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For a one-component superfluid with nonzero order-parameter magnitude on a closed loop, the phase defines a map from the loop into the circle. Its winding number isThe fundamental group ${\pi }_1(S^1)=\mathbb{Z}$ means that smooth deformations preserving nonzero amplitude cannot change this integer. With superfluid velocity ${\mathbf{v}}_s=\hslash \nabla \theta /m_B$, it gives quantized circulation $\oint {\mathbf{v}}_s\cdot d\mathbf{l}=nh/m_B$. A vortex core permits the order parameter to vanish and carries the winding; a superfluid phase slip changes a loop's winding when a core crosses it. For paired neutral fermions, the relevant condensed-object mass is the pair mass.

For charged pairs, gauge-covariant phase rigidity gives fluxoid and flux quantization in a superconductor:The magnitude of the pair flux quantum is $h/(2e)$. Magnetic flux alone approaches an integer multiple of this value when the circulating-velocity term on a bulk contour is negligible. The topological condition is not a claim that an arbitrary finite loop has exactly quantized magnetic flux independent of its screening current.

There are also momentum-space invariants, distinct from real-space vortex winding. A fully gapped two-dimensional Bogoliubov--de Gennes Hamiltonian can have a Chern number obtained from its occupied quasiparticle bands' Berry curvature. In a chiral paired model it can protect edge modes while the bulk gap and relevant symmetry class are maintained. Such an invariant can change only when those assumptions fail, for example through a bulk gap closing. A nodal d-wave state requires a different analysis; it is not automatically a fully gapped Chern superconductor.

A complex order parameter captures both the amplitude and phase rigidity of a superfluid. A local isotropic Ginzburg-Landau theory writes its free-energy functional, to fourth order in amplitude and second order in gradients, asThe uniform minimum has $\Psi =0$ for $\alpha >0$ and $|{\Psi }_0{|}^2=-\alpha /\beta$ for $\alpha <0$. Its phase remains arbitrary under a global $U(1)$ transformation. Minimizing with respect to ${\Psi }^{\ast }$ gives $-K{\nabla }^2\Psi +\alpha \Psi +\beta |\Psi {|}^2\Psi =0$. The phenomenological description specifies the allowed structures and coefficients; microscopic input is needed to identify the normalization of $\Psi$ and calculate those coefficients. A static functional by itself does not fix the dynamical equation or mode frequencies.

For a dilute three-dimensional Bose gas with short-range repulsion and particle mass $m_B$, the microscopic low-energy Hamiltonian isThe interaction parameter is expressed through the low-energy s-wave scattering length $a_s$, rather than an unrenormalized bare delta potential. In a symmetry-selected description, $\Psi =⟨\hat{\psi }⟩$ and $|\Psi {|}^2=n_0$ is condensate density. Equivalently, fixed-number off-diagonal long-range order identifies the macroscopically occupied natural orbital and its amplitude without a nonzero one-point expectation.

Evaluating $\hat{H}-\mu \hat{N}$ in a coherent-field condensate variational state gives the Ginzburg-Landau functional of a dilute Bose gas, with the mean-field microscopic identificationsIts stationary variational equation is the Gross–Pitaevskii equation. The time-dependent variational action additionally contains $i\hslash {\Psi }^{\ast }{\partial }_t\Psi$, givingThis term expresses density-phase conjugacy. It is the missing microscopic dynamical input; a purely dissipative relaxation law would describe a different approximation.

Bogoliubov theory writes $\hat{\psi }=\Psi +\delta \hat{\psi }$ and retains quadratic fluctuations. In a uniform condensate, $\mu =g_B{n}_0$ and the fluctuation matrix isIts eigenvalues are $\pm \sqrt{{ϵ}_k({ϵ}_k+2g_B{n}_0)}$. The positive branch is the gapless sound mode, continuously crossing over to particle-like excitations at larger $k$. The same result follows by linearizing the time-dependent Gross–Pitaevskii equation about its uniform solution. Thus the microscopic condensate amplitude, scattering interaction and mass determine the mean-field static coefficients and the collective mode. Quadratic quantum fluctuations also produce quantum depletion of a Bose condensate, which a pure coherent-state product omits.

