Statistical Distribution Functions

Prerequisites: Statistics. Full path (2 pages)

• Intuitive
• Concrete
• Abstract

Intuitive

In particle physics distribution functions describe the energies of particles in a system depending on its temperature. Even in systems which appear to be at a single temperature, such as the particles in air, some will be moving faster or slower than others. Understanding the distribution function is important for evaluating microscopic and macroscopic properties of a thermodynamic system.

Concrete

Three of the most important examples of distribution function are the Maxwell-Boltzmann, Fermi-Dirac and Bose-Einstein distributions; the former is a classical distribution, appropriate for identical but distinguishable particles, and the latter two are quantum distributions, used for fermions and bosons respectively.

• https://physics.stackexchange.com/questions/210552/what-really-is-a-dirac-delta-function/210557#210557

Abstract

All three distributions follow from the grand canonical equilibrium state. For a system with Hamiltonian $H$ and particle number $N$ on Fock space, at inverse temperature $\beta = 1/(k_BT)$ and chemical potential $\mu$, the density matrix is

$$\rho = \frac{1}{Z}\, e^{-\beta(H-\mu N)}, \qquad Z = \mathrm{Tr}\, e^{-\beta(H-\mu N)} .$$

For non-interacting particles $H-\mu N = \sum_k(\epsilon_k-\mu)\,n_k$ with occupation number operators $n_k = a_k^\dagger a_k$, and $Z$ factorises over single-particle modes. With $x_k = \beta(\epsilon_k-\mu)$, the fermionic sum runs over $n_k\in\{0,1\}$ and the bosonic one over $n_k\in\{0,1,2,\ldots\}$:

$$Z_{\rm FD} = \prod_k\left(1+e^{-x_k}\right), \qquad Z_{\rm BE} = \prod_k\frac{1}{1-e^{-x_k}} ,$$

where the bosonic sum converges only if $\mu<\epsilon_k$ for all $k$. Then $\langle n_k\rangle = -\partial\ln Z/\partial x_k$ gives

$$\langle n_k\rangle = \frac{1}{e^{\beta(\epsilon_k-\mu)}\pm 1} ,$$

with $+$ for fermions and $-$ for bosons. Both reduce to the Maxwell-Boltzmann form $e^{-\beta(\epsilon_k-\mu)}$ when $e^{\beta(\epsilon_k-\mu)}\gg 1$, i.e. when all occupation numbers are small. When the bosonic $\mu$ approaches the lowest single-particle energy from below, the occupation of that mode can become macroscopic, which is Bose-Einstein condensation.

For infinite systems the trace does not exist and equilibrium states are characterised instead by the KMS condition. Let $\alpha_t(B) = e^{iKt}Be^{-iKt}$ with $K = H-\mu N$ (in units with $\hbar=1$). A state $\omega$ is a $(\beta,\alpha)$-KMS state if, for suitable observables, $\omega(A\,\alpha_{i\beta}(B)) = \omega(BA)$. For a finite system the Gibbs state satisfies this by cyclicity of the trace. The KMS condition alone determines the free occupation numbers. For a single mode $\alpha_t(a^\dagger) = e^{i(\epsilon-\mu)t}a^\dagger$, so taking $A = a$, $B = a^\dagger$ gives

$$e^{-\beta(\epsilon-\mu)}\,\omega(aa^\dagger) = \omega(a^\dagger a) .$$

With the canonical anticommutation relation $aa^\dagger = 1 - a^\dagger a$ this yields the Fermi-Dirac distribution, and with the commutation relation $aa^\dagger = 1 + a^\dagger a$ the Bose-Einstein distribution. The resulting states are quasi-free (Gaussian): all higher correlation functions follow from the two-point function by Wick's theorem.

In the kinetic description a distribution function is a density $f(x,p)$ on one-particle phase space, and the equilibrium distributions above are the stationary solutions of the Boltzmann equation, with the quantum statistics entering through the factors $(1\mp f)$ in the collision term. In the relativistic case $\epsilon$ is replaced by $u_\mu p^\mu$ for a fluid with four-velocity $u^\mu$ (up to the sign convention of the metric), which makes $f$ a Lorentz scalar.

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1. Calculus
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