# Degeneracy of a 2-dimensional isotropic Harmonic Oscillator

1. Oct 20, 2015

### silverwhale

1. The problem statement, all variables and given/known data
The Hamiltonian is given by:
$$H = \frac{1}{2} \sum_{i=1,2}[p_i^2 + q_i^2]$$
We define the following operators:
$$J = \frac{1}{2} (a_1^+ a_1 + a_2^+ a_2)$$
$$J_1 = \frac{1}{2} (a_2^+ a_1 + a_1^+ a_2)$$
$$J = \frac{i}{2} (a_2^+ a_1 - a_1^+ a_2)$$
$$J = \frac{1}{2} (a_1^+ a_1 - a_2^+ a_2)$$

I have shown previosly that $$\textbf{J}^2=J_1^2 + J_2^2 + J_3^2 = J(J+1)$$

The question: Using this result (the one above), discuss what is the degeneracy of the eigenvalues of H.
2. Relevant equations
$$H = 2J + 1$$
$$\textbf{J}^2=J_1^2 + J_2^2 + J_3^2 = J(J+1)$$

3. The attempt at a solution
I thought just saying that $$[H, \textbf{J}^2] = 0$$ would be enough to say that the eigenvalues of H are degenerate with respect to the eigenstates of $$\textbf{J}^2$$. But I am not sure if this is enough, can I calculate the degeneracy of the eigenvalues of H just using the equations found above?

2. Oct 24, 2015

### MisterX

One makes an analogy to angular momentum. So one does something like
$$J_{\pm} = J_1 \pm i J_2$$
and find that
$$[J_+, J_3] = J_+$$
$$[J_-, J_3] = -J_-$$
One also finds that $[J_3, J^2] = 0$ so one can give the eigenvalues of $J^2$ and $J_3$ a separate quantum number, so the eigenvalues of the two operators are $j(j+1)$ and $m$, respectively.
So from this we obtain $$J_3J_+\mid j m \rangle = (m+1) J_+\mid j m \rangle$$
$$J_+\mid j m \rangle = \alpha_{jm+}\mid j m+1 \rangle$$
and similiarly for $J_-$. Also since $[J_\pm, J^2] = 0$, one gets that $J_\pm\mid j m \rangle$ has the same energy eigenvalue as $\mid j m \rangle$
To find the degeneracy one calculates $\left| J_\pm\mid j m \rangle\right|^2 = |\alpha_{jm\pm}|^2$ and notice that if one tries to exceed a particular range of $m$ one just gets $\alpha=0$. So there is a finite range of $m$ with the same energy. Let me know if that helps.

Last edited: Oct 24, 2015