---
title: Quantum Statistics — Bose-Einstein and Fermi-Dirac
module: Quantum Statistics
moduleNumber: 7
lessonNumber: 1
order: 701
summary: >
  Quantum particles of the same kind are genuinely indistinguishable: no label
  survives an overlap of their wave functions. Counting states with that
  constraint replaces the Boltzmann distribution with two quantum laws — the
  Bose-Einstein distribution for integer-spin particles, which clump into shared
  states, and the Fermi-Dirac distribution for half-integer-spin particles,
  which exclude one another. Both reduce to Boltzmann in the dilute, hot limit,
  and a de Broglie criterion says exactly when.
topics: [Quantum Statistics]
sources:
  - book: Tipler & Llewellyn
    ref: "Ch. 8 Statistical Physics; §8-2 Quantum Statistics (Bose-Einstein and Fermi-Dirac Distributions, Comparison, Density of States)"
  - book: Tipler & Llewellyn
    ref: "Ch. 7 Atomic Physics; §7-6 Schrödinger Equation for Two or More Particles"
draft: false
---

The [classical distribution](/statistical-mechanics/foundations/classical-statistics-and-equipartition)
treated the particles of a gas as identical but distinguishable — the same as
one another, yet trackable through a collision, like billiard balls with numbers
painted on their sides. Quantum mechanics denies the numbers. The wave function
of a particle has finite extent, and when two identical particles pass within a
[de Broglie wavelength](/statistical-mechanics/quantum-statistics/the-classical-limit-and-quantum-concentration) of
each other their wave functions overlap, after which no measurement can say
which emerging particle was which. Identical quantum particles are
**indistinguishable**, and rebuilding the statistics on that fact changes the
occupation law and, with it, the low-temperature behavior of every gas.

$$
% caption: Two identical particles approach within a de Broglie wavelength and
% their wave functions overlap; the emerging particles cannot be matched to the
% incoming ones, so labeling them is meaningless.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\fill[acc!12, draw=acc, dashed] (0,0) circle (18pt);
\node[text=acc, font=\scriptsize] at (0,0) {overlap};
\draw[->, black, thick] (-2.6,0.7) -- (-0.7,0.2);
\draw[->, black, thick] (-2.6,-0.7) -- (-0.7,-0.2);
\node[anchor=east, font=\scriptsize] at (-2.6,0.7) {particle 1};
\node[anchor=east, font=\scriptsize] at (-2.6,-0.7) {particle 2};
\draw[->, black, thick] (0.7,0.2) -- (2.6,0.7);
\draw[->, black, thick] (0.7,-0.2) -- (2.6,-0.7);
\node[anchor=west, font=\scriptsize] at (2.6,0.7) {which one?};
\node[anchor=west, font=\scriptsize] at (2.6,-0.7) {which one?};
\end{tikzpicture}
$$

## Symmetry of the two-particle wave function

The origin of the two quantum statistics is the symmetry of the wave function
under exchange of two identical particles, developed for multielectron
atoms. For two
particles, one in state $n$ and one in state $m$, the two single-particle
product solutions

$$
\psi_{nm}(1,2) = \psi_n(1)\psi_m(2), \qquad
\psi_{nm}(2,1) = \psi_n(2)\psi_m(1)
$$

are distinct only if the particles can be told apart. For indistinguishable
particles the physical states are the symmetric and antisymmetric combinations,

$$
\psi_S = \tfrac{1}{\sqrt{2}}\big[\psi_n(1)\psi_m(2) + \psi_n(2)\psi_m(1)\big],
\qquad
\psi_A = \tfrac{1}{\sqrt{2}}\big[\psi_n(1)\psi_m(2) - \psi_n(2)\psi_m(1)\big].
$$

