Grand Canonical Ensemble for Ideal Quantum Gases

An ideal quantum gas is a crowd of non-interacting indistinguishable particles whose only correlation comes from the symmetrization postulate: a many-body state is fixed once you say how many particles sit in each single-particle level, the occupation numbers $\{n_\varepsilon\}$ . At fixed particle number this is awkward — the sum over microstates must respect $\sum_\varepsilon n_\varepsilon = N$ , and that one constraint entangles every level with every other. The grand canonical ensemble cuts the knot. Let the gas trade particles with a reservoir at chemical potential $\mu$ ; the fixed- $N$ constraint becomes a Lagrange multiplier, and the levels decouple. We take from the quantum-mechanics track exactly one input and no more. The article aq-sm-occupation-number-representation (built on QM's aq-qm-occupation-number-fock-space) establishes that a many-body state of identical particles is nothing but a tally $\{n_\varepsilon\}$ , with $n_\varepsilon\in\{0,1\}$ for fermions and $n_\varepsilon\in\{0,1,2,\dots\}$ for bosons. That counting rule is the whole of the quantum microscopy we need. What this article owns is the statistical machinery that turns those tallies into thermodynamics. The decisive structural fact is that energy and number are both additive over levels, $E=\sum_\varepsilon\varepsilon\,n_\varepsilon$ and $N=\sum_\varepsilon n_\varepsilon$ , so the grand canonical weight $e^{-\beta(E-\mu N)}$ factorizes into a product of single-level pieces. The grand partition function becomes $\mathcal{Z}=\prod_\varepsilon\mathcal{Z}_\varepsilon$ , and each factor is an elementary sum — a two-term binomial $1+ze^{-\beta\varepsilon}$ for fermions, a geometric series $1/(1-ze^{-\beta\varepsilon})$ for bosons. From this one factorization the mean occupation follows by a single logarithmic derivative, $\langle n_\varepsilon\rangle = 1/(e^{\beta(\varepsilon-\mu)}\pm1)$ , the upper sign fermionic and the lower bosonic. This article sets up that framework and delivers the general $\langle n_\varepsilon\rangle$ ; the two specialized distributions and everything they imply — the Bose $-1$ that seeds condensation, the Fermi $+1$ that caps occupation at one — are worked out in aq-sm-bose-einstein-distribution and aq-sm-fermi-dirac-distribution.

Key takeaways

The grand canonical ensemble fixes $T$ and $\mu$ and lets the gas exchange energy and particles with a reservoir; for an ideal quantum gas it is the natural framework because the symmetrization constraint couples particle number to occupation.
Because the Hamiltonian is a sum over independent single-particle modes, the grand partition function factorizes: $\mathcal{Z}=\prod_\varepsilon\mathcal{Z}_\varepsilon$ , one independent open subsystem per level.
Each single-level sum is elementary: a two-term binomial $1+ze^{-\beta\varepsilon}$ for fermions (occupations $\{0,1\}$ ), a geometric series $1/(1-ze^{-\beta\varepsilon})$ for bosons (occupations $\{0,1,2,\dots\}$ ).
One logarithmic derivative $\langle n_\varepsilon\rangle = z\,\partial_z\ln\mathcal{Z}_\varepsilon$ gives both quantum distributions at once: $\langle n_\varepsilon\rangle = 1/(e^{\beta(\varepsilon-\mu)}\pm1)$ .
The Bose geometric series converges only for $\mu<\varepsilon_0$ (the ground-level energy) — a mathematical ceiling that becomes the physics of Bose-Einstein condensation; the Fermi sum has no such ceiling.
Key idea: for non-interacting indistinguishable particles the grand canonical weight $e^{-\beta(\varepsilon-\mu)n}$ factorizes over single-particle levels, so each level is an independent open subsystem; both quantum distributions differ by exactly one sign, flipped by whether the occupation sum is a two-term binomial or an unbounded geometric series.

The core insight

The grand canonical weight of an ideal quantum gas factorizes over single-particle levels: each level is an independent open subsystem in equilibrium with the particle reservoir. Summing over its allowed occupations — a two-term binomial $1+x$ for fermions, an unbounded geometric series $1/(1-x)$ for bosons, with $x=ze^{-\beta\varepsilon}$ — and differentiating once gives $\langle n_\varepsilon\rangle=x/(1\pm x)=1/(e^{\beta(\varepsilon-\mu)}\pm1)$ . Bose and Fermi statistics are one calculation separated by a single sign.

