Quantum Statistical Fluctuations in Occupation Numbers

The Bose-Einstein and Fermi-Dirac distributions answer "how full is a level on average?" This article answers the sharper question hiding behind the average: how much does the occupation of a single level wobble around its mean — and does that wobble reveal which statistics the particles obey? It does, decisively. In the grand canonical ensemble every single-particle level is an independent little system trading particles with a reservoir, so its occupation variance follows from one derivative and lands on a single compact law, $\langle(\Delta n_k)^2\rangle=\langle n_k\rangle(1\pm\langle n_k\rangle)$ , with the upper sign for bosons and the lower for fermions. That one sign carries the whole difference between the quantum statistics, read as noise. Bosons are noisier than a classical Poisson gas — they bunch, piling into the same state so the variance overshoots the mean by $\langle n_k\rangle^2$ . Fermions are quieter — they antibunch, because a level holds only $0$ or $1$ particle and a certainly-filled or certainly-empty level cannot fluctuate at all. The Maxwell-Boltzmann gas sits exactly between, with Poisson statistics where variance equals mean. These occupation variances are not an abstraction: they are the equal-state limit of the second-order coherence $g^{(2)}$ measured in Hanbury Brown-Twiss experiments, and the mesoscopic shot noise measured in a quantum point contact. Photon bunching (1956), and later atomic bunching for bosonic $^4$ He against antibunching for fermionic $^3$ He, directly confirm the formula derived here — the thermodynamic shadow of the symmetrization postulate, made audible as noise.

Key takeaways

In the grand canonical ensemble each single-particle level is independent, so its occupation variance is $\langle(\Delta n_k)^2\rangle=\langle n_k\rangle(1\pm\langle n_k\rangle)$ — upper sign bosons, lower sign fermions.
The relative (Fano) fluctuation $\langle(\Delta n_k)^2\rangle/\langle n_k\rangle=1\pm\langle n_k\rangle$ sits above the Poisson value $1$ for bosons (bunching) and below it for fermions (antibunching).
The means $\langle n_k\rangle$ are owned by aq-sm-bose-einstein-distribution and aq-sm-fermi-dirac-distribution; this article owns the variance and its statistics-selecting sign.
Fermionic noise vanishes for filled ( $\langle n\rangle=1$ ) and empty ( $\langle n\rangle=0$ ) levels and is maximal ( $=\tfrac14$ ) at half-filling $\varepsilon_k=\mu$ ; bosonic noise grows without bound as a mode fills.
The single-mode second-order coherence is $g^{(2)}=2$ for thermal bosons (bunching) and $g^{(2)}=0$ for fermions (antibunching) — the Hanbury Brown-Twiss signature.
Key idea: because the grand partition function factorizes over levels, each variance is one derivative, $\langle(\Delta n_k)^2\rangle=k_BT\,\partial\langle n_k\rangle/\partial\mu=\langle n_k\rangle(1\pm\langle n_k\rangle)$ — the $+$ (bosons, bunching) versus $-$ (fermions, antibunching) is the entire statistical content of the noise.

The core insight

In the grand canonical ensemble a level's occupation variance is a single logarithmic derivative of its partition function, and it collapses to $\langle(\Delta n_k)^2\rangle=\langle n_k\rangle(1\pm\langle n_k\rangle)$ . The sign is the statistics: $+$ makes bosons super-Poissonian (bunching, variance overshoots the mean), $-$ makes fermions sub-Poissonian (antibunching, a Bernoulli level that cannot fluctuate when full or empty). The classical gas is the sign-blind Poisson case in between.