The antisymmetric $\psi_A$ describes particles that obey the exclusion
principle, the **fermions**; the symmetric $\psi_S$ describes particles that do
not, the **bosons**. Put both particles into the same state $n$. The symmetric
combination becomes

$$
\psi_{BE} = \tfrac{1}{\sqrt{2}}\big[\psi_n(1)\psi_n(2) + \psi_n(2)\psi_n(1)\big]
= \sqrt{2}\,\psi_n(1)\psi_n(2),
$$

whose probability density is $2\psi_n^\ast(1)\psi_n^\ast(2)\psi_n(1)\psi_n(2)$ —
_twice_ the classical value. The antisymmetric combination becomes

$$
\psi_{FD} = \tfrac{1}{\sqrt{2}}\big[\psi_n(1)\psi_n(2) - \psi_n(2)\psi_n(1)\big] = 0.
$$

Two statements follow, and they are the whole physical content of quantum
statistics:[^tl-symm]

- **Bosons cluster.** The presence of a boson in a state _raises_ the
  probability that another identical boson occupies the same state, as if
  identical bosons attract. Two in one state are twice as likely as for
  classical particles.
- **Fermions exclude.** The probability that two identical fermions occupy the
  same state is zero, as if identical fermions repel. This is the exclusion
  principle, read off the antisymmetry.

$$
% caption: Two identical particles sharing a single-particle state. Classical
% and boson particles may pile into one level (bosons with enhanced
% probability); two fermions never share a level and must split across two.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% distinguishable
\foreach \dx/\lab in {0/{distinguishable}} {
  \draw[black, thick] (\dx+0,0) -- (\dx+1.4,0);
  \draw[black, thick] (\dx+0,0.9) -- (\dx+1.4,0.9);
  \fill[black] (\dx+0.5,0.13) circle (2.6pt);
  \fill[black] (\dx+0.9,0.13) circle (2.6pt);
  \node[font=\scriptsize, anchor=north] at (\dx+0.7,-0.15) {\lab};
  \node[font=\scriptsize] at (\dx+0.7,1.35) {may share};
}
% bosons
\begin{scope}[xshift=3.2cm]
  \draw[black, thick] (0,0) -- (1.4,0);
  \draw[black, thick] (0,0.9) -- (1.4,0.9);
  \fill[acc] (0.5,0.13) circle (2.6pt);
  \fill[acc] (0.9,0.13) circle (2.6pt);
  \node[font=\scriptsize, anchor=north] at (0.7,-0.15) {bosons};
  \node[text=acc, font=\scriptsize] at (0.7,1.35) {clustering};
\end{scope}
% fermions
\begin{scope}[xshift=6.4cm]
  \draw[black, thick] (0,0) -- (1.4,0);
  \draw[black, thick] (0,0.9) -- (1.4,0.9);
  \fill[black] (0.7,0.13) circle (2.6pt);
  \fill[black] (0.7,1.03) circle (2.6pt);
  \node[font=\scriptsize, anchor=north] at (0.7,-0.15) {fermions};
  \node[text=black, font=\scriptsize] at (0.7,1.35) {one per state};
\end{scope}
\end{tikzpicture}
$$

## The three distributions

Bose (1924), then Einstein, built the distribution for integer-spin particles;
Fermi and Dirac built the one for half-integer spin. Both keep the Boltzmann
factor $e^{-E/kT}$ but change the denominator.

> **Definition (Bose-Einstein distribution).** For integer-spin particles the
> probability that a state of energy $E$ is occupied is
> $$
> f_{BE}(E) = \frac{1}{e^{\alpha}\,e^{E/kT} - 1},
> $$
> with $\alpha$ a system-dependent normalization constant.[^tl-be]

> **Definition (Fermi-Dirac distribution).** For half-integer-spin particles,
> $$
> f_{FD}(E) = \frac{1}{e^{\alpha}\,e^{E/kT} + 1}.
> $$

Writing the Boltzmann normalization $A = e^{-\alpha}$ puts the classical law in
the same form,

$$
f_{B}(E) = \frac{1}{e^{\alpha}\,e^{E/kT}},
$$

and the three differ only by the term added to the denominator: $-1$ for
bosons, $0$ for the classical gas, $+1$ for fermions. The $-1$ is the boson
clustering; the $+1$ is the fermion exclusion. Everything about the low-energy
behavior of a quantum gas turns on that one digit.

