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In three dimensions an ideal Bose gas can saturate its excited-state population at chemical potential zero, with the remaining particles forming a Bose-Einstein condensate. The transition follows from the Bose-Einstein distribution integrated against its density of states; below the ideal-gas transition, $n_0/n=1-(T/T_c{)}^{3/2}$. Condensation alone does not supply the finite Landau critical velocity of an interacting superfluid: the ideal quadratic spectrum has ${\inf }_{p}E(p)/p=0$.

By contrast, a weak attractive interaction near a Fermi surface produces the Cooper instability and the BCS gap equation. Its logarithmic enhancement permits pairing at arbitrarily weak attraction in the ideal model. The low-temperature excitations have $E=\sqrt{{\xi }^2+{\Delta }^2}$, and in weak coupling $k_B{T}_c=(e^{{\gamma }_E}/\pi )\Delta (0)\simeq 0.567\Delta (0)$. This transition is controlled by pair formation when phase fluctuations are small; deeply bound pairs instead motivate a Bose-condensation description with their own density and effective mass.

For the given ellipsoidal dispersion, set $q_x=k_x$, $q_y=k_y$ and $q_z=k_z/\sqrt{\gamma }$. Then $\epsilon ={\hslash }^2{q}^2/(2m)$ while the integration measure is $d^{3}k=\sqrt{\gamma }{d}^{3}q$. Thus the anisotropic ideal-gas density of states per unit volume and per internal state isThe full continuum momentum range is assumed; this is an anisotropic three-dimensional gas, not yet a discrete stack of two-dimensional bands.

At the Bose transition, substitute $x=\epsilon /(k_{B}T)$:The integral is $\Gamma (3/2)\zeta (3/2)\simeq 2.315$. Solving for the Bose-Einstein condensation temperature for an anisotropic effective mass givesThe heavier $z$-direction effective mass increases the available excited-state count and lowers the condensation temperature at fixed number density.

For a balanced two-spin Fermi gas, let $n_F$ denote total fermion number density and take $N_s$ to be the single-spin density of states used in the conventional gap equation. Counting the filled ellipsoid givesIf a spin-summed density of states is used instead, the coupling convention in the supplied integral must change accordingly; inserting a factor two into only one side would change the exponent incorrectly.

For ${\epsilon }_c\ll {\epsilon }_F$, replace the density of states across the interaction window by $N_s({\epsilon }_F,\gamma )$. Writing ${\lambda }_0=|g|N_{s0}$, the zero-temperature integral becomesand thereforeThe second expression additionally requires ${\lambda }_0{\gamma }^{1/3}\ll 1$. The explicit density dependence is ${\lambda }_0=|g|m(3{\pi }^2{n}_F{)}^{1/3}/(2{\pi }^2{\hslash }^2)$. This mass-anisotropy enhancement of a weak-coupling BCS gap initially raises the pairing scale, while the ideal Bose condensation scale decreases as ${\gamma }^{-1/3}$.

Large anisotropy weakens transport and coherence in the heavy direction. The separation of pair formation and phase ordering then matters, so a large mean-field gap need not imply a large actual $T_c$. A qualitative sketch has $T_c$ initially rising with the pairing scale and eventually turning down when coherence is the smaller scale. If preformed fermion pairs behave as bosons, use $n_B=n_F/2$ and pair mass about $2m$; in this continuum model this gives $k_B{T}^{\mathrm{BEC}}\simeq 0.218{\epsilon }_{F0}{\gamma }^{-1/3}$.

The turnover cannot be located reliably by taking the weak-coupling formulas to $\gamma \to \infty$: ${\epsilon }_F$ falls, ${\lambda }_0{\gamma }^{1/3}$ grows, the narrow-window assumption fails and the pair chemical potential and mass require a self-consistent treatment. The smooth curve in the figure is explicitly an illustrative interpolation of two extrapolated scales, not such a calculation. If actual decoupled two-dimensional layers are approached, ordinary ideal-gas finite-temperature condensation is absent and interacting phase order requires a different analysis, including the Berezinskii–Kosterlitz–Thouless transition when appropriate.