Where this lives today

This factorized grand-canonical calculation is the workhorse of condensed-matter and atomic physics. Applied to electrons with a Fermi–Dirac occupation and a $\sqrt{\varepsilon}$ density of states it gives the free-electron model of metals — the Fermi energy, the linear electronic heat capacity, Pauli paramagnetism, and the entire physics of the Fermi surface probed by quantum-oscillation experiments. Applied to bosons it predicts Bose–Einstein condensation, realized in 1995 in dilute atomic vapors and now a routine platform for quantum simulation, atom interferometry, and studies of superfluidity. With $\mu=0$ it reproduces Planck radiation and Debye phonons. The same $\Omega=-k_BT\ln\mathcal{Z}$ is what modern many-body codes evaluate when they compute the equation of state of a neutron-star crust or a trapped ultracold gas.

The physical picture

Picture the spectrum of single-particle levels as a row of boxes, each connected to a huge reservoir held at chemical potential $\mu$ . Because the particles do not interact, what happens in one box is utterly independent of the others — the reservoir mediates no correlation between levels. The whole gas is therefore a collection of independent subsystems: two-state boxes for fermions (empty or singly filled), many-state boxes for bosons (any occupancy). Equilibrium statistical mechanics of the gas is nothing but equilibrium statistical mechanics of one box, repeated over the spectrum. The chemical potential is the "price" of adding one particle. A level with $\varepsilon<\mu$ is cheap, so it is heavily occupied; a level with $\varepsilon>\mu$ is expensive, so it is nearly empty; the transition has width $\sim k_BT$ . Fermions hit a ceiling of one particle per box, so their occupation can never exceed unity. Bosons have no ceiling — which is exactly why $\mu$ must stay below the ground-level energy for a Bose gas, or the lowest box would demand an infinite, and then negative, occupation.

Figure

One factorized calculation, two sign choices: fermion binomial box versus boson geometric box

The whole of quantum statistics for an ideal gas is one box calculation done twice: a two-term binomial (fermion) or an unbounded geometric series (boson). The two answers differ by a single sign.

Figure

The general occupation curve: FD saturates at 1, BE diverges as ε to μ, MB is the shared classical tail

The punchline of the framework: the general $\langle n_\varepsilon\rangle=1/(e^{\beta(\varepsilon-\mu)}\pm1)$ carries both statistics. The $+1$ pins the Fermi curve below one; the $-1$ lets the Bose curve blow up as $\varepsilon\to\mu$ — the mathematical seed of condensation.

The core equations

For non-interacting particles the total energy and number are additive over levels, so the grand canonical Boltzmann weight of a whole configuration is a product of one-level weights. Summing over all configurations therefore factorizes into a product of independent single-level sums — the step that fails at fixed $N$ , where the constraint $\sum_\varepsilon n_\varepsilon=N$ ties the levels together.

\mathcal{Z} = \sum_{\{n_\varepsilon\}} e^{-\beta\sum_\varepsilon (\varepsilon-\mu)\,n_\varepsilon} = \prod_\varepsilon \mathcal{Z}_\varepsilon, \qquad \mathcal{Z}_\varepsilon = \sum_{n} \left(z\,e^{-\beta\varepsilon}\right)^{n} \qquad (1)

where: $\mathcal{Z}$ — grand partition function; $\{n_\varepsilon\}$ — occupation configuration over single-particle levels; $\beta=1/(k_BT)$ — inverse temperature; $\mu$ — chemical potential set by the reservoir; $z=e^{\beta\mu}$ — the fugacity; $\varepsilon$ — single-particle level energy; $\mathcal{Z}_\varepsilon$ — single-level grand partition function. Physical reading: the gas is a warehouse of independent boxes, one per level; the fugacity $z$ is the reservoir's "willingness to supply" a particle. The occupation range is the only place the quantum statistics enters. Fermions allow $n\in\{0,1\}$ (a finite binomial), bosons allow $n\in\{0,1,2,\dots\}$ (a geometric series that converges only when its ratio is below one).

