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I have a box with $x,y,z$ all ranging from 0 to $l$. It has $V(x)$=0 inside and =$\infty$ outside. By extending the 1D Schrodinger equation, I have that the allowed energy eigenvalues are $\hbar^2\pi^2\over2ml^2$$(n_1^2+n_2^2+n_3^2)$. What is the degeneracy of the 1st excited energy level? By "1st excited energy level" does that mean 1 of the $n_k$'s, say $n_1$, =2 while $n_2=n_3=1$? Or does it mean all 3 dimensions are in their 1st excited state -- $n_k=2$ $ \forall k\in${$1,2,3$}? Also, how does one find the degeneracy? I am guessing that it is 3? Thanks for any help.

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Degeneracy of a level means many different wave functions that give the same E. In your case it is not sufficient to write E, you need to prove that for this E there is only one wave function.

Consider, for example, a case with $n_1=2, n_2=1, n_3 = 1$. The same value of energy can be found in the states $n_1=1, n_2=2, n_3 = 1$ and in $n_1=1, n_2=1, n_3 = 2$ which have different wave functions.

The ground state is thus not degenerate but all excited ones are.

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Thanks, Vladimir. I understand what degeneracy is, in principle. What I am unclear about is what qualifies as the first excited state -- if only one dimension is excited or whether all 3 are. Also, I don't know how many degeneracies there are. – G. K. Sep 3 '11 at 16:39
Any state with $E > E_0$ is an excited one. In your case it is any state with $n > 1$, not obligatory with all 1D modes excited. For the first excited state there are three equal energies with different 1D excited modes, so the degeneracy degree is 3. For higher excited states there might be more combinations of $n$s that give the same sum of their squares. – Vladimir Kalitvianski Sep 3 '11 at 17:11
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Small correction: most excited states are degenerate, but not all. For instance, the state $(n_1,n_2,n_3)=(2,2,2)$ is nondegenerate. Most triples $(n,n,n)$ are degenerate -- e.g. $(3,3,3)$ is degenerate with $(1,1,5)$. A quick check with Mathematica seems to indicate that $(n,n,n)$ is nondegenerate if and only if $n$ is a power of 2. At least that's true for the first couple of hundred $n$'s, but I don't know a proof that it always is. – Ted Bunn Sep 3 '11 at 18:32
Yes, Ted, you are right. I did not check the higher numbers. – Vladimir Kalitvianski Sep 3 '11 at 18:42
Thanks, @Ted. Why is (2,2,2) nondegenerate? Can't we have, say (1,2,3)? I must be missing sth here... – G. K. Sep 3 '11 at 22:06
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