| | Boltzmann | Bose-Einstein | Fermi-Dirac |
| --- | --- | --- | --- |
| Applies to | distinguishable | integer spin | half-integer spin |
| Exchange symmetry | none | symmetric $\psi_S$ | antisymmetric $\psi_A$ |
| Denominator term | $0$ | $-1$ | $+1$ |
| Occupancy per state | unlimited | unlimited, enhanced | at most one |
| Example | dilute gas | $^4$He, photons | electrons, protons |

Setting $\alpha = 0$ and comparing over energies from $0$ to $5kT$ makes the
ordering visible. At every energy the boson curve lies above the classical one,
which lies above the fermion one:

$$
f_{FD}(E) < f_{B}(E) < f_{BE}(E).
$$

$$
% caption: The three occupation functions at alpha equal to zero. The
% Bose-Einstein curve diverges toward low energy, the Fermi-Dirac curve
% saturates at one-half, and all three merge above about five kT.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.6,0) node[right] {E in units of kT};
\draw[->, black] (0,0) -- (0,3.2) node[above] {f(E)};
\foreach \x/\lab in {1/1, 2/2, 3/3, 4/4, 5/5} \node[font=\scriptsize, anchor=north] at (\x,-0.03) {\lab};
% BE
\draw[acc, very thick, smooth] plot coordinates
  {(0.45,3.05)(0.6,2.4)(0.8,1.75)(1,1.35)(1.5,0.72)(2,0.42)(2.5,0.25)(3,0.155)(3.5,0.096)(4,0.06)(4.5,0.038)(5,0.024)(5.5,0.015)};
% B
\draw[black, thick, densely dashed, smooth] plot coordinates
  {(0,3.0)(0.5,1.82)(1,1.1)(1.5,0.67)(2,0.41)(2.5,0.25)(3,0.15)(3.5,0.091)(4,0.055)(4.5,0.033)(5,0.02)(5.5,0.012)};
% FD (scaled ×3 for visibility: f runs 0..0.5, plot 3×)
\draw[black!70, thick, densely dotted, smooth] plot coordinates
  {(0,1.5)(0.5,1.13)(1,0.81)(1.5,0.55)(2,0.36)(2.5,0.23)(3,0.142)(3.5,0.087)(4,0.053)(4.5,0.032)(5,0.02)(5.5,0.012)};
\node[text=acc, font=\scriptsize, anchor=west] at (0.9,1.9) {Bose-Einstein};
\node[text=black, font=\scriptsize, anchor=west] at (1.7,0.95) {Boltzmann};
\node[text=black!70, font=\scriptsize, anchor=west] at (2.4,0.55) {Fermi-Dirac};
\end{tikzpicture}
$$

The two quantum curves both approach the classical one when
$e^{\alpha}\,e^{E/kT} \gg 1$, that is when the occupancy is much less than one.
Then the $\pm 1$ is negligible against the large denominator and
$f_{BE} \approx f_{B} \approx f_{FD} \ll 1$. Quantum statistics matters only
where states are appreciably occupied — at low energy, low temperature, or high
density.

## When the classical distribution is valid

The condition "occupancy much less than one" has a physical reading in terms of
wave-function overlap. Identical particles become indistinguishable when their
de Broglie waves overlap, and they can be treated classically only when the
typical de Broglie wavelength is small compared with the average spacing
$\langle d \rangle$ between particles,

$$
\lambda \ll \langle d \rangle.
$$

Using the thermal momentum $p = \sqrt{2m\langle E \rangle} = \sqrt{3mkT}$ from
$\langle E \rangle = \tfrac{3}{2}kT$ gives the thermal de Broglie wavelength
$\lambda = h/\sqrt{3mkT}$, and with $\langle d \rangle = (V/N)^{1/3}$ the
criterion becomes

> **Definition (Classical validity criterion).** The Boltzmann distribution
> describes a gas of number density $N/V$ when
> $$
> \frac{N}{V}\,\frac{h^3}{(3mkT)^{3/2}} \ll 1,
> $$
> which requires low density, high temperature, or large particle mass.[^tl-crit]

The same quantity appears, up to a numerical factor, as $e^{-\alpha}$ in the
normalization below, so the criterion is equivalent to $e^{-\alpha} \ll 1$, i.e.
$e^{\alpha} \gg 1$.

