Bose-Einstein Distribution Calculator

The Bose-Einstein distribution tells you, on average, how many bosons sit in a quantum state of energy E when the system is at temperature T and chemical potential μ.

n̄(E) = 1 / (exp((E - μ) / k_B T) - 1)

Bosons are particles with integer spin: photons, phonons (lattice vibrations), helium-4 nuclei, the Higgs. Unlike fermions, there is no Pauli rule, so any number of bosons can pile into the same state. That minus one in the denominator, where Fermi-Dirac has a plus one, is the whole reason lasers and Bose-Einstein condensates exist.

Three regimes the formula lives in:

When (E - μ) ≫ k_B T, the exponential is huge and n̄ ≈ exp(-(E - μ) / k_B T). That is Maxwell-Boltzmann, the classical limit. High-energy states are empty, and quantum statistics don’t matter.

When (E - μ) ≈ k_B T, you’re in the genuinely quantum regime. Bose-Einstein gives noticeably more occupation than Maxwell-Boltzmann, about 58% more at E - μ = k_B T exactly. This regime is where blackbody radiation lives, and it is why Planck’s law differs from the Rayleigh-Jeans classical prediction.

When (E - μ) → 0⁺, the denominator collapses toward zero and n̄ diverges. That is the signature of Bose-Einstein condensation. As you cool a dilute boson gas, μ climbs toward the ground-state energy from below, and at the critical temperature a macroscopic number of atoms suddenly pour into the single lowest state. Cornell, Wieman, and Ketterle won the 2001 Nobel Prize for observing this directly in rubidium-87 cooled to 170 nanokelvin.

For photons, μ = 0. Photons can be created and destroyed freely, which forces zero chemical potential. Set μ = 0 in this calculator and the result is exactly the number of photons per mode in a thermal cavity at temperature T, which is the Planck radiation law in disguise.

Worked example: sunlight

Green light at 2.5 eV, at the Sun’s surface temperature of 5778 K, with μ = 0.

k_B T = 8.617 × 10⁻⁵ × 5778 = 0.4979 eV
x = (E - μ) / k_B T = 2.5 / 0.4979 = 5.021
n̄ = 1 / (e^5.021 - 1) = 0.006642

Fewer than seven photons per thousand modes. That is what “the Sun is not very bright in the green if you count per mode” means, and it is also why x above 5 is a good working definition of the classical regime: the quantum enhancement here is only 0.7%, so Maxwell-Boltzmann would have given you essentially the same answer.

Those three numbers are what the input boxes suggest, so type them in and you get exactly this back.

Units to know. k_B is the Boltzmann constant. In SI, k_B = 1.381 × 10⁻²³ J/K. In electron-volts (much friendlier for atomic-scale problems), k_B = 8.617 × 10⁻⁵ eV/K. This calculator does the unit conversion for you. Pick eV or Joules for the energy and just enter the number. T is always in Kelvin (room temperature is 293 K, the Sun’s surface is 5778 K, the cosmic microwave background is 2.725 K).