Fermi Dirac distribution derivation Does anybody understand how my lecturer is normalising the probability distribution at the end to achieve the Fermi Dirac distribution? I don’t understand how he gets 0x1 or the denominator at all.

 A: As far, as I understand, your lecturer considers an isolated system build from a reservoir and an electron state. Then the number of states of the isolated system, corresponding to the electron state being empty, is equal to $\Omega_0$. And the number of states of the isolated system, corresponding to the electron state being occupied, is equal to $\Omega$. According to the microcanonical distribution probability of electron state being occupied is
$$
p = \frac{\Omega}{\Omega+\Omega_0} = \frac1{\frac{\Omega_0}{\Omega}+1} = \frac1{\exp\left[(\varepsilon-\mu)/k_B T\right]+1}
$$
Upd. It can also be understood in the following way. Non-normalized statistical probabilities, as they are understood after Boltzmann, are
$$
P_0 = \Omega_0, \quad P = \Omega.
$$
Hence, true normalized probabilities are
$$
p = \frac{P}{P+P_0}=\frac{\Omega}{\Omega+\Omega_0},\qquad p_0 = \frac{P_0}{P+P_0}=\frac{\Omega_0}{\Omega+\Omega_0}.
$$
A: They seem to just have the sum over N=0 and N=1 in the nominator.
EDIT: So we need to normalize probability $p(\epsilon)=A\exp\left(-\frac{(\varepsilon-\mu)N}{k_B T}\right)$. The probability of $N$ being either 0 or 1 is unity, so $0\cdot A\exp\left(-\frac{(\varepsilon-\mu)\cdot 0}{k_B T}\right)+1\cdot A\exp\left(-\frac{(\varepsilon-\mu)\cdot 1}{k_B T}\right)=1$, so you can find the normalization constant $A$.
