Degeneracy of 2 Dimensional Harmonic Oscillator If we consider a particle in a 2 dimensional harmonic oscillator potential with Hamiltonian $$H = \frac{\mathbf{p}^2}{2m} + \frac{m w^2 \textbf{r}^2}{2}$$ it can be shown that the energy levels are given by $$E_{n_x,n_y} = \hbar \omega (n_x + n_y + 1) = \hbar \omega (n + 1)$$ where $n = n_x + n_y$. Is it then true that the n$^{\text{th}}$ energy level has degeneracy $n-1$ for $n \geq 2$, and 1 for $0 \leq n \leq 1$? 
How common is this scenario where it is possible to calculate the degeneracy of a "general" or "n$^\text{th}$" energy level? How common is this in more complicated quantum systems?
 A: In the case of the n-dimensional harmonic oscillator, possibly the most elegant method is to recognize that the set of states with total number $m$ of excitation span the irrep $(m,0,\ldots,0)$ of $su(n)$.  Thus the degeneracy is the dimension of this irrep.


*

*For the 2D oscillator and $su(2)$ this is just $m+1$,

*For the 3D oscillator and $su(3)$ this is $\frac{1}{2}(m+1)(m+2)$

*For the 4D oscillator and $su(4)$ this is $\frac{1}{3!}(m+1)(m+2)(m+3)$ 
etc.

A: Yes that's correct, and in general it's very common to be able to count. For more info look into the mircocanonical density of states - it's very closely related to the idea of entropy (i.e. entropy is related to the number of degeneracies in a system). 
A: If you ignore the group theoretical implications, the number operator eigenstates are simply
$$
|n_1,n_2,\dots,n_l\rangle,
$$
with the restriction
$$
\sum_{j=1}^l n_j = N.
$$
This is because the energy of $l$ uncoupled oscillators is
$$
E_l = N
$$
plus a constant. For fixed $N$, the degeneracy space is simply how many of these $N$ excitations (actually bosons!) you can distribute over $l$-levels. This is the typical combinatorial problem with replacement, since any number of bosons can fit in any state out of $l$ possible. Hence the degeneracy is
$$
\binom{l+N-1}{N}.
$$
