quantum mechanics - Fermi-Dirac distribution derivation?

I am trying to derive the Fermi-Dirac statistics using density matrix formalism. I know that

$$= Tr \rho A.$$

So I started from

$$= Tr \rho n(\epsilon_i)=\frac {1}{Z} \sum e^{-\beta \epsilon_i n_i}n_i=\frac {1}{Z} e^{-\beta \epsilon_i}. $$

In the last passage I used the pauli principle ($n_i=0,1$). Now to derive the correct Fermi-Dirac distribution I have to use for $Z=1 +e^{-\beta \epsilon_i}$. Why I have not to use the general form of

$$Z=\prod_i (1 +e^{-\beta \epsilon_i})~?$$

Can anybody give me a good explanation?

Answer

The derivation of the Fermi-Dirac distribution using the density matrix formalism proceeds as follows:

The setup.

We assume that the single-particle hamiltonian has a discrete spectrum, so the single-particle energy eigenstates are labeled by an index $i$ which runs over some finite or countably infinite index set $I$. A basis for the Hilbert space of the system is the occupation number basis $$\begin{align} |\mathbf n\rangle = |n_0, n_1, \dots\rangle \end{align}$$ where $n_i$ denotes the number of particles occupying the single-particle energy eigenstate $i$. For a system of non-interacting identical fermions, the set $\mathscr N_-$ of admissible occupation sequences $\mathbf n$ consists of those sequences with each $n_i$ equal to either $0$ or $1$. Let $H$ be the hamiltonian for such a system, and let $N$ be the number operator, then we have $$\begin{align} H|\mathbf n\rangle = \left(\sum_{i\in I}n_i\epsilon_i\right)|\mathbf n\rangle, \qquad N|\mathbf n\rangle = \left(\sum_{i\in I} n_i\right) |\mathbf n\rangle \end{align}$$ where $\epsilon_i$ is the energy of eigenstate $i$. We can also define an observable $N_i$ which tells us the occupation number of the $i^\mathrm{th}$ single-particle energy state; $$\begin{align} N_i|\mathbf n\rangle = n_i|\mathbf n\rangle \end{align}$$

Note that we are attempting to determine the ensemble average occupation number of the $j^\mathrm{th}$ energy eigenstate. In the density matrix formalism, this is given by $$\begin{align} \langle n_j\rangle =\mathrm{tr}(\rho N_i) \end{align}$$ where $$\begin{align} \rho = \frac{e^{-\beta(H-\mu N)}}{Z}, \qquad Z = \mathrm {tr}\big(e^{-\beta(H-\mu N)}\big) \end{align}$$

The proof.

1. Show that $$\begin{align} Z = \sum_{\mathbf n\in \mathscr N_-}\prod_{i\in I}x_i^{n_i} \end{align}$$ where $x_j = e^{-\beta(\epsilon_j-\mu)}$, the sum is over admissible sequences $\mathbf n$ of occupation numbers of single-particle energy states, and the product is over indices $i$ labeling an orthonormal basis of single particle energy eigenstates.


2. Show that the ensemble average occupation number of the $j^\mathrm{th}$ state can be computed as follows: $$\begin{align} \langle n_j\rangle = x_j\frac{\partial}{\partial x_j}\ln Z \end{align}$$


3. Show that the product and the sum in the partition function can be "exchanged" to give $$\begin{align} Z = \prod_{i\in I}\sum_{n=0}^1 x_i^n \end{align}$$ where the product is now over single-particle energy eigenstates, and the sum is over admissible occupation numbers of a single-particle state.



4. Combine the results of steps 2 and 3 to show that $$\begin{align} \langle n_j\rangle = \frac{1}{e^{\beta(\epsilon_j-\mu)}+1} \end{align}$$ which is the desired result.