Bose–Einstein statistics

Bose–Einstein statistics describes the equilibrium distribution of indistinguishable bosons, particles with integer spin that are not subject to the Pauli exclusion principle. Because any number of bosons may occupy a single quantum state, their statistical behavior differs profoundly from that of fermions. Photons, phonons, and ultracold bosonic atoms are the most important physical examples.

In equilibrium, the distribution of bosons in phase space depends only on temperature and (where allowed) chemical potential. For photons in thermal radiation fields, chemical equilibrium with matter forces the chemical potential to vanish, \( \mu = 0 \). For massive bosons, \( \mu \) varies with temperature and density but always satisfies \( \mu \le \epsilon_{0} \), the ground-state energy.

The fundamental description of a bosonic gas begins not with the occupation of discrete energy levels, but with the phase-space distribution function, which specifies how many particles occupy each differential volume of phase space.

The statistical mechanics of bosons gives the number of particles per phase-space cell of volume \( h^{3} \) as

$$ f = \frac{2}{h^{3}} \, \frac{1}{e^{h\nu/(kT)} - 1}. $$

Here the factor of 2 accounts for two photon polarization states. For massive bosons this factor is replaced by the appropriate spin degeneracy \( g \). This function expresses a key difference between classical and quantum statistics: the denominator can become small, allowing large occupation of low-energy states.

For photons, the variable \( \nu \) labels frequency, and this same distribution underlies the Planck spectrum of Blackbody radiation. For massive bosons, the energy is instead \( \epsilon = p^{2}/(2m) \), but the statistical form is identical:

$$ f = \frac{g}{h^{3}} \, \frac{1}{e^{(\epsilon-\mu)/(kT)} - 1}. $$

The form of \( f \) therefore unifies the physics of blackbody photons and ideal Bose gases.

The phase-space distribution function determines the mean number of bosons occupying each single-particle quantum state. Integrating \( f \) over a phase-space cell associated with state \( i \) leads to the Bose–Einstein occupation number

$$ \bar{n}_i = \frac{1}{e^{(\epsilon_i - \mu)/(kT)} - 1}. $$

This quantity plays the same role for massive bosons that the photon distribution function plays for radiation. Summing over all states gives

$$ N = \sum_i \bar{n}_i, \qquad E = \sum_i \epsilon_i \bar{n}_i. $$

In the continuum limit with density of states \( g(\epsilon) \),

$$ n = \frac{N}{V} = \int_0^\infty \frac{g(\epsilon)}{e^{(\epsilon-\mu)/(kT)} - 1} \, d\epsilon. $$

The crucial feature is the “\(-1\)” in the denominator; its absence in the classical Boltzmann limit suppresses large occupation at low energy.

When the exponent is large,

$$ e^{(\epsilon-\mu)/(kT)} \gg 1, $$

the Bose–Einstein distribution reduces to the Maxwell–Boltzmann form

$$ \bar{n}_i \approx e^{-(\epsilon_i-\mu)/(kT)}, $$

and the phase-space distribution becomes

$$ f \approx \frac{g}{h^{3}} \, e^{-(\epsilon-\mu)/(kT)}. $$

Quantum statistical effects appear when the thermal de Broglie wavelength

$$ \lambda_{\mathrm{th}} = \sqrt{\frac{2\pi\hbar^{2}}{m kT}} $$

becomes comparable to the mean particle spacing. The transition from classical to quantum behavior occurs when

$$ n \lambda_{\mathrm{th}}^{3} \gtrsim 1. $$

Below this threshold, the large-occupation low-energy behavior of bosons becomes unavoidable.

For a uniform ideal Bose gas, the chemical potential satisfies \( \mu \le \epsilon_{0} \). As the gas is cooled at fixed density, \( \mu \to \epsilon_{0} \), and the excited states are no longer capable of holding all particles. The excess particles accumulate in the ground state, producing macroscopic occupation,

$$ \bar{n}_{0} \gg 1. $$

This is Bose–Einstein condensation, a striking manifestation of the Bose–Einstein distribution function. The condensation phenomenon is therefore the direct consequence of the form of \( f \) and the associated occupation numbers \( \bar{n}_i \): large phase-space density implies large ground-state occupation.

In contrast, for photons the chemical potential is fixed to zero, and instead of condensing, photons redistribute themselves to form the blackbody spectrum.

To analyze how the Bose–Einstein distribution behaves, it is useful to define dimensionless variables

$$ x = \frac{\epsilon}{kT}, \qquad a = -\frac{\mu}{kT}, $$

so that

$$ \bar{n}(x; a) = \frac{1}{e^{x+a} - 1}. $$

Small \( a \) (that is, large \( \mu \)) enhances the low-energy peak of the distribution. The special case \( a = 0 \) corresponds exactly to the photon distribution function in blackbody radiation.

• The phase-space distribution \( f = g h^{-3} [\,e^{(\epsilon-\mu)/(kT)} - 1\,]^{-1} \) is the foundational expression from which all Bose–Einstein behavior follows.
• Large low-energy occupation arises because the denominator may become small as \( \mu \to \epsilon_{0} \).
• Classical statistics emerges only when \( e^{(\epsilon-\mu)/(kT)} \gg 1 \), suppressing large occupation.
• Quantum degeneracy appears when \( n \lambda_{\mathrm{th}}^{3} \gtrsim 1 \), forcing the use of Bose–Einstein statistics.
• Bose–Einstein condensation results directly from the inability of excited states to hold all particles as \( \mu \to \epsilon_0 \).
• For photons, the condition \( \mu = 0 \) transforms the Bose–Einstein distribution into the Planck spectrum of blackbody radiation.

• Derive the phase-space distribution \( f = g h^{-3} [e^{(\epsilon-\mu)/(kT)} - 1]^{-1} \) from the grand canonical ensemble.
• Show how the occupation number \( \bar{n}_i \) emerges from integrating \( f \) over a state’s phase-space cell.
• Demonstrate explicitly how the Maxwell–Boltzmann limit arises from the Bose–Einstein distribution.
• Using \( n \lambda_{\mathrm{th}}^{3} \sim 1 \), estimate the temperature at which a dilute Bose gas becomes quantum degenerate.
• Explain how varying \( a = -\mu/(kT) \) alters the shape of the occupation curves in the interactive figure.