Answer to Question #89304 in Molecular Physics | Thermodynamics for Shivam Nishad

Question

Starting from the expression for thermodynamic probability for a Fermi system, obtain expression for its distribution function.

Answer 1

Consider a system of non-interacting particles. Energy in the states of a single particle are denoted as {E₁, E₂,...}, then a system with corresponding {n₁,n₂,...} numbers of particles in the states has energy

$$E = \sum_{i=1} n_i E_i$$

thus the probability for the system to be in state with numbers {n₁,n₂,...}:

$$P \propto \exp\bigg(- \frac{\sum_{i=1} n_i E_i}{kT}\bigg)= \exp\bigg(- \frac{n_1 E_1}{kT}\bigg) \cdot\exp\bigg(- \frac{\sum_{i=2} n_i E_i}{kT}\bigg)$$

So, the probability to have N particle in the arbitrary chosen first state regardless the state of the rest of the system:

$$P(n_1 = N) \propto \exp\bigg(-\frac{NE_1}{kT}\bigg) \sum_{\{n_i\}} \exp\bigg(-\frac{\sum_{i=2} n_i E_i}{kT}\bigg)$$

here sum over {n_i} means sum over all possible sequences of {n₂,n₃,...}. This sum is just a numerical factor and does not depend on N, so

$$P(n_1 = N) =\frac{\exp(-NE_1/kT)}{\exp(-0 E_1/kT) + \exp(-1 E_1/kT)}$$

where we use the information that for Fermi statistic N can be only 0 or 1. The average number of particles in this state is

$$\langle N \rangle = \sum_{N=0,1} N P(n = N) = \frac{0 + \exp(-E/kT)}{1+\exp(-E/kT)}$$

or

$$\langle N \rangle = \frac{1}{\exp\frac{E}{kT}+1}$$

If there is a chemical potential, the distribution turns to

$$\langle N \rangle = \frac{1}{\exp\frac{E-\mu}{kT}+1}$$