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?

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. — Cosmas Zachos, Nov 09 '18 at 15:36
I forgot to take into account the cases where $n_x = 0$ or $n_y=0$. Thanks! — , Nov 09 '18 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. — Siham Salman, May 28 '23 at 07:45

score 6 · Answer 1 · answered Mar 25 '18 at 03:01

6

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.

answered Mar 25 '18 at 03:01

ZeroTheHero

45,515

3

Perhaps for the benefit of future users, you might use N for su(N) and provide the general formula for the degeneracy $\frac{(m+N-1)}{(N-1)! m!}$ ? – Cosmas Zachos Nov 09 '18 at 20:04

score 2 · Accepted Answer · answered Mar 25 '18 at 01:15

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).

score 0 · Answer 3 · answered Feb 24 '22 at 14:41

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

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

Degeneracy of 2 Dimensional Harmonic Oscillator

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