Bose–Einstein Distribution

Purpose

For an independent bosonic mode of one-particle energy $\epsilon_i$, the grand-canonical equilibrium occupation is

$$b_i \equiv \langle\hat n_i\rangle = \frac{1}{ e^{\beta(\epsilon_i-\mu)}-1 }, \qquad \beta = \frac{1}{k_{\mathrm B}T}.$$

A bosonic number measurement returns

$$n_i=0,1,2,\ldots,$$

whereas $b_i$ is the ensemble mean of those outcomes. Unlike a fermionic mean occupation, it has no upper bound.

The canonical derivation and interpretation are at Bose–Einstein Statistics. This card collects calculation forms, convergence conditions, limiting regimes, and condensate bookkeeping.

At a glance

Quantity	Formula
Dimensionless energy	$x_i=\beta(\epsilon_i-\mu)>0$
Mean occupation	$b_i=(e^{x_i}-1)^{-1}$
One-mode partition factor	$\mathcal Z_i=(1-e^{-x_i})^{-1}$
Number probability	$P_i(n)=(1-e^{-x_i})e^{-nx_i}$
Number variance	$\operatorname{Var}(n_i)=b_i(1+b_i)$
Energy derivative	$\partial b/\partial\epsilon=-\beta b(1+b)$
Chemical-potential derivative	$\partial b/\partial\mu=\beta b(1+b)$
Dilute classical limit	$b(\epsilon)\simeq e^{-\beta(\epsilon-\mu)}$
Highly occupied limit	$b(\epsilon)\simeq k_{\mathrm B}T/(\epsilon-\mu)$
Finite-system convergence	$\mu<\epsilon_0$

Here $\epsilon_0=\min_i\epsilon_i$. If the energy zero is shifted by $C$, both $\epsilon_i$ and $\mu$ must be shifted by $C$; all occupations then stay unchanged.

One-mode geometric law

For the grand-energy contribution

$$\hat K_i = (\epsilon_i-\mu)\hat n_i,$$

define

$$q_i = e^{-\beta(\epsilon_i-\mu)}.$$

Convergence requires

$$0\leq q_i<1.$$

The one-mode partition factor is the geometric sum

$$\mathcal Z_i = \sum_{n=0}^{\infty} q_i^n = \frac{1}{1-q_i},$$

and the normalized counting law is

$$P_i(n) = (1-q_i)q_i^n, \qquad n=0,1,2,\ldots.$$

Its mean is

$$b_i = \frac{q_i}{1-q_i} = \frac{1}{e^{\beta(\epsilon_i-\mu)}-1}.$$

The formula gives a mean occupation per complete mode. If a level has $g_i$ independent modes at the same energy,

$$\langle N_i\rangle = g_i b_i.$$

Degeneracy multiplies the number of modes; it does not modify the single-mode denominator.

Chemical-potential constraint

Every bosonic geometric series must converge:

$$\epsilon_i-\mu>0 \quad \text{for all }i.$$

Therefore a finite ideal Bose system requires

$$\mu<\epsilon_0.$$

If one chooses $\epsilon_0=0$, this becomes

$$\mu<0, \qquad z=e^{\beta\mu}<1.$$

The sign of $\mu$ by itself is not invariant under an energy-zero shift. The meaningful statement is always $\mu<\epsilon_0$.

For particles whose number is conserved, such as trapped bosonic atoms, the number equation determines $\mu$. For equilibrium excitations whose number is not conserved, the associated chemical potential is ordinarily fixed to zero. Photons in blackbody equilibrium and harmonic-crystal phonons are the standard examples:

$$\mu_\gamma=0, \qquad \mu_{\mathrm{ph}}=0.$$

This does not imply that every bosonic species has zero chemical potential. An effective chemical potential for driven or approximately conserved excitations requires a timescale and conservation-law justification.

Equivalent and stable forms

With

$$x=\beta(\epsilon-\mu)>0,$$

the occupation can be written as

$$b(x) = \frac{1}{e^x-1} = \frac{e^{-x}}{1-e^{-x}} = \frac12 \left[ \coth\left(\frac{x}{2}\right)-1 \right].$$

For large $x$, the $e^{-x}$ form avoids overflow. For small $x$, direct subtraction in $e^x-1$ loses precision; use an exponential-minus-one routine or the series

$$b(x) = \frac1x - \frac12 + \frac{x}{12} - \frac{x^3}{720} + O(x^5).$$

The leading term is the classical-field or Rayleigh–Jeans approximation. The subleading $-1/2$ is often needed when estimating its error.