**Helium in the atmosphere and in the liquid.** Helium atoms have spin $0$ and
are bosons. In the atmosphere the number density is
$N/V = 1.41 \times 10^{20}\ \mathrm{m^{-3}}$ at $T = 273\ \mathrm{K}$, and

$$
\frac{N}{V}\,\frac{h^3}{(3mkT)^{3/2}} = 6.3 \times 10^{-11} \ll 1,
$$

so atmospheric helium is classical. Liquid helium at its boiling point
$T = 4.2\ \mathrm{K}$ has $N/V = 1.87 \times 10^{28}\ \mathrm{m^{-3}}$, and the
same expression is $4.39$, not small. The Boltzmann distribution fails for
liquid helium; the Bose-Einstein distribution is required, which is the doorway
to [superfluidity and condensation](/statistical-mechanics/bose-systems/bose-einstein-condensation-and-the-fermion-gas).[^tl-hecrit]

$$
% caption: Two regimes for a gas. Hot and dilute: de Broglie wavelengths are
% small against the spacing, packets do not overlap, particles are effectively
% distinguishable. Cold and dense: wavelengths exceed the spacing, packets
% overlap, and quantum statistics governs.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% left: classical regime
\draw[black] (-0.2,-0.2) rectangle (3.2,2.4);
\node[font=\footnotesize, anchor=south] at (1.5,2.45) {hot, dilute};
\foreach \x/\y in {0.5/0.5, 2.3/0.4, 1.2/1.5, 2.6/1.8, 0.7/2.0} {\fill[black] (\x,\y) circle (1.6pt); \draw[black] (\x,\y) circle (3.2pt);}
\node[font=\scriptsize, anchor=north] at (1.5,-0.25) {packets separate};
% right: quantum regime
\begin{scope}[xshift=4.6cm]
\draw[black] (-0.2,-0.2) rectangle (3.2,2.4);
\node[font=\footnotesize, anchor=south] at (1.5,2.45) {cold, dense};
\foreach \x/\y in {0.7/0.7, 1.5/0.7, 2.3/0.7, 1.1/1.5, 1.9/1.5, 1.5/1.1} {\fill[acc!25, draw=acc] (\x,\y) circle (7pt);}
\foreach \x/\y in {0.7/0.7, 1.5/0.7, 2.3/0.7, 1.1/1.5, 1.9/1.5, 1.5/1.1} \fill[black] (\x,\y) circle (1.6pt);
\node[font=\scriptsize, anchor=north] at (1.5,-0.25) {packets overlap};
\end{scope}
\end{tikzpicture}
$$

## Density of states in a box

To turn an occupation probability $f(E)$ into an actual particle count
$n(E) = g(E)f(E)$, and to fix $\alpha$ by normalization, the density of states
$g(E)$ is needed. Confine $N$ particles in a cube of side $L$ and treat it as a
three-dimensional infinite square
well. The energy levels
are

$$
E = E_0(n_x^2 + n_y^2 + n_z^2), \qquad E_0 = \frac{h^2}{8mL^2},
$$

with positive integers $n_x, n_y, n_z$. Each triple is a lattice point in an
abstract "quantum-number space," and $E = E_0(n_x^2+n_y^2+n_z^2)$ is the
equation of a sphere of radius $R = (E/E_0)^{1/2}$. Because the quantum numbers
are all positive, the allowed points fill one octant.

