Bose-Einstein and Fermi-Dirac Occupation Numbers

Derivation

Bose-Einstein and Fermi-Dirac Occupation Numbers

D-240 Home PU-302 Threads matter · chance · symmetry Depends on Grand-Canonical Ensemble and the Grand Potential

Statement

For a system of indistinguishable non-interacting particles in thermal and diffusive contact with a reservoir at temperature \(T\) (with \(\beta = 1/k_{\mathrm B}T\)) and chemical potential \(\mu\), the mean number of particles occupying a single-particle energy level of energy \(\varepsilon\) is \(\displaystyle \langle n \rangle = \frac{1}{e^{\beta(\varepsilon-\mu)} \mp 1}\), where the upper sign (\(-1\)) holds for bosons (Bose-Einstein statistics) and the lower sign (\(+1\)) for fermions (Fermi-Dirac statistics). Both reduce to the Maxwell-Boltzmann form \(\langle n\rangle \approx e^{-\beta(\varepsilon-\mu)}\) when \(\langle n\rangle \ll 1\).

Why it matters

These two occupation numbers are the load-bearing result of quantum statistical mechanics. Every equilibrium property of an ideal quantum gas — the pressure of a white dwarf's electrons, the blackbody spectrum, the heat capacity of metals, the onset of Bose-Einstein condensation — is obtained by summing the appropriate \(\langle n\rangle\) over the single-particle spectrum. The derivation shows that the entire difference between the two great families of particles collapses to a single sign, fixed by the exchange symmetry of the many-body wavefunction.

The mode-by-mode grand-canonical argument is also a template: because the modes are statistically independent once we allow particle number to fluctuate, an intractable many-body counting problem factorises into a product of trivial single-mode partition functions. This is why the grand-canonical ensemble, not the canonical one, is the natural language of quantum gases.

Assumptions

Non-interacting particles.If particles interact, the many-body energy is not a sum \(\sum_k n_k \varepsilon_k\) over fixed single-particle levels; the modes no longer decouple and the grand partition function does not factorise, so no closed-form \(\langle n\rangle\) exists without further approximation. A fixed, discrete single-particle spectrum \(\{\varepsilon_k\}\).Without well-defined stationary single-particle states (e.g. in a strongly disordered or time-dependent potential) the labelling by mode occupation \(n_k\) is ill-defined and the ensemble cannot be organised mode-by-mode. Indistinguishable particles with definite exchange symmetry.If particles were distinguishable, one would count microstates by which particle sits where and recover Maxwell-Boltzmann exactly; the \(\mp 1\) arises solely from the boson/fermion restriction on allowed occupations. The reservoir fixes both \(T\) and \(\mu\), and equilibrium exists.If \(\mu\) is not well-defined (open, driven, or non-equilibrium steady states) the grand-canonical weight \(e^{-\beta(\varepsilon-\mu)n}\) has no meaning and the result does not apply. The grand partition function converges.For bosons, convergence of the geometric series requires \(\mu < \varepsilon_k\) for every mode, i.e. \(\mu\) below the ground-state energy; if violated the single-mode sum diverges and the formula breaks (this is the edge of Bose-Einstein condensation).

Derivation

\[ \hat H = \sum_k \varepsilon_k \, \hat n_k, \qquad \hat N = \sum_k \hat n_k \]

For non-interacting particles the total energy and total number are additive over occupation-number eigenstates \(|\{n_k\}\rangle\) of the single-particle modes; this is the definition of an ideal gas in second-quantised form. A

\[ \mathcal Z = \sum_{\{n_k\}} e^{-\beta\left(E_{\{n_k\}} - \mu N_{\{n_k\}}\right)} = \sum_{\{n_k\}} \exp\!\left[-\beta \sum_k (\varepsilon_k - \mu)\, n_k\right] \]

The grand partition function sums the Gibbs weight \(e^{-\beta(E-\mu N)}\) over all microstates; a microstate of an ideal gas is fully specified by the occupation list \(\{n_k\}\). (Prior result: grand-canonical ensemble and fugacity.) A

\[ \mathcal Z = \sum_{n_1}\sum_{n_2}\cdots \prod_k e^{-\beta(\varepsilon_k-\mu)n_k} = \prod_k \left( \sum_{n_k} e^{-\beta(\varepsilon_k-\mu)n_k} \right) \equiv \prod_k \mathcal Z_k \]

Because the exponent is a sum over independent modes and each \(n_k\) is summed independently, the multiple sum of a product factorises into a product of sums. This factorisation is the whole payoff of working grand-canonically. B

\[ \text{Bosons: } n_k \in \{0,1,2,\dots\}, \qquad \text{Fermions: } n_k \in \{0,1\} \]