Where this lives today

Occupation-number fluctuations are the working currency of quantum optics and mesoscopic physics. In quantum optics the single-mode $g^{(2)}$ classifies light: $g^{(2)}=2$ for thermal sources, $g^{(2)}=1$ for lasers, and $g^{(2)}<1$ — impossible for classical fields — is the modern definition of nonclassical antibunched light. In mesoscopic conductors, electrons partitioned at a quantum point contact carry current noise with the Pauli factor $T(1-T)$ , the transport avatar of $\langle n\rangle(1-\langle n\rangle)$ ; the resulting sub-Poissonian Fano factor is measured routinely and was used to detect the fractional charge $e/3$ of fractional-quantum-Hall quasiparticles. The matter-wave versions are equally direct: cold-atom Hanbury Brown-Twiss experiments show bunching for bosonic isotopes and antibunching for fermionic ones — a laboratory reading of the sign in Eq. (3). Summed over levels, the same variances give the total number fluctuation $\langle\Delta N^2\rangle=k_BT N^2\kappa_T/V$ , tying counting statistics to the isothermal compressibility. And because variance and mean are locked by $1\pm\langle n\rangle$ , measuring fluctuation power calibrates temperature without knowing the coupling — the basis of primary noise thermometry and of the sensitivity limits of quantum-limited amplifiers.

The physical picture

Picture one single-particle level as a small box trading particles with a large reservoir at chemical potential $\mu$ and temperature $T$ . For a fermion the box is a switch: it holds $0$ or $1$ particle, a biased coin with "heads" probability $\langle n\rangle$ . Its spread is largest when the coin is fair ( $\langle n\rangle=\tfrac12$ , at the Fermi surface) and exactly zero when the outcome is certain — a level deep below $\mu$ (always full) or far above (always empty) does not fluctuate. For a boson the box has no ceiling, and identical bosons "like company": the more particles already present, the more the distribution spreads, adding the extra $+\langle n\rangle^2$ that classical particles never have. The hero figure below is the punchline: plot the Fano factor $\langle(\Delta n)^2\rangle/\langle n\rangle=1\pm\langle n\rangle$ against the mean occupation. Three curves fan out from the same Poisson value $1$ at low occupancy — the boson line rises (bunching), the classical line stays flat (Poisson), and the fermion line falls to zero at $\langle n\rangle=1$ (antibunching). The sign of a single term splits the plane into "louder than classical" and "quieter than classical."

Figure

Fano factor versus mean occupation: bosons bunch above the Poisson line, fermions antibunch below it

The punchline: at low occupancy all three statistics collapse to Poisson ( $1$ ); as levels fill, the single sign $\pm\langle n\rangle$ pushes bosons up (louder) and fermions down (quieter), reaching the extremes $g^{(2)}=2$ and $g^{(2)}=0$ .

Compare three counting distributions for one level: fermions are Bernoulli (two outcomes), classical particles are Poisson (mean equals variance), bosons are geometric (a long tail). The second figure lays them side by side, so the variance ordering $\langle n\rangle(1-\langle n\rangle)<\langle n\rangle<\langle n\rangle(1+\langle n\rangle)$ becomes the visible shape of each distribution.

Figure

Single-level occupation distributions: Bernoulli fermion, Poisson classical, geometric boson

Fermions repel (two-valued, narrow), classical particles ignore each other (Poisson), bosons cluster (fat geometric tail). The variance is literally the width of these pictures.

The factor $(1\pm\langle n\rangle)$ is the same factor that governs transition rates: $(1+n)$ for stimulated bosonic emission and $(1-n)$ for fermionic Pauli blocking. Fluctuation and response are two faces of one quantity — the fluctuation-dissipation theorem with the chemical potential as conjugate field. And because the grand partition function factorizes over levels, distinct levels are statistically independent, making the total number variance a simple sum of per-level variances — a simplification special to the grand canonical ensemble that fails the moment you fix $N$ exactly.

The core equations

The occupations owned by the distribution articles are the input. For bosons and fermions the mean occupation of level $k$ is a single expression with a sign, $-$ for bosons and $+$ for fermions in the denominator.

\langle n_k\rangle=\frac{1}{e^{\beta(\varepsilon_k-\mu)}\mp1} \qquad (1)

where: $\langle n_k\rangle$ — mean occupation of single-particle level $k$ ; $\beta=1/k_BT$ ; $\varepsilon_k$ — level energy; $\mu$ — chemical potential; the denominator sign is $-$ for bosons, $+$ for fermions. Physical reading: the Bose case can diverge as $\varepsilon_k\to\mu^+$ (owned by aq-sm-bose-einstein-distribution), the Fermi case is bounded in $[0,1]$ (owned by aq-sm-fermi-dirac-distribution); this article takes both as given and computes their fluctuations. Both statistics come from one single-level grand partition function, whose occupancy sum runs to infinity for bosons and stops at $n=1$ for fermions. Writing $x_k=e^{-\beta(\varepsilon_k-\mu)}$ , the two cases are a geometric sum and a two-term sum, compactly one formula.

