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Each electromagnetic normal mode is a quantum harmonic oscillator with $E_n=(n+1/2)h\nu$. For $x=h\nu /(k_{B}T)$ its canonical partition function and occupation probabilities areThe geometric distribution is normalized and gives thermal mode energy $h\nu /(e^x-1)$. The zero-point term $h\nu /2$ is not part of the thermally emitted black-body intensity. The equilibrium photon chemical potential is zero because photon number is not conserved.

Counting two transverse polarizations gives $8\pi {\nu }^2/c^3$ modes per volume per frequency interval. Thus the spectral energy density is $u_{\nu }=8\pi h{\nu }^3/[c^3(e^x-1)]$. In an isotropic radiation field, spectral radiance is $B_{\nu }=c{u}_{\nu }/(4\pi )$. Transforming the spectral measure, $B_{\lambda }=B_{\nu }|d\nu /d\lambda |$, givesHere intensity means radiance per wavelength and solid angle; the hemispheric spectral exitance of a black surface is $M_{\lambda }=\pi B_{\lambda }$. The extra Jacobian is essential: the wavelength peak is not found by simply converting the frequency peak using $\lambda =c/\nu$.

At short wavelength the exponential suppresses radiation. At long wavelength the Rayleigh-Jeans limit is $B_{\lambda }\simeq 2c{k}_{B}T/{\lambda }^4$; using classical equipartition at every frequency instead would produce the ultraviolet catastrophe. The wavelength maximum solves $5(1-e^{-x})=x$, with nonzero root $x=4.9651$, yielding Wien's displacement law. For the specified temperature,Integrating the hemispheric flux gives the Stefan-Boltzmann law, $M=\sigma T^4\simeq 3.01\times 10^{-6}\mathrm{W}{\mathrm{m}}^{-2}$. The plot uses the specified ideal temperature, rather than substituting a later measured value. The physical black-body description of the cosmic microwave background is tested by comparing the sky spectrum with an absolute reference blackbody and separating instrumental and Galactic contributions.^[ref]

A diatomic molecule has one internal bond-stretching vibrational coordinate. Near its equilibrium separation, approximate the bond by a spring of stiffness $K$ and use the reduced mass $\mu =m_1{m}_2/(m_1+m_2)$. Its frequency is $\omega =\sqrt{K/\mu }$ and its quantum harmonic oscillator levels are ${ϵ}_r=(r+1/2)\hslash \omega$.

For independent molecules in thermal equilibrium, set $x=\hslash \omega /(k_{B}T)={\theta }_v/T$. The single-molecule canonical partition function is a geometric series:Using $U=-\partial \log (z_v^N)/\partial \beta$, where $\beta =1/(k_{B}T)$, givesPer mole replace $N{k}_B$ by $R$. The zero-point energy is present in $U_v$ but is temperature-independent and contributes nothing to the heat capacity.

For $T\ll {\theta }_v$, occupation of the first excited vibrational state is exponentially suppressed: $C_{V,v}\simeq N{k}_B{x}^2{e}^{-x}\to 0$. The vibrational degree of freedom is frozen out. For $T\gg {\theta }_v$, $C_{V,v}\to N{k}_B$. This recovers classical equipartition: the quadratic kinetic term and quadratic elastic term each contribute $k_B/2$ per molecule.

The vibrational heat capacity of a diatomic gas must be distinguished from the total heat capacity. In a dilute gas with rotations already active and vibration still frozen, translation contributes $3R/2$ and rotation $R$, so molar $C_V\simeq 5R/2$. Once vibration is active it adds $R$, giving $C_V\simeq 7R/2$. For an ideal gas of a fixed number of intact molecules, $C_P=C_V+R$, so the corresponding values are $7R/2$ and $9R/2$. Very low temperatures also freeze rotations, while sufficiently high temperatures bring anharmonicity, dissociation or electronic excitations; the harmonic plateau is restricted to the range where those effects are negligible.

The expression is the canonical partition function for a system in thermal equilibrium with a large heat bath at fixed temperature $T$. The system exchanges energy with the bath, while its particle number and external mechanical parameters, such as volume, are fixed. The bath is large enough that its temperature is not appreciably changed by this exchange.

