Fermi-Dirac statistics

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In statistical mechanics, Fermi-Dirac statistics determines the statistical distribution of fermions over the energy states for a system in thermal equilibrium. In other words, it is a probability of a given energy level to be occupied by a fermion. Fermions are particles which are indistinguishable and obey the Pauli exclusion principle, i.e., no more than one particle may occupy the same quantum state at the same time. Statistical thermodynamics is used to describe the behaviour of large numbers of particles. A collection of non-interacting fermions is called a Fermi gas.

F-D statistics was introduced in 1926 by Enrico Fermi and Paul Dirac and applied in 1927 by Arnold Sommerfeld to electrons in metals.

The expected number of particles in an energy state i for F-D statistics is

<math>

n_i = \frac{g_i}{e^{\left(\epsilon_i-\mu\right) / k T} + 1} </math>

where:

n_i is the number of particles in state i

ε_i is the energy of state i

g_i is the degeneracy of state i (the number of states with energy ε_i )

μ is the chemical potential. Sometimes the Fermi energy E_F is used instead, as a low-temperature approximation.

k is Boltzmann's constant

T is absolute temperature

1 Which distribution to use
2 Brief derivation
3 A more thorough derivation
4 See also

[edit]

Which distribution to use

Template:Physics/ParticleDistributions

[edit]

Brief derivation

Consider a two-level system, in which the excited state is an energy <math>\mathbf{\epsilon}</math> above the ground state. Because we are dealing with fermions, only one particle can occupy one energy level (quantum state). In other words, the energy contribution is either zero or <math>\mathbf{\epsilon}</math>. The partition function can be written

where <math>i</math> sums over all possible energy levels, <math>g_i</math> is the degeneracy of the <math>i</math>^th energy level, i.e. number of ways of getting/having that energy. As we have established before, there are only two possible energies: 0 and <math>\mathbf{\epsilon}</math>, both non-degenerate (<math>g_i = 1</math>). So for this system the partition function is

<math>Z = \sum_{n=0}^1 e^{-n\epsilon/k T} = 1 + e^{-\epsilon/k T}</math>.

In general, probability of being in an energy state <math>i</math> is given by

<math>P(E_i) = g_i \frac{e^{-E_i/k T}}{Z}</math>.

So the probability of the energy level to be occupied by a particle, or the probability of a particle having energy <math>\mathbf{\epsilon}</math> is

<math>\bar{n} = \frac{e^{-\epsilon/k T}}{Z} = \frac{e^{-\epsilon/k T}}{1 + e^{-\epsilon/k T}} = \frac{1}{e^{\epsilon/k T}+1}</math>

One can also use the standardized formula for <math>\bar{n}</math>

<math>\bar{n} = -{1\over Z\epsilon}\partial_{\beta}Z = -{-\epsilon e^{-\epsilon/k T}\over\epsilon\left(1 + e^{-\epsilon/k T}\right)} = \frac{1}{e^{\epsilon/k T}+1}</math>

For massive particles, all fermions are massive, zero energy is unachievable, we thus alter the formula to reflect that fact

<math>\bar{n} = \frac{1}{e^{\left(\epsilon-\mu\right)/k T}+1}</math>

where <math>\mathbf{\mu}</math> is the chemical potential of the system. At zero temperature, the chemical potential is exactly equal to the Fermi energy, <math>E_F</math>. For systems well below the Fermi temperature, it is often sufficient to use <math>\mathbf{\mu}</math> ≈ <math>E_F</math>. This formula is the Fermi-Dirac distribution.

[edit]

A more thorough derivation

Say there are two fermions placed in a system with four energy levels. There are six possible arrangements of such a system, which are shown in the diagram below.

   ε₁   ε₂   ε₃   ε₄
A  *    *
B  *         *
C  *              *
D       *    *
E       *         *
F            *    *

Each of these arrangements is called a microstate of the system. The ergodic hypothesis states that at thermal equilibrium, each of these microstates will be equally likely, subject to the constraints that there be a fixed total energy and a fixed number of particles.

Depending on the values of the energy for each state, it may be that total energy for some of these six combinations is the same as others. Indeed, if we assume that the energies are multiples of some fixed value ε, the energies of each of the microstates become:

A: 3ε

B: 4ε

C: 5ε

D: 5ε

E: 6ε

F: 7ε

So if we know that the system has an energy of 5ε, we can conclude that it will be equally likely that it is in state C or state D. Note that if the particles were distinguishable (the classical case), there would be twelve microstates altogether, rather than six.

Now suppose we have a number of energy levels, labelled by index i , each level having energy ε_i and containing a total of n_i particles. Suppose each level contains g_i distinct sublevels, all of which have the same energy, and which are distinguishable. For example, two particles may have different momenta, in which case they are distinguishable from each other, yet they can still have the same energy. The value of g_i associated with level i is called the "degeneracy" of that energy level. The Pauli exclusion principle states that only one fermion can occupy any such sublevel.

Let w(n, g) be the number of ways of distributing n particles among the g sublevels of an energy level. Its clear that there are g ways of putting one particle into a level with g sublevels, so that w(1, g) = g which we will write as:

<math>

w(1,g)=\frac{g!}{1!(g-1)!} </math>

We can distribute 2 particles in g sublevels by putting one in the first sublevel and then distributing the remaining n − 1 particles in the remaining g − 1 sublevels, or we could put one in the second sublevel and then distribute the remaining n − 1 particles in the remaining g − 2 sublevels, etc. so that w'(2, g) = w(1, g − 1) + w(1,g − 2) + ... + w(1, 1) or

<math>

w(2,g)=\sum_{k=1}^{g-1}w(1,g-k) = \sum_{k=1}^{g-1}\frac{(g-k)!}{1!(g-k-1)!}=\frac{g!}{2!(g-2)!} </math>

where we have used the following theorem involving binomial coefficients:

<math>

\sum_{k=n}^g \frac{k!}{n!(k-n)!}=\frac{(k+1)!}{(n+1)!(k-n)!} </math>

Continuing this process, we can see that w(n, g) is just a binomial coefficient

<math>

w(n,g)=\frac{g!}{n!(g-n)!} </math>

The number of ways that a set of occupation numbers n_i can be realized is the product of the ways that each individual energy level can be populated:

<math>

W = \prod_i w(n_i,g_i) = \prod_i \frac{g_i!}{n_i!(g_i-n_i)!} </math>

Following the same procedure used in deriving the Maxwell-Boltzmann distribution, we wish to find the set of n_i for which W is maximised, subject to the constraint that there be a fixed number of particles, and a fixed energy. We constrain our solution using Lagrange multipliers forming the function:

<math>

f(n_i)=\ln(W)+\alpha(N-\sum n_i)+\beta(E-\sum n_i \epsilon_i) </math>

Again, using Stirling's approximation for the factorials and taking the derivative with respect to n_i, and setting the result to zero and solving for n_i yields the Fermi-Dirac population numbers:

<math>

n_i = \frac{g_i}{e^{\alpha+\beta \epsilon_i}+1} </math>

It can be shown thermodynamically that β = 1/kT where k is Boltzmann's constant and T is the temperature, and that α = -μ/kT where μ is the chemical potential, so that finally:

<math>

n_i = \frac{g_i}{e^{(\epsilon_i-\mu)/kT}+1} </math>

Note that the above formula is sometimes written:

<math>

n_i = \frac{g_i}{e^{\epsilon_i/kT}/z+1} </math>

where <math>z=exp(\mu/kT)</math> is the absolute activity.

[edit]

Fermi-Dirac statistics

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Contents

Which distribution to use

Brief derivation

A more thorough derivation

See also

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