\Xi_k=\big(1\mp x_k\big)^{\mp1},\qquad x_k=e^{-\beta(\varepsilon_k-\mu)} \qquad (2)

where: $\Xi_k$ — grand partition function of level $k$ alone; the upper sign (boson) gives $\Xi_k=(1-x_k)^{-1}$ (geometric series, convergent for $\mu<\varepsilon_k$ ), the lower (fermion) gives $\Xi_k=1+x_k$ (only $n=0,1$ ). Physical reading: the single stopped term — the difference between " $\dots+x^2+x^3+\dots$ " and "stop at $x$ " — is the entire origin of the sign in every result below. The central object of this article is the occupation variance. It is the second cumulant, obtained from the same $\Xi_k$ , and it collapses to a strikingly simple product of the mean with $1\pm\langle n_k\rangle$ .

\langle(\Delta n_k)^2\rangle=\langle n_k\rangle\big(1\pm\langle n_k\rangle\big) \qquad (3)

where: $\langle(\Delta n_k)^2\rangle=\langle n_k^2\rangle-\langle n_k\rangle^2$ is the variance; $+$ for bosons (bunching), $-$ for fermions (antibunching). Physical reading: the leading $\langle n_k\rangle$ is the classical Poisson noise; the correction $\pm\langle n_k\rangle^2$ is the purely quantum-statistical excess or deficit that the sign selects. Equation (3) is a fluctuation-response (fluctuation-dissipation) relation in disguise: the variance equals the temperature times the response of the mean occupation to the chemical potential. This is why measuring noise measures a susceptibility.

\langle(\Delta n_k)^2\rangle=k_BT\,\frac{\partial\langle n_k\rangle}{\partial\mu}=-k_BT\,\frac{\partial\langle n_k\rangle}{\partial\varepsilon_k} \qquad (4)

where: $\partial\langle n_k\rangle/\partial\mu$ — isothermal response of the occupation to the chemical potential. Physical reading: because $\langle n_k\rangle$ depends on $\varepsilon_k$ and $\mu$ only through $\varepsilon_k-\mu$ , the two derivatives are equal and opposite; the noise is the slope of the occupation curve. Normalizing by the mean gives the Fano factor $F_k\equiv\langle(\Delta n_k)^2\rangle/\langle n_k\rangle=1\pm\langle n_k\rangle$ : exactly one for Poisson, above one for bunching ( $F\to\infty$ as a boson mode fills), below one for antibunching ( $F\to0$ as a fermion mode fills). Because the grand partition function factorizes, $\Xi=\prod_k\Xi_k$ , different levels are uncorrelated, so the total particle-number variance is just the sum of single-level variances — exact only in the grand canonical ensemble, where cross-covariances vanish for $k\neq k'$ .

\langle\Delta N^2\rangle=\sum_k\langle n_k\rangle\big(1\pm\langle n_k\rangle\big) \qquad (5)

where: $N=\sum_k n_k$ — total particle number; $\langle\Delta N^2\rangle$ — its variance. Physical reading: number fluctuations add level by level and connect to the isothermal compressibility $\kappa_T$ .

Derivations

Derivation AMean and variance from the single-level grand partition function

The goal is to show that both the mean occupation and its variance are logarithmic derivatives of the single-level grand partition function $\Xi_k$ , and that carrying this through for the boson (geometric) and fermion (two-term) sums produces exactly $\langle(\Delta n_k)^2\rangle=\langle n_k\rangle(1\pm\langle n_k\rangle)$ . The route is to treat each level as an independent system in the variable $\alpha_k=\beta(\varepsilon_k-\mu)$ , recognize $\ln\Xi_k$ as a cumulant generating function, and read off the first two cumulants. Watch the sign bookkeeping: one derivative in $\alpha_k$ carries a minus, and the boson/fermion split enters only through whether the occupancy sum stops at $n=1$ .

