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For noninteracting particles in a three-dimensional isotropic quantum harmonic oscillator, measure the one-particle ground energy as zero. The single-particle partition function is $Z_1=(1-e^{-\beta \hslash \omega }{)}^{-3}$, from the product of three geometric series. In the dilute Maxwell–Boltzmann statistics regime, $\ln \Xi =e^{\beta \mu }{Z}_1$, so the grand potential is the displayed expression and $N=e^{\beta \mu }{Z}_1$. For $k_{B}T\gg \hslash \omega$,The same approximation follows from integrating the density of states $g(E)\simeq E^2/[2(\hslash \omega {)}^3]$ against $e^{-\beta E}$. Extrapolating the classical formula until the fugacity is one gives the scale of Bose condensation in a three-dimensional harmonic trap, but the quantitative transition temperature requires the Bose occupation law rather than Maxwell-Boltzmann statistics.

A two-component Nambu spinor $({\psi }_{k\uparrow },{\bar{\psi }}_{-k\downarrow })$ for every k accounts for both spin species once after relabelling. Its Grassmann integral contributes one functional determinant. A fully doubled spin-particle-hole basis instead has redundancy and requires one half in the trace logarithm or a corresponding Pfaffian convention. A reliable check is the zero-gap limit: the grand potential must become $-2{\beta }^{-1}{\sum }_k\ln (1+e^{-\beta {\xi }_k})$, with exactly two spin states and no invented vacuum offset.

For noninteracting fermions with density of states $g$, define ${\Phi }_0(x)=-{\int }_0^x(x-y)g(y)dy$. The grand potential is ${\int }_0^{\infty }g(ϵ)F(ϵ)dϵ$, where $F=k_{B}T\ln (1-n_F)$ has derivative $F^{\prime }=n_F$. Integrating by parts twice gives the title expression, since ${\Phi }_0(0)={\Phi }_0^{\prime }(0)=0$ and upper boundary terms vanish for ordinary polynomially growing densities of states. The Fermi window therefore thermally averages the zero-temperature grand potential as a function of trial Fermi energy.

For one internal species, the grand canonical partition function of a classical ideal gas satisfies $\ln \Xi =zV/{\lambda }^3$, with quantum gas fugacity $z=e^{\beta \mu }$ and thermal de Broglie wavelength $\lambda =\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$. Differentiating the grand potential gives $N=zV/{\lambda }^3$, hence $\mu =k_{B}T\ln (n{\lambda }^3)$. An internal partition factor must be included consistently if internal states are available.

Differentiating the grand partition function at fixed temperature and volume gives $⟨N⟩=-{\partial }_{\mu }\Phi$ and $\mathrm{Var}N=-k_{B}T{\partial }_{\mu }^2\Phi =k_{B}T{\partial }_{\mu }⟨N⟩$. It follows because derivatives of the logarithm of the state sum generate cumulants of particle number. Nonnegative variance makes the grand potential concave in chemical potential in a stable grand-canonical description.

For a uniform exchange field with cost $N{M}^2/(2U)$ and spin-band energies ${ϵ}_k-\sigma M$, the grand potential is even in M. Its curvature at zero field is $N[U^{-1}-{\chi }_0]$, with ${\chi }_0=\int \nu (-n_F^{\prime })dϵ$ and spin-summed density of states per site. Negative curvature proves local paramagnetic instability. It does not by itself exclude a first-order transition before that instability, nor determine the stable finite-field amplitude.

Assume $A>0$ and $\alpha >0$, so the excitation energy increases with momentum and vanishes in the ground state. With one internal state, the number of states in a spherical momentum shell is $V4\pi k^{2}dk/(2\pi {)}^3$. Since $k=(ϵ/A{)}^{1/\alpha }/\hslash$,An internal degeneracy multiplies this density of states by its degeneracy factor.

