Consider a particle subject to a central force **F** = -k**r** directed
towards the origin and proportional to the distance away from the origin. Then

U(r) = ½kr^{2} = ½mω^{2}r^{2}, with ω^{2}
= k/m and **F** = - **∇**U(**r**).

The Hamiltonian is

H = ½P^{2}/m
+ ½mω^{2}R^{2} = ½(P_{x}^{2} + P_{y}^{2}
+ P_{z}^{2})/m
+ ½mω^{2}(X^{2} + Y^{2} + Z^{2})

= 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 E_{0 }are
degenerate. E_{0} = (3/2)ħω is not degenerate.

For the three-dimensional
isotropic harmonic
oscillator the energy eigenvalues are E = (n + 3/2)ħω,
with n = n_{1} + n_{2} + n_{3}, where n_{1}, n_{2},
n_{3} 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 = n_{1}+ n_{2}+ n_{3}, with n_{i}= 0,1,2, ... .

For a given n choose a particular n_{1}. Then n_{2}+ n_{3}= n - n_{1}.

There are n - n_{1}+ 1 possible pairs {n_{2},n_{3}}. n_{2}can take on the values 0 to n-1, and for each n_{2}the value of n_{3}is fixed. The degree of degeneracy therefore is

g_{n}= ∑_{n1=0}^{n}(n - n_{1}+ 1) = ∑_{n1=0}^{n}(n + 1) - ∑_{n1=0}^{n}n_{1}= (n + 1)(n + 1) - ½n(n + 1) = ½n(n + 1)(n + 2).