The system has allowed energy levels ${ϵ}_j$. The index $j$ labels distinct energy values, and the integer degeneracy factor $g_j$ counts distinct quantum states with that same energy. $k_B$ is the Boltzmann constant, and $\exp [-{ϵ}_j/(k_{B}T)]$ is a state's Boltzmann factor. The dimensionless sum $Z$ normalizes the probabilities:Thus the degeneracy multiplies the probability of occupying the level, not the probability assigned to each individual state within it.

The canonical model is not restricted to a classical ideal gas. It applies to a quantum rotor, for example, provided the sum counts all of that system's allowed states with their proper degeneracies. For a many-particle system, ${ϵ}_j$ would be the energy of a many-particle state, with any quantum-statistical restrictions already built into the state count.

Let $\beta =(k_{B}T{)}^{-1}$, and sum over all $2^N$ configurations satisfying the periodic identification. The canonical partition function and finite-size free energy per site areThe plus sign multiplying $J$ in the energy means that $J>0$ favors opposite neighboring spins: this is an antiferromagnetic Ising model. Even $N$ permits an alternating configuration to close without frustration. The finite-size formula does not require taking a thermodynamic limit; that limit is taken only after evaluating $Z_N$.

Let $\beta =(k_{B}T{)}^{-1}$ and sum over all $2^{2L}$ configurations of the Ising spins. The canonical partition function and Helmholtz free energy per site areBoth spins in each cell count as sites. In particular the denominator is $2L$, rather than $L$.

The mean spin magnetization per site is the Boltzmann distribution averageEquivalently introduce a uniform magnetic field through $E_h=E-h{\sum }_j({\sigma }_j+{\tau }_j)$ and differentiate:This convention absorbs the microscopic magnetic moment into $h$. Simultaneous reversal of every spin leaves $E$ invariant and changes the sign of the numerator, so $⟨m⟩=0$ at zero field for finite $L$.

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.

Entropy curves and reversible cooling. For a fixed $N$ and a reference pair $(T_{\ast },a_{\ast })$, the dependence can be written without logarithms of dimensional quantities:At each fixed $a$, the entropy increases logarithmically and is concave downward in $T$; decreasing $a$ shifts the curve upward. Its formal classical divergence as $T\to 0$ marks failure of classical statistics, not a physical negative-infinite entropy.

For a reversible adiabatic change, $S$ is constant at fixed $N$, soHence reducing $a$ lowers the temperature: this is isentropic decompression of a spherical linear trap. The gas does work on the control that weakens the trap. Equivalently $dU=⟨\partial H/\partial a⟩da=N⟨r⟩da$, with $⟨r⟩=3k_{B}T/a$; inserting $U=9N{k}_{B}T/2$ gives the same relation. Reversibility requires an equilibrium sequence without heat input or entropy production. Because $Z_1$ stays constant on this adiabat, $N/Z_1$ also stays constant: this mechanism cools the cloud but does not by itself increase quantum degeneracy.

Chemical potential and quantum threshold. From the extensive free energy,Thus the quantum gas fugacity is $z=e^{\mu /(k_{B}T)}\simeq N/Z_1$. Classical statistics requires $z\ll 1$; quantum effects become important as the chemical potential comes within order $k_{B}T$ of the lowest one-particle energy. With the semiclassical zero at the potential minimum, the quantum degeneracy threshold in a spherical linear trap is consequentlyThere are no numerical values of $m,T,a$ in the paper, so this parameter-dependent estimate is the requested atom number. It also follows from the peak density: $n(0)=N/[8\pi (k_{B}T/a{)}^3]$, and degeneracy starts when $n(0){\lambda }_T^3$ is order one. The criterion is a crossover estimate, not an assertion that the classical formula remains valid at $\mu =0$. For spinless bosons the continuum excited-state saturation is more specifically $N_c=\zeta (9/2)Z_1$, while for fermions the onset is occupation constrained by exclusion. At temperatures comparable with the quantum trap spacing, the actual ground energy must be included consistently rather than identified exactly with the classical minimum.

Particles within each species are indistinguishable, while the two species are physically distinguishable and their numbers are fixed. Independence makes the canonical partition function factor by species:Since $Z_1\propto T^{3/2}$ at fixed volume, $U=-{\partial }_{\beta }\ln Z=3N{k}_{B}T/2$. The canonical entropy is thereforeFor large species populations, Stirling's approximation gives the equivalent Sackur-Tetrode equation formFor structureless particles with the same translational partition function, $Z_1=V/{\lambda }^3$, where $\lambda$ is their common thermal de Broglie wavelength. More generally the supplied common constant factor in $Z_1$ should be retained in the entropy.