These identifications are controlled for a dilute weakly interacting gas away from a fluctuation-dominated critical region. They are not a universal statement that the exact finite-temperature GL coefficient is always $-\mu$ or that its quartic coefficient never receives corrections. Integrating out noncondensed and short-wavelength modes produces effective temperature-dependent coefficients and can change their near-critical behavior. The phase rigidity described by $K$ and the thermodynamic order-parameter expansion remain the shared organizing structure.

For BCS theory, the microscopic order parameter is instead an anomalous pair amplitude, commonly normalized as an energy gap $\Delta \propto ⟨{\hat{\psi }}_{\downarrow }{\hat{\psi }}_{\uparrow }⟩$. It is not an ordinary single-electron field expectation or automatically the square root of a pair number density. Integrating out fermions and expanding in small $\Delta$ and slow gradients near $T_c$ gives the microscopic derivation of Ginzburg-Landau theory. The BCS pairing-field effective action supplies its coefficients through normal-state pair-response functions and thermal fermion sums.

For example, a clean isotropic three-dimensional weak-coupling s-wave model, with $N(0)$ the density of states per spin at the Fermi energy and $\Delta$ measured in energy units, giveswith microscopic Ginzburg-Landau coefficients from BCS theoryThe numerical factors arise from fermionic Matsubara sums and the isotropic velocity average; they depend on the stated gap and density-of-states conventions. Rescaling the order parameter rescales all three coefficients. The expansion applies near the transition with sufficiently slow spatial variation; disorder, anisotropic pairing and strong coupling alter the coefficients or require additional components. Microscopic derivation also fixes the pair charge in the covariant gradient, producing the superconducting screening current and the Meissner effect.^[ref]

The neutral Bose and charged Fermi theories therefore share a complex order parameter and a phase-stiffness functional, while deriving that field from different microscopic objects. The Bogoliubov transformation mixes particle and hole fluctuations of the bosonic condensate in one case and paired fermionic excitations in the other. The static GL limit organizes long-distance behavior; the full microscopic theories retain excitation spectra and temperature dependence beyond that local near-transition expansion.

The superconducting order is a coherent pair amplitude, schematically $\Delta \propto ⟨c_{\downarrow }{c}_{\uparrow }⟩=|\Delta |e^{i\varphi }$. Under a global fermion phase change $c\mapsto e^{i\alpha }c$, it acquires phase $2\alpha$. A symmetry-broken thermodynamic description selects a pair phase. This must be distinguished from electromagnetic gauge freedom: a local gauge transformation changes the phase and potentials together and is a redundancy of the description, not an observable symmetry that the material literally violates.

A conventional centrosymmetric s-wave superconductor usually has an even-parity orbital pair state and a spin singlet state, consistent with the antisymmetry of two fermions. Its ordinary zero-field equilibrium state preserves time reversal. The electromagnetic coupling makes phase dynamics and screening part of the Anderson-Higgs mechanism in a superconductor; the Meissner effect is a measurable consequence. The condensate phase description does not imply that electric charge is no longer conserved.

In an unconventional superconductor, the gap can transform nontrivially under the crystal point group or have other nonstandard pairing structure. A common example is d-wave superconductivity, with sign changes and possible nodes in the gap. This does not automatically imply an additional observable fourfold-rotation breaking: a rotation that reverses the gap's sign can be accompanied by a gauge phase change. A nematic superconducting state instead breaks an actual crystal rotation symmetry in its gauge-invariant responses.

A chiral combination such as $p_x+i{p}_y$ can break time reversal, while a spin-triplet order parameter can also distinguish spin directions when the applicable spin-rotation and spin-orbit constraints permit it. Helium-3 superfluidity illustrates paired-fermion order with richer spin and orbital structure. These possibilities must be tested rather than equated: unconventional pairing does not by itself mean spin-triplet pairing, time-reversal breaking, or nematicity. Gap angular dependence, phase-sensitive interference, spin response and spontaneous magnetic signatures probe different symmetry questions. The phase-sensitive test of d-wave pairing is more specific than inferring a gap sign merely from its magnitude.

There is a state-assumption issue. For a number-conserving Gaussian quantum state with no anomalous contractions, put ${\psi }_i=\psi ({\mathbf{r}}_i)$, $n_i=⟨{\psi }_i^{†}{\psi }_i⟩$ and the one-body density matrix $g_{ij}=⟨{\psi }_i^{†}{\psi }_j⟩$. Wick's theorem pairs each creation operator with an annihilation operator. The direct pairing gives $n_1{n}_2$; the exchanged pairing gives $ϵg_{12}{g}_{21}$, with $ϵ=+1$ for bosons and $-1$ for fermions. Hencenormal ordering removes the equal-position delta-function self-counting term. Fermionic occupation-number product states are Slater determinants and obey this Wick theorem. Independent thermal bosonic modes also obey it, with $N_{\alpha }$ understood as mean occupations.

