Degeneracy of 2 Dimensional Harmonic Oscillator

Question

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?

Answer 1

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.

Answer 2

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).

Answer 3

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

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