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(P_x² + P_y² + P_z²)/m + ½mω²(X² + Y² + Z²)
= H_x + H_y + H_z.
The state space E can be written as a tensor product space, E = E_x ⊗ E_y ⊗ E_z .
H_x acts in E_x , H_y acts in E_y , and H_z acts in E_z . We know the eigenfunctions if H_i in E_i.
H_i|Φ_ni> = (n_i + ½)ħω|Φ_ni>.
{|Φ_ni>} is an orthonormal basis for E_i.
{|Ψ_nx,ny,nz> =  |Φ_nx>⊗|Φ_ny>⊗|Φ_nz>} is an orthonormal basis for E .
We have H|Ψ_n1,n2,n3> = (n_x + n_y + n_z + 3/2)ħω|Ψ_nx,ny,nz>.
The energy levels of the three-dimensional harmonic oscillator are denoted by E_n = (n_x + n_y + n_z + 3/2)ħω, with n a non-negative integer, n = n_x + n_y + n_z. 
All energies except E₀ are degenerate.  E₀ = (3/2)ħω is not degenerate.

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 n_i  =  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
  g_n = ∑_n1=0ⁿ(n - n₁ + 1) = ∑_n1=0ⁿ(n + 1)  - ∑_n1=0ⁿ n₁ = (n + 1)(n + 1) - ½n(n + 1) = ½n(n + 1)(n + 2).