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Sheet 2, Question 5

Hello all! Above is my question. I am fine all the way up to the final part about the degeneracy. I find counting degeneracies quite difficult, and this is no exception! I have really no idea why the trace is as that power series, and don't know why the degeneracies are as given.

By the standard power series / binomial expansion, I get $$ d_n = {(-1)^n \over n!} \prod_{m=0}^{n-1}(-D-m) = {1 \over n!} \prod_{m=0}^{n-1}(D+m).$$ (Empty sum is identity, as usual.) I can't work out why this is the degeneracy (I know it is related to $n \choose r$, but not sure how to use that...). I can't see any reason why this should be the coefficients in the power series expansion.

Also, while I'm on the topic of degeneracies, how can I show that the degeneracy of the $n-$th excited state of the 3D harmonic oscillator ($E_{n_1,n_2,n_3} = \hbar \omega (n_1 + n_2 + n_3 + 3/2))$ is $(n+1)(n+2)/2$ (which I know is equal to $\sum_{r=0}^{n+1}r$). I guess it's just the number of ways that three (positive) integers can add up to $n$, but I don't know how to prove that.

Thank you in advance for your help!
(Please note: I'm going away shortly, so if I don't respond to your answer, then I'm not ignoring it, I simply haven't seen it yet; once I see any responses, I shall reply asap! Thanks!)
(Oh, and please no-one try to move this to physics.SE - this is a maths questions; I am not asking what the physical significance of this is, just the maths behind it - it's linear algebra really! Sorry to be blunt, but I had someone move one over incorrectly before - rather annoying! Thanks!)

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When you have $D$ oscillators the state will be $| n\rangle = | n_1, n_2, \cdots, n_D\rangle$ and the trace you want:

$$ Tr(x^{\mathcal{N}}) = \sum_{n} \langle n|x^{\mathcal{N}}|n\rangle = \sum_{n_1, \cdots, n_D =0}^{\infty} \langle n_1, \cdots, n_D|x^{\mathcal{N}}| n_1,\cdots, n_D \rangle $$ The second sum is actually (all number operators commute with each other)

$$ \sum_{n_1, \cdots, n_D = 0}^{\infty} x^{n_1 + \cdots +n_D } = \sum_{n_1 =0}^{\infty}x^{n_1} \sum_{n_2=0}^{\infty}x^{n_2} \cdots \sum_{n_D =0}^{\infty}x^{n_D} = (1-x)^{-D} $$ In the last step one uses the result derived in (a).

The degeneracy of the operator $\mathcal{N}$ is the amount of all such partitions of the integer $n$ into the sum $n_1 +n_2 + \cdots + n_D$, this is a standard combinatioral problem (solved usually by introducing "separations"), the answer is

$$ d_n = \binom{n+D-1}{n} $$ One can see this number is exactly the coefficent in the trace expansion by noting that the one can recast the sum into the form

$$ \sum_{n_1, \cdots, n_D = 0}^{\infty} x^{n_1 + \cdots +n_D } = \sum_{n=0}^{\infty} \sum_{n_1 + \cdots + n_D =n}x^{n} = \sum_{n=0}^{\infty} d_n x^n $$ (all we have done is classify the terms of the sum according to their powers)

Your approach seems to make use of the generating function for $d_n$, which should also work.

For the 3D harmonic oscillator you have $D=3$ so you get

$$ d_3 = \binom{n+3-1}{n} = \binom{n+2}{n} = \frac{(n+1)(n+2)}{2} $$

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  • $\begingroup$ Thank you very much! (Sorry for the slow reply - I was away camping for a week!) $\endgroup$ – Sam T Aug 9 '14 at 8:03
  • $\begingroup$ I'm also out today, but tomorrow I'll go through this carefully on the paper to check everything! :) - I think the key issue for me was that I didn't know that the number of integer partitions is as you claimed; can you explain why that is? $\endgroup$ – Sam T Aug 9 '14 at 8:04
  • $\begingroup$ Sure, think of the integer $n$ as a collection of $n$ balls: $oo\cdots o$ and place $D-1$ "separators" like bars $|$, so that the partition of say 8 = 4 + 1+ 3 has e.g. $n=8$, $D=3$ so it has two separators and looks like: $oooo|o|ooo$. It becomes clear that the number of such partitions will be the number of ways of placing the balls and the separators, they are $n+D-1$ objects, so you have $(n+D-1)!$, but then balls are indistuinguishable, and so are bars, hence one must divide by $n!$ and by $(D-1)!$ to account for this. $\endgroup$ – Rogelio Molina Aug 11 '14 at 3:03
  • $\begingroup$ That's very clear, thank you. :) $\endgroup$ – Sam T Aug 11 '14 at 7:26

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