A bosonic Fock product state does not generally obey the asserted identity. The bosonic number-state corrections to Gaussian density correlations follow from $⟨a_{\alpha }^{†2}{a}_{\alpha }^2⟩=N_{\alpha }(N_{\alpha }-1)$ and giveFor one occupied mode this is $N(N-1)|{\phi }_1{|}^2|{\phi }_2{|}^2$, whereas the uncorrected Gaussian expression gives $2N^2|{\phi }_1{|}^2|{\phi }_2{|}^2$. This counterexample disproves the asserted identity for arbitrary bosonic number-product states. The formulas below give both the Gaussian result, with its additional state assumption, and the exact bosonic number-state alternative. The ensemble distinction is also explicit in a later primary treatment of ideal-gas correlations.^[ref]

The three-point density correlation of an ideal quantum gas follows first for a number-conserving Gaussian quantum state. There are six pairings, one for each permutation of three annihilation operators. The identity gives the product of densities; each transposition has sign $ϵ$; both three-cycles have positive sign. ThusThe last two terms are complex conjugates. They equal $2\mathrm{Re}(g_{12}{g}_{23}{g}_{31})$, not an additional signed pair term. This is the permanent of a matrix $(g_{ij})$ for bosons and its determinant for fermions. Wick's theorem supplies the six terms only with the state assumption stated in part (a).

For completeness an exact bosonic Fock-state alternative can be given without that assumption. Let $\mathbf{\ell }$ be a mode-occupation decrement with ${\sum }_{\alpha }{\ell }_{\alpha }=3$. Define $A_{\mathbf{\ell }}(1,2,3)$ as the sum of ${\prod }_{j=1}^3{\phi }_{{\alpha }_j}({\mathbf{r}}_j)$ over all distinct ordered triples having these mode multiplicities. Acting three annihilation fields on the number state leaves mutually orthogonal states for different $\mathbf{\ell }$. ThereforeHere $N^{\underline{\ell }}=N!/(N-\ell )!$, with zero when $\ell >N$. This expression includes repeated-mode occupation corrections absent from the Gaussian permanent; for a single mode it reduces to $N(N-1)(N-2){\prod }_j|\phi ({\mathbf{r}}_j){|}^2$.

Choose orthonormal, flux-normalized incoming and outgoing channels and include all channels of this lossless scattering problem. Conservation of probability requires $\Vert \beta {\Vert }^2=\Vert \alpha {\Vert }^2$ for every input vector. With $\beta =S\alpha$,For a square channel matrix this also gives $S{S}^{†}=I$. A submatrix obtained by omitting loss channels need not be unitary, but that is not the stated two-channel model.

The given real Hadamard matrix is symmetric, unitary and its own inverse. HenceThis same transformation preserves the canonical commutation relations for bosons and anticommutation relations for fermions. To extract probabilities one must use normalized number states, including their factorial factors, even when the supplied input expression is unnormalized.

For identical fermions, $(b_j^{†}{)}^2=0$ and $b_2^{†}{b}_1^{†}=-b_1^{†}{b}_2^{†}$. ThereforeThe minus sign is a global phase for this output. The only occupancy is $(n_1,n_2)=(1,1)$, with probability one. Double occupancy of either specified single-particle mode is excluded; no extra internal states are present in this model.

In second quantization, expand the annihilating quantized field in a complete orthonormal basis of one-particle states, $\hat{\psi }(x)={\sum }_{\alpha }{\phi }_{\alpha }(x)a_{\alpha }$. At equal times, the canonical commutation relation for bosons and the canonical anticommutation relations for fermions are, respectively,andHere $\delta$ is a Dirac delta distribution. Multiplying by the basis wave functions and integrating gives $[a_{\alpha },a_{\beta }^{†}]={\delta }_{\alpha \beta }$ for bosons, or $\{a_{\alpha },a_{\beta }^{†}\}={\delta }_{\alpha \beta }$ for fermions; the two-annihilator and two-creator brackets vanish in the corresponding convention.