The Bose-Einstein distribution is $\bar{n}(ϵ)=[e^{(ϵ-\mu )/(k_{B}T)}-1{]}^{-1}$. For a finite nonnegative occupation of the zero-energy state, its denominator must be positive, requiring $\mu <0$. At Bose-Einstein condensation the thermodynamic-limit chemical potential approaches $0$ from below, and the ground state must be counted separately; therefore “$\mu <0$” is a finite-system condition, while the condensed limit has $\mu =0$.

In a box whose shape is fixed, each momentum quantum number scales as $V^{-1/3}$, so each energy level scales as $V^{-\alpha /3}$. Differentiating the grand potential $\Omega =k_{B}T{\sum }_j\log (1-e^{-({ϵ}_j-\mu )/(k_{B}T)})$ at fixed $T,\mu$ givesThe zero-energy condensate contributes neither pressure nor kinetic energy.

The number outside the ground state isAt $\mu \uparrow 0$, its low-energy integrand is proportional to ${ϵ}^{3/\alpha -2}$. The integral is finite precisely when $3/\alpha >1$, soFor these exponents, the maximum excited population isThis follows by expanding the Bose denominator in positive exponentials and integrating each term. It tends to zero as $T\downarrow 0$, forcing excess particles into the ground state at fixed nonzero total number. At $\alpha =3$ the divergence is logarithmic; for $\alpha >3$ it is stronger, so there is no positive-temperature thermodynamic-limit condensation of this homogeneous gas.

Photons have $\alpha =1$ and two polarizations, but in ordinary blackbody equilibrium their number is not conserved and $\mu =0$ at every temperature. Cooling reduces their number rather than forcing an excess into a ground state, so an ordinary equilibrium photon gas does not condense. Systems imposing an effective photon-number constraint and an appropriate confined spectrum are different from the unconstrained photon gas considered here.

At $U=0$, use fermionic Matsubara frequencies ${\omega }_r=(2r+1)\pi /\beta$. The fermionic Gaussian integral has inverse propagator $-i{\omega }_r+{\xi }_{\mathbf{k}}$. With its thermal normalization, the determinant for one single-particle level is $1+e^{-\beta \xi }$, as also follows from its empty and occupied states. Two spin species therefore giveThe expression labeled Helmholtz free energy in the PDF is the grand potential, since the trace contains $H-\mu \hat{N}$. The thermodynamic Helmholtz free energy at a prescribed electron number $N_e$ is $F={\Omega }_0+\mu N_e$, with $\mu$ fixed by $N_e=2{\sum }_{\mathbf{k}}{n}_F({ϵ}_{\mathbf{k}})$. The Matsubara determinant requires a normalization prescription; a raw infinite sum of logarithms is not an ordinary convergent sum.

The $-\mu \psi$ term means that the generator is the grand potential, rather than just the physical energy. With boundary variations vanishing, the Hamiltonian functional isIts functional derivative isso the Gross–Pitaevskii equation is $i\hslash {\psi }_t=\delta H/\delta {\psi }^{\ast }$. Treating $\psi$ and ${\psi }^{\ast }$ as independent variables in this variation gives the factor $g$, not $2g$.

For repulsive interactions $g>0$ and $\mu >0$, complete the square in the local energy:The gradient contribution is nonnegative, so the homogeneous ground state has arbitrary constant complex argument andThe infinite-volume constant may be subtracted to define a relative grand potential; it does not affect the functional derivative. At fixed particle number one instead minimizes the physical energy $H+\mu N$, with $\mu$ as its Lagrange multiplier. For $g>0$, $\mu \le 0$ the grand-potential minimum over $n\ge 0$ is the vacuum $n=0$. For $g<0$ the displayed unconstrained functional is unbounded below; the positive-density stable uniform state assumed in the following parts requires repulsion.

Suppress tildes in the normalized equation and let $X=x-vt$. A Gross–Pitaevskii solitary wave stationary in these translating coordinates obeysThe boundary fixes the bulk complex argument and describes a localized disturbance with zero net vortex winding number; a vortex pair is possible. No extra plane-wave factor is needed when only the coordinates are changed. A full Galilean transformation of the field would additionally change the bulk flow and its complex argument convention.

