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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?

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    $\begingroup$ 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. $\endgroup$ Nov 9, 2018 at 15:36
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    $\begingroup$ I forgot to take into account the cases where $n_x = 0$ or $n_y=0$. Thanks! $\endgroup$
    – user154080
    Nov 9, 2018 at 20:02
  • $\begingroup$ degenercy of nth state for 2D harmonic oscillator is given by; d(n)=n+1 where n is the principle quantum number. $\endgroup$ May 28, 2023 at 7:45

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

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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

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

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