\mathcal{Z}_\varepsilon = \begin{cases} 1 + z\,e^{-\beta\varepsilon} & \text{(Fermi, } n\in\{0,1\}) \\[4pt] \dfrac{1}{1 - z\,e^{-\beta\varepsilon}} & \text{(Bose, } n\in\{0,1,2,\dots\},\ z\,e^{-\beta\varepsilon}<1) \end{cases} \qquad (2)

where: $z\,e^{-\beta\varepsilon}=e^{-\beta(\varepsilon-\mu)}$ — the per-particle grand-canonical weight of the level. Physical reading: the fermion box is a coin (in or out); the boson box is an unlimited stack whose geometric mean fill grows without bound as its weight approaches one. The grand potential collects the logs, and its derivative with respect to a level's energy is that level's mean occupation. Taking $z\,\partial_z\ln\mathcal{Z}_\varepsilon$ brings a factor of $n$ down inside the sum, which is exactly the average.

\Omega = -k_BT\ln\mathcal{Z} = \mp\,k_BT\sum_\varepsilon \ln\!\left(1\pm z\,e^{-\beta\varepsilon}\right), \qquad \langle n_\varepsilon\rangle = z\,\frac{\partial\ln\mathcal{Z}_\varepsilon}{\partial z} = \frac{1}{e^{\beta(\varepsilon-\mu)}\pm 1} \qquad (3)

where: $\Omega=E-TS-\mu N=-PV$ — the grand potential; the upper sign is Fermi, the lower Bose. Physical reading: one master formula holds both quantum distributions, distinguished by a single $\pm$ . The number that fixes $\mu$ is the sum of these occupations, $N=\sum_\varepsilon\langle n_\varepsilon\rangle$ , and the pressure is $PV=-\Omega$ — no further work. The fluctuations follow from a second derivative and expose the statistics in a different way: the variance of a level's occupation is $\langle n\rangle$ times $(1\mp\langle n\rangle)$ , sub-Poissonian for fermions (a filled level cannot fluctuate up), super-Poissonian for bosons (they bunch).

\mathrm{Var}(n_\varepsilon) = k_BT\,\frac{\partial\langle n_\varepsilon\rangle}{\partial\mu} = \langle n_\varepsilon\rangle\left(1\mp\langle n_\varepsilon\rangle\right) \qquad (4)

where: the upper sign ( $-$ in $1\mp$ ) is Fermi (variance vanishes at $\langle n\rangle=0$ and $1$ ), the lower ( $+$ ) is Bose (variance grows quadratically — bunching). Physical reading: fermions damp their own fluctuations by exclusion; bosons amplify theirs by sociability.

Derivations

Derivation ABoth quantum distributions from one factorized grand-canonical sum

The goal is to obtain the mean occupation $\langle n_\varepsilon\rangle$ for fermions and bosons in a single unified calculation, showing that the two statistics are one algebraic problem separated only by the range of the occupation sum. We use exactly one quantum input — that a many-body state is a tally $\{n_\varepsilon\}$ with $n_\varepsilon\in\{0,1\}$ (fermions) or $\{0,1,2,\dots\}$ (bosons), established by aq-sm-occupation-number-representation — and derive everything else from the additivity of energy and number over levels. This is the structural result the whole chapter of quantum gases rests on.

Step 1. Write the grand-canonical probability of a whole configuration. In the ensemble at $(T,\mu)$ the weight of a configuration $\{n_\varepsilon\}$ is $e^{-\beta\sum_\varepsilon(\varepsilon-\mu)n_\varepsilon}/\mathcal{Z}$ , because energy $E=\sum_\varepsilon\varepsilon n_\varepsilon$ and number $N=\sum_\varepsilon n_\varepsilon$ are both additive over levels. Why: non-interacting particles carry no cross-level energy, so the exponent is a plain sum over modes.

Step 2. Factorize. The exponential of a sum is a product, $e^{-\beta\sum_\varepsilon(\varepsilon-\mu)n_\varepsilon}=\prod_\varepsilon e^{-\beta(\varepsilon-\mu)n_\varepsilon}$ , and summing each occupation independently distributes the total sum through the product: $\mathcal{Z}=\prod_\varepsilon\mathcal{Z}_\varepsilon$ with $\mathcal{Z}_\varepsilon=\sum_n x^n$ and $x=e^{-\beta(\varepsilon-\mu)}=z\,e^{-\beta\varepsilon}$ . Why: the occupations are unconstrained in the grand ensemble, so the levels are genuinely independent — the step that fails at fixed $N$ .