Subtract the uniform background from the grand potential. A convenient dimensionless energy and the renormalized momentum of a condensate areThe subtraction in $\mathbf{p}$ removes the troublesome background complex argument gradient. In particular, for the usual subsonic localized far field $\psi -1=O(r^{-1})$, $\nabla \psi =O(r^{-2})$, and $1-|\psi {|}^2=O(r^{-2})$, the energy density is $O(r^{-4})$ and the momentum integrand is $O(r^{-3})$, both integrable in two dimensions. Merely writing $\psi \to 1$ without sufficient decay is not by itself a convergence proof. The common solitary-wave branch is subsonic: linearizing this normalization gives ${\omega }^2=k^2/2+k^4/4$ and speed of sound $1/\sqrt{2}$.

Integration by parts, with decaying variations and their boundary terms vanishing, givesThus the translating equation is precisely $\delta (E-v{p}_x)=0$. Along a differentiable family ${\psi }_{\lambda }$ with fixed bulk normalization, evaluate this stationarity on ${\partial }_{\lambda }\psi$ to obtainwhere $d{p}_x/d\lambda \ne 0$. This energy–momentum slope of a solitary wave does not require differentiating the Lagrange multiplier $v$ inside the field variation. At a turning point the parametrized identity remains true, while $E$ need not be a single-valued function of $p_x$ locally. The chosen factors make $i{\psi }_t=\delta E/\delta {\psi }^{\ast }$, consistent with the normalized equation; changing the normalization of $E$ alone would change the claimed slope.

Differentiating the grand potential givesA reversible adiabatic change in an isolated gas keeps $S$ and $N$ fixed. To infer constant $y$, one must check that this ratio is one-to-one, rather than merely assuming it.

Put $q=y/k$, $n(x)=1/(1+e^{x-q})$, $F_j={\int }_0^{\infty }{x}^{j}n(x)dx$ and $M_j={\int }_0^{\infty }{x}^{a+j}n(x)(1-n(x))dx$ for $j=0,1,2$. Integration by parts gives $M_1=b{F}_a$ and $M_2=(b+1)F_{a+1}$, andThe strict inequality is the Cauchy-Schwarz inequality for the positive weight $x^{a}n(x)(1-n(x))$, since $x$ is not constant on its support. This proves strict monotonicity of Fermi gas entropy per particle and hence $\mu /T$ is constant. Then $N=kV{T}^b{f}^{\prime }(y)$ gives $V{T}^b$ constant, and $P=k{T}^{b+1}f(y)$ givesThis adiabatic scaling of a Fermi gas with power-law density of states concerns positive temperature; the zero-temperature limit is a separate limiting case.

For noninteracting fermions, each single-particle level has occupation $n_i\in \{0,1\}$ by the Pauli exclusion principle. A microstate has $E={\sum }_i{ϵ}_i{n}_i$, $N={\sum }_i{n}_i$, so its grand-canonical weight is the product of its independent level weights. The grand potential is $\Omega =-kT\log \mathcal{Z}$; the particle number, entropy and pressure follow from $N=-{\partial }_{\mu }\Omega$, $S=-{\partial }_T\Omega$ and $P=-{\partial }_V\Omega$. This occupation description is the basis for the product, continuum and adiabatic calculations.

Summing separately over the two allowed occupations of every level givesThusThe factorization uses the absence of interactions; each level, including any degeneracy labels, is counted separately.

The grand potential is

For an ultrarelativistic particle, $\epsilon =pc$. Including the two spin states of an electron, counting wavevectors in a sphere gives the density of statesThe grand canonical partition function of the ideal Fermi gas is thereforeSince the grand potential satisfies $pV=k_{B}T\log \mathcal{Z}$, integration by parts givesThusThis is the equation of state of an ideal ultrarelativistic gas.