$$
% caption: States of a particle in a cubic box are lattice points in
% quantum-number space; a fixed energy is a sphere of radius proportional to the
% square root of E, and the allowed positive quantum numbers fill one octant of
% it.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (4.2,0) node[right] {$n_x$};
\draw[->, black] (0,0) -- (0,4.0) node[above] {$n_y$};
\draw[->, black] (0,0) -- (-1.9,-1.5) node[below left] {$n_z$};
% quarter arc
\draw[acc, thick] (3.4,0) arc (0:90:3.4);
\fill[acc!10] (0,0) -- (3.4,0) arc (0:90:3.4) -- cycle;
% lattice dots inside the arc (n_x^2 + n_y^2 within R^2)
\foreach \pt in {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} \fill[acc] \pt circle (1.8pt);
\node[text=acc, anchor=west, font=\scriptsize] at (2.4,2.7) {larger E, larger radius};
\end{tikzpicture}
$$

Counting lattice points inside the octant and differentiating gives the density
of states,[^tl-dos]

$$
g(E) = \frac{4\pi\sqrt{2}\,m^{3/2}V}{h^3}\,E^{1/2},
$$

with $V = L^3$. The $E^{1/2}$ dependence is the same factor that shaped the
Maxwell energy distribution. For electrons, each spatial state holds two — spin
up and spin down — so the electron density of states is doubled,

$$
g_e(E) = \frac{4\pi(2m_e)^{3/2}V}{h^3}\,E^{1/2}.
$$

## Fixing the normalization constant

The constant $\alpha$ follows from requiring that the counts add up to $N$. For
a classical electron gas,

$$
N = \int_0^\infty g_e(E)\,f_B(E)\,\d E
  = e^{-\alpha}\,\frac{4\pi(2m_e)^{3/2}V}{h^3}\int_0^\infty E^{1/2}e^{-E/kT}\,\d E,
$$

and evaluating the standard integral gives

$$
e^{-\alpha} = \frac{N h^3}{2(2\pi m_e kT)^{3/2}\,V}.
$$

The right side is, up to the numerical factor, exactly the quantity in the
classical-validity criterion. So the two conditions coincide: the Boltzmann
distribution is valid precisely when $e^{-\alpha} \ll 1$, which is when the gas
is dilute and hot enough that the states it occupies are, on average, nearly
empty. Where $e^{-\alpha}$ approaches or exceeds one — cold, dense, or light
particles — the $\pm 1$ in the denominator dominates, and the gas becomes a
[Bose-Einstein condensate or a degenerate Fermi
gas](/statistical-mechanics/bose-systems/bose-einstein-condensation-and-the-fermion-gas) depending on the spin.

[^tl-symm]: **Tipler & Llewellyn**, _Modern Physics_, §8-2 (Comparison of the Distribution Functions), Eqs. 8-27 to 8-33 — symmetric and antisymmetric two-particle wave functions, the doubled same-state probability for bosons, and the vanishing one for fermions.
[^tl-be]: **Tipler & Llewellyn**, _Modern Physics_, §8-2 (Bose-Einstein and Fermi-Dirac Distributions), Eqs. 8-24 to 8-26 — the two quantum distributions and the Boltzmann law rewritten with $A = e^{-\alpha}$, differing only by the added $\mp 1$.
[^tl-crit]: **Tipler & Llewellyn**, _Modern Physics_, §8-2, Eqs. 8-34 to 8-36 — the overlap criterion $\lambda \ll \langle d \rangle$ and the resulting density-temperature condition for Boltzmann validity.
[^tl-hecrit]: **Tipler & Llewellyn**, _Modern Physics_, §8-2, Example 8-6 (Statistical Distribution of He in the Atmosphere) — atmospheric helium is classical, liquid helium is not.
[^tl-dos]: **Tipler & Llewellyn**, _Modern Physics_, §8-2 (Density of States), Eqs. 8-38 to 8-44 — the octant-of-a-sphere count in quantum-number space, the $E^{1/2}$ density of states, the spin-doubled electron form, and the normalization giving $e^{-\alpha}$.