# Grand partition function substate average number of particles

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

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

I am interested in obtaining the average $$n_1$$.

WRONG APPROACH:

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

• 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. Jul 28 at 12:00
• 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 $\gamma$? Jul 28 at 14:21
• @RogerVadim, thanks, there was a typo that I fixed and $\delta$ is not a good variable, see edit Jul 28 at 17:06
• @BySymmetry, this is much clearer way to formulate the problem, which is now mentioned in the question. Thank you! Jul 28 at 17:07

## 1 Answer

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*}

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