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?

• No, it is not true. In 2 d, the degeneracy is n +1, as @ZeroTheHero 's answer details. Check the n =4 case to reassure yourself. read up on your Jordan map. Commented Nov 9, 2018 at 15:36
• I forgot to take into account the cases where $n_x = 0$ or $n_y=0$. Thanks!
– user154080
Commented Nov 9, 2018 at 20:02
• degenercy of nth state for 2D harmonic oscillator is given by; d(n)=n+1 where n is the principle quantum number. Commented May 28, 2023 at 7:45

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.
• Perhaps for the benefit of future users, you might use N for su(N) and provide the general formula for the degeneracy $\frac{(m+N-1)}{(N-1)! m!}$ ? Commented Nov 9, 2018 at 20:04

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).

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

$$\binom{l+N-1}{N}.$$