Put $z=e^{\beta \mu }$, with $\beta =1/(k_{B}T)$. For a single quantum state of energy $ϵ$, dilute Maxwell–Boltzmann statistics gives $\ln {\Xi }_{ϵ}\simeq z{e}^{-\beta ϵ}$. For bosons this is the leading expansion of $-\ln (1-z{e}^{-\beta ϵ})$; the occupation must be small for that expansion. Thus the grand potential isWith the ground-state energy set to zero, a dilute classical gas requires $e^{\beta \mu }\ll 1$, giving negative $\mu$. This sign is relative to the chosen energy zero. For an ideal Bose gas, $\mu$ approaches the ground energy from below at condensation, so it approaches zero with this convention.

The printed degeneracy formula requires $n_x,n_y,n_z=0,1,2,\dots$, rather than strictly positive integers. With zero-point energy omitted these are the appropriate oscillator quantum numbers. The stars and bars count of three nonnegative integers summing to $n$ givesThere is one ground state at $n=0$. If the integers were strictly positive, the count would instead be $\left(\genfrac{}{}{0ex}{}{n-1}{2}\right)$ for $n\ge 3$, inconsistent with both the printed formula and its energy reference.

For large $n$, $d_n\simeq n^2/2$. Consecutive energy shells are separated by $\hslash \omega$, giving the continuum density of statesUse this when $k_{B}T\gg \hslash \omega$, and integrate the single-state grand potential over energy:because the integral is $2/{\beta }^3$. Particle number follows by differentiating at fixed temperature and trap frequency:These are the high-temperature expressions for the classical grand potential in a harmonic trap.

Extrapolating to $\mu =0$ gives the requested transition scaleIt is an order-one estimate because Maxwell-Boltzmann occupations cease to be accurate near condensation. The Bose-Einstein distribution gives the more precise ideal-trap result $N_{\mathrm{ex}}^{\max }=\zeta (3)(k_{B}T/\hslash \omega {)}^3$, and hence $T_c=(\hslash \omega /k_B)[N/\zeta (3){]}^{1/3}$, about $0.94$ times the estimate. Excess particles populate the ground state after the excited-state capacity saturates: Bose condensation in a three-dimensional harmonic trap.

The observation of Bose-Einstein condensation in rubidium in 1995 implemented laser precooling followed by radio-frequency evaporation from a magnetic trap. Imaging the released cloud revealed a narrow anisotropic component within a broader thermal population. This provides an experimental connection to the population-counting argument; the measured expansion also reflects interactions and imaging resolution, not just the ideal-trap density of states.^[ref]

Ensemble and applications. A grand canonical ensemble describes a system exchanging both energy and particles with a large reservoir. Its fixed control parameters are $T,V,\mu$, while microstates with different $N$ have probabilitiesWeak coupling allows the system's energy levels to remain meaningful; equilibrium fixes the reservoir temperature and chemical potential. The grand potential is $\Phi =-k_{B}T\ln \Xi =U-TS-\mu N$, with $N$ now the mean population. It generates entropy, pressure and mean particle number by differentiation. This ensemble is convenient for gases in contact with particle reservoirs, independently occupied quantum levels and surface adsorption; its particle-number fluctuations are part of the description, not an error in counting.

Classical single-level limit. For the single-level grand partition function in the dilute limit, put $x_k=e^{-\beta ({ϵ}_k-\mu )}$. Empty and singly occupied levels contribute $1$ and $x_k$. In the dilute classical limit $x_k\ll 1$, higher occupancies contribute only at order $x_k^2$ or beyond:This truncation does not impose the Pauli principle on a classical gas. A fermion state has exactly $1+x_k$, a boson state has $(1-x_k{)}^{-1}$, and Maxwell-Boltzmann occupancy weights give $e^{x_k}$; they all agree to this order. In particular, $\ln {\Xi }_k=x_k+O(x_k^2)$, with the classical Maxwell-Boltzmann result exactly $x_k$.

grand potential and chemical potential. With $z=e^{\beta \mu }$, summing the single-level logarithms and using the stated density of states givesSubstitute $ϵ=k_{B}T{x}^2$, for whichCombining this with the density-of-states prefactor gives $\ln \Xi =zV/{\lambda }^3$. ThusDifferentiating at fixed $T,V$,Similarly $p=-({\partial }_V\Phi {)}_{T,\mu }=k_{B}Tz/{\lambda }^3$, so $pV=N{k}_{B}T$. This is the grand-canonical chemical potential of a classical ideal gas. The condition $N{\lambda }^3/V\ll 1$ states the classical regime. The supplied density of states counts one internal state; extra internal degeneracies would have to be included consistently rather than inserted at only one stage.