The cuprates make the distinction between pairing and phase coherence relevant: they are layered and have anisotropic motion, but their strong correlations, lattice band structure and predominantly d-wave superconductivity are outside this attractive continuum $s$-wave model. Anisotropy by itself is therefore not an explanation or quantitative fit of cuprate transition temperatures.

The electron-phonon interaction lets one electron polarize the ionic lattice and another interact with that delayed deformation. Phonon-mediated electron attraction acts over energy transfers below a characteristic phonon scale. Pairing additionally requires a net attractive interaction in some allowed pairing channel after screened Coulomb repulsion is included. In the conventional weak-coupling situation, retardation separates the low phonon scale from the much larger electronic scale, reducing the effective repulsion. The Cooper instability then makes a Fermi surface unstable to an arbitrarily weak net attraction in the idealized channel; finite temperature cuts off its logarithmic enhancement.

In BCS theory, a momentum-independent pairing amplitude within the cutoff obeys a self-consistency equation containing $\tanh (E/2k_{B}T)/(2E)$, with quasiparticle energy $E=\sqrt{{\xi }^2+|\Delta {|}^2}$. At a continuous transition, take $\Delta \to 0$. A constant effective density of states and coupling combine into $\lambda$, while the band bottom truncates the negative-energy side at $-\sigma$, with $\sigma =\min (\mu ,\theta )$. This gives the supplied linearized gap equation. Its integrand is even, so withit becomes $1/\lambda =[I(\theta ,T_c)+I(\sigma ,T_c)]/2$. If both cutoffs are much larger than $k_B{T}_c$, the given large-cutoff integral gives $I(a,T_c)\simeq \ln (1.13a/(k_B{T}_c))$. Consequently the asymmetric-cutoff BCS transition temperature isFor $\mu >\theta$, this reduces to $1.13\theta e^{-1/\lambda }$; for $0<\mu <\theta$, it is $1.13\sqrt{\theta \mu }{e}^{-1/\lambda }$. The second limit requires $k_B{T}_c\ll \mu$ as well as $k_B{T}_c\ll \theta$. Weak $\lambda$ at fixed nonzero $\mu /\theta$ supplies that limit, but taking $\mu \to 0$ afterward need not preserve it.

For a three-dimensional spin-degenerate parabolic electron band in the degenerate regime,DefineHolding the other model parameters fixed givesThus the log-log slope is $1/3$ below the cutoff-density scale and zero above it. For the specified parameters,This density crossover of a fixed-coupling BCS cutoff model is the requested prescribed model. A real three-dimensional parabolic band also has a changing density of states, so keeping $\lambda$ constant as density varies is an additional imposed assumption, not a consequence of that dispersion.

For noninteracting spinless bosonic pairs, mass $m_B=2m$ and density $n_B=n/2$, homogeneous ideal Bose gas condensation givesUsing the supplied integral $I_B={\int }_0^{\infty }{x}^{1/2}/(e^x-1)dx\simeq 2.3$ gives $C_B=\frac{1}{2}[2/(3I_B){]}^{2/3}\simeq 0.219$. Its slight difference from the more precise $0.218$ is consistent with the rounded integral. Write $a=T_0/(\theta /k_B)=1.13e^{-1/\lambda }\simeq 0.0403$. On the low-density branch,Since $a/C_B<1$, no part of the high-density plateau exceeds $T_B$. The formal crossing condition isThe crossing occurs near $T_c\simeq 7.4\mathrm{K}$ and $T_F\simeq 34\mathrm{K}$.