Derivatives and fluctuations

Differentiation gives

$$\frac{\partial b}{\partial\epsilon} = - \beta b(1+b), \qquad \frac{\partial b}{\partial\mu} = \beta b(1+b).$$

At fixed $\mu$,

$$\frac{\partial b}{\partial T} = \frac{\epsilon-\mu}{ k_{\mathrm B}T^2 } b(1+b).$$

For the geometric probability law,

$$\operatorname{Var}(n_i) = \langle n_i^2\rangle - \langle n_i\rangle^2 = b_i(1+b_i).$$

Thus

$$\operatorname{Var}(n_i) = b_i+b_i^2 > b_i$$

whenever $b_i>0$. The variance exceeds the Poisson value with the same mean, an equilibrium manifestation of bosonic bunching.

The derivative and variance combine into

$$\frac{\partial b_i}{\partial\mu} = \beta \operatorname{Var}(n_i).$$

For independent ideal modes in the grand-canonical ensemble,

$$\operatorname{Var}(N) = \sum_i b_i(1+b_i) = k_{\mathrm B}T \frac{\partial\langle N\rangle}{\partial\mu}.$$

Exactly fixed particle number correlates the modes, so this grand-canonical sum is not a canonical-ensemble fluctuation formula. Condensate-number fluctuations are especially ensemble sensitive.

Dilute and highly occupied limits

In the dilute regime,

$$x=\beta(\epsilon-\mu)\gg1,$$

and

$$b(x) = \frac{e^{-x}}{1-e^{-x}} = \sum_{\ell=1}^{\infty} e^{-\ell x}.$$

Therefore

$$b(x) = e^{-x} + e^{-2x} + \cdots.$$

The first term is the Maxwell–Boltzmann occupation. The positive higher terms are the bosonic statistical enhancement.

In the opposite regime,

$$0<x\ll1,$$

the leading occupation is

$$b(x) \simeq \frac1x = \frac{k_{\mathrm B}T}{\epsilon-\mu}.$$

This approximation is useful for highly occupied low-frequency modes, but it cannot be extended to arbitrarily high energy: doing so removes the quantum exponential cutoff and can create an ultraviolet divergence.

Thermodynamic sums

For independent modes,

$$\hat H = \sum_i \epsilon_i\hat n_i, \qquad \hat N = \sum_i \hat n_i,$$

the mean number and excitation energy are

$$\langle N\rangle = \sum_i b_i, \qquad U = \sum_i\epsilon_i b_i.$$

Any mode-independent zero-point contribution must be added separately when the physical Hamiltonian contains it. It does not alter the occupation probabilities.

The grand potential is

$$\Omega = k_{\mathrm B}T \sum_i \ln \left[ 1- e^{-\beta(\epsilon_i-\mu)} \right],$$

and the entropy is

$$S = k_{\mathrm B} \sum_i \left[ (1+b_i)\ln(1+b_i) - b_i\ln b_i \right].$$

These formulas assume independent thermal modes. Coherent, squeezed, and number states are bosonic states but do not have this geometric thermal counting law merely because their excitations are bosons.

Density-of-states form

Let $D(\epsilon)$ be the total one-particle density of states, including the physical volume and every degeneracy not already counted. For noncondensed modes,

$$N_{\mathrm{ex}} = \int_{\epsilon_0^+}^{\infty} D(\epsilon) \frac{ d\epsilon }{ e^{\beta(\epsilon-\mu)}-1 },$$

and

$$U_{\mathrm{ex}} = \int_{\epsilon_0^+}^{\infty} \epsilon D(\epsilon) \frac{ d\epsilon }{ e^{\beta(\epsilon-\mu)}-1 }.$$

The notation $\epsilon_0^+$ emphasizes that a discrete lowest mode must be kept separate when its occupation can be macroscopic. Replacing the whole spectrum by a continuum integral can erase the condensate mode.

At a bosonic boundary,

$$\mu\to\epsilon_0^-.$$

If near the spectral edge

$$D(\epsilon) \propto (\epsilon-\epsilon_0)^{s-1},$$

then the number integrand behaves as

$$D(\epsilon)b(\epsilon) \propto (\epsilon-\epsilon_0)^{s-2}.$$

The excited-state capacity is finite at the lower endpoint exactly when

$$s>1.$$

For a homogeneous system in $d$ dimensions with dispersion $\epsilon-\epsilon_0\propto k^r$, one has $s=d/r$. The ideal-gas continuum criterion is therefore

$$d>r.$$

This criterion is spectrum specific. Finite traps, lattices, interactions, and low-dimensional phase fluctuations require separate analysis.

