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Let $\Delta =\hslash \Omega$ and choose the nondegenerate energy levels as 0 and $\Delta$. For N independent identical units, $Z=(1+e^{-\beta \Delta }{)}^N$; set N=1 for a single two-level system. Differentiating its logarithm givesAt low temperature, $U\sim N\Delta e^{-x}$ and $C_V\sim N{k}_B{x}^2{e}^{-x}$: the upper level is frozen out. At high temperature, the two levels approach equal populations, giving $U\to N\Delta /2$ while $C_V\sim N{k}_B{x}^2/4\to 0$. The energy therefore rises monotonically to a plateau, while the heat capacity has a Schottky peak:Unequal degeneracies would change the populations and peak; the usual equal-degeneracy model is stated here. Shifting both energy levels changes the energy baseline but not heat capacity.

Equipartition gives mean energy $k_{B}T/2$ for each independent quadratic term in a classical canonical Hamiltonian. For example, $p^2/(2m)$ and $k{x}^2/2$ each contribute this amount in a harmonic oscillator. It is a thermal average, not an instantaneous equality; constraints, nonquadratic terms or quantum level spacings large compared with $k_{B}T$ require a different calculation.

The fluctuation-dissipation theorem connects equilibrium fluctuations with the dissipative part of linear response. If an applied force couples as $-f(t)x$, use response to $e^{-i\omega t}$ and define the two-sided angular-frequency spectrum $S_{xx}(\omega )=\int e^{i\omega t}⟨\delta x(t)\delta x(0)⟩dt$. The classical relation isFor a viscously damped particle, $m\dot{v}=-\zeta v+\xi (t)$, the corresponding white-noise force satisfies $⟨\xi (t)\xi (t^{\prime })⟩=2\zeta k_{B}T\delta (t-t^{\prime })$. Noise replenishes energy lost by friction; their strengths cannot be varied independently at fixed equilibrium temperature.

Electrically, the one-sided voltage-noise spectrum per hertz of an equilibrium resistor is $4k_{B}TR$. A capacitor in parallel filters it toThe same equilibrium voltage variance follows from equipartition. Increasing R makes the spectrum taller and narrower without changing its integrated area. This thermodynamic connection between dissipative conductance and thermal electrical fluctuations is the content of the classical noise calculation.^[ref]

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.

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.

Let $E$ denote the fluctuating microscopic energy and $U=⟨E⟩$ its mean, at fixed $N,V$. In a canonical ensemble with $\beta =1/(k_{B}T)$,Since $dT/d\beta =-k_B{T}^2$, this givesHere $\Delta U$ denotes an energy fluctuation about the mean, as in the question. The relation is not restricted to ideal gases when the Hamiltonian operator has no explicit temperature dependence. Alternatively, the stated equilibrium free-energy curvature gives a Gaussian energy weight proportional to $\exp [-(\Delta U{)}^2/(2k_B{T}^2{C}_V)]$, with the same variance.

For the relative energy fluctuation of a monatomic ideal gas, $U=3N{k}_{B}T/2$ and $C_V=3N{k}_B/2$, henceTaking room conditions as approximately $T=300$ K and $p=10^5$ Pa, with $V=10^{-6}$ m${}^3$, givesThe mean energy is about $0.15$ J and its standard deviation about $2.49\times 10^{-11}$ J. The tiny relative fluctuation explains why macroscopic equilibrium internal energy behaves as a sharply defined thermodynamic quantity. The estimate uses a thermal reservoir, not a particle reservoir; allowing $N$ itself to fluctuate changes the relevant fluctuation problem.

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.

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]

Landau description. Introduce a macroscopic order parameter and expand an appropriate free energy analytically in that parameter, retaining terms permitted by the physical symmetries. Stable equilibrium minimizes this free energy globally; a local minimum with larger free energy is metastable. Coefficient changes with temperature can exchange the preferred minima. A continuous emergence from zero can produce a continuous transition, while competing separated minima can produce a first-order transition. Stability requires a positive highest retained even-power term. This local mean-field description neglects spatial fluctuations and nucleation dynamics unless gradient and dynamical terms are added; near a critical point those omissions can matter.

For this temperature-dependent cubic-quartic Landau free energy, abbreviate $B=a/(12\pi )>0$ and $A(T)=c(T^2-T_0^2)$, so $F=A{\varphi }^2/2-BT{\varphi }^3+\lambda {\varphi }^4/4$ and $\gamma =2B^2/(\lambda c)$.

