I have to calculate the grand canonical partition function of a system of hypothetical particles, wherein each single-particle quantum state can be occupied by up to 3 particles.

Obviously, this is a sort of joke, referring to fermions (with a maximum of 2 particles per state) and bosons (unlimited particles per state). It is assumed that these hypothetical particles do not interact with each other.

So I tried viewing each single-particle quantum state as a separate grand canonical ensemble, following the approach on https://en.wikipedia.org/wiki/Fermi%E2%80%93Dirac_statistics

At chemical potential $\mu$ and temperature $T$, where the energy of the state is $\epsilon$, I get: \begin{equation} \mathcal{Z} = \sum_{n=0}^{3}{\exp{\left(\frac{n(\mu-\epsilon)}{k_B T}\right)} = \frac{1-\exp{\left(4\frac{\mu-\epsilon}{k_B T}\right)}}{1-\exp{\left(\frac{\mu-\epsilon}{k_B T}\right)}}} \end{equation} where I used the finite geometric progression.

Now I also have to determine the average occupation number $\langle n_i \rangle$ for a state with energy $\epsilon_i$ at temperature $T=0$.

In general, we have \begin{equation} \langle n_i \rangle = k_B T \frac{\partial \ln{\mathcal{Z}}}{\partial \mu} \end{equation}

which yields me $\langle n_i \rangle =2-\frac{1}{1+\exp(x)}+\tanh(x)$ where I defined $x=\frac{\mu-\epsilon_i}{k_B T}$. (I used Wolfram Mathematica for simplifying the algebra.)

Clearly at $T=0$ this expression is ill-defined, but by taking the limit $T\rightarrow 0$ we see that $\langle n_i\rangle=0$ if $\epsilon_i>\mu$, $\langle n_i\rangle=3/2$ if $\epsilon_i=\mu$ and $\langle n_i\rangle=3$ if $\epsilon_i<\mu$, correct?

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    $\begingroup$ That final result sounds fine to me; you wouldn't have any particles if the 'cost' to having one is infinite. $\endgroup$
    – knzhou
    May 19, 2018 at 21:49
  • $\begingroup$ @Sylorinnis, If you write $\epsilon_i$, you assume at least two states for the system, so you have to differentiate not $\mathcal{Z}_i$, but the total partition function which is the product of $\mathcal{Z}_i$: $\mathcal{Z}=\prod \mathcal{Z}_i $ $\endgroup$ May 20, 2018 at 18:54
  • $\begingroup$ @AlekseyDruggist I want to calculate the avarage number of particles in the state with energy $\epsilon_i$, this state is a grand canonical ensemble on its own, so I can simply differentiate its own partition function, right? This is also the approach followed on en.wikipedia.org/wiki/… $\endgroup$
    – Sylorinnis
    May 20, 2018 at 21:28
  • $\begingroup$ @Sylorinnis, you are right, I meant total mean number of particles in the system $<N>$ $\endgroup$ May 20, 2018 at 21:48
  • $\begingroup$ If the particles are non-interacting, this seems as a bit of overkill - e.g., why use "finite geometric progression) to sum four terms? And using Wolfram for simple differentiations (instead of doing is a good way to get simple results expressed by not-so-transparent expressions (clearly the case here). $\endgroup$ Jan 7, 2022 at 16:12

1 Answer 1


Your formulas seem correct to me. But you really can not justify the $\mu <\epsilon$ condition in this case. In my opinion $\mu$ can take any value from $-\infty$ to $+\infty$ in this problem. At fixed temperature $T>0$ it follows from your formula that $\left<n\right> = 0$ at $\mu = -\infty$ and $\left<n\right> = 3$ at $\mu = +\infty$. These are correct limiting cases. At $T = 0$ we also have $\left<n\right> = 3$ if $\mu > \epsilon$ and $\left<n\right> = \frac{3}{2}$ if $\mu = \epsilon$.

The $\mu < \epsilon$ condition is a must only for an ideal Bose gas.

  • $\begingroup$ Thanks, I updated the post with these considerations, and I'm pretty sure I got it solved now. $\endgroup$
    – Sylorinnis
    May 20, 2018 at 21:37

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