Calculating degeneracy of the energy levels of a 2D harmonic oscillator

[sukmeov]

Homework Statement

I calculated the energies for decoupled oscillators to be E_n_1 = 3 ħω(n_1+1/2) and E_n_2 = ħω(n_2 +1/2) and so the total energy of the 2D harmonic oscillator is E = ħω(3n_1+n_2 +2). What's the degeneracy for each energy level?

Relevant Equations

none... just counting

Too dim for this kind of combinatorics. Could anyone refer me to/ explain a general way of approaching these without having to think :D. Thanks.

 

[sukmeov]

Wait... If n=3n_1 + n_2 then is it just floor(n/3) +1?

 

[hutchphd]

Make some kind of graph or ladder diagram. Then just look. This one seems hardly worth the effort...

 

[sukmeov]

Yeah. Think I was being silly. A general method might be useful for 3 or more summands... I posted an attempt above. Do you think it is correct?

 

[hutchphd]

I don't think so. If n=4 specify the 2 degenerate states by $(n_1,n_2)$.
Make a little matrix table of degeneracy with $(n_1,n_2)$, you'll see

 

[sukmeov]

4= 3x0+4= 3x1 +1, so degeneracy is 2. floor(4/3)+1=2... am I missing something?

 

[hutchphd]

You know I made a little table. You are in fact correct! Apologies for the brain fade.

 

[vanhees71]

Shouldn't the energy eigenvalues be
$$E_{n_1,n_2}=\hbar \omega (n_1+n_2+1)$$
with $n_1,n_2 \in \mathbb{N}_0$?

 

[hutchphd]

Shouldn't the energy eigenvalues be
En1,n2=ℏω(n1+n2+1)

I don't think the problem is isotropic. But I don't know of any system where this is a good model (other than pedagogy).

 

[vanhees71]

Well, but then you have
$$E_{n_1,n_2}=\hbar (\omega_1 n_1 + \omega_2 n_2 +1)$$
with different frequencies for the normal modes of the plane oscillator. I'm still puzzled about where the formula for the energy eigenvalues in #1 comes from.

 

[hutchphd]

I assumed they just defined $\omega _1= 3\omega_2$

Clearly it is artificial. As I consider it, are there subtleties to such "accidental" degeneracies? Everything has a finite energy width in practice.

 

[sukmeov]

Oscillator with diagonal potential (1 0
0 9).

 

[vanhees71]

Argh. My fault. Of course, if you want to have degeneracy at all $\omega_1$ and $\omega_2$ must be "commensurable", i.e., $\omega_1/\omega_2 \in \mathbb{Q}$. Now it makes sense!