The apparent low-density excess should not be read as a reliable transition above the ideal condensation scale of preformed pairs. On this extrapolation $k_B{T}_c/\mu =a{u}^{-1/3}$ grows without bound, violating the degenerate-gas and large-lower-cutoff approximations. It reaches one already near $7.5\times 10^{21}{\mathrm{m}}^{-3}$; well before using an arbitrarily small density, the chemical potential, pairing, density of states and pair fluctuations need a self-consistent treatment. The asymptotic formula and ideal-boson curve compare two stipulated models, rather than solving the full BCS-BEC crossover.

Condensate depletion means particles outside a macroscopically occupied bosonic single-particle mode. Interactions can deplete that mode even at zero temperature: the Bogoliubov transformation mixes creation and annihilation operators, so its quasiparticle vacuum contains particles at nonzero momentum. A weak homogeneous three-dimensional gas with repulsive scattering length $a_s$ has leading quantum depletion of a Bose condensateThermal quasiparticles add another contribution. Depletion and normal-fluid density are distinct: a Galilean-invariant fluid can have quantum depletion while retaining its full superfluid mass response at zero temperature.

For the fermionic pairing problem, mean-field decoupling of the reduced BCS pairing Hamiltonian introduces the anomalous amplitude $\Delta$. Each paired Nambu block has diagonal entries ${\xi }_{\mathbf{k}},-{\xi }_{\mathbf{k}}$ and off-diagonal pairing entries of magnitude $\Delta$. A phase convention with negative off-diagonal entries gives eigenvalues $\pm E_{\mathbf{k}}$, where $E_{\mathbf{k}}=\sqrt{{\xi }_{\mathbf{k}}^2+{\Delta }^2}$. Its canonical Bogoliubov rotation yields the BCS coherence factorsThe partner transformation has the complementary sign required by fermionic anticommutation. At zero temperature all positive-energy BCS quasiparticle modes are empty. Expanding the supplied electron operator therefore givesSince $E_{\mathbf{k}}^2-{\xi }_{\mathbf{k}}^2={\Delta }^2$,The electron occupation is close to one far below the normal-state Fermi surface, equals one half at $\xi =0$, and decays toward zero above it. Pairing smooths the Fermi step over an energy range of order $\Delta$. The positive-energy quasiparticle occupation is instead zero everywhere at this temperature; the electron sea and the quasiparticle vacuum are different descriptions of the same state. At finite temperature, $n_{\mathbf{k}\sigma }=[1-(\xi /E)\tanh (E/(2k_{B}T))]/2$ in the same mean-field model.

For a three-dimensional parabolic normal band, the total density of states including both spins is $D(\mu )=3N/(2\mu )$. In the weak-gap limit, replace it by this constant across the narrow pairing region and use $\mu \simeq E_F$. The number of electrons above the normal-state Fermi surface isThe integral is $[\xi -\sqrt{{\xi }^2+{\Delta }^2}{]}_0^{\infty }/2=\Delta /2$. Therefore the BCS population above the normal-state Fermi surface isThe constant-density-of-states approximation and a small gap relative to the chemical potential are essential. This is the question's explicitly defined fermionic population measure; it is not the fraction of electrons outside a single bosonic condensate orbital.

For an order-of-magnitude mercury estimate, use $T_c\simeq 4.2\mathrm{K}$ and the weak-coupling scale ${\Delta }_0\simeq 1.76k_B{T}_c\simeq 0.64\mathrm{meV}$, with a representative metallic $\mu \simeq 7\mathrm{eV}$. These input approximations give $\Delta N/N\sim 7\times 10^{-5}$, of order $10^{-4}$. The use of a weak-coupling gap is an estimate, not an assertion of mercury's exact measured gap.

