring out the
*r ^{l}* exp(-

The polynomials we have been calling *F* are Laguerre polynomials
(see for example, Abramowitz & Stegun, §22 p. 771 or
Szegö Ch.V).

Here is the overall solution:

Here is a fancier display of energy levels as a function of *l*