Step 3. Do the two sums. For fermions the exclusion truncates the sum after two terms, $\mathcal{Z}_\varepsilon=\sum_{n=0}^{1}x^n=1+x$ . For bosons the sum is a geometric series, $\mathcal{Z}_\varepsilon=\sum_{n=0}^{\infty}x^n=1/(1-x)$ , convergent only if $x<1$ , i.e. $\mu<\varepsilon$ . Why: the whole quantum-statistical content is here — a finite binomial versus an infinite geometric series, one algebraic step apart.

Step 4. Extract the mean by a logarithmic derivative. Because $x=z\,e^{-\beta\varepsilon}$ , the operator $z\,\partial_z$ acts as $x\,\partial_x$ , and $x\,\partial_x\ln\mathcal{Z}_\varepsilon$ brings a factor $n$ down inside the sum to give $\langle n\rangle$ . Fermi: $x\,\partial_x\ln(1+x)=x/(1+x)$ . Bose: $x\,\partial_x\ln\!\big(1/(1-x)\big)=x/(1-x)$ . Why: the log-derivative is the generating-function trick for the first moment.

Step 5. Rearrange to standard form. In both cases $\dfrac{x}{1\pm x}=\dfrac{1}{x^{-1}\pm1}=\dfrac{1}{e^{\beta(\varepsilon-\mu)}\pm1}$ , the master occupation formula, $+$ Fermi, $-$ Bose. The two distributions are literally the same expression with one sign flipped. Why: multiplying numerator and denominator by $e^{\beta(\varepsilon-\mu)}$ is the only algebra needed, and it treats both signs identically. Verified: the geometric and binomial sums, the log-derivative $\langle n\rangle=x/(1\pm x)$ , the standard-form rewrite, and the unified $s=\pm1$ selection are all checked symbolically in verify_grand-canonical-ideal-quantum-gases.py.

Go deeper · Derivation B · 2 minThe Bose convergence bound is the chemical-potential ceiling

Derivation BThe Bose convergence bound is the chemical-potential ceiling

The goal is the structural result this article owns: that the mathematical requirement for the boson geometric series to converge — nothing more than $|x|<1$ — is identical to the physical ceiling $\mu<\varepsilon_0$ on the chemical potential of an ideal Bose gas, and that this ceiling is precisely where the ground-level occupation diverges and Bose-Einstein condensation begins. Fermions have no analogous bound because their sum is finite for any argument. We prove the equivalence and locate the divergence, then hand the condensation physics itself to aq-sm-bose-einstein-condensation-ideal-gas.

Step 1. State the convergence condition. The single-level boson sum $\mathcal{Z}_\varepsilon=\sum_{n\ge0}x^n$ converges to $1/(1-x)$ only when its ratio $x=z\,e^{-\beta\varepsilon}<1$ . Applied to every level, the tightest condition is at the lowest energy $\varepsilon_0$ , where $x$ is largest. Why: a geometric series with ratio $\ge1$ diverges term by term, so the partition function is undefined.

Step 2. Translate to the chemical potential. The condition $z\,e^{-\beta\varepsilon_0}<1$ is $e^{\beta(\mu-\varepsilon_0)}<1$ , i.e. $\mu<\varepsilon_0$ . Why: taking the log of the fugacity condition converts a bound on $z$ into a bound on $\mu$ .

Step 3. Show the ceiling is a genuine divergence, not a bookkeeping choice. As $\mu\to\varepsilon_0^-$ the ground-level occupation $\langle n_0\rangle=1/(e^{\beta(\varepsilon_0-\mu)}-1)$ grows without bound; numerically, at $\beta=1$ and $\varepsilon_0=0$ it climbs through $\langle n_0\rangle\approx 9.5,\,99.5,\,999.5$ as $\mu\to0^-$ by factors of ten. Push $\mu$ past $\varepsilon_0$ and the formula returns a negative "occupation" — the unmistakable sign that the ensemble has left its domain. Why: the pole of $1/(e^{\beta(\varepsilon_0-\mu)}-1)$ at $\mu=\varepsilon_0$ is the ground state demanding a macroscopic population, which is condensation.