For a copper oxide, an illustrative gap $20\mathrm{meV}$ and effective chemical-potential scale $0.5\mathrm{eV}$ give about three per cent in this formula. Choosing $0.2$–$1\mathrm{eV}$ instead gives roughly $7.5$–$1.5$ per cent. These are explicit example inputs: the paper specifies no compound, doping or band parameters from which a unique number follows. A real cuprate's anisotropic gap, quasi-two-dimensional band and strong correlations require a band-resolved occupation integral, so the reliable contrast is the much larger gap-to-electronic-energy scale, rather than a universal cuprate percentage.

Assume three-dimensional isotropic gapless excitations, zero chemical potential and one independent branch. Momentum-state counting givesSubstituting $x=ϵ/(k_{B}T)$ into the Bose energy integral shows $U/V\propto T^{1+3/n}$ and $C/V\propto T^{3/n}$. The observed exponent therefore gives $n=2$.

For the quadratic-boson heat-capacity coefficient, the required Bose integral isThe equality follows by expanding $(e^x-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term. HenceThe supplied units make its $C$ a heat capacity per volume. Equating its coefficient to $220\mathrm{J}{\mathrm{K}}^{-5/2}{\mathrm{m}}^{-3}$ givesEquivalently $ϵ=p^2/(2m_{\ast })$ has $m_{\ast }=1.09\times 10^{-28}\mathrm{kg}$. For $q_b$ identical branches, the measured coefficient fixes $q_b/{\alpha }^{3/2}$, so instead $\alpha =(4.58\times 10^{27}\mathrm{k}{\mathrm{g}}^{-1})q_b^{2/3}$. The one-branch convention is necessary for the quoted numerical value. The supplied integral with $x^4$ is not the moment arising for the exponent determined here.

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 $1\ll X\ll 2{\pi }^2\mu /\Delta$.

A black body absorbs incident electromagnetic radiation and, in equilibrium, emits a spectrum determined by temperature alone. A cavity with a small aperture approximates this ideal. Photon number is not conserved in thermal equilibrium, so its chemical potential is zero. Combining the Bose occupation with the two-polarization electromagnetic mode density gives the spectral energy densityChanging variable to $x=\hslash \omega /(k_{B}T)$ shows directly why the total energy density is proportional to $T^4$:Three powers of T arise from the number of thermally populated modes and one from their characteristic energy. Isotropic radiation incident on a surface has outward hemispherical flux $cu/4$, givingThe factor 1/4 is the hemispherical angular average; multiplying energy density by c alone would overcount. The spectrum's frequency scale grows with T. Classical equipartition over all modes would instead produce the Rayleigh-Jeans ultraviolet divergence; the Planck occupation suppresses high-frequency modes. Real materials can have wavelength-dependent emissivity, and a body's net radiative exchange also includes absorbed environmental radiation.

An equation of state relates equilibrium pressure, temperature and density. The ideal-gas form $p=n{k}_{B}T$ neglects intermolecular forces and excluded volume. A low-density real-gas expansion iswhere the virial coefficients encode interaction effects and generally depend on temperature.

A simple phenomenological example is the molar Van der Waals equationHere v is molar volume, b represents excluded volume and the attractive interaction lowers pressure through a. Its dilute expansion gives $pv/(RT)=1+[b-a/(RT)]/v+\cdots$. It recovers the ideal-gas limit but is not a quantitatively universal equation for all real gases.

Below its critical temperature, the raw isotherm contains a region with $(\partial p/\partial v{)}_T>0$, corresponding to negative isothermal compressibility and mechanical instability. Equilibrium phase separation replaces that segment by a coexistence pressure satisfying equal chemical potentials, equivalently the Maxwell equal-area construction. The critical point follows from $p_v=p_{vv}=0$:The distinction between an unstable homogeneous branch and a stable two-phase state is essential when interpreting a real-gas equation of state.