Step 1. Set $\alpha_k=\beta(\varepsilon_k-\mu)$ and $x_k=e^{-\alpha_k}$ . The probability of occupation $n$ is $P(n)=x_k^{\,n}/\Xi_k$ , so the mean is $\langle n_k\rangle=\sum_n n\,x_k^{\,n}/\Xi_k=-\partial_{\alpha_k}\ln\Xi_k$ . Why: each power of $n$ is brought down by $-\partial_{\alpha_k}$ acting on $x_k^{\,n}=e^{-\alpha_k n}$ , and dividing by $\Xi_k$ turns the derivative of $\Xi_k$ into a derivative of its logarithm.

Step 2. The second cumulant is the second derivative: $\partial_{\alpha_k}^2\ln\Xi_k=\langle n_k^2\rangle-\langle n_k\rangle^2=\langle(\Delta n_k)^2\rangle$ . Why: $\ln\Xi_k$ generates cumulants in $\alpha_k$ ; its first derivative gives the mean (with a sign), its second gives the variance directly.

Step 3. Bosons: $\Xi_k=(1-x_k)^{-1}$ , so $\ln\Xi_k=-\ln(1-x_k)$ and $\langle n_k\rangle=-\partial_{\alpha_k}\ln\Xi_k=x_k/(1-x_k)=1/(e^{\alpha_k}-1)$ , the Bose-Einstein form. The variance is $\partial_{\alpha_k}^2\ln\Xi_k=x_k/(1-x_k)^2$ . Since $1+\langle n_k\rangle=1/(1-x_k)$ , this equals $\langle n_k\rangle(1+\langle n_k\rangle)$ . Why: the extra factor $1/(1-x_k)$ in the variance is precisely $1+\langle n_k\rangle$ , producing the super-Poissonian bunching excess.

Step 4. Fermions: $\Xi_k=1+x_k$ , so $\ln\Xi_k=\ln(1+x_k)$ and $\langle n_k\rangle=x_k/(1+x_k)=1/(e^{\alpha_k}+1)$ , the Fermi-Dirac form. The variance is $x_k/(1+x_k)^2=\langle n_k\rangle(1-\langle n_k\rangle)$ , because $1-\langle n_k\rangle=1/(1+x_k)$ . Why: stopping the sum at $n=1$ flips the sign of the quadratic term, turning the excess into a deficit — the antibunching that caps the variance at $\tfrac14$ .

Step 5. Cross-check with fluctuation-response: since $\langle n_k\rangle$ depends only on $\alpha_k$ , $k_BT\,\partial_\mu\langle n_k\rangle=-\partial_{\alpha_k}\langle n_k\rangle$ , and differentiating $1/(e^{\alpha_k}\mp1)$ returns $\langle n_k\rangle(1\pm\langle n_k\rangle)$ for both signs. Why: this confirms Eq. (4): the noise is a susceptibility, the same quantity from a completely independent route. Verified: the geometric and two-term $\Xi_k$ , both means, both variances, and the fluctuation-response identity $-\partial_{\alpha}\langle n\rangle=\langle n\rangle(1\pm\langle n\rangle)$ are checked symbolically in verify_occupation-number-fluctuations.py.

Go deeper · Derivation B · 1 minFermionic occupation as a Bernoulli variable and the vanishing at full or empty

Derivation BFermionic occupation as a Bernoulli variable and the vanishing at full or empty

The goal is to prove the fermionic variance $\langle n_k\rangle(1-\langle n_k\rangle)$ a second way — with no calculus at all — by recognizing that a fermion level is a two-valued random variable, and to extract from that the two facts that define antibunching: the noise is maximal ( $=\tfrac14$ ) at half-filling and vanishes exactly for a certainly-full or certainly-empty level. The route uses only the exclusion cap $n_k\in\{0,1\}$ (owned microscopically by aq-qm-pauli-exclusion-principle) and the algebra of a biased coin, so it makes clear that antibunching is a direct consequence of the Pauli principle, independent of any thermodynamic detail.

