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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 $Z$?

Qmechanic
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Andrea
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1 Answers1

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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...

However I am stuck when trying to compute $Z$

...

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

$$ Z = \sum_{n, m}e^{-\beta E_{n, m}} = e^{-\beta \hbar \omega}\sum_{n, m}\Big(e^{-\beta\hbar \omega}\Big)^n\Big(e^{-\beta\hbar \omega}\Big)^m $$

The two sums are already unconstrained. There is no energy constraint to worry about because we are working at a fixed temperature not a fixed energy.

So, the sums are independent and therefore almost trivial. (N.b., In an attempt to preserve my own sanity I will choose units such that $\hbar=k_B=1$.) $$ Z = e^{-\beta \omega}\sum_{n=0}^1{(e^{-\beta\omega})}^n \sum_{m=0}^1{(e^{-\beta\omega})}^m $$ $$ =e^{-\beta \omega}(1+e^{-\beta\omega})(1+e^{-\beta\omega}) $$ $$ e^{-\beta \omega}\left({1+e^{-\beta\omega}}\right)^2\;. $$

And we have $$ \ln Z = -\beta\omega + 2\ln(1+e^{-\beta\omega}) = -\beta F\;. $$

hft
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