Consider a particle subject to a central force F = -kr directed towards the origin and proportional to the distance away from the origin.  Then
U(r) = ½kr2 = ½mω2r2, with ω2 = k/m and F = - U(r).

The Hamiltonian is
H = ½P2/m + ½mω2R2 = ½(Px2 + Py2 + Pz2)/m + ½mω2(X2 + Y2 + Z2)
= Hx + Hy + Hz.
The state space E can be written as a tensor product space, E = Ex ⊗ Ey ⊗ Ez .
Hx acts in Ex , Hy acts in Ey , and Hz acts in Ez . We know the eigenfunctions if Hi in Ei.
Hini> = (ni + ½)ħω|Φni>.
{|Φni>} is an orthonormal basis for Ei.
{|Ψnx,ny,nz> =  |Φnx>⊗|Φny>⊗|Φnz>} is an orthonormal basis for E .
We have H|Ψn1,n2,n3> = (nx + ny + nz + 3/2)ħω|Ψnx,ny,nz>.
The energy levels of the three-dimensional harmonic oscillator are denoted by En = (nx + ny + nz + 3/2)ħω, with n a non-negative integer, n = nx + ny + nz
All energies except E0 are degenerate.  E0 = (3/2)ħω is not degenerate.

Problem:

For the three-dimensional isotropic harmonic oscillator the energy eigenvalues are E = (n + 3/2)ħω, with n = n1 + n2 + n3, where n1, n2, n3 are the numbers of quanta associated with oscillations along the Cartesian axes.  Derive a formula for the degeneracy of the quantum state n, for spinless particles confined in this potential.

• Solution:
We have n = n1 + n2 + n3, with ni  =  0,1,2, ... .
For a given n choose a particular n1. Then n2 + n3 = n - n1.
There are n - n1 + 1 possible pairs {n2,n3}.  n2 can take on the values 0 to n-1, and for each n2 the value of n3 is fixed.  The degree of degeneracy therefore is
gn = ∑n1=0n(n - n1 + 1) = ∑n1=0n(n + 1)  - ∑n1=0n n1 = (n + 1)(n + 1) - ½n(n + 1) = ½n(n + 1)(n + 2).