Step 1. Because Pauli exclusion allows only $n_k\in\{0,1\}$ , the variable is idempotent: $n_k^2=n_k$ . Let $p=\langle n_k\rangle$ be the probability the level is filled. Why: a two-valued $0/1$ variable squares to itself, which is the algebraic heart of everything that follows.

Step 2. Take expectations: $\langle n_k^2\rangle=\langle n_k\rangle=p$ . Why: idempotency means the second moment equals the first, so no separate computation of $\langle n_k^2\rangle$ is needed.

Step 3. Therefore $\langle(\Delta n_k)^2\rangle=\langle n_k^2\rangle-\langle n_k\rangle^2=p-p^2=p(1-p)$ , identical to the Fermi-Dirac variance from Step 4 of the previous derivation. Why: the whole antibunching factor $1-\langle n_k\rangle$ drops out of the Bernoulli structure with no reference to temperature or spectrum.

Step 4. Maximize: $\tfrac{d}{dp}\,p(1-p)=1-2p=0$ gives $p=\tfrac12$ , where the variance is $\tfrac14$ ; at $p=0$ or $p=1$ it is zero. Why: a certainly-empty or certainly-full level has a definite occupation and cannot fluctuate, while the fair-coin point $p=\tfrac12$ (the Fermi surface, $\varepsilon_k=\mu$ ) is where a fermion level is noisiest. Verified: the idempotency $n^2=n$ , the Bernoulli variance $p(1-p)$ , the maximum $\tfrac14$ at $p=\tfrac12$ , and the vanishing at $p=0,1$ are checked in verify_occupation-number-fluctuations.py.

Go deeper · Derivation C · 1 minBunching, the second-order coherence, and the classical Poisson limit

Derivation CBunching, the second-order coherence, and the classical Poisson limit

The goal is to compute the single-mode second-order coherence $g^{(2)}=\langle n(n-1)\rangle/\langle n\rangle^2$ for both statistics — obtaining $g^{(2)}=2$ for thermal bosons and $g^{(2)}=0$ for fermions — and then to show that all three statistics collapse to Poisson noise in the dilute limit $\langle n\rangle\ll1$ . The route evaluates $\langle n(n-1)\rangle$ from the variance via $\langle n(n-1)\rangle=\langle(\Delta n)^2\rangle+\langle n\rangle^2-\langle n\rangle$ , then normalizes. Watch that the single-mode $g^{(2)}$ is a fixed number, independent of $\langle n\rangle$ , while the excess-over-Poisson form makes the $\pm$ structure explicit.

Step 1. Rearranging the variance definition, $\langle n(n-1)\rangle=\langle n^2\rangle-\langle n\rangle=\langle(\Delta n)^2\rangle+\langle n\rangle^2-\langle n\rangle$ . Why: $\langle n(n-1)\rangle$ counts ordered pairs of particles in the mode, the quantity a coincidence detector registers.

Step 2. Bosons: substitute $\langle(\Delta n)^2\rangle=\langle n\rangle(1+\langle n\rangle)$ to get $\langle n(n-1)\rangle=\langle n\rangle(1+\langle n\rangle)+\langle n\rangle^2-\langle n\rangle=2\langle n\rangle^2$ , hence $g^{(2)}=2$ . Why: a single thermal (chaotic) mode is exactly twice as likely to be found doubly occupied as independent particles would be — textbook Hanbury Brown-Twiss bunching.

Step 3. Fermions: substitute $\langle(\Delta n)^2\rangle=\langle n\rangle(1-\langle n\rangle)$ to get $\langle n(n-1)\rangle=\langle n\rangle(1-\langle n\rangle)+\langle n\rangle^2-\langle n\rangle=0$ , hence $g^{(2)}=0$ . Why: a level cannot hold two fermions, so the ordered-pair count is identically zero — perfect antibunching.