Local behavior at the origin. To determine constrained order-parameter boundary stability, note that $F^{\prime }(0)=0$ and $F^{\prime \prime }(0)=c(T^2-T_0^2)$, the origin is a local minimum for $T>T_0$ and a local maximum for $0\le T<T_0$. At $T=T_0$, the quadratic term vanishes, but $F=-B{T}_0{\varphi }^3+O({\varphi }^4)<0$ for small positive $\varphi$. ThereforeThe equality case uses the stipulated domain $\varphi \ge 0$. On an unrestricted real line, the nonzero cubic term would instead make it a stationary inflection point.

Global stability in both inline cases. For $\varphi >0$, complete the square in $F/{\varphi }^2$:The quartic coefficient guarantees a bounded global minimum. A negative value exists precisely when the minimum of this quadratic factor is negative. Hence case a, $\gamma >1$, has a nonzero globally stable phase with $F<0$ for every physical temperature. Case b, $0<\gamma <1$, has such a phase exactly forAt $T=0$, the square's centre is at the boundary, but its negative constant still gives $F<0$ for small positive $\varphi$, so this endpoint is included.

At $T=T^{\ast }$,The two global minima $\varphi =0,{\varphi }^{\ast }$ have equal free energy, and $F^{\prime \prime }({\varphi }^{\ast })=\lambda ({\varphi }^{\ast }{)}^2/2>0$. The coexistence barrier of a cubic-quartic Landau transition has its maximum at ${\varphi }^{\ast }/2$, with barrier height $\lambda ({\varphi }^{\ast }{)}^4/64$.

For locating the extrema away from coexistence,Retain only real positive roots; when both are positive, ${\varphi }_{-}$ is a local maximum and ${\varphi }_{+}$ a local minimum. For $T<T_0$, only ${\varphi }_{+}$ is positive. If $\gamma <8/9$, the pair first appears on cooling at $T_{\mathrm{sp}}=T_0/\sqrt{1-9\gamma /8}>T^{\ast }$; if $8/9\le \gamma <1$, a higher-free-energy nonzero minimum already exists at arbitrarily high temperature. These local-stability limits must not be mistaken for the global coexistence condition. The boundary $\gamma =1$ also favors a nonzero minimum at every finite temperature, but has no finite $T^{\ast }$.

Cooling and transition temperature. For $\gamma <1$, equilibrium starts at the zero minimum at high temperature and switches to the nonzero minimum at $T^{\ast }$. The jump is finite, $0\to {\varphi }^{\ast }>0$, and there is a barrier between the coexisting states. This is a first-order transition. The latent heat of a temperature-dependent cubic-quartic Landau transition follows from differentiating the minimized free energy: the heat released on cooling isIf $F$ is an energy density rather than total energy, $L$ is likewise a density. An actual system can supercool below $T^{\ast }$ before nucleation, so $T^{\ast }$ is the equilibrium transition temperature.

For thermal symmetry nonrestoration in a cubic-quartic model, with $\gamma >1$, the equilibrium global minimum is nonzero already at arbitrarily high temperature. Following it does not force a transition. A system initially prepared in the metastable zero minimum can still leave it by nucleation while $T>T_0$; that minimum loses local stability at $T_0$. This is the sense in which a change is possible rather than inevitable. There is no equilibrium coexistence temperature for $\gamma >1$ in this model. A kinetic transition can depend on preparation and nucleation dynamics; $T_0$ is the limiting loss of local stability of the zero state, not a second coexistence temperature. Continuing $T^{\ast }$ to an imaginary value does not define a physical transition temperature.

Mass window in the specified electroweak phase transition model. For the mass window in a cubic electroweak transition model, put $x=m_H^2/m_0^2>0$. The supplied ratios implyThe two strict conditions therefore requireSince $x(A+x)$ is increasing for $x>0$, the finite window isUsing the specified $A$ and $m_0$ givesAt $m_H=134$ u, $x=0.3581801$, $\gamma \simeq 0.0030033<1$ but ${\varphi }^{\ast }/T^{\ast }\simeq 0.1481<1$. Thus 134 u lies outside the required window, failing the order-parameter-strength condition. This does not remove the polynomial model's finite first-order jump; it fails the additional requirement ${\varphi }^{\ast }>T^{\ast }$. The result is a bound within the specified simplified model, not a model-independent Higgs-mass bound.

Take $\mu >0$ to be the magnitude of either spin projection's magnetic moment, so the independent magnetic levels are $E_{\pm }=\mp \mu B$. Their partition function and magnetic energy arewhere $\beta =1/(k_{B}T)$ and $x=\beta \mu B$. The canonical entropy is consequentlyThis independent-spin magnetic entropy counts the two orientation states of each independent moment; translational entropy and interactions are excluded.

