Proceeding exactly as before we have the dimensionless form

Since the total Hamiltonian can be written as
the sum of Hamiltonians (one just in *x'* the other
just in *y'*, each exactly the same form as in the
1-d oscillator), the solution is just the product of
1-d oscillator wavefunctions (one just in *x'* the other
just in *y'*):

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

Here's what the probability density looks like for *n _{x}*=16 and

In this display the bright parts of the image are where the particle is likely to be found. Here is a expanded version of the same:

Note the 4 nodal lines perpendicular to *y* and the
16 nodal lines perpendicular to *x*.

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 polar coordinates
(*r'*-).

Changing coordinates changes the Schrödinger's equation to:

Seeking a solution where the dependence on *r'* factors
from the dependence on ("separation of variables"),
we find:

The solution to the equation is easy:

Our "boundary condition" is that if we go round the origin exactly once, the value of the wavefunction is exactly the same, i.e., the wavefunction must be the same at and +2:

Hence *m* must be an integer.

This leaves us with the *r'* differential equation
(given below in three equivalent forms):

We have to work harder to solve this differential equation!
Start by seeing how the 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
*m*^{2}/*r'*^{2} term.
If we try a power-law solution *r'*^{n}

we find *n*=|*m*|.

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

=*r'*^{|m|}
exp(-½*r' ^{2}*)

So the next step is to
rewrite Schrödinger's equation for *G(r')*.

So:

(In the above we've used *m* where we actually mean
|*m*|.)

If we now try to write *G* 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' ^{m}* 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'*=2(2*n _{r}*+|

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

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

Here is the overall solution:

Here is what the probability density looks like for *n _{r}*=10,

Note the solution is circle-symmetric and has 10 nodal circles.

Here is what the probability density looks like for *n _{r}*=5,

Note the solution is circle-symmetric and has 5 nodal circles, and that the particle is unlikely to be found exactly at the origin.

Note that in producing these circle-symmetric solutions, we have
produced *complex* rather than real wavefunctions....
that factor of exp(i*m*). This factor
cancels out of our ^{*}
probability density (so the probability density is circle-symmetric),
but if we were to look at just the real part of
we would find oscillation as a function of
. Oscillation indicates momentum, so these
solutions have angular momentum...in fact *m*
angular momentum in the "*z*" direction (i.e., perpendicular to the
plane).

Because the angular dependence is "simple" we can usefully plot the
wavefunction just as a function of *r'*. Here are
"stacked wavefunction" plots for *m*=0,1,3:

The red line is the classical "effective potential"=
*m ^{2}/r'^{2}*+