One adsorption site. Take the free-vapour energy reference as zero and the binding energy as $-ϵ$. A site is either empty or singly occupied; the latter has vibrational partition function $z_s(T)$. Its adsorption-site grand partition function and mean occupation areThe chemical potential of an adsorbed species on equivalent sites follows by inverting the occupation ratioThe factor $1-\theta$ is the vacant-site probability. The binding reference and the energies included in $z_s$ must be chosen consistently, so that vibrational zero-point energy is not counted twice.

Isotherm. The vibrational factor in a Langmuir equilibrium constant is fixed by adsorption equilibrium, ${\mu }_s={\mu }_{\mathrm{vap}}$. Neglecting vapour internal degrees of freedom and writing its number density as $p/(k_{B}T)$ gives $e^{\beta {\mu }_{\mathrm{vap}}}=p{\lambda }^3/(k_{B}T)$. ThereforeThis is the Langmuir isotherm: initially linear, increasing monotonically and concave down, approaching unit coverage. Half coverage occurs at $p_{1/2}=1/b_s(T)$. The coefficient has inverse-pressure units. It assumes equivalent independent sites, at most one molecule per site and no lateral interaction or multilayer adsorption. The molecular mass appropriate to the vapour determines $\lambda$.

A real uptake measurement must separate adsorbed molecules from free gas in the apparatus. Langmuir's original mica experiment compared a mica-containing bulb with a similar blank and inferred uptake from calibrated gas volumes and pressures. His stated blank contribution and bath-temperature drift limited the inferred adsorption; a saturating curve alone does not establish ideal equivalent sites.^[ref]

An statistical ensemble specifies probabilities of microstates under particular macroscopic constraints. It represents statistical information about a system; the reservoir constraints of equilibrium ensembles specify what it can exchange with its surroundings.

The thermodynamic potentials generate mean quantities and response functions. For example, canonical energy variance is $⟨(\Delta E{)}^2⟩=k_B{T}^2{C}_V$. With short-range interactions and stable bulk phases, relative fluctuations usually vanish for large systems and these ensembles give equivalent bulk predictions after matching their mean variables. Finite systems, constraints and long-range interactions require greater care; the ensembles retain different fluctuating variables.^[ref]

Classical translational statistics. Write the number density as $n=N/V$ and the thermal de Broglie wavelength as $\lambda =\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$. Quantum exchange is negligible when wave packets rarely overlap:The equivalent state-counting statement is that the mean occupation of each thermally relevant one-particle state is much smaller than one. The equations in this question count one internal species; with $g$ equally populated spin states the relevant density per species is $n/g$.

Let $z=e^{\beta \mu }$ be the dimensionless quantum gas fugacity, with the lowest one-particle energy chosen as zero. Both quantum distributions can be writtenIf $z\ll 1$, then $n_{\eta }(E)\simeq z{e}^{-\beta E}\ll 1$ for all $E\ge 0$, recovering Maxwell–Boltzmann statistics. Conversely, low occupation even at the spectrum's bottom requires $z/(1-z)\ll 1$ for bosons or $z/(1+z)\ll 1$ for fermions; either condition impliesA single high-energy state can be sparsely occupied even in a degenerate gas, so the condition must hold across the relevant spectrum, not just for one arbitrarily chosen energy. Also, classical translational statistics does not imply that all internal excitation energies or other quantum effects can be ignored.