Along the magnetic-entropy demagnetization adiabat, constant magnetic entropy fixes $x$, hence $B/T$ is constant. Lowering the field therefore gives the ideal adiabatic demagnetization relationA useful cycle first magnetizes isothermally while releasing heat to a reservoir, then thermally isolates the sample and lowers the field. Cooling the entire sample requires its other heat capacities and spin-lattice equilibration to be accounted for. Magnetic interactions, residual fields and heat leaks prevent extrapolating the independent-moment formula to an attainable zero temperature.^[ref]

For a complete single-particle state, the Fermi-Dirac distribution isHere ${\mu }_{\mathrm{ch}}$ denotes the chemical potential, distinct from the magnetic-moment symbol in (b). At zero temperature it approaches the Fermi energy $E_F$ and the occupation has a sharp edge. At $T\ll T_F=E_F/k_B$, thermal changes occur within an energy interval of order $k_{B}T$ about that edge.

The Fermi-edge heat-capacity counting argument uses the fact that only order $g(E_F)k_{B}T$ electrons can be rearranged because states well below the edge are blocked by the Pauli exclusion principle. Each active excitation carries energy of order $k_{B}T$, so $U(T)-U(0)\sim g(E_F)(k_{B}T{)}^2$ and $C_{V,e}\propto T$. For a smooth density of states, the Sommerfeld expansion givesHere $g$ is the total single-particle density of states, including spin. In a three-dimensional free-electron gas, $g(E_F)=3N/(2E_F)$, giving $C_{V,e}=({\pi }^2/2)N{k}_B(T/T_F)$. This is the low-temperature normal-state electronic contribution, not the lattice heat capacity or a description across a superconducting transition.^[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]

Measure energies from the single-particle ground state, and assume the three-dimensional ideal Bose gas in the large-particle-number limit. The harmonic-trap chemical-potential pinning discussed here uses a chemical potential satisfying ${\mu }_{\mathrm{ch}}\le 0$. On cooling, it approaches zero; below the condensation temperature it remains at zero to leading thermodynamic order. Above that temperature it is negative. The detailed negative-chemical-potential branch shape is determined by the density of states; the figure uses the harmonic trap analysed below. The finite-size condensate chemical potential stays slightly below zero rather than being exactly pinned there.

A Bose-Einstein condensate has macroscopic occupation of a single quantum state. A dilute atomic gas needs low temperature and high phase-space density while remaining metastable against ordinary liquid or solid formation. Laser cooling followed by evaporative cooling of atoms in a weak trap can achieve the required regime. For the trapped ideal gas the quantitative threshold is derived below; a uniform ideal gas has $n{\lambda }_T^3=\zeta (3/2)$ at onset. An experimental signature is a narrow low-momentum component appearing within the broader thermal distribution after time-of-flight expansion, as observed in magnetically trapped rubidium-87.^[ref]

Spinless cubic-box wave-vector counting. Impose periodic boundary conditions on a cube of side $L$. The allowed vectors are $\mathbf{k}=(2\pi /L)(n_x,n_y,n_z)$ with integer components. Each state occupies $(2\pi /L{)}^3$ in wave-vector space. For a spinless particle, counting the spherical shell givesThere is no spin multiplicity. This is continuum shell counting; a finite box has discrete levels. Using $ϵ={\hslash }^2{k}^2/(2m)$ gives the free-particle density of states per unit energy and volume

Semiclassical density of states in a central potential. At position $\mathbf{r}$, the available kinetic energy is $ϵ-U(r)$. The local-density approximation integrates the preceding free-particle result over the classically allowed region:This form assumes the allowed radial region is the connected interval $0\le r\le r_{ϵ}$. A nonmonotonic potential would require integrating all allowed shells.

For the isotropic harmonic trap, $r_{ϵ}=\sqrt{2ϵ/(m{\omega }^2)}$. Substituting $r=r_{ϵ}\sin \vartheta$ givesConsequentlyThe exact oscillator ground energy is $3\hslash \omega /2$ above the potential minimum. The harmonic-trap energy-zero convention subtracts that constant, while the displayed density of states is the leading semiclassical continuum expression, neglecting the finite-level corrections produced by that shift. The discrete ground state must be counted separately.

