# Isotropic harmonic potential degeneracy of two identical particles

1. Apr 8, 2012

### hpar

Hello Everyone,

I've come across the following question and can't seem to get the right answer out:

1. The problem statement, all variables and given/known data

Two identical particles are in an isotropic harmonic potential. Show that, if the particles do not interact and there are no spin-orbit forces the degeneracies of the three lowest energy values are 1, 12, 39 if particles have spin 1/2, and 6, 27, 99 if the particles have spin 1.

2. Relevant equations

In an isotropic 3D harmonic oscillator we can write the energy levels as:

E = (n+3/2)$\hbar$$\omega$
where n = nx + ny + nz

3. The attempt at a solution

Now, in the first case, particles have spin 1/2, we have fermions, where particles with same quantum numbers can't coexist in same state due to the antisymmetric wavefunction, unless it's the spin part of the wavefunction that is antisymmetric.

now, our first energy level is when n = 0. Here, the spatial degeneracy is 1. Now, we can have two identical particles of spin 1/2 in that state as we have two possibilities for the spin. ie: we have two state at that energy so the degeneracy is two. and two particles of spin 1/2 can fill that state.
For the next value of n, n=1, we have three possible degeneracies.
0 0 1, 0 1 0, 1 0 0 for nx ny and nz. and we have two spin states. So spatial degeneracy is 3, spin degeneracy 2, so total states at the energy level is 3*2 = 6 different states at that energy. Degeneracy is 6. And so on. (we can derive general formula for such a degeneracy). Now my reasoning must obviously be wrong as I don't get the values stated in the question.

I would apply the same reasoning to spin 1 particles, but multiply by three since we have three possible spin states: -1, 0 and 1.

Can anybody help me out and point out the error in my reasoning in this problem?