Write $z=e^{\beta (\mu -ϵ)}$. For one nondegenerate fermionic quantum state, the allowed occupations 0 and 1 give ${\Xi }_F=1+z$. For a bosonic state, summing all nonnegative occupations gives ${\Xi }_B=(1-z{)}^{-1}$, requiring $z<1$ or $\mu <ϵ$. Differentiating the logarithm,The interacting point defect has two distinct spin states, not one fermionic state. Its four microstates have grand-canonical weights $1,z,z,q{z}^2$, with $q=e^{-\beta U}$. The first two single-electron weights are separate up/down spin possibilities; the double state has one electron of each spin. ThusFor $U=0$, $\Xi =(1+z{)}^2$ and $\bar{n}=2z/(1+z)$. The sketch is a single smooth step from 0 to 2, centred on $\mu =ϵ$, where $\bar{n}=1$. As $T\to 0$ its width tends to zero.

For positive U with $\beta U\gg 1$, the lowest grand energies are successively $0$, $ϵ-\mu$ and $2ϵ+U-2\mu$. Away from the two crossings,Finite temperature smooths each step over an energy range of order $k_{B}T$. Near the lower step double occupation is exponentially suppressed, so $\bar{n}\simeq 2z/(1+2z)$. Near the upper step put $y=e^{\beta (\mu -ϵ-U)}$ and neglect the empty state, giving $\bar{n}\simeq 1+y/(2+y)$. Spin degeneracy shifts the half-step positions to $ϵ-k_{B}T\ln 2$ and $ϵ+U+k_{B}T\ln 2$.

For a grand-canonical probability $p_s\propto e^{-\beta (E_s-\mu n_s)}$, differentiation gives ${\partial }_{\mu }{p}_s=\beta (n_s-\bar{n})p_s$. Multiplying by $n_s$ and summing provesFor this defect, direct state counting givesAt $\mu =ϵ$, $z=1$ and thereforeThe same limit gives $\bar{n}=2/3$: empty, up and down states have equal weights, while the double state is suppressed. Replacing this degeneracy by a two-outcome equal-weight model would incorrectly give variance $1/4$ at this chemical potential.

Across the full chemical-potential range the variance has two well-separated narrow peaks and tends to zero on the 0, 1 and 2 plateaux. The limiting shapes are $2z/(1+2z{)}^2$ near the lower step and $2y/(2+y{)}^2$ near the upper one. Each peak has height $1/4$, at the shifted half-step position; the variance at the unshifted crossings is instead $2/9$.

At $T\ll T_F$, Pauli exclusion restricts thermally changed occupations to an energy interval of order $k_{B}T$ about the Fermi energy. The affected fraction is of order $T/T_F$; their typical excitation energy is of order $k_{B}T$, so the thermal energy correction is proportional to $T^2$ and the heat capacity is linear in $T$.

More precisely, with $g(E_F)$ the total density of one-particle states per unit energy, including spin,For a three-dimensional ideal gas with quadratic dispersion, $g(E_F)=3N/(2E_F)$, henceThe fixed-particle-number Sommerfeld expansion gives $U=3N{E}_F/5+({\pi }^2/4)N(k_{B}T{)}^2/E_F+\cdots$. The temperature dependence of the chemical potential must be included when obtaining this expression; holding it artificially fixed changes the intermediate energy correction.

For a single quantum state, the equilibrium occupations areBoth reduce to $e^{(\mu -ϵ)/(k_{B}T)}$ when $ϵ-\mu \gg k_{B}T$.

For one three-dimensional quadratic branch, the density of states per unit volume isWith ${\lambda }_T=\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$, changing integration variable to $x=ϵ/(k_{B}T)$ givesA degeneracy factor multiplies both densities if additional equivalent branches are present. The first expression is the noncondensed density: a macroscopically occupied ground state must be counted separately.