Step 4. Classical limit: in $\langle(\Delta n_k)^2\rangle=\langle n_k\rangle(1\pm\langle n_k\rangle)$ the quadratic term is negligible when $\langle n_k\rangle\ll1$ , leaving $\langle(\Delta n_k)^2\rangle\to\langle n_k\rangle$ for both signs — Poisson statistics, mean equal to variance. Why: when levels are sparsely occupied ( $n\lambda_T^3\ll1$ ) double occupancy is rare, so the distinction between "pile up" and "exclude" never comes into play and quantum statistics become invisible. Verified: $g^{(2)}=2$ (boson) and $g^{(2)}=0$ (fermion) from $\langle n(n-1)\rangle$ , and the Poisson reduction of the variance as $\langle n\rangle\to0$ , are checked in verify_occupation-number-fluctuations.py.

Limiting cases

The single formula $\langle(\Delta n_k)^2\rangle=\langle n_k\rangle(1\pm\langle n_k\rangle)$ interpolates among the statistics and across the filling of a level.

Case	Result	Meaning
Bosons (bunching)	$\langle(\Delta n_k)^2\rangle=\langle n_k\rangle(1+\langle n_k\rangle)$	Super-Poissonian; the variance exceeds the mean by $\langle n_k\rangle^2$ . As a mode macroscopically fills, $\Delta n_k\to\langle n_k\rangle$ — relative noise stays order one, never vanishing
Fermions (antibunching)	$\langle(\Delta n_k)^2\rangle=\langle n_k\rangle(1-\langle n_k\rangle)$	Sub-Poissonian Bernoulli noise, maximal $=\tfrac14$ at $\langle n_k\rangle=\tfrac12$ (the Fermi surface), vanishing for filled or empty levels
Classical / Maxwell-Boltzmann	$\langle(\Delta n_k)^2\rangle\to\langle n_k\rangle\ (\langle n_k\rangle\ll1)$	The $\pm\langle n_k\rangle^2$ correction is negligible: Poisson, mean $=$ variance, statistics-blind — the dilute limit $n\lambda_T^3\ll1$
Second-order coherence	$g^{(2)}=2$ (boson), $0$ (fermion), $1$ (Poisson)	Single-mode $g^{(2)}$ is a fixed number: bunched thermal light, perfectly antibunched fermions, coherent/classical in between

Common trapIndependence of levels is a grand-canonical privilege, not a law of nature

The clean sum $\langle\Delta N^2\rangle=\sum_k\langle n_k\rangle(1\pm\langle n_k\rangle)$ and the statistical independence of different levels hold in the grand canonical ensemble, where the total particle number is free to fluctuate. In the canonical or microcanonical ensemble the constraint $\sum_k n_k=N$ correlates the occupations: they are no longer independent, and each single-level variance is corrected by terms of order $1/N$ . For extensive quantities in the thermodynamic limit the ensembles agree, but for the Bose-Einstein condensate's ground-level occupation the grand canonical variance is anomalously huge — the notorious "grand-canonical catastrophe" that forces a canonical treatment of the condensate. Do not carry the additive formula into a fixed- $N$ problem without checking the correlations.

Worked examples

Example 1 — Enumerate a two-state fermion system and read off the antibunching.

Step 1. Setup. Take a single spinless fermion level and put it in contact with a reservoir at $\alpha=\beta(\varepsilon-\mu)$ . The only microstates are $n=0$ and $n=1$ , with grand weights $1$ and $x=e^{-\alpha}$ . Choose $\alpha=0$ (level exactly at the chemical potential, $\varepsilon=\mu$ ), so $x=1$ .

Step 2. Probabilities. $\Xi=1+x=2$ , so $P(0)=1/2$ and $P(1)=1/2$ — a fair coin. The mean is $\langle n\rangle=0\cdot\tfrac12+1\cdot\tfrac12=\mathbf{0.5}$ .

Step 3. Variance by direct enumeration. $\langle n^2\rangle=0^2\cdot\tfrac12+1^2\cdot\tfrac12=\tfrac12$ , so $\langle(\Delta n)^2\rangle=\tfrac12-(\tfrac12)^2=\mathbf{0.25}$ . This matches $\langle n\rangle(1-\langle n\rangle)=\tfrac12\cdot\tfrac12=\tfrac14$ exactly — the maximal fermionic fluctuation, at half-filling.

