Fermi-Dirac distribution derivation?

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

$$<A>= Tr \rho A.$$

So I started from

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

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?

• This might help: physics.stackexchange.com/q/18576 Commented Mar 1, 2014 at 12:56
• You are using the canonical ensemble, which means that you can only get Fermi Dirac statistics after making some approximations. Your derivation will probably also be long and ugly. You can keep using the density matrix formalism but consider switching to Fock space and the grand canonical ensemble, where Fermi Dirac statistics are exactly derivable in about two lines. Commented Mar 1, 2014 at 14:29
• @m.mybo: I'm likely missing something (I'm not an expert at all in Stat Mech) but wouldn't a state with no fermions have $\epsilon_0=0$ and hence both forms of the partition function are equivalent? Commented Mar 1, 2014 at 15:42

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.