## Physics Friday 59

Let us consider the isotropic three-dimensional quantum harmonic oscillator: we have  (where the particle mass is now m0 to prevent confusion later). In cartesian coordinates, this becomes:

where Hx, Hy, and Hz are each the hamiltonian for a one-dimensional harmonic oscillator, in the x, y, and z directions respectively. Thus, the energy eigenstates will be products of eigenstates of these three; we have energy levels
. Thus, we have energies , with degeneracy equal to the number of solutions of , which, via combinatorics, is . (Only the ground state n=0 is non-degenerate.)

Now, suppose instead we consider spherical coordinates. As  depends on r only (spherical symmetry), we see that under separation of variables, the angular components will be given by the spherical harmonics  (see here and here). We will then have . With this in place, our Schrödinger equation becomes, for the radial component:

Defining , we see that this differential equation simplifies to
.
Now, we rescale the radial coordinate by defining dimensionless . Then the above equation becomes
, where  is the rescaled u(r). We can expect that far from the origin, we should have something like a Gaussian. If we attempt the substitution  (where k is a constant for which we will find later a value convenient for solving the equation), then this becomes, after eliminating a factor of ,
.
We see that this simplifies greatly if , so that , and then
.
Lastly, defining , and , we get
,
which is the associated Laguerre differential equation

with parameters  and .
The solutions are associated Laguerre functions , , and to be physically valid, we require that  be a non-negative integer. Thus ln, and they are both odd or both even integers. Thus, for even n, l can take values 0,2,4,…,n-2,n; and for odd n, l can take values 1,3,5,…,n-2,n. Adding the fact that m is an integer with possible values –l,-l+1,…,l-1,l; we have 2l+1 different possibilities for m. So, for even n, we have
 different possible angular momentum states; and for odd n, there are
 angular momentum states; the same formula as for even n. Looking back at the start of this post, we note that this formula matches the result for the degeneracies of the energy states we found using cartesian coordinates.