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Say I have two identical (fermionic) non-interacting particles in a 1D harmonic oscillator. I would like to compute the entropy of the system as the temperature $T$ varies, for which I need the partition function of this system, then I would calculate the entropy like

$$ F = -k_B T \log(Z) \Longrightarrow S = -\frac{\partial F}{\partial T} $$

However I am stuck when trying to compute $Z$, can anyone help?

My attempt

The particles are non-interacting, therefore the energy of the system is given by $E_{n, m} = E_n + E_m= \hbar \omega(1+n+m)$ so to compute the (canonical) partition function we need to compute

$$ Z = \sum_{n, m}e^{-\beta E_{n, m}} = e^{-\beta \hbar \omega}\sum_{n, m}\Big(e^{-\beta\hbar \omega}\Big)^n\Big(e^{-\beta\hbar \omega}\Big)^m = e^{-\beta \hbar \omega}\sum_n\Big(e^{-\beta\hbar \omega}\Big)^n\Big(e^{-\beta\hbar \omega}\Big)^n + e^{-\beta \hbar \omega}\sum_{n<m}\Big(e^{-\beta\hbar \omega}\Big)^n\Big(e^{-\beta\hbar \omega}\Big)^m = \frac{e^{-\beta \hbar \omega}}{1-e^{-2\beta \hbar \omega}} + \sum_{n<m}\Big(e^{-\beta\hbar \omega}\Big)^n\Big(e^{-\beta\hbar \omega}\Big)^m $$

at this point I am stuck. How can I compute the second part of $Z$?

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    $\begingroup$ If you must evaluate $Z$ without noting it's the square of the one-particle case, the $n<m$ double sum you're stuck on is$$\sum_ne^{-n\beta\hbar\omega}\sum_{m\ge n+1}e^{-m\beta\hbar\omega},$$which is easily rewritten (by evaluating the inner sum) as a geometric series. $\endgroup$
    – J.G.
    Apr 10, 2022 at 14:09
  • $\begingroup$ Possibly related, check also the links to the papers therein. $\endgroup$
    – Tobias Fünke
    Apr 10, 2022 at 14:33
  • $\begingroup$ You might be interested in this. $\endgroup$
    – Tobias Fünke
    Apr 10, 2022 at 17:19

1 Answer 1

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Assuming Bosonic statistics, we write:

$$ Z=\sum_{n=m} e^{-\beta E_{n,n}}+ \frac12 \sum_{n\neq m} e^{-\beta E_{n,m}} = \frac12 \sum_{n=m} e^{-\beta E_{n,n}}+ \frac12 \sum_{n, m} e^{-\beta E_{n,m}} $$

The second sum is unconstrained now.

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