Quantum harmonic oscillators - grand partition function

  • Thread starter Heirot
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Homework Statement



Calculate the grand partition function for a system of [tex]N[/tex] noninteracting quantum mechanical oscillators, all of which have the same natural frequency [tex]\omega_0[/tex]. 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
[tex]E(\{n_i\})=\frac{N}{2}\hbar\omega_0+\hbar\omega_0\sum_{i=1}^Nn_i[/tex]
where [tex]n_i \geq 0[/tex] is the number of phonons in the i-th harmonic oscillator. For a given
[tex]s=\sum_{i=1}^Nn_i[/tex]
the grand partition function is
[tex]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)}[/tex]
The function [tex]g(s)[/tex] 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.
[tex]g(s)=\frac{(s+N-1)!}{s!}[/tex]
On the other hand, for Bose statistics, the oscillators (boxes) are now indistinguishable and one has
[tex]g(s)=\frac{(s+N-1)!}{s!(N-1)!}[/tex]

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

Thank you.
 

Answers and Replies

  • #2
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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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