Let's say I have a grand partition function with two states $\epsilon_1$ and $\epsilon_2$:

\begin{equation} Z = \sum\limits_i \exp\left[-\beta\left(\epsilon_1 n_1+\epsilon_2 n_2 - \mu n_i\right)\right], \end{equation} where $n_1+n_2 = n_i$. Better formulation is in the comment of @By Symmetry.

I am interested in obtaining the average $n_1$.


I can rewrite my partition function as: \begin{equation} Z = \sum\limits_i \exp\left[-\beta\left(\epsilon_1 n_{1i}+\epsilon_2 \delta^{-1} n_{1i} - \mu (1+\delta^{-1}) n_{1i}\right)\right], \end{equation} where I used $\delta = n_1/n_2$.

If I take partial derivative over $\mu(1+\delta^{-1})$ this does NOT give me the average number of particles in the state $\epsilon_1$, because there is $\delta^{-1}$ in front of the $\epsilon_2$ term. Is there a way around it?

EDIT: $\delta$ is a bad variable as @Roger Vadim mentioned

  • 3
    $\begingroup$ I do not understand the first equation - what is $n_i$ here? Then it seems like you are summing over different values of $n_1$ and $n_2$, but treating their ratio, $\delta$, as if it were a constant. $\endgroup$ Jul 28, 2021 at 12:00
  • 2
    $\begingroup$ Am I correct in thinking the your first equation should be understood as $Z = \sum_\gamma \exp\left[-\beta(\epsilon_1 n_{1\gamma} + \epsilon_2 n_{2\gamma} - \mu (n_{1\gamma}+n_{2\gamma}))\right]$ where $n_{i\gamma}$ is the number of particles in energy state $i\in \{1,2\}$ in global system state $\gamma$? $\endgroup$ Jul 28, 2021 at 14:21
  • $\begingroup$ @RogerVadim, thanks, there was a typo that I fixed and $\delta$ is not a good variable, see edit $\endgroup$ Jul 28, 2021 at 17:06
  • $\begingroup$ @BySymmetry, this is much clearer way to formulate the problem, which is now mentioned in the question. Thank you! $\endgroup$ Jul 28, 2021 at 17:07

1 Answer 1


Add the multiplicity to your equation, given $n = n_1 + n_2$:

\begin{align*} Z &= \sum_{n=0}\,\,\,\sum_{n1=0, n2=n-n1}^n \frac{n!}{n_1! n_2!} e^{-\beta\left\{\epsilon_1 n_1+\epsilon_2 n_2 - \mu (n_1+n_2)\right\}}.\\ &= \sum_n \,\sum_{n_1=0}^n \frac{n!}{n_1! n_2!} e^{-\left\{\beta\epsilon_1 \,n_1+\beta\epsilon_2\, n_2 - \beta\mu (n_1+n_2)\right\}}\\ &= \sum_n\,\sum_{n1, n2} \frac{n!}{n_1! n_2!} e^{-\left\{\beta (\epsilon_1 - \mu) \,n_1+\beta (\epsilon_2-\mu)\, n_2\right\}}\\ &= \sum_{n=0}^{\infty}\left\{ e^{-\beta (\epsilon_1 - \mu) } + e^{-\beta (\epsilon_2 - \mu) }\right\}^n = \sum_n z_1^n = \frac{1}{1-z_1} \end{align*} where $z_1 = e^{-\beta (\epsilon_1 - \mu) } + e^{-\beta (\epsilon_2 - \mu) }$, and $\ln Z = - \ln \left(1 - z_1 \right).$

The average number $\bar n_1$ \begin{align*} \bar n_1 &= \frac{1}{Z} \sum_n \, \sum_{n_1=0}^n n_1\,\frac{n!}{n_1! n_2!} e^{-\beta\left\{\epsilon_1 n_1+\epsilon_2 n_2 - \mu (n_1+n_2)\right\}}\\ &= \frac{-1}{Z}\frac{\partial}{\partial (\beta \epsilon_1)}\sum_n \, \sum_{n_1=0}^n \frac{n!}{n_1! n_2!} e^{-\beta\left\{\epsilon_1 n_1+\epsilon_2 n_2 - \mu (n_1+n_2)\right\}} \\ &= \frac{-1}{Z}\frac{\partial Z}{\partial (\beta \epsilon_1)} = -\frac{\partial \ln Z}{\partial (\beta \epsilon_1)}\\ &= + \frac{\partial}{\partial (\beta \epsilon_1)} \ln \left(1 - z_1 \right)\\ &= \frac{\partial}{\partial (\beta \epsilon_1)}\ln \left\{1- e^{-\beta (\epsilon_1 - \mu) } - e^{-\beta (\epsilon_2 - \mu) }\right\}\\ &= \frac{e^{-\beta (\epsilon_1 - \mu) } }{1 -e^{-\beta (\epsilon_1 - \mu) } - e^{-\beta (\epsilon_2 - \mu) } }\\ \end{align*}

  • $\begingroup$ My last version strayed into canonical ensemble. I had corrected to be in grand canonical. $\endgroup$
    – ytlu
    Jul 28, 2021 at 23:38

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