Quantum Statistics — Bose-Einstein and Fermi-Dirac

The classical distribution 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 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.

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

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

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

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

Two statements follow, and they are the whole physical content of quantum statistics:¹

• 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.

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.

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

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.

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:

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 $⟨d⟩$ between particles,

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

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

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.⁴

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

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.

Counting lattice points inside the octant and differentiating gives the density of states,⁵

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,

Fixing the normalization constant

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

and evaluating the standard integral gives

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 depending on the spin.