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I was working on the 3D isotropic harmonic oscillator and I found that the energies are given by:

$$E=\hbar\omega(n_x+n_y+n_z+3/2)$$

Which has a degeneracy of $\tfrac12(n+1)(n+2)$. However, when dealing with the anisotropic case, I'm not sure if there's a degeneracy in energies. For example, for the 2D anisotropic harmonic oscillator with frequencies $\omega$ and $3\omega$, I found the energies to be:

$$E=\hbar\omega (n_x +3n_y +2)$$

For example here, for $n=1$ we can either have $n_x=1$ so $E_1=3\hbar\omega$ or $n_y=1$ so $E_2=5\hbar\omega$. For $n=2$, we can have $n_x=2$ so $E_2=4\hbar\omega$ or $n_y=2$ so $E_2=8\hbar\omega$ or $n_x=1,n_2=1$ so $E_2=6\hbar\omega$. So there seems to be no degeneracy, but I'm not sure as this is the first time I've found this problem.

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The states

  • $n_x=3$, $n_y=0$
  • $n_x=0$, $n_y=1$

are degenerate in energy. $\tag{QED}$


More generally, any harmonic oscillator of the form $$ E = \hbar \omega_1 n_1 + \hbar \omega_2 n_2 $$ will be degenerate if $\displaystyle \frac{\omega_1}{\omega_2} \in \mathbb Q$. It is an important exercise to prove that that is the case and to calculate the degeneracies in both 2D and 3D.

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