Quantum harmonic oscillators - grand partition function

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The discussion focuses on calculating the grand partition function for a system of N noninteracting quantum harmonic oscillators with the same natural frequency, considering both Boltzmann and Bose statistics. The energy of the system is expressed as a function of phonon numbers, leading to the formulation of the grand partition function. For Boltzmann statistics, the degeneracy function g(s) accounts for distinguishable oscillators, while for Bose statistics, it reflects indistinguishable oscillators. The participant seeks confirmation on the correctness of their reasoning regarding the calculations of g(s) and the possibility of summing the series into closed-form expressions. The example provided illustrates the difference in counting configurations between the two statistical cases.
Heirot
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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.

Homework Equations





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?
 

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