Grand partition function, density and pressure. Noninteracting particles have independent occupations of each one-particle state. Put $x_{\mathbf{k}}=z{e}^{-\beta {\epsilon }_{\mathbf{k}}}$. Summing the allowed occupations givesThereforeThe grand potential is ${\Omega }_G=-k_{B}T\ln \Xi =-pV$ for this homogeneous gas. Replacing the sum by $V\int d^{3}k/(2\pi {)}^3$ and using ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ givesThe mean number is $N=z{\partial }_z\ln \Xi$, givingFor bosons, the grand sum requires $z<1$ when the ground energy is zero. A macroscopic condensate at the limiting $z\to 1$ must be separated from the continuum integral in the condensed phase. The following small-$z$ calculation is in the normal, nondegenerate regime, so that complication does not enter.

Density expansion and inversion. Expand the occupation as a geometric series:The Gaussian wavevector integral for its $\ell$th term isIt follows thatFor the inverse series, put $y=n{\lambda }^3$ and try $z=y+c{y}^2+O(y^3)$. Substitution gives $0=(c+\eta /2^{3/2})y^2+O(y^3)$, henceThe leading equality $n{\lambda }^3\simeq z$ directly connects the density and occupancy criteria for classical behavior.

Pressure and its first quantum correction. Expanding the logarithm in the pressure expression gives one extra factor $1/\ell$ compared with the density series:Insert the inverse series. To second order in $y=n{\lambda }^3$,ThusThe first quantum correction at fixed density and temperature isIt is negative for bosons and positive for fermions, and vanishes in the classical density limit.

Virial coefficient. Define the number-density virial expansion byThe coefficient $B_j$ has units ${\mathrm{volume}}^{j-1}$ in this convention. Comparing terms givesThis is the exchange contribution to the second virial coefficient. It arises without a pair-interaction potential; the underlying gas is noninteracting. A molar-density convention would multiply this $B_2$ by the Avogadro constant. For $g$ equally populated internal species, exchange occurs within each species and the coefficient in terms of total density is smaller by a factor $g$.

Why the two signs differ. Bosonic exchange symmetry enhances the probability of nearby identical bosons, whereas fermionic exchange symmetry suppresses nearby identical fermions in the same internal state. In the dilute regime, the normalized equal-time pair detection correlation isThe exchange part follows by squaring the Gaussian one-particle coherence factor $e^{-\pi r^2/{\lambda }^2}$. At zero separation this gives $g^{(2)}(0)=2$ for bosons and $0$ for one-component fermions. Integrating the exchange pair weight provides a direct coefficient check:Bunching therefore acts like an effective statistical attraction in the dilute equation of state, lowering the pressure needed at fixed density. Exclusion acts like an effective statistical repulsion, raising it. Neither description asserts an actual interparticle force. Different internal fermion species do not exclude each other from a common spatial location in the same way, explaining the importance of the species convention.

The regulated bare coupling $g_0$ is matched to the physical two-body scattering length. In the uniform one-loop grand potential this matching supplies the subtraction that makes $\frac{1}{2}\int d^{3}k/(2\pi {)}^3[E_k-{\epsilon }_k-\mu +{\mu }^2/(2{\epsilon }_k)]$ ultraviolet finite. Merely using the physical coupling in an unregulated zero-point integral would retain a divergent answer. The contact description and its cutoff cannot be extended to arbitrary momenta independently of that matching.

Subtracting the normal-state grand potential fixes the reference constant in the uniform BCS pairing-field effective action. Here ${\omega }_n=(2n+1)\pi /\beta$ has energy units and $E_k^2={\xi }_k^2+|\Delta {|}^2$. The ratio cancels a field-independent determinant phase and produces a convergent high-frequency sum under the same momentum regulator. Differentiation gives the BCS gap equation. For a fixed attractive pairing shell, this is a finite regulated model; an unrestricted continuum contact interaction additionally needs ultraviolet renormalization. Its weak-coupling condensation density $-{\nu }_F{\Delta }^2/2$ uses a single-spin density of states per volume.