Condensate population. Below the transition, use ${\mu }_{\mathrm{ch}}\simeq 0$ for the excited states. The saturation of harmonic-trap excited states follows from the supplied integral:where $\zeta$ is the Riemann zeta function and $\zeta (3)\simeq 1.202$. Setting $N_{\mathrm{ex}}(T_0)=N$ gives the harmonic-trap critical-temperature prefactor:Above $T_0$ there is no macroscopic ground-state fraction in this approximation. For completeness, above the transition the quantum gas fugacity $z=e^{{\mu }_{\mathrm{ch}}/(k_{B}T)}<1$ satisfies $N=(k_{B}T/\hslash \omega {)}^3{\mathrm{Li}}_3(z)$; the polylogarithm ${\mathrm{Li}}_3$ therefore defines the plotted negative branch. A finite trapped gas has rounded behaviour, weak interactions shift the threshold, and the continuum thermal population is inaccurate when $k_{B}T$ is comparable to or below $\hslash \omega$. The power law is intended for the large-$N$ condensed regime, not an exact discrete-spectrum formula at arbitrarily small temperature.^[ref]

For the ideal dilute Maxwell-Boltzmann gas implicit in a single-particle partition function, indistinguishability gives $Z_N=Z_1^N/N!$. With Stirling's approximation,Differentiate at fixed $T,V$:This is a thermodynamic-limit ideal-gas result; arbitrary interacting classical gases do not have $Z_N=Z_1^N/N!$.

Equilibrium constraints and the Gibbs minimum. The open-system equilibrium contacts allow exchange of heat, volume and matter. For unconstrained contacts without external-field or interface corrections, equilibrium requires equal temperatures, pressures and each exchangeable species' chemical potentials across the contacts. These conditions follow by maximizing total entropy under conserved total energy, volume and species amounts.

The next minimization statement fixes the system's particle number. For a system in a reservoir at $T_0,p_0$, with $N$ fixed and only pressure-volume work, the reservoir entropy change givesFor candidate states at the imposed $T=T_0$, $p=p_0$, the expression in parentheses is $G$. Thus the Fixed-particle-number Gibbs minimum selects stable equilibrium at fixed $T,p,N$, with stationary first variation and nonnegative allowed second variation. If particle exchange remains unconstrained, $N$ is not fixed and this minimization statement must be replaced by one incorporating the particle reservoir; it is not a justification for minimizing $G$ over arbitrary $N$.

The simple-compressible-system first law givesFor a homogeneous extensive single-component phase, the Single-component Gibbs Euler relation follows from $U=TS-pV+{\mu }_{\mathrm{ch}}N$, yielding $G={\mu }_{\mathrm{ch}}N$. For a mixture, the corresponding identities are $G={\sum }_i{\mu }_i{N}_i$ and $dG=-SdT+Vdp+{\sum }_i{\mu }_{i}d{N}_i$. Surface effects or other nonextensive contributions require additional terms.^[ref]

Reaction equilibrium. Let $d\xi$ be an allowed reaction extent for the forward process, so $d{N}_A=-d\xi$ and $d{N}_B=2d\xi$. At fixed $T,p$,For an interior equilibrium allowing both reaction directions, the coefficient vanishes:At fixed $T,V$, minimizing $F$ gives the same condition. The reaction formula-unit conservation gives $M=N_A+N_B/2$, not the total number $N_A+N_B$ of gas particles. This conservation law replaces fixed total particle number for the reacting mixture.

Using the stipulated single-particle functions,The number-based dissociation equilibrium ratio is thereforeThis number-based quantity has a factor of volume; the corresponding concentration ratio $n_B^2/n_A=K_N/V$ depends on temperature alone in the stated ideal model. A standard activity-based equilibrium constant would additionally specify reference states. No bond-energy factor is added to the functions given in the paper.

Define the dissociated fraction of original A formula units by $\alpha =N_B/(2M)$. Then $N_A=M(1-\alpha )$ and $N_B=2M\alpha$, soThe dissociation fraction versus conserved density follows because the left side increases monotonically from zero to infinity on $0<\alpha <1$. At fixed $V,T$, increasing $M$ decreases $q$, hence decreases dissociation. Equivalently, at fixed $M,T$ compression raises $M/V$ and favours A. Association produces fewer independent translating particles; at high density the translational-entropy advantage of the two-particle side is smaller, so equilibrium shifts toward the one-particle side.

Put $\xi =ϵ-{ϵ}_F$ and $\Delta =|\Delta |>0$. The BCS quasiparticle spectrum has $E=\sqrt{{\xi }^2+{\Delta }^2}$, so on either inverse branchConservation of states on one branch gives $D_{\mathrm{branch}}(E)dE=g({ϵ}_F)|d\xi |$. Consequently the single-branch expression requested in the source isThere are no such states for $0<E<\Delta$ in this ideal fully gapped model. Near the edge the density of states diverges as $(E-\Delta {)}^{-1/2}$, an integrable singularity; at large $E/\Delta$ it approaches its normal-state value.