The mode number operator and total particle number operator areThe bosonic occupation numbers are $0,1,2,\dots$. For fermions, $a_{\alpha }^2=0$ and the canonical anticommutation relations imply $n_{\alpha }^2=n_{\alpha }$, so the fermionic occupation numbers are only $0$ and $1$, as required by the Pauli exclusion principle.

Keeping only two modes later is a projection: the projected field bracket is ${\sum }_{j=1}^2{\phi }_j(x){\phi }_j^{\ast }(x^{\prime })$, rather than the full Dirac delta distribution.

For independent bosonic modes, the canonical commutation relation isFor independent fermionic modes, the canonical anticommutation relations arewhere $[A,B]=AB-BA$ and $\{A,B\}=AB+BA$. The creation operator is the adjoint of the annihilation operator. In either case the mode's number operator iswith $a_i=b_i$ or $c_i$ as appropriate. It measures the mode's occupation number. Its possible eigenvalues areFor fermions, $c_i^2=0$ implies $n_i^2=c_i^{†}(1-c_i^{†}{c}_i)c_i=n_i$, so the number operator is a projection. This expresses the exclusion of double occupation of one fermionic mode. The total particle number is $N={\sum }_i{n}_i$.

Fix the mode ordering with the $b^{†}$ operator before the $c^{†}$ operator, as in the given state. In the sector with one fermion in each subsystem, use the ordered basisThe state vector is $v=(1/2,1/\sqrt{2},0,1/2{)}^T$, whose squared norm is one. Thus the global density matrix is the rank-one projectorIt has trace one and satisfies ${\rho }_{BC}^2={\rho }_{BC}$, so the global state is pure.

Take the partial trace over $B$. The conditional vectors in $C$ are $v_1=(1/2,1/\sqrt{2}{)}^T$ for $B_1$ and $v_2=(0,1/2{)}^T$ for $B_2$. Therefore the reduced density matrix isIts determinant is $1/16$, giving eigenvaluesBoth are positive; in particular, $\mathrm{Tr}{\rho }_C^2=7/8<1$, so the reduced state is mixed despite global purity. The entanglement entropy is obtained by diagonalizing this reduced matrix and evaluatingFor a globally pure bipartite state this equals the entropy of either reduced subsystem. No numerical evaluation is needed.

Since every quasiparticle energy is positive, the ground state is the vacuum of all the ${\alpha }_p$ annihilation operators, including ${\alpha }_0=c_0$. Its zero-momentum mode is therefore empty. For each positive momentum, the inverse Bogoliubov transformation givesBoth annihilate the normalized pair state $(u_p+i{v}_p{c}_p^{†}{c}_{-p}^{†})|0⟩$. For example, applying ${\alpha }_p$ produces the two cancelling terms $i{u}_p{v}_p{c}_{-p}^{†}|0⟩$. ThusThe factors have even fermion parity and commute for disjoint momentum pairs. Each normalized factor is a superposition of zero and two fermions, whose pair-occupation probability is $v_p^2$. ThereforeThere is no zero-mode contribution. For fixed $L$, the sum converges since $v_p^2=O(p^{-4})$.

A normalized simple excited state is $|k⟩={\alpha }_k^{†}|\Omega ⟩$ for $k>0$. Its excitation energy is $E_k$, and its total momentum is $k$: both terms in ${\alpha }_k^{†}=u_k{c}_k^{†}+i{v}_k{c}_{-k}$ add momentum $k$. Acting on the $k,-k$ pair givesThe affected pair now contains exactly one fermion instead of mean number $2v_k^2$; all other pairs are unchanged. ConsequentlyThis is the fermion number of a Bogoliubov excitation. Its change in mean fermion number lies between zero and one for $\mu >0$, with the value one at $\Delta =0$. The Bogoliubov quasiparticle coherently mixes adding a fermion and removing the opposite-momentum fermion; it is not an eigen-excitation of the original fermion number operator. Indeed,so fermion number is not conserved when $\Delta \ne 0$, while fermion parity is conserved. The generally noninteger change is an expectation value in a coherent paired background, rather than a fractional eigenvalue of the fermion number operator. In the weak-pairing limit at fixed $k$, it tends to one, as expected for an ordinary added fermion.