Writing $x_i=N_i/N$ separates the total-number contribution from the composition term:The last term is the entropy of mixing relative to separated ideal components at the same temperature and pressure. This is the entropy of a two-species classical ideal gas. Artificially labeling otherwise identical chemical atoms would not justify dividing by separate factorials or adding a mixing entropy. The pure-species limit is obtained by $x\ln x\to 0$.

Canonical probabilities. A canonical ensemble describes a system of fixed particle number and fixed external parameters such as volume, weakly coupled to a much larger thermal reservoir. They exchange energy and reach thermal equilibrium, while the reservoir's temperature stays effectively constant. The interaction energy is negligible in the state counting. The combined isolated system is described by equal probabilities for its accessible microstates.

If its total energy is $E_{\mathrm{tot}}$, the probability of a particular system microstate $i$ is proportional to the reservoir multiplicity at $E_{\mathrm{tot}}-E_i$. Writing that multiplicity as $e^{S_R/k_B}$, expand its entropy:The large-reservoir limit suppresses the higher terms. Normalization therefore givesThe sum counts individual microstates; when grouping them by energy, their state degeneracies must be included. This $Z$ is the canonical partition function.

Classical orientational heat capacity. Hold the magnetic field fixed. Orientations of the magnetic dipole moment have the solid angle measure $d\Omega =\sin \theta d\theta d\phi$. With $x=\beta mB$, the angular partition function isChoosing a normalized angular measure would remove $4\pi$, with no effect on the energy derivatives. Differentiating at fixed $B$,The angular average $⟨\cos \theta ⟩=\coth x-1/x$ is the Langevin function. Since $dx/dT=-x/T$,This is the heat capacity of a classical magnetic dipole, in $\mathrm{J}{\mathrm{K}}^{-1}$ for one magnetic moment. The word specific here refers to that single moment; $N$ independent magnetic moments have $N$ times this heat capacity. At $B=0$ the energy is independent of orientation and $C_B=0$, obtained by the continuous $x\to 0$ limit. At fixed nonzero $B$, the classical low-temperature limit instead gives $C_B\to k_B$.

Equipartition and its proof. The equipartition theorem states that an independent quadratic term $a{q}^2/2$ in a classical Hamiltonian, with $a>0$, contributes $k_{B}T/2$ to its mean energy. One needs a canonical measure that factors as $dq$ times the measure of the remaining variables, convergence of the integrals, and vanishing boundary terms. The coordinate must not be restricted by a thermally important hard boundary or a coupled constraint.

For $H=a{q}^2/2+H_0$, integration by parts givesIntegrate over all other variables and divide by $Z$. It follows thatThe same reasoning applies to quadratic momenta among the canonical variables. It is the number of quadratic energy contributions that is counted, not an unrestricted count of coordinate labels.

Two soft orientational directions. When $x\gg 1$, the moment is concentrated near alignment. Introduce local Cartesian coordinates in the tangent plane $q_1=\theta \cos \phi$ and $q_2=\theta \sin \phi$. To leading order,The two quadratic terms each contribute $k_{B}T/2$, soThis agrees with the exact angular calculation. Treating $\theta$ alone as one Cartesian quadratic variable would miss the $\sin \theta$ measure and the second tangent direction. There are no extra rotational kinetic terms in the stipulated Hamiltonian; adding such terms would be a different model.

Quantum spin one. The spin projections are $m_s=-1,0,1$, with $S_z=\hslash m_s$. Put $\Delta =\hslash |\gamma |B>0$, where $\gamma$ is the gyromagnetic ratio. The three energies are $-\Delta ,0,+\Delta$, irrespective of the sign of $\gamma$. Relative to the ground state they are $0,\Delta ,2\Delta$. Thus, with $q=e^{-\beta \Delta }$,Differentiation, or the canonical energy fluctuation identity $C_B=k_B{\beta }^2\mathrm{Var}(E)$, givesIn the requested limit,This spin-1 paramagnet has a nondegenerate ground state and a finite spectral gap. Excitations are exponentially rare, so its heat capacity tends to zero. Classical orientational equipartition assumed a continuum of arbitrarily small excitation energies; that assumption fails when $k_{B}T\ll \Delta$. This is a change in the accessible state spectrum, not a violation of the equipartition theorem within its classical domain.

