1

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 $n_1$.

WRONG APPROACH:

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

stinglikeabeer
  • 380
  • 3
    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. – Roger V. Jul 28 '21 at 12:00
  • 2
    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 $\gamma$? – By Symmetry Jul 28 '21 at 14:21
  • @RogerVadim, thanks, there was a typo that I fixed and $\delta$ is not a good variable, see edit – stinglikeabeer Jul 28 '21 at 17:06
  • @BySymmetry, this is much clearer way to formulate the problem, which is now mentioned in the question. Thank you! – stinglikeabeer Jul 28 '21 at 17:07

1 Answers1

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*}

ytlu
  • 4,196
  • My last version strayed into canonical ensemble. I had corrected to be in grand canonical. – ytlu Jul 28 '21 at 23:38