The Pauli exclusion principle for fermions caps each mode at one particle; symmetric bosonic states admit any non-negative occupation. This single input distinguishes the two statistics. A

\[ \mathcal Z_k^{\text{FD}} = \sum_{n=0}^{1} e^{-\beta(\varepsilon_k-\mu)n} = 1 + e^{-\beta(\varepsilon_k-\mu)} \]

The fermionic single-mode sum has just two terms. A

\[ \mathcal Z_k^{\text{BE}} = \sum_{n=0}^{\infty} \left(e^{-\beta(\varepsilon_k-\mu)}\right)^{n} = \frac{1}{1 - e^{-\beta(\varepsilon_k-\mu)}} \]

The bosonic single-mode sum is geometric with ratio \(e^{-\beta(\varepsilon_k-\mu)}\); it converges iff that ratio is \(<1\), i.e. \(\mu < \varepsilon_k\). B

\[ \langle n_k \rangle = \frac{1}{\mathcal Z_k}\sum_{n} n\, e^{-\beta(\varepsilon_k-\mu)n} = -\frac{1}{\beta}\,\frac{\partial \ln \mathcal Z_k}{\partial \varepsilon_k} \]

The mean occupation is the ensemble average of \(n_k\); differentiating \(\ln\mathcal Z_k\) with respect to \(\varepsilon_k\) brings down a factor \(-\beta n\), reproducing the average. This is the standard "differentiate the log of the partition function" trick. B

\[ \langle n_k \rangle^{\text{FD}} = -\frac{1}{\beta}\frac{\partial}{\partial \varepsilon_k}\ln\!\left(1 + e^{-\beta(\varepsilon_k-\mu)}\right) = \frac{e^{-\beta(\varepsilon_k-\mu)}}{1 + e^{-\beta(\varepsilon_k-\mu)}} \]

Direct differentiation of the two-term fermionic log; the chain rule supplies \(-\beta e^{-\beta(\varepsilon_k-\mu)}\) in the numerator, cancelling the \(-1/\beta\). A

\[ \langle n_k \rangle^{\text{FD}} = \frac{1}{e^{\beta(\varepsilon_k-\mu)} + 1} \]

Multiply numerator and denominator by \(e^{+\beta(\varepsilon_k-\mu)}\) to clear the negative exponent; purely algebraic rearrangement. A

\[ \langle n_k \rangle^{\text{BE}} = -\frac{1}{\beta}\frac{\partial}{\partial \varepsilon_k}\Big[-\ln\!\left(1 - e^{-\beta(\varepsilon_k-\mu)}\right)\Big] = \frac{e^{-\beta(\varepsilon_k-\mu)}}{1 - e^{-\beta(\varepsilon_k-\mu)}} \]

Same differentiation applied to the bosonic log; note the internal minus sign of \(\ln(1-x)\) flips to give a positive result. B

\[ \langle n_k \rangle^{\text{BE}} = \frac{1}{e^{\beta(\varepsilon_k-\mu)} - 1} \]

Again multiply top and bottom by \(e^{+\beta(\varepsilon_k-\mu)}\). The two results differ only in the sign of the \(1\), which we now unify. A

\[ \boxed{\;\langle n_k \rangle = \frac{1}{e^{\beta(\varepsilon_k-\mu)} \mp 1}\;}\qquad \begin{cases} - & \text{bosons}\\ + & \text{fermions}\end{cases} \]

Combining steps 9 and 11: the entire content of quantum statistics for ideal gases is the single sign in the denominator, traceable to the allowed range of \(n_k\) fixed by exchange symmetry in step 4. Dropping the mode label \(k\) gives the level-by-level statement. A

Result

\[ \langle n \rangle = \frac{1}{e^{\beta(\varepsilon-\mu)} \mp 1}, \qquad \beta = \frac{1}{k_{\mathrm B}T} \]

Reading. A single-particle level at energy \(\varepsilon\) is, on average, occupied by \(\langle n\rangle\) particles. When \(\varepsilon \gg \mu\) the exponential dominates and both statistics give a small, Boltzmann-like occupation. When \(\varepsilon \to \mu\) the denominators diverge in opposite ways: bosons pile up without bound (\(\langle n\rangle \to \infty\)), while fermions saturate at \(\langle n\rangle^{\text{FD}} \to 1/2\) exactly at \(\varepsilon=\mu\) and approach \(1\) below it — the Pauli ceiling. The chemical potential \(\mu\) is the energy at which one more particle costs no net free energy; it is fixed by requiring \(\sum_k \langle n_k\rangle = N\).