An additive-looking energy expression does not by itself establish two noninteracting subsystems. Their allowed microstates must also form a Cartesian product of independently available states. Here the second label's allowed values depend on the first. The four allowed pairs and their energies areThe canonical partition function, with $\beta =(k_{B}T{)}^{-1}$, is thereforeThe conditional sum over $m_J$ is $2\cosh (\beta ϵJ)$ and depends on $J$; it cannot be pulled out as the partition function of an independent second subsystem. Summing independently over $J\in \{1,2\}$ and all four values $m\in \{-2,-1,1,2\}$ would invent four forbidden pairs and give eight states instead of four.

One can make the state labels independent by writing $s=m_J/J=\pm 1$, but then $E(J,s)=ϵJ(1+s)$ contains a coupling between them. For genuinely additive independent energies, the difference between the $s=+1$ and $s=-1$ states would be independent of $J$; here it is $2ϵJ$. Hence the proposed $J$ and $m_J$ subsystems cannot be treated as noninteracting independent subsystems. The literal two-value reading of the PDF's endpoint notation has been used. If that notation were instead intended as shorthand for full angular-momentum multiplets, including every integer between the endpoints, the eight allowed states would giveThe conclusion about the failure of independent-subsystem factorization is unchanged. The two state-counting conventions should not be mixed.

Put $\beta =1/(k_{B}T)$ and define the thermal de Broglie wavelength of species $\alpha$ byIgnoring the internal degrees of freedom as stipulated, each molecule's translational canonical partition function is $V/{\lambda }_{\alpha }^3$. The two species do not interact, so their partition functions multiply; the treatment of indistinguishable particles requires a separate factorial for each species, not a factorial of the total particle number. ThusUsing $F=-k_{B}T\ln Z$ and Stirling's approximation gives the extensive Helmholtz free energyThese are ideal gas expressions, valid for $n_{\alpha }{\lambda }_{\alpha }^3\ll 1$, with $n_{\alpha }=N_{\alpha }/V$. The energy zero has been chosen so that separated CO and oxygen molecules have zero internal reference energy.

The given binding energy $ϵ$ belongs to the formation of two carbon-dioxide molecules. Relative to the reference used in (a), the internal energy offset of one carbon-dioxide molecule is therefore $-ϵ/2$, not $-ϵ$. Its mass is $m_c=m_{\mathrm{CO}}+m_{{\mathrm{O}}_2}/2$ and its thermal de Broglie wavelength is ${\lambda }_c=h/\sqrt{2\pi m_c{k}_{B}T}$, where the subscript $c$ denotes $\mathrm{C}{\mathrm{O}}_2$.

The single-molecule canonical partition function is $(V/{\lambda }_c^3)e^{\beta ϵ/2}$. For $N_c$ indistinguishable noninteracting molecules,The corresponding Helmholtz free energy, using Stirling's approximation, isThe negative sign in the last term expresses the stabilizing binding energy and ensures that low temperature favors association.

For $N$ indistinguishable classical particles the canonical partition function is an integral over phase space,The factor $1/N!$ removes overcounting under particle permutations, and $h^{3N}$ makes the classical phase-space measure dimensionless. The pair potential depends only on positions, so all momentum integrations separate. The three-dimensional Gaussian integral for each particle givesThe grand canonical partition function is $\Xi ={\sum }_{N=0}^{\infty }{e}^{\beta \mu N}{Z}_N$, where chemical potential $\mu$ controls the mean number of particles. ThereforeFor $N=0$, the empty products and configurational integral equal one. The factor $1/2$ in the interaction prevents double counting of unordered particle pairs.

For a single-particle state of energy $ϵ$, put $z=e^{\mu /(k_{B}T)}$. The equilibrium mean occupancies areThe Maxwell-Boltzmann distribution is the dilute classical limit. The Fermi-Dirac distribution applies to identical fermions with at most one particle in each complete single-particle state by the Pauli exclusion principle. The Bose-Einstein distribution applies to identical bosons with no such upper occupation bound. The chemical potential fixes total particle number; these are mean state occupancies, not three separately normalized single-particle probability densities. For a number-conserving ideal Bose gas, $\mu$ cannot exceed the lowest single-particle energy.