Step 4. Contrast the fermion case. The Fermi sum $1+x$ converges for any $x\ge0$ , so $\mu$ (the Fermi level) may sit anywhere — above or below any level — and $\langle n\rangle=1/(e^{\beta(\varepsilon-\mu)}+1)$ stays safely in $[0,1]$ . There is no ceiling and no condensation. Why: a two-term sum has no convergence requirement, so exclusion removes the very instability that defines the Bose gas. Verified: the geometric convergence branch, the $\mu<\varepsilon_0$ translation, the monotone divergence of $\langle n_0\rangle$ as $\mu\to0^-$ , the negative-occupation flag for $\mu>\varepsilon_0$ , and the Fermi $\langle n\rangle\in[0,1]$ bound are checked in verify_grand-canonical-ideal-quantum-gases.py.

Limiting cases

With $z=e^{\beta\mu}$ and $x_\varepsilon=z\,e^{-\beta\varepsilon}$ , the mean occupation $\langle n_\varepsilon\rangle=x_\varepsilon/(1\pm x_\varepsilon)$ interpolates between regimes:

Regime	Result	Meaning
Classical (non-degenerate)	$z\to0:\ \langle n_\varepsilon\rangle\approx z\,e^{-\beta\varepsilon}\ll1$	All occupations tiny; Bose and Fermi merge into Maxwell–Boltzmann (aq-sm-maxwell-boltzmann-classical-limit). Holds when $n\lambda^3\ll1$ .
Degenerate Fermi gas	$\beta(\varepsilon-\mu)\to-\infty:\ \langle n_\varepsilon\rangle\to1$	Levels below $\mu$ filled to the exclusion ceiling; at $T=0$ a sharp Fermi surface at $\varepsilon=\mu=\varepsilon_F$ .
Degenerate Bose gas	$\mu\to\varepsilon_0^-:\ \langle n_{\varepsilon_0}\rangle\to\infty$	Ground-level occupation diverges; a macroscopic condensate forms. Requires $\mu\le\varepsilon_0$ always.
High-energy tail	$\varepsilon-\mu\gg k_BT:\ \langle n_\varepsilon\rangle\approx e^{-\beta(\varepsilon-\mu)}$	Both statistics decay with the Boltzmann factor, whatever is happening elsewhere in the spectrum.
Photon / phonon gas	$\mu=0$ (number not conserved): $\langle n_\varepsilon\rangle=1/(e^{\beta\varepsilon}-1)$	The Bose result with zero chemical potential — Planck radiation and Debye phonons (aq-sm-photon-gas-planck-radiation).

Common trapThe factorization needs the grand ensemble, and μ is not a free knob

The product $\mathcal{Z}=\prod_\varepsilon\mathcal{Z}_\varepsilon$ holds only in the grand canonical ensemble. In the canonical ensemble the constraint $\sum_\varepsilon n_\varepsilon=N$ couples the modes and the partition function does not factorize — this is the entire reason we go grand-canonical for quantum gases. A second trap: $\mu$ is not an independent parameter you dial freely. It is fixed implicitly by the number equation $N=\sum_\varepsilon\langle n_\varepsilon\rangle$ , and for bosons it is bounded above by $\varepsilon_0$ . Treating $\mu$ as free and setting it above $\varepsilon_0$ produces negative "occupations" — a signal that condensation has taken over, not a value to be used.

Worked examples

Example 1 — Occupation of a level near the chemical potential, both statistics.

Step 1. Set the scale. At $T=300\,\mathrm{K}$ , $k_BT=1.381\times10^{-23}\times300=4.14\times10^{-21}\,\mathrm{J}=\mathbf{0.02585\ eV}$ .

Step 2. Fermi level $0.1\,\mathrm{eV}$ below $\mu$ . Then $\beta(\varepsilon-\mu)=-0.1/0.02585=-3.868$ , so $\langle n\rangle=1/(e^{-3.868}+1)=1/(0.0209+1)=\mathbf{0.980}$ — nearly full, as a level well below the Fermi level should be.

Step 3. Fermi level $0.1\,\mathrm{eV}$ above $\mu$ . Now $\beta(\varepsilon-\mu)=+3.868$ and $\langle n\rangle=1/(e^{3.868}+1)=1/(47.85+1)=\mathbf{0.0205}$ — nearly empty. The two answers add to $\approx1$ , the particle–hole symmetry about $\mu$ .

Step 4. A boson level $0.01\,\mathrm{eV}$ above $\mu$ (bosons need $\varepsilon>\mu$ ). Here $\beta(\varepsilon-\mu)=0.3868$ , so $\langle n\rangle=1/(e^{0.3868}-1)=1/(0.4722)=\mathbf{2.12}$ — already above one, and it would run to infinity as the gap closes. Same framework, opposite sign, wildly different behavior.