Units check. The argument \(\beta(\varepsilon-\mu)\) is (energy)/(energy) = dimensionless, as an exponent must be. The exponential and the \(\mp 1\) are therefore dimensionless, so \(\langle n\rangle\) is a pure number — a count of particles — as required.

Limiting cases

Non-degenerate / classical limit (\(e^{\beta(\varepsilon-\mu)} \gg 1\), i.e. \(\langle n\rangle \ll 1\)): the \(\mp 1\) is negligible and both reduce to \(\langle n\rangle \approx e^{-\beta(\varepsilon-\mu)} = e^{\beta\mu}e^{-\beta\varepsilon}\), the Maxwell-Boltzmann distribution with fugacity \(z=e^{\beta\mu}\).
Fermion, \(T\to 0\): \(\langle n\rangle^{\text{FD}} \to 1\) for \(\varepsilon<\mu\) and \(0\) for \(\varepsilon>\mu\) — a sharp step at the Fermi energy \(\varepsilon_F = \mu(T{=}0)\).
Boson with \(\varepsilon=\mu\): occupation diverges; a macroscopic fraction condenses into the ground mode (Bose-Einstein condensation).
Photon gas (\(\mu=0\), massless bosons): \(\langle n\rangle = 1/(e^{\beta\varepsilon}-1)\), giving the Planck blackbody law when multiplied by the photon density of states.
High energy tail (\(\varepsilon-\mu \gg k_{\mathrm B}T\)): both distributions decay as \(e^{-\beta(\varepsilon-\mu)}\); quantum and classical predictions become indistinguishable.

Breaks when

Interactions matter. For a real gas near condensation or in a dense electron liquid, the energy is not \(\sum_k n_k\varepsilon_k\); mode factorisation (step 3) fails and \(\langle n\rangle\) acquires interaction corrections (e.g. Landau Fermi-liquid quasiparticle shifts).
Bosons with \(\mu \ge \varepsilon_0\). The geometric series in step 6 diverges once \(\mu\) reaches the ground-state energy; the formula gives a negative or infinite occupation and must be replaced by explicit treatment of the condensate plus the excited-state continuum.
Small systems / large fluctuations. The grand-canonical result is the mean; when the number of accessible modes is tiny, relative number fluctuations \(\sim 1/\sqrt{N}\) are not negligible and the mean occupation is not the whole story (canonical and grand-canonical ensembles cease to be equivalent).
No equilibrium chemical potential. Driven, pumped, or transient systems (a laser mode, a shock front) have no single \(\mu\); the Gibbs weight is undefined and one needs a kinetic (Boltzmann/quantum-Boltzmann) description instead.

Failure modes

Sign confusion. Writing \(+1\) for bosons and \(-1\) for fermions — exactly backwards. Anchor it: fermions add the \(1\) (their occupation is bounded, the \(+1\) keeps \(\langle n\rangle \le 1\)); bosons subtract.
Forgetting the density of states. Treating \(\langle n\rangle\) as the number of particles in an energy interval. It is the occupation of a single mode; physical particle counts require \(\int \langle n(\varepsilon)\rangle\, g(\varepsilon)\, d\varepsilon\).
Setting \(\mu=0\) for a massive gas. \(\mu=0\) holds only for particles with no conservation law (photons, phonons). For atoms or electrons \(\mu\) is fixed by the number constraint and is generally nonzero and temperature-dependent.
Confusing \(\mu\) with the Fermi energy at finite \(T\). \(\varepsilon_F\equiv\mu(T{=}0)\); at finite temperature \(\mu(T)\neq\varepsilon_F\), and for a 3D free-electron gas \(\mu\) actually decreases with \(T\).
Applying Maxwell-Boltzmann inside the degenerate regime. Using \(e^{-\beta(\varepsilon-\mu)}\) when \(\langle n\rangle\) is not \(\ll 1\) — e.g. for conduction electrons at room temperature, where the gas is strongly degenerate and MB is qualitatively wrong.
Double-counting spin. Omitting the \((2s+1)\) spin degeneracy that multiplies the density of states, or conversely folding it into \(\langle n\rangle\) itself.

Discussion

The derivation isolates why the grand-canonical ensemble is indispensable for quantum gases. In the canonical ensemble the global constraint \(\sum_k n_k = N\) couples all the modes, so the sum over microstates does not factorise. Trading the fixed \(N\) for a fixed \(\mu\) — allowing number to fluctuate against a reservoir — decouples the modes and turns an interlocking combinatorial problem into a product of one-line geometric or two-term sums. The price, a fluctuating \(N\), is negligible in the thermodynamic limit, where the ensembles are equivalent.