Step 4. Interpret. Move the level far below $\mu$ ( $\alpha\to-\infty$ , $x\to\infty$ ): $P(1)\to1$ , $\langle n\rangle\to1$ , and the variance $\langle n\rangle(1-\langle n\rangle)\to0$ . A certainly-filled level does not fluctuate. The noise lives only in the thin shell near $\varepsilon=\mu$ — the same active window that gives metals their linear-in- $T$ heat capacity.

Example 2 — Thermal versus coherent light: same brightness, different noise.

Step 1. Setup. A single optical mode is populated to mean occupation $\langle n\rangle=1$ (one photon on average per mode). Compare a thermal (chaotic) source, whose mode obeys Bose-Einstein statistics, with an ideal laser, whose mode is coherent (Poisson).

Step 2. Thermal mode. Variance $\langle(\Delta n)^2\rangle=\langle n\rangle(1+\langle n\rangle)=1\cdot(1+1)=\mathbf{2}$ , so the standard deviation is $\sqrt2\approx\mathbf{1.41}$ — larger than the mean itself. The second-order coherence is $g^{(2)}=2$ : photons arrive in bunches.

Step 3. Coherent (laser) mode. Poisson statistics give $\langle(\Delta n)^2\rangle=\langle n\rangle=\mathbf{1}$ , standard deviation $\mathbf{1.0}$ , and $g^{(2)}=1$ : photons arrive at random with no bunching.

Step 4. Interpret. Both sources have the same mean brightness, but the thermal source is measurably noisier — its variance is twice the Poisson value. This excess is exactly the extra $+\langle n\rangle^2$ in the bosonic formula, and it is what Hanbury Brown and Twiss detected in 1956 as photon bunching, using the correlation to measure a star's angular size.

Origins · 1 minpascual jordan and how this was discovered

pascual jordan

1902–1980 · Pascual Jordan, co-founder of matrix mechanics with Born and Heisenberg, built the occupation-number (second-quantized) formalism in which a many-body state is specified by the set of level occupations. It is precisely in this representation that the occupation of a mode becomes a fluctuating quantum variable — bosonic modes with unbounded n, fermionic modes with n in {0,1}. Jordan's 1927–28 work with Klein and Wigner introduced the boson and fermion field operators whose number-operator variances are the formulas of this article.

History

1926

— Bose-Einstein and Fermi-Dirac mean occupations established, fixing the

\langle n\rangle

this article varies about

1927

— Jordan and Klein quantize the boson field; occupation number becomes an operator

1928

— Jordan and Wigner introduce fermionic anticommutators, enforcing

n\in\{0,1\}

1956

— Hanbury Brown and Twiss measure photon bunching (

g^{(2)}=2

) and use it to size the star Sirius

2005

— Cold-atom experiments observe bunching for bosonic

^4

He and antibunching for fermionic

^3

2007

— Sub-Poissonian atom-number fluctuations imaged directly in degenerate Fermi gases

Going further

Texts: Pathria & Beale, Statistical Mechanics (3rd ed.), §6.1–6.3 (mean occupations and fluctuations); Kardar, Statistical Physics of Particles, Ch. 7; Reif, Fundamentals of Statistical and Thermal Physics, §9.5–9.6; Landau & Lifshitz, Statistical Physics Part 1, §54–56; Sethna, Statistical Mechanics, Ch. 7.
Builds on: Grand Canonical Ensemble for Ideal Quantum Gases · Occupation Number Representation · Bose-Einstein Distribution · Fermi-Dirac Distribution · Symmetrization Postulate and Wavefunction Symmetry. The means $\langle n_k\rangle$ are owned by the distribution articles; this article owns their variances.
Leads to: Hanbury Brown–Twiss correlations (measured $g^{(2)}$ ) · Maxwell-Boltzmann Classical Limit (the Poisson watershed) · Pauli Exclusion Principle (the microscopic $n\in\{0,1\}$ behind antibunching) · Quantum Concentration and the Classicality Criterion.
Provenance: AllQFT `sm` export (Pathria & Beale, §6.1–6.3), aq-sm-occupation-number-fluctuations, ref STAT.QSTAT.OCCFL.1215. Every variance, sign, coherence value, and worked number python-verified (14 checks, all PASS): the geometric/two-term $\Xi_k$ , $\langle n\rangle(1\pm\langle n\rangle)$ for both signs, the fluctuation-response identity, $g^{(2)}=2$ (boson) and $0$ (fermion), the half-filled maximum $\tfrac14$ , and the Poisson dilute limit.