Example 2 — The Fermi energy of copper as the ground state of the framework.

Step 1. Setup. Copper has one conduction electron per atom, number density $n=8.5\times10^{28}\,\mathrm{m^{-3}}$ ; electrons are spin- $\tfrac12$ fermions of mass $m=9.11\times10^{-31}\,\mathrm{kg}$ . At $T=0$ the Fermi–Dirac occupation is a step: every level up to $\varepsilon_F=\mu(0)$ is filled, all above empty.

Step 2. Filling the sphere. With the free-particle density of states, the number equation $N=\int_0^{\varepsilon_F}g(\varepsilon)\,d\varepsilon$ inverts to $\varepsilon_F=\dfrac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ (the density-of-states machinery is owned by aq-sm-density-of-states-energy; we use its result).

Step 3. Plug in. $(3\pi^2 n)^{2/3}=(3\pi^2\times8.5\times10^{28})^{2/3}=1.81\times10^{19}\,\mathrm{m^{-2}}$ , and $\hbar^2/2m=6.10\times10^{-39}\,\mathrm{J\,m^2}$ , so $\varepsilon_F=6.10\times10^{-39}\times1.81\times10^{19}=1.13\times10^{-18}\,\mathrm{J}=\mathbf{7.05\ eV}$ — matching the measured copper value of $7.0\,\mathrm{eV}$ .

Step 4. The consequence. The Fermi temperature is $T_F=\varepsilon_F/k_B=\mathbf{8.2\times10^{4}\ K}$ . Room temperature is $T/T_F\approx0.0037\ll1$ : the electron gas is strongly degenerate, the Fermi–Dirac step barely smeared. This is why metals have a linear-in- $T$ electronic heat capacity — worked out in aq-sm-electronic-heat-capacity-metals.

Origins · 1 minfermi gas and how this was discovered

fermi gas

ideal quantum gas concept · The ideal quantum gas — non-interacting indistinguishable particles whose only correlation is the symmetrization postulate — is the arena this article sets up. Its grand-canonical treatment rests on Josiah Willard Gibbs's 1902 ensemble language, then acquires quantum content through Bose and Einstein (1924–25) for bosons and Fermi and Dirac (1926) for fermions. The decisive move is to relax the fixed-number constraint: only then do the single-particle levels decouple, turning an intractable coupled sum into one independent-box calculation per level. That factorization is why a gas of quantum particles is tractable at all, and it is the launch pad for the Bose–Einstein and Fermi–Dirac distributions derived next.

History

Origins · 1 minhow this was discovered

History

1900

— Planck fits blackbody radiation, implicitly using the Bose factor for photons

1902

— Gibbs formulates the grand canonical ensemble in Elementary Principles in Statistical Mechanics

1924

— Bose derives Planck's law by counting photon states, Einstein extends the idea to material gases

1926

— Fermi and, independently, Dirac introduce the antisymmetric statistics for particles obeying exclusion

1927

— Sommerfeld applies Fermi–Dirac statistics to the free-electron gas in metals

1938

— London links Bose–Einstein condensation to the superfluidity of liquid helium-4

1995

— Bose–Einstein condensation realized in dilute rubidium and sodium vapors

Going further

Texts: Pathria & Beale, Statistical Mechanics (3rd ed.), Ch. 6; Kardar, Statistical Physics of Particles, Ch. 7; Reif, Fundamentals of Statistical and Thermal Physics, Ch. 9; Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 7; Landau & Lifshitz, Statistical Physics Part 1, §53–56.
The quantum input: the occupation-tally description of many-body states is owned by aq-sm-occupation-number-representation (built on the QM article aq-qm-occupation-number-fock-space); this article supplies the statistical machinery, not the microscopic one.
Leads to: the two specialized distributions aq-sm-bose-einstein-distribution and aq-sm-fermi-dirac-distribution; the classical crossover aq-sm-maxwell-boltzmann-classical-limit; the degeneracy parameter aq-sm-quantum-concentration-classicality-criterion; and downstream, degenerate Fermi and Bose gases in SM5.
Provenance: AllQFT `sm` export (Pathria & Beale), aq-sm-grand-canonical-ideal-quantum-gases, ref STAT.QSTAT.GCIQG.1204. Every distribution, sign, limit, and the Cu Fermi energy ( $7.05\,\mathrm{eV}$ ) are python-verified (22 checks, all PASS).