Physically, the \(\mp 1\) encodes quantum indistinguishability. Distinguishable classical particles would be counted by which particle occupies which state, giving Maxwell-Boltzmann. Indistinguishability erases those labels; what remains is the exchange symmetry of the wavefunction, symmetric for bosons (any occupation) and antisymmetric for fermions (at most one). The bosonic \(-1\) reflects an effective statistical attraction — bunching, the same effect behind stimulated emission and the Hanbury Brown-Twiss correlation — while the fermionic \(+1\) reflects statistical repulsion, the Pauli pressure that supports white dwarfs and neutron stars.

The chemical potential is the quiet workhorse. It is not a free parameter but is pinned by the number constraint \(N=\sum_k\langle n_k\rangle\); its temperature dependence governs everything from the electronic heat capacity's linear-in-\(T\) law (only states within \(\sim k_{\mathrm B}T\) of \(\mu\) can be excited) to the critical temperature of condensation (where \(\mu\) reaches the ground state). Both distributions share the same \(\mu\); it is the sign, not the chemical potential, that separates the physics.

At a deeper level, the two statistics are the only possibilities in three dimensions because the configuration space of \(N\) indistinguishable particles has fundamental group \(S_N\), whose one-dimensional representations are just the trivial (symmetric) and sign (antisymmetric) ones. In two dimensions the relevant group is the braid group, admitting a continuum of anyonic statistics interpolating between Bose and Fermi — and the clean \(\mp 1\) derived here is then replaced by fractional generalisations. The occupation formula is thus a low-dimensional shadow of the topology of particle exchange.

Common misconceptions. The distribution does not say "particles prefer low energy" in isolation — it is the product \(\langle n(\varepsilon)\rangle\,g(\varepsilon)\) that determines where particles actually sit, and a rising density of states can put the population peak well above \(\mu\). Also, \(\langle n\rangle\) is a mean over a fluctuating ensemble, not a fixed integer occupation of any given microstate.

Worked examples

\[ \text{Conduction electron 0.10 eV above } \mu \text{ in copper at } T=300\text{ K} \]

Fermions, so use Fermi-Dirac. Symbolic form first, then numbers.

\[ \langle n\rangle^{\text{FD}} = \frac{1}{e^{(\varepsilon-\mu)/k_{\mathrm B}T} + 1} \]

Write \(\beta(\varepsilon-\mu)=(\varepsilon-\mu)/k_{\mathrm B}T\).

\[ k_{\mathrm B}T = (8.617\times10^{-5}\ \text{eV/K})(300\ \text{K}) = 0.02585\ \text{eV} \]

Thermal energy scale in eV for convenience.

\[ \frac{\varepsilon-\mu}{k_{\mathrm B}T} = \frac{0.100}{0.02585} = 3.868, \qquad e^{3.868} = 47.9 \]

Dimensionless exponent, then exponentiate.

\[ \langle n\rangle^{\text{FD}} = \frac{1}{47.9 + 1} = 0.0204 \approx 2.0\% \]

Reading. A state 0.1 eV (about 4 \(k_{\mathrm B}T\)) above the Fermi level is only 2% occupied at room temperature — consistent with the sharp Fermi step barely softened by thermal smearing.

\[ \text{Photon mode at } \lambda = 500\ \text{nm in a cavity at } T = 3000\text{ K} \]

Massless bosons with \(\mu=0\); use Bose-Einstein.

\[ \langle n\rangle^{\text{BE}} = \frac{1}{e^{\varepsilon/k_{\mathrm B}T} - 1}, \qquad \varepsilon = \frac{hc}{\lambda} \]

Set \(\mu=0\) (photon number not conserved) and use the photon energy.

\[ \varepsilon = \frac{(6.626\times10^{-34})(3.00\times10^{8})}{5.00\times10^{-7}} = 3.98\times10^{-19}\ \text{J} = 2.48\ \text{eV} \]

Photon energy in J and eV.

\[ k_{\mathrm B}T = (1.381\times10^{-23})(3000) = 4.14\times10^{-20}\ \text{J} = 0.2585\ \text{eV} \]

Thermal energy at 3000 K.

\[ \frac{\varepsilon}{k_{\mathrm B}T} = \frac{2.48}{0.2585} = 9.59, \qquad e^{9.59} = 1.46\times10^{4} \]

Dimensionless exponent and its exponential.

\[ \langle n\rangle^{\text{BE}} = \frac{1}{1.46\times10^{4} - 1} = 6.8\times10^{-5} \]

Reading. Each 500 nm mode of a 3000 K cavity holds far less than one photon on average; visible modes are in the Wien (Boltzmann) tail, where \(\langle n\rangle\approx e^{-\varepsilon/k_{\mathrm B}T}\) and the \(-1\) is irrelevant. This is why an incandescent lamp at 3000 K is dim in the blue.