Why does one sign flip carry the entire difference between bosons and fermions?

Everything traces to whether the single-level occupancy sum stops. For bosons $\Xi_k=1+x+x^2+\cdots=(1-x)^{-1}$ ; for fermions Pauli exclusion truncates it at $\Xi_k=1+x$ . That one stopped term flips the sign of the quadratic correction: the variance is $x/(1-x)^2=\langle n\rangle(1+\langle n\rangle)$ for bosons and $x/(1+x)^2=\langle n\rangle(1-\langle n\rangle)$ for fermions. The leading Poisson piece $\langle n\rangle$ is shared; only the $\pm\langle n\rangle^2$ excess distinguishes them, and it is that stopped term made quantitative.

Is the occupation variance the same as the second-order coherence g(2)?

They are closely related but not identical. The variance $\langle(\Delta n)^2\rangle=\langle n^2\rangle-\langle n\rangle^2$ measures the spread of a single mode's occupation. The single-mode $g^{(2)}=\langle n(n-1)\rangle/\langle n\rangle^2$ counts ordered pairs of particles and is a normalized, dimensionless number ( $2$ thermal, $0$ fermionic, $1$ coherent). You can build one from the other: $\langle n(n-1)\rangle=\langle(\Delta n)^2\rangle+\langle n\rangle^2-\langle n\rangle$ . The variance grows with $\langle n\rangle$ ; the single-mode $g^{(2)}$ does not, which is why $g^{(2)}$ is the cleaner statistics label.

Does this article re-derive the Bose-Einstein and Fermi-Dirac distributions?

No. The mean occupations $\langle n_k\rangle=1/(e^{\beta(\varepsilon_k-\mu)}\mp1)$ are owned by aq-sm-bose-einstein-distribution and aq-sm-fermi-dirac-distribution. This article takes those means as input and computes what fluctuates around them — the variance and the coherence. The two derivations here re-obtain the means only as a by-product of the same logarithmic-derivative machinery, so the article is self-contained, but the physical ownership of the distributions rests with the distribution articles.

Why does the total number fluctuation add level by level here but not at fixed N?

In the grand canonical ensemble the grand partition function factorizes, $\Xi=\prod_k\Xi_k$ , which makes the occupations of distinct levels statistically independent — their covariances vanish, so variances add. In the canonical or microcanonical ensemble the total $N=\sum_k n_k$ is fixed, so if one level gains a particle another must lose one: the occupations are anticorrelated. That constraint modifies each single-level variance by terms of order $1/N$ and breaks the simple sum, most dramatically for the Bose condensate mode.

How does fermionic occupation noise become measurable shot noise in a wire?

When electrons hit a quantum point contact, each channel is a fermionic mode with transmission probability $T$ . Whether an electron is transmitted is a Bernoulli event, so the partitioned current carries variance $\propto T(1-T)$ — exactly the shape of $\langle n\rangle(1-\langle n\rangle)$ . The measured noise is sub-Poissonian, with a Fano factor below one that directly reports the channel transmissions. Push $T\to1$ (a perfectly open channel) and the noise vanishes: a certainly-transmitted electron, like a certainly-filled level, does not fluctuate.

Is antibunching a purely quantum effect with no classical analogue?

The bosonic bunching value $g^{(2)}=2$ can be understood semiclassically as intensity fluctuations of a chaotic wave, so it is not exclusively quantum. Antibunching, $g^{(2)}<1$ , is different: it is forbidden for any classical field, where $g^{(2)}\ge1$ always. A fermionic mode's $g^{(2)}=0$ has no classical description — it is a direct signature of the discreteness and exclusion of quanta, which is why antibunching, not bunching, is the operational definition of nonclassical light and of genuine Fermi statistics.