Three-dimensional uniform gas

For a free nonrelativistic gas with $\epsilon_0=0$, define

$$\lambda_T = \sqrt{ \frac{ 2\pi\hbar^2 }{ mk_{\mathrm B}T } }, \qquad z=e^{\beta\mu}<1.$$

Using the Bose function

$$g_\nu(z) = \sum_{\ell=1}^{\infty} \frac{z^\ell}{\ell^\nu},$$

the excited-state density is

$$n_{\mathrm{ex}} = \frac{g}{\lambda_T^3} g_{3/2}(z),$$

where $g$ counts independent internal components sharing the same spectrum and chemical potential.

At the condensation boundary $z\to1^-$,

$$n_{\mathrm{ex}}^{\max} = \frac{g}{\lambda_T^3} \zeta\left(\frac32\right).$$

For fixed total density $n$, the ideal critical temperature is

$$T_c = \frac{2\pi\hbar^2}{ mk_{\mathrm B} } \left[ \frac{ n }{ g\zeta(3/2) } \right]^{2/3}.$$

Below this ideal thermodynamic-limit boundary,

$$\frac{N_0}{N} = 1- \left( \frac{T}{T_c} \right)^{3/2}.$$

These last two formulas belong specifically to a uniform, three-dimensional, quadratic, noninteracting gas. They are not universal Bose–Einstein distribution identities.

Separating a condensate mode

For a discrete ground mode,

$$N = N_0 + N_{\mathrm{ex}},$$

with

$$N_0 = \frac{1}{ e^{\beta(\epsilon_0-\mu)}-1 }.$$

For every finite grand-canonical system, $\mu$ remains strictly below $\epsilon_0$. In the fixed-density thermodynamic limit, one may have

$$\mu\to\epsilon_0^-, \qquad N_0=O(N),$$

while the excited-state integral saturates.

Do not substitute $\mu=\epsilon_0$ into the finite ground-mode formula; it would diverge. In a condensed-phase calculation, isolate $N_0$ and use the number constraint. The limits of infinite volume, fixed density, and $\mu\to\epsilon_0$ must be stated in the correct order.

Condensation is macroscopic occupation of a one-particle state, more generally an extensive eigenvalue of the one-body density matrix. It is not automatically equivalent to superfluidity, and the condensate orbital need not be a zero-momentum plane wave in a trap or interacting system.

Oscillator, photon, and phonon forms

For an equilibrium oscillator-like mode with excitation energy

$$\epsilon=\hbar\omega, \qquad \mu=0,$$

the occupation is

$$\bar n_\omega = \frac{1}{ e^{\beta\hbar\omega}-1 }.$$

Its thermal energy, excluding zero point, is

$$U_{\mathrm{th}} = \hbar\omega \bar n_\omega,$$

and its heat capacity is

$$C = k_{\mathrm B} \left( \beta\hbar\omega \right)^2 \frac{ e^{\beta\hbar\omega} }{ \left( e^{\beta\hbar\omega}-1 \right)^2 }.$$

For photons, multiplication by the electromagnetic density of states and two polarizations produces Planck’s radiation law. For phonons, sum over wavevector and branch labels. The mode occupation alone is not yet an energy or spectral density.

Worked numerical check

For a nonconserved oscillator mode with

$$\hbar\omega = 3k_{\mathrm B}T,$$

the occupation is

$$\bar n_\omega = \frac{1}{e^3-1} \simeq 0.0524.$$

The classical approximation would give $1/3$, which is far too large because $x=3$ is not a highly occupied regime. Conversely, at $x=0.1$,

$$b(0.1) \simeq 9.508,$$

while the leading Rayleigh–Jeans value is $10$. These limits provide simple numerical sign and scale checks.

Assumptions and scope

The distribution is exact for a grand-canonical ensemble of independent bosonic modes with convergent one-mode sums. It can also describe quasiparticles when a controlled quadratic theory identifies the relevant mode energies and conserved charges.

Additional care is required for:

• a macroscopically occupied condensate mode;
• finite fixed-$N$ systems and condensate fluctuations;
• interacting particles, whose bare-mode occupations need not be geometric;
• Bogoliubov quasiparticles, where coherence factors relate quasiparticle and particle occupations;
• driven, pumped, or lossy modes outside thermal equilibrium;
• low-dimensional systems and spectra with singular or discrete density-of-states structure.