Problems

At what energy relative to \(\mu\) is a fermionic state exactly half-occupied, for any \(T\)?
Solution
Set \(\langle n\rangle^{\text{FD}}=1/2\): \(e^{\beta(\varepsilon-\mu)}+1 = 2\), so \(e^{\beta(\varepsilon-\mu)}=1\), giving \(\beta(\varepsilon-\mu)=0\), i.e. \(\varepsilon=\mu\). The chemical potential is precisely the half-occupancy energy at every temperature — a temperature-independent landmark of the Fermi-Dirac distribution.
Show that the Fermi-Dirac and Bose-Einstein occupations are related to the "hole" occupation \(1-\langle n\rangle^{\text{FD}}\) by a particle-hole symmetry. Specifically, evaluate \(1-\langle n\rangle^{\text{FD}}(\varepsilon)\) and express it as a Fermi function.
Solution
\(1-\dfrac{1}{e^{x}+1} = \dfrac{e^{x}}{e^{x}+1} = \dfrac{1}{1+e^{-x}}\) with \(x=\beta(\varepsilon-\mu)\). Thus \(1-\langle n\rangle^{\text{FD}}(\varepsilon) = \langle n\rangle^{\text{FD}}(2\mu-\varepsilon)\): the hole occupation at energy \(\varepsilon\) equals the electron occupation at the mirror energy \(2\mu-\varepsilon\). The distribution is antisymmetric about \(\mu\) — the basis of the electron-hole symmetry in semiconductors.
A cavity at \(T=5800\) K (solar surface). Compute \(\langle n\rangle\) for a photon mode at \(\lambda=1000\) nm (near-IR).
Solution
\(\varepsilon = hc/\lambda = (6.626\times10^{-34})(3.00\times10^{8})/(1.00\times10^{-6}) = 1.99\times10^{-19}\,\text{J}=1.24\,\text{eV}\). \(k_{\mathrm B}T=(8.617\times10^{-5})(5800)=0.500\,\text{eV}\). Ratio \(=1.24/0.500=2.48\); \(e^{2.48}=11.9\). \(\langle n\rangle = 1/(11.9-1)=0.0917\approx 0.092\). Roughly a tenth of a photon per mode — noticeably larger than the visible example, since near-IR sits lower on the Wien tail.
For a 3D ideal Fermi gas, argue qualitatively why \(\mu(T)\) must decrease as \(T\) rises from zero (assuming fixed \(N\)).
Solution
The particle number is \(N=\int_0^\infty g(\varepsilon)\langle n\rangle^{\text{FD}}\,d\varepsilon\) with \(g(\varepsilon)\propto\sqrt{\varepsilon}\) in 3D. Thermal broadening of the step promotes electrons from just below \(\mu\) to just above. Because \(g(\varepsilon)\) is larger above \(\mu\) than below (it rises with \(\varepsilon\)), symmetric broadening in energy would over-count particles unless \(\mu\) drops to compensate. Keeping \(N\) fixed therefore forces \(\mu(T)<\varepsilon_F\), and to leading order \(\mu(T)\approx\varepsilon_F\big[1-\frac{\pi^2}{12}(k_{\mathrm B}T/\varepsilon_F)^2\big]\) (Sommerfeld expansion).
Two identical bosonic modes each at energy \(\varepsilon-\mu = k_{\mathrm B}T\). Find the mean total occupation and compare with the classical (Maxwell-Boltzmann) prediction.
Solution
Per mode: \(\langle n\rangle^{\text{BE}}=1/(e^{1}-1)=1/(2.718-1)=1/1.718=0.582\). Two modes: total \(=1.164\). Classical: \(\langle n\rangle^{\text{MB}}=e^{-1}=0.368\) per mode, total \(0.736\). The bosonic occupation exceeds the classical value by a factor \(0.582/0.368=1.58\) — the statistical bunching enhancement. At this energy the gas is already mildly degenerate (\(\langle n\rangle\) not \(\ll 1\)), so the \(-1\) matters and MB is a poor approximation.