Use thermal de Broglie wavelength ${\lambda }_T=h/\sqrt{2\pi m{k}_{B}T}$ and spacing $a=n^{-1/3}$. Degeneracy onset from thermal wavelength occurs atThe numerical factor depends on the convention for a typical momentum, but the scale ${\hslash }^2{n}^{2/3}/(m{k}_B)$ does not. For fermions the Fermi temperature $T_F={\hslash }^2(3{\pi }^{2}n{)}^{2/3}/(2m{k}_B)$ gives a comparable scale for spin degeneracy two.

Representative estimates, with the assumed densities stated explicitly, are:

Here $u$ is the atomic mass constant. Thus the characteristic orders are $10^5\mathrm{K}$ for electrons, sub-microkelvin for a dilute heavy atomic gas, and $10^{12}\mathrm{K}$ for neutrons. There is no density-independent quantum-degeneracy temperature for atomic gases: for example, taking $n=2.5\times 10^{25}{\mathrm{m}}^{-3}$ and mass $4u$ instead gives about $0.065\mathrm{K}$. These are illustrative input estimates, not values supplied by the paper. Denser neutron-star cores require relativistic and interaction corrections; the estimate identifies the degeneracy scale rather than a precise stellar temperature.

Let $\beta =1/(k_{B}T)$. Each independent fermion orbital is either empty or occupied once, so its grand partition factor is $1+e^{-\beta ({ϵ}_j-\mu )}$. Multiplying factors and taking the logarithm givesSince $1-n_F=[1+e^{-\beta (ϵ-\mu )}{]}^{-1}$, this is the requested form. Spin or other orbital degeneracy is already included in $g$.

For the Fermi-window convolution of the grand potential, defineDirect differentiation gives $F^{\prime }=+n_F$. The supplied PDF hint has the opposite sign; using that sign would contradict its grand-potential formula. Integrating twice with $A(0)={\Phi }_0(0)=0$ and vanishing upper boundary terms,At zero temperature $-{\partial }_{ϵ}{n}_F\to \delta (ϵ-\mu )$, confirming ${\Phi }_0(\mu )=\Phi (0,\mu )$.

Use the source's implicit unit normalization of the constant density of states; restoring a constant baseline $g_0$ multiplies the following potentials by $g_0$. For the sinusoidal density of states, direct integration givesPhysical nonnegative density of states requires $|\alpha |\le 1$. Write $t=k_{B}T$ and $y=ϵ-\mu$. The Fermi windowis even, with full-line moments $\int wdy=1$, $\int ywdy=0$ and $\int y^{2}wdy={\pi }^2{t}^2/3$. The last integral follows, for example, by expanding $e^x/(1+e^x{)}^2$ on $x>0$ into its alternating exponential series. When $t\ll \mu$, extending the lower integration limit to $-\infty$ introduces only exponentially small band-edge corrections. The supplied Fourier transform givesThe odd sine term vanishes. Substitution provesThe thermal damping of a sinusoidal density of states is therefore $X/\sinh X$ in the full oscillatory potential. The requested ratio instead concerns the thermal change $\Phi (T,\mu )-{\Phi }_0(\mu )$. Its oscillatory thermal-change amplitude ratio isFor positive $X$ it decreases smoothly from the limit $|\alpha |$ to zero:At exactly $T=0$ both thermal changes vanish; the quoted value is their limiting ratio. The amplitude is the envelope in chemical potential. At a particular fixed $\mu$, the signed contribution ratio has $\alpha \cos (2\pi \mu /\Delta )$ in place of $|\alpha |$; at an oscillation node the actual oscillatory contribution is zero. The large-$X$ behaviour still requires $k_{B}T\ll \mu$, so the useful range is $1\ll X\ll 2{\pi }^2\mu /\Delta$.

The Fermi-Dirac distribution gives the mean occupancy of a single fermion state:Its maximum occupancy of one expresses the Pauli exclusion principle. A spatial orbital with two spin states can hold two electrons because they are different single-particle states. At zero temperature, the occupation is a step at the Fermi energy $E_F$; at low nonzero temperature, occupation changes appreciably only in an energy shell of width of order $k_{B}T$ about the chemical potential.

Consequently, most conduction electrons cannot absorb small thermal energies into already occupied states. Only a fraction of order $T/T_F$ participates, where $T_F=E_F/k_B$. For a normal metal with a smooth density of states $g(E_F)$, including both spin states and using the total-system normalization, the Sommerfeld expansion at fixed particle number givesand thereforeFor a three-dimensional free electron gas, $g(E_F)=3N/(2E_F)$, so $C_{\mathrm{el}}=({\pi }^2/2)N{k}_B(T/T_F)$. This is much smaller than a classical equipartition estimate at $T\ll T_F$. The chemical potential must adjust with temperature to maintain $N$; treating it as fixed while differentiating an energy integral gives the wrong fixed-$N$ calculation in general.

