2
$\begingroup$

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 $Z$?

$\endgroup$
3
  • 1
    $\begingroup$ 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.
    Commented Apr 10, 2022 at 14:09
  • $\begingroup$ Possibly related, check also the links to the papers therein. $\endgroup$ Commented Apr 10, 2022 at 14:33
  • $\begingroup$ You might be interested in this. $\endgroup$ Commented Apr 10, 2022 at 17:19

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.