Why go grand-canonical instead of just fixing N?

Because at fixed $N$ the constraint $\sum_\varepsilon n_\varepsilon=N$ links every level to every other, and the partition function does not factorize — you would face a genuinely coupled combinatorial sum. Relaxing $N$ and introducing $\mu$ turns the constraint into a Lagrange multiplier that acts level by level, so each level becomes an independent open subsystem and the whole calculation collapses to one box done once. The ensembles agree in the thermodynamic limit, so nothing physical is lost.

Is the chemical potential something I choose?

No. In practice $N$ , $V$ , and $T$ are given, and $\mu$ is whatever value makes $N=\sum_\varepsilon\langle n_\varepsilon\rangle$ come out right. You solve that implicit equation for $\mu(N,T,V)$ . For fermions $\mu$ can be anywhere (it becomes the Fermi energy at $T=0$ ); for bosons it is pinned below $\varepsilon_0$ and, at the condensation point, sticks essentially at $\varepsilon_0$ while the surplus particles pour into the ground state.

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

It derives the general occupation $\langle n_\varepsilon\rangle=1/(e^{\beta(\varepsilon-\mu)}\pm1)$ and the factorization that produces it, which is the shared backbone of both. The dedicated treatments — the Bose $-1$ with its condensation ceiling, the Fermi $+1$ with its exclusion cap and $T=0$ step — are the subjects of aq-sm-bose-einstein-distribution and aq-sm-fermi-dirac-distribution. Think of this article as the launch pad and those two as the specializations.

Where does the density of states come in?

It does not appear in the single-level occupation $\langle n_\varepsilon\rangle$ , which is per level. It enters when you sum over levels to get the total number or energy, $N=\int g(\varepsilon)\langle n_\varepsilon\rangle\,d\varepsilon$ . The function $g(\varepsilon)$ that converts a sum over discrete levels into an energy integral is owned by aq-sm-density-of-states-energy and aq-sm-density-of-states-k-space; the number equation that fixes $\mu$ (and the Fermi energy in Example 2) uses it.

Why do the fluctuations differ for bosons and fermions?

Both come from the same identity $\mathrm{Var}(n)=k_BT\,\partial\langle n\rangle/\partial\mu=\langle n\rangle(1\mp\langle n\rangle)$ , but the sign flips the story. For fermions ( $1-\langle n\rangle$ ) a nearly empty or nearly full level cannot fluctuate — exclusion forbids a second particle — so the variance is sub-Poissonian and vanishes at both ends. For bosons ( $1+\langle n\rangle$ ) the variance exceeds the Poisson value by $\langle n\rangle^2$ : bosons bunch, arriving in correlated groups, as measured directly in the Hanbury Brown–Twiss experiment.

Does the Gibbs 1/N! come out of this, or is it inserted by hand?

It comes out. In the classical limit $z\to0$ , the grand potential collapses to $\Omega\to-k_BT\,z\,Z_1$ , and converting back to the Helmholtz free energy gives $Z_{\mathrm{can}}=Z_1^N/N!$ automatically, with the $N!$ appearing as $N\ln N-N$ from Stirling. The factorial that classical statistical mechanics has to postulate to fix extensivity is here a consequence of counting indistinguishable occupations — the subject of aq-sm-gibbs-correction-boltzmann-counting.

What sets the boundary between quantum and classical behavior?

The fugacity $z=e^{\beta\mu}$ , equivalently the phase-space density $n\lambda^3$ where $\lambda$ is the thermal de Broglie wavelength. When $z\to0$ (equivalently $n\lambda^3\ll1$ ), occupations are tiny, the $\pm1$ is negligible, and both quantum distributions collapse to Maxwell–Boltzmann. Degeneracy — the regime where quantum statistics matters — sets in when $z$ becomes of order one, i.e. $n\lambda^3\sim1$ . The dimensionless number $n\lambda^3$ is developed in aq-sm-quantum-concentration-classicality-criterion.

Grand Canonical Ensemble for Ideal Quantum Gases

Where this lives today

The physical picture

The core equations

Derivations

Limiting cases

Worked examples

History

Going further

Before this

Where this leads