Bose-Einstein condensation occurs when the excited states can no longer hold all particles and the ground-state occupation becomes macroscopic. At fixed total density,For field-induced triplon condensation, write $\Delta =\nu n-\eta (H-H_g)$, so $ϵ(k)={\hslash }^2{k}^2/(2m)+\Delta$. At fixed reference $n$, the three requested dispersions have gaps $\nu n+\eta H_g$, $\nu n$ and zero respectively. Their curvature is unchanged. Because the actual thermodynamic chemical potential is zero, the Bose distribution requires $\Delta \ge 0$. At $H_c=H_g+\nu n/\eta$, the minimum reaches zero and the noncondensed population saturates, permitting macroscopic occupation.

For the triplon condensation field scaling, the critical density at onset is $n_T=\zeta (3/2)/{\lambda }_T^3$. ThereforeThe low-temperature phase boundary rises from $(T,H)=(0,H_g)$; the condensed phase lies on the higher-field side.

For $H>H_c$, simply continuing the gap below zero would give unphysical Bose occupations. For the condensate density in a shifted-parabola triplon model, added condensate density keeps the minimum at zero: $\nu (n_0+n_T)=\eta (H-H_g)$. At fixed temperature this giveswith $n_0=0$ below onset. At zero temperature it becomes $n_0=\eta (H-H_g)/\nu$ within that model.

This prefactor is conditional on continuing the stated single-density mean-field shift into the condensed phase. A microscopic interacting Bose condensate has collective Bogoliubov excitations $E_k=\sqrt{{ϵ}_k^0({ϵ}_k^0+2g{n}_0)}$, with ${ϵ}_k^0={\hslash }^2{k}^2/(2m)$, which are linear at small $k$. The normal-state shift alone does not fix the condensed interaction closure or identify $g$ with $\nu$. The gap closes and the condensate density grows linearly with excess field in the simple mean-field description; the detailed collective dispersion needs that additional interaction model. A dilute-boson treatment of field-induced magnetic ordering makes this normal/condensed distinction explicitly.^[ref]

Differentiate the Euler relation for a single-component simple compressible system:The thermodynamic fundamental relation for the same system is $dU=TdS-pdV+\mu dN$. Subtraction givesDividing by $N$, with $s=S/N$ and $v=V/N$, proves the Gibbs-Duhem equationThe underlying Euler relation assumes extensive bulk thermodynamics and no additional work coordinates. The result says that the intensive quantities are constrained; the chemical potential cannot be varied independently of temperature and pressure within one homogeneous phase.

Bose-Einstein condensation is macroscopic occupation of the lowest one-particle state. It is a change in occupation statistics, not ordinary liquid formation and not a requirement that all atoms occupy one spatial point. For ideal bosons,Set the lowest one-particle energy to zero. Then $\mu \le 0$, and the population of excited states in a uniform three-dimensional box isHere $g_s(z)={\sum }_{j\ge 1}{z}^j/j^s$. As $\mu$ approaches zero, the excited-state population saturates at $V\zeta (3/2)/{\lambda }^3$. Any excess particles occupy the ground state. In the thermodynamic limit,Thus the onset occurs when the thermal de Broglie wavelength is comparable to the mean interparticle spacing. At finite $N$, the chemical potential remains slightly below the ground-state energy; setting it equal to zero is a thermodynamic-limit statement.

The 1995 rubidium experiment used magnetic confinement and evaporative cooling of atoms. Absorption images after trap manipulation revealed a narrow, anisotropically expanding component superposed on the broader thermal distribution; the new component grew as cooling proceeded. The thermal component supplied an internal comparison through the same imaging optics. Trap decompression and release geometry mattered, and finite imaging resolution and interactions limited direct comparison with an ideal-gas ground state.^[ref]

A narrow cloud alone is not sufficient evidence. One should check its emergence across a controlled cooling sequence, atom-number and temperature calibration, the thermal background and the predicted expansion anisotropy. These are tests of an occupation transition rather than a simple focusing artifact. The box formula above is not the exact critical-temperature or condensate-fraction formula for the trapped experiment: the trap changes the density of states, and interactions also matter. Condensation and superfluid response are related but distinct properties; the former definition is ground-state occupation.

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.