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I am reading Section 2.4 of Equilibrium Statistical Physics by Plischke and Bergersen, where they briefly discuss a system of noninteracting fermions in the grand canonical ensemble, and I am having some trouble understanding how to derive their expression for the mean occupation number of each state.

They index each single particle state by a wave vector $\mathbf k$ and a spin index $\sigma$, but let us write $\gamma = (\mathbf k, \sigma)$ for short. The contribution of one such state to the grand partition function is

$$\sum_{n=0}^1 \exp\{-n \beta (E_\gamma - \mu)\} = 1 + \exp\{-\beta (E_\gamma - \mu)\} \tag 1$$

and hence

$$Z_G = \prod_\gamma [1 + \exp\{-\beta (E_\gamma - \mu)\}]. \tag 2$$

They then state, without derivation, that the mean total number of particles is given by

$$\langle N \rangle = \sum_\gamma \frac{1}{\exp\{\beta (E_\gamma - \mu)\} + 1}, \tag 3$$

and that the mean occupation number of state $\gamma$ is

$$\langle n_\gamma \rangle = \frac{1}{\exp\{\beta (E_\gamma - \mu)\} + 1}. \tag 4$$

I have managed to derive (3) as follows. First I note that I can expand the product above to obtain

$$Z_G = \sum_{N=0}^\infty \sum_{\langle \gamma_1, \dotsc, \gamma_N \rangle} \exp\left\{ -\beta \left( \sum_{i=1}^N E_{\gamma_i} - N \mu \right) \right\}, \tag 5$$

where I use the notation $\langle \gamma_1, \dotsc, \gamma_N \rangle$ to mean that the inner sum is taken over all combinations of $N$ state indices. (For $N=0$ we must define the inner sum to be 1.) It is then clear that each term in the sum is proportional to the probability of the corresponding configuration. Hence we can use it to compute expectation values in the usual way, and thus obtain

$$\langle N \rangle = \frac{1}{Z_G} \sum_{N=0}^\infty \sum_{\langle \gamma_1, \dotsc, \gamma_N \rangle} N \exp\left\{ -\beta \left( \sum_{i=1}^N E_{\gamma_i} - N \mu \right) \right\} = \beta^{-1} \frac{\partial \ln Z_G}{\partial \mu}. \tag 6$$

Then (3) follows easily from (6). I have tried to do something similar to derive (4), but with little progress. Could anyone show how to do it, or at least give some pointers?

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  • $\begingroup$ Derivation of eq.(3) from eq. (6) is correct. What about a derivative of $\ln Z_G$ wrt $\beta E_{\gamma_i}$ for the average occupation number? $\endgroup$ Sep 5, 2023 at 15:33
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    $\begingroup$ I think I have figured it out with the help of your comment @GiorgioP-DoomsdayClockIsAt-90. Thank you! I will write down my computations as an answer and credit you, if that sound okay? $\endgroup$
    – ummg
    Sep 5, 2023 at 15:49
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    $\begingroup$ It's ok. Just a curiosity: do Plischke and Bergersen suggest such a procedure somewhere in their book? $\endgroup$ Sep 5, 2023 at 21:59
  • $\begingroup$ Not really. In general it is my impression that their book lets the reader work out a lot of stuff on their own. I find it quite a difficult read, but the lecturer likes the book. The exercise problems are pretty hard as well. $\endgroup$
    – ummg
    Sep 5, 2023 at 22:07

1 Answer 1

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With the help of a comment from user GiorgioP-DoomsdayClockIsAt-90 I was able to figure it out. Consider a configuration $\langle \gamma_1, \dotsc, \gamma_N \rangle$ and let $\gamma$ be some particular single-particle state. The occupation number $n_\gamma$ is equal to 1 if $\gamma = \gamma_i$ for some $i \in \{1, \dotsc N\}$, and 0 otherwise. (Note that there is at most one $i$ such that $\gamma = \gamma_i$.) Hence we can write

$$n_\gamma |_{\langle \gamma_1, \dotsc, \gamma_N \rangle} = \sum_{i=1}^N \delta_{\gamma\gamma_i}$$

where $\delta_{ab}$ is the Kronecker delta. We then form the expectation $\langle n_\gamma \rangle$ much like how we formed (6) from (5) above, thus obtaining

$$ \begin{split} \langle n_\gamma \rangle &= \frac{1}{Z_G} \sum_{N=0}^\infty \sum_{\langle \gamma_1, \dotsc, \gamma_N \rangle} n_\gamma |_{\langle \gamma_1, \dotsc, \gamma_N \rangle} \exp\left\{ -\beta \left( \sum_{i=1}^N E_{\gamma_i} - N \mu \right) \right\} \\ &= \frac{1}{Z_G} \sum_{N=0}^\infty \sum_{\langle \gamma_1, \dotsc, \gamma_N \rangle} \exp\left\{ -\beta \left( \sum_{i=1}^N E_{\gamma_i} - N \mu \right) \right\} \sum_{j=1}^N \delta_{\gamma\gamma_j} \\ &= \frac{1}{Z_G} \sum_{N=0}^\infty \sum_{\langle \gamma_1, \dotsc, \gamma_N \rangle} \exp\left\{ -\beta \left( \sum_{i=1}^N E_{\gamma_i} - N \mu \right) \right\} \frac{\partial}{\partial E_\gamma} \sum_{j=1}^N E_{\gamma_j} \\ &= -\frac{1}{\beta} \frac{\partial \ln Z_G}{\partial E_\gamma} \\ &= \frac{1}{\exp\{\beta (E_\gamma - \mu)\} + 1}. \end{split} $$

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    $\begingroup$ The whole point of the grand canonical ensemble is that the number of particles in each energy state is independent of all the others. In the canonical ensemble, you fix the temperature rather than the energy, so that when calculating $\langle E_{\gamma}\rangle$, you do not need to know how much energy is tied up in the other modes. The grand canonical ensemble applies the same to particle number, fixing $\mu$ instead of $N$; $\langle n_{\gamma}\rangle$ is then determined by $\mu$, without any direct limit on the total $N$. This makes ${\cal Z}$ a product, with one factor for each $\gamma$. $\endgroup$
    – Buzz
    Sep 5, 2023 at 18:39

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