Take $\mathbf{k}$ to be a wavevector and count periodic-box states with cell volume $(2\pi {)}^3/V$ in momentum space, parameterized by $\mathbf{k}=\mathbf{p}/\hslash$. Write $g_s$ for internal degeneracy. The number of states in a spherical shell is $g_{s}V4\pi k^{2}dk/(2\pi {)}^3$. Since $ϵ=\alpha k$,Thus the one-particle canonical partition function isHere $\beta =1/(k_{B}T)$, $g_s=1$ for one nondegenerate branch, and $g_s=2$ for the spin-half case below. The integral uses ${\int }_0^{\infty }{u}^2{e}^{-u}du=2$. For classical indistinguishable particles, $Z_N=Z_1^N/N!$, givingThe factor three in the energy arises from the linear dispersion relation and three-dimensional state counting; the usual nonrelativistic quadratic-dispersion value $3N{k}_{B}T/2$ does not apply.

For spin-half fermions at $T=0$, all states up to the Fermi energy are occupied. With $g_s=2$,Integrating the occupied energy states givesThese are the corresponding ultrarelativistic ideal Fermi gas formulae with general slope $\alpha$. The nonzero ground-state internal energy and degeneracy pressure reflect the Pauli exclusion principle: particles cannot all occupy the lowest-momentum state. Formally extrapolating the classical equations gives $U,p\to 0$, but Maxwell–Boltzmann statistics is not valid at fixed positive $n$ as $T\to 0$.

At large $T$, the number equation gives $z\simeq N/Z_1\ll 1$, so expanding the Fermi-Dirac distribution recovers the Maxwell-Boltzmann distribution. A convenient crossover scale is the Fermi temperature,For $T\gg T_F$, both gases have $p\simeq n{k}_{B}T$. At fixed $n$, the Fermi curve begins at $p_0=n{ϵ}_F/4$, has zero initial slope, lies above the classical line, and approaches it at high temperature. For example its low-temperature expansion is $p/(n{ϵ}_F)=1/4+({\pi }^2/6)(T/T_F{)}^2+\cdots$. The plot evaluates the finite-temperature occupation integrals at fixed number density.

If the source's symbol $\mathbf{k}$ is interpreted literally as momentum rather than a wavevector, the state-space cell is $h^3/V$. In that convention replace $\alpha$ in the wavevector-based coefficients and energy scales above by $\hslash \alpha$. The dimensionless thermodynamics and the relation $p=U/(3V)$ are unchanged.

The ferromagnetic Potts model on a labelled ring has $Q$ possible site states, $J>0$ and ${\sigma }_{L+1}={\sigma }_1$. Its spin-chain transfer matrix is $T=(e^{\beta J}-1)I+P$, with $P$ the $Q$-dimensional all-ones matrix. The canonical partition function is $(e^{\beta J}+Q-1{)}^L+(Q-1)(e^{\beta J}-1{)}^L$. The convention counts two bonds when $L=2$.

A quantum Gibbs state is the density operator of a canonical ensemble with inverse temperature $\beta =1/(k_{B}T)$. It exists when the canonical partition function $Z=\mathrm{Tr}(e^{-\beta H})$ is finite. The spectral theorem assigns probabilities $e^{-\beta E_j}/Z$ to an orthonormal eigenbasis of the Hamiltonian. An additive constant in the Hamiltonian cancels between numerator and denominator. For $H=-\mu B{\sigma }_z$, the exponential of a Pauli vector, continued to an imaginary argument, gives ${\rho }_{\beta }=[I+\tanh (\beta \mu B){\sigma }_z]/2$.

Write the linear rigid rotor energy eigenvalues as $E_J=k_{\mathrm{B}}{\theta }_{r}J(J+1)$, where ${\theta }_r={\hslash }^2/(2I{k}_{\mathrm{B}})=hB/k_{\mathrm{B}}$ is the rotational temperature scale, $I$ the moment of inertia, $B$ the frequency-valued rotational constant, and $k_{\mathrm{B}}$ the Boltzmann constant. Put $t=T/{\theta }_r$ and ${ϵ}_J=J(J+1)$. Define sums that include rotational multiplicity but exclude nuclear-spin multiplicity:The canonical partition function for one equilibrating internal rotor is $q_{\mathrm{eq}}=q_{\mathrm{p}}+3q_{\mathrm{o}}$, so its equilibrium orthohydrogen fraction isFor $t\ll 1$, the lowest allowed levels give $q_{\mathrm{p}}\simeq 1$ and $q_{\mathrm{o}}\simeq 3e^{-2/t}$, hence $x_{\mathrm{o}}^{\mathrm{eq}}\simeq 9e^{-2/t}/(1+9e^{-2/t})\to 0$. For $t\gg 1$, the densely populated even and odd ladders have equal leading sums, $q_{\mathrm{p}}\sim q_{\mathrm{o}}\sim t/2$, hence $x_{\mathrm{o}}^{\mathrm{eq}}\to 3/4$. The factor nine belongs to the low-temperature two-level approximation, and the factor three to the high-temperature spin multiplicities.