Common mistakes

• Treating $b_i$ as a probability distribution over mode labels rather than a mean occupation of one mode.
• Setting $\mu=\epsilon_0$ inside a finite-system grand partition function.
• Saying simply that bosonic $\mu$ is negative without declaring the energy zero.
• Assigning $\mu=0$ to conserved massive bosons whose number equation determines it.
• Assuming every bosonic system condenses merely because one mode can hold many particles.
• Including the condensate ground mode in a continuum density-of-states integral.
• Double-counting spin, polarization, or branch degeneracy.
• Calling condensate fraction and superfluid fraction the same observable.
• Applying the Rayleigh–Jeans form into the ultraviolet.
• Applying independent grand-canonical mode fluctuations to an exactly fixed-$N$ condensate.
• Using the ideal occupation unchanged for strongly interacting or nonequilibrium bosons.

Exercises

1. Mean and variance of the geometric law

Starting from $P(n)=(1-q)q^n$, derive the mean and variance.

Solution

For $0\leq q<1$,

$$\sum_{n=0}^{\infty} q^n = \frac{1}{1-q}.$$

Applying $q\,d/dq$ gives

$$\langle n\rangle = (1-q) \sum_{n=0}^{\infty} nq^n = \frac{q}{1-q} \equiv b.$$

A second application gives

$$\langle n^2\rangle = \frac{q(1+q)}{(1-q)^2}.$$

Therefore

$$\operatorname{Var}(n) = \langle n^2\rangle-\langle n\rangle^2 = \frac{q}{(1-q)^2} = b(1+b).$$

2. Invert the occupation

Find $\mu$ for a mode of energy $\epsilon$ with mean occupation $b$. Evaluate the result for $b=9$.

Solution

From

$$b = \frac{1}{ e^{\beta(\epsilon-\mu)}-1 },$$

one obtains

$$e^{\beta(\epsilon-\mu)} = 1+\frac1b.$$

Hence

$$\mu = \epsilon - k_{\mathrm B}T \ln \left( 1+\frac1b \right).$$

For $b=9$,

$$\mu = \epsilon - k_{\mathrm B}T \ln\left(\frac{10}{9}\right),$$

which is strictly below $\epsilon$, as convergence requires.

3. Oscillator heat capacity

Differentiate the thermal energy $U_{\mathrm{th}}=\hbar\omega/(e^{\beta\hbar\omega}-1)$ and recover the single-mode heat capacity.

Solution

Let $x=\hbar\omega/(k_{\mathrm B}T)$. Then

$$U_{\mathrm{th}} = \hbar\omega \frac{1}{e^x-1}, \qquad \frac{dx}{dT} = - \frac{x}{T}.$$

Therefore

$$\begin{aligned} C &= \frac{dU_{\mathrm{th}}}{dT} \\ &= \hbar\omega \frac{x}{T} \frac{e^x}{(e^x-1)^2} \\ &= k_{\mathrm B}x^2 \frac{e^x}{(e^x-1)^2}. \end{aligned}$$

The zero-point energy, if included, is temperature independent and does not contribute.

4. Excited-state saturation criterion

Suppose $D(\epsilon)\propto(\epsilon-\epsilon_0)^{s-1}$ near the lowest energy. Determine when the excited-state number remains finite as $\mu\to\epsilon_0^-$.

Solution

Near the boundary,

$$b(\epsilon) \simeq \frac{ k_{\mathrm B}T }{ \epsilon-\epsilon_0 }.$$

Thus

$$D(\epsilon)b(\epsilon) \propto (\epsilon-\epsilon_0)^{s-2}.$$

The lower-endpoint integral

$$\int_0^\Delta x^{s-2}\,dx$$

converges exactly when

$$s-2>-1,$$

or

$$s>1.$$

Only then can the excited states saturate and force an extensive occupation outside the continuum contribution in the ideal thermodynamic limit.

Canonical explanations

• Bose–Einstein Statistics
• Occupation Numbers
• Ideal Bose Gas
• Bose–Einstein Condensation
• Chemical Potential
• Grand-Canonical Ensemble
• Bosonic Fock Space
• Phonons as Many-Body Excitations

Related lookup pages

• Bose Gas Formula Sheet
• Quantum Gas Formula Sheet
• Fermi–Dirac Distribution
• Bose Gas Model Card
• Statistical Mechanics Checklist

References

• R. K. Pathria and P. D. Beale, Statistical Mechanics, 4th ed., Academic Press, 2021.
• K. Huang, Statistical Mechanics, 2nd ed., Wiley, 1987.
• C. J. Pethick and H. Smith, Bose–Einstein Condensation in Dilute Gases, 2nd ed., Cambridge University Press, 2008.
• L. Pitaevskii and S. Stringari, Bose–Einstein Condensation and Superfluidity, Oxford University Press, 2016.