Isotropic harmonic potential degeneracy of two identical particles

[hpar]

Hello Everyone,

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

Homework Statement

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.

Homework 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

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?

 

[mark726]

Hello!

Your reasoning is on the right track, but there are a few key points that you are missing.

Firstly, in the case of spin 1/2 particles, you are correct that the spatial degeneracy is 1 for the n=0 energy level. However, the spin degeneracy is not 2. Remember that for fermions, the total wavefunction must be antisymmetric. This means that the spin part of the wavefunction must be symmetric. Therefore, for two particles with spin 1/2, the only possible spin states are both spin up or both spin down. This gives us a spin degeneracy of 1, not 2.

For the next energy level, n=1, you correctly identified that there are 3 possible spatial states (0 0 1, 0 1 0, 1 0 0) and 2 possible spin states (both spin up or both spin down). However, these are not all distinct states. In fact, the state (0 0 1) with both particles in the same spatial state and the same spin state is not allowed due to the Pauli exclusion principle. This leaves us with only 2 possible states at this energy level, giving a total degeneracy of 2.

You are correct that for spin 1 particles, the spin degeneracy will be 3 instead of 1. However, you must also take into account that there are now 3 possible spatial states for each n value, not just 1. This means that for n=0, there are 3 possible states with a total degeneracy of 3. For n=1, there are 9 possible states with a total degeneracy of 9.

I hope this helps clarify the reasoning behind the given degeneracies. Let me know if you have any further questions. Good luck!