# Quantum harmonic oscillators - grand partition function

## Homework Statement

Calculate the grand partition function for a system of $$N$$ noninteracting quantum mechanical oscillators, all of which have the same natural frequency $$\omega_0$$. Do this for the following cases: (i) Boltzmann statistics; (ii) Bose statistics.

## The Attempt at a Solution

The energies of the system are given by
$$E(\{n_i\})=\frac{N}{2}\hbar\omega_0+\hbar\omega_0\sum_{i=1}^Nn_i$$
where $$n_i \geq 0$$ is the number of phonons in the i-th harmonic oscillator. For a given
$$s=\sum_{i=1}^Nn_i$$
the grand partition function is
$$Z_G(\beta,\mu)=e^{-\beta\frac{N\hbar\omega_0}{2}}\sum_{s=0}^{\infty}g(s)e^{-\beta s (\hbar \omega_0 - \mu)}$$
The function $$g(s)$$ represents density of states (degeneracy) of the bosonic system, and I have a hard time calculating it.

For Boltzmann statistics, the oscillators are distinguishable and the degeneracy should be equal to the number of ways one can partition s identical objects into N different boxes, e.g.
$$g(s)=\frac{(s+N-1)!}{s!}$$
On the other hand, for Bose statistics, the oscillators (boxes) are now indistinguishable and one has
$$g(s)=\frac{(s+N-1)!}{s!(N-1)!}$$

My question is, is this reasoning correct? If so, can I sum the series into closed-form expressions?

Thank you.

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It looks as though I was wrong in the last post. E.g. for N=3 oscillators and s=3 phonons, one has:

Boltzmann case:
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10 different possibilities, but only

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3 in Bose Einstein case.

Is this reasoning correct?