A frozen mixture requires two separately normalized probability distributions,Multiply these by the prescribed fractions $1-x$ and $x$, respectively. The nuclear-spin factor three cancels from the conditional orthohydrogen distribution. For fixed counts $N_{\mathrm{o}}$ and $N_{\mathrm{p}}$, the internal part of the many-molecule partition function is proportional to $q_{\mathrm{p}}^{N_{\mathrm{p}}}(3q_{\mathrm{o}}{)}^{N_{\mathrm{o}}}$, with combinatorial factors independent of temperature. Its logarithm yields the two conditional means weighted by the fixed counts. Using $q_{\mathrm{p}}+3q_{\mathrm{o}}$ instead would silently permit conversion and change the physical constraint.

For a linear rigid rotor, rotational temperature expresses the rotational energy constant in kelvins: $E_J=k_B{\theta }_{r}J(J+1)$, where $I$ is the perpendicular moment of inertia. With all integer $J\ge 0$ allowed and degeneracies $2J+1$, its rotational canonical partition function is $Z={\sum }_J(2J+1)e^{-{\theta }_{r}J(J+1)/T}$. It approaches $1+3e^{-2{\theta }_r/T}$ at low temperature and $T/{\theta }_r$ at high temperature. The latter limit gives mean rotational energy $k_{B}T$ and heat capacity $k_B$ per rotor.

For a quantum harmonic oscillator with $\hslash =1$, $\omega >0$ and inverse temperature $\beta >0$, the canonical partition function is ${\sum }_{r\ge 0}{e}^{-\beta \omega (r+1/2)}$. The geometric series gives the displayed formula. Keeping the zero-point energy is essential to its prefactor.

The trace of imaginary-time evolution is the canonical partition function of a quantum system. In a coordinate basis it is the integral of the diagonal Euclidean path integral kernel. Inserting a symmetry operator instead gives a weighted thermal trace.

The thermodynamic limit describes a sequence of larger systems, not a single enormous specimen: particle number $N$ and volume $V$ tend to infinity with number density $n=N/V$ fixed. Hold the chosen intensive and extensive thermodynamic quantities such as temperature $T$ and an applied magnetic field fixed along that sequence. In the canonical ensemble the controls are $T,N,V$; in the grand canonical ensemble one instead fixes $T$ and chemical potential, and the limiting number density is a result, possibly with a jump at phase coexistence. These are different specifications of a limiting experiment.

The useful limit is usually an intensive and extensive thermodynamic quantities normalization of the Helmholtz free energy, not a finite limit of the total extensive quantity itself:Here $Z_{N,V}$ is the canonical partition function, $k_B$ is the Boltzmann constant, $f$ is free-energy density, and $a$ is Helmholtz free energy per particle. Thus $F_{N,V}=Vf+o(V)$ when the limit exists; finite systems need not be exactly extensive quantities because of the surface correction to thermodynamic free energy.

Specify the interactions and the growing containers too. For a conventional fluid, use a fixed stable pair potential with a tempered pair potential tail, and containers whose boundary layer of any fixed thickness occupies a vanishing fraction of the volume; uniformly enlarged cubes work, a tube of fixed cross-section does not give the same bulk limit. For a finite-range Ising model, keep the lattice spacing and bond coupling fixed while enlarging all intended spatial directions. Ordinary bounded wall interactions or periodic boundary conditions can then give the same bulk free-energy density, although boundary conditions can select different spontaneous symmetry breaking states. This is a set of hypotheses, not a theorem that every interaction has a boundary-independent limit.^[ref]

The Curie–Weiss model below deliberately uses a different limiting prescription: its all-to-all coupling is divided by $N$. Without that factor its interaction energy would grow as $N^2$, rather than as an extensive quantity. The model has no spatial wall or correlation length to compare with a container; it is an exactly summable illustration of finite-size rounding, not a quantitative model of the fluid experiment below.