The lattice adds a Debye contribution $C_{\mathrm{ph}}={\beta }_D{T}^3$ at temperatures well below the Debye temperature. HenceA plot of measured $C/T$ against $T^2$ can separate the electronic intercept from the phonon slope. calorimetry must account for the heat capacity of the sample holder, heat leaks, thermal equilibration and thermometer calibration. A magnetic impurity contribution or a superconducting transition can invalidate this two-term fit; the prediction is for the normal metallic state in an appropriate low-temperature interval. Differences from a free-electron coefficient can reveal quasiparticle effective-mass enhancement or nontrivial electronic structure rather than a failure of Fermi statistics.

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]

In the grand canonical ensemble, the system exchanges both energy and particles with a reservoir at fixed temperature $T$ and chemical potential $\mu$. The probability of a particular many-body microstate $i$, with energy $E_i$ and particle number $N_i$, is its Boltzmann weight $e^{-\beta (E_i-\mu N_i)}$ divided by the grand canonical partition function $\Xi$, where $\beta =1/(k_{B}T)$. The denominator normalizes ${\sum }_i{p}_i=1$; a degeneracy is accounted for by counting distinct microstates, rather than changing an individual state's probability.

For noninteracting particles, specify a microstate by the occupation numbers $n_r$ of one-particle levels of energy ${ϵ}_r$. Then $E-\mu N={\sum }_r({ϵ}_r-\mu )n_r$, so the Gibbs probability factorizes between levels. Put $x_r=e^{-\beta ({ϵ}_r-\mu )}$.

For fermions, the Pauli exclusion principle permits $n_r=0,1$ in each complete one-particle state, giving ${\Xi }_r=1+x_r$. Its mean occupancy isthe Fermi-Dirac distribution.

For bosons, $n_r=0,1,2,\dots$, so the geometric sum is ${\Xi }_r=1/(1-x_r)$ when $x_r<1$. Differentiating that sum givesthe Bose-Einstein distribution. Finite-system convergence requires $\mu <{ϵ}_{\min }$; condensation is the limiting approach to the lowest energy with macroscopic ground-state occupation. Both quantum distributions reduce to $e^{-\beta ({ϵ}_r-\mu )}$ when occupation is small. The quantities derived here are mean occupancies of states, rather than normalized probabilities over the set of one-particle energies.

One particle. Use the classical phase-space normalization of one quantum state per volume $h^3$, with no unspecified internal degeneracy. For $a>0$ and $\beta =1/(k_{B}T)$, the canonical partition function factorizes into momentum and position integrals:The Gaussian momentum integral is $(2\pi m{k}_{B}T{)}^{3/2}$. In spherical coordinates, put $t=\beta ar$:Therefore the classical ideal gas in a spherical linear trap hasHere $a$ is a force, in $\mathrm{J}{\mathrm{m}}^{-1}$, so $k_{B}T/a$ is a length and $Z_1$ is dimensionless. A nonconfining $a\le 0$ would not give this finite spatial integral. As required by the classical phase-space approximation, the temperature must exceed the trap's discrete level scale.

Indistinguishable dilute particles. Independent labeled particles would give $Z_1^N$. In the classical dilute regime, division by $N!$ removes the overcounting of permutations of identical particles:This is the Maxwell-Boltzmann limit, not an exact quantum canonical partition function at high occupation. Exchange effects cannot be captured merely by continuing this formula beyond its dilute regime.

Free energy, energy and entropy. The Helmholtz free energy isWith leading Stirling's approximation, $\ln N!\simeq N\ln N-N$, this becomesAt fixed trap strength, ${\partial }_T\ln Z_1=9/(2T)$. Differentiating $F$ gives the requested entropy,The $11/2$ is $1+9/2$: the first term comes from indistinguishability through Stirling's formula, and the second from the temperature dependence of $Z_1$. Without the large-$N$ approximation, $S=k_B[N\ln Z_1-\ln N!+(9/2)N]$. Independently,The kinetic contribution is $3N{k}_{B}T/2$ and the linear-trap potential contribution is $3N{k}_{B}T$, also obtained from generalized equipartition.