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For a positive quadratic bending operator $K$ with a real orthonormal basis of eigenfunctions, the equipartition theorem gives $C(x,y)=k_{B}T{\sum }_n{W}_n(x)W_n(y)/{\mu }_n$, where $K{W}_n={\mu }_n{W}_n$. This is the inverse-operator Green function multiplied by Boltzmann constant and temperature. Unconstrained zero-energy filament modes prevent a normalizable canonical ensemble, and negative modes signal an unstable quadratic model.

For both ends clamped and $K=A{\partial }_x^4$, the diagonal is $C(x,x)=k_{B}T{x}^3(L-x{)}^3/(3A{L}^3)$. This differs from a clamped-free tip variance. The center variance is $k_{B}T{L}^3/(192A)$.

The canonical ensemble assigns the density matrix $e^{-\beta H}/Z$, with $Z=\mathrm{Tr}{e}^{-\beta H}$. Its expectation value of an observable is the displayed normalized trace. In imaginary time, cyclicity of this trace gives periodic bosonic correlation functions; it does not make the imaginary-time Heisenberg operator itself periodic.

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.

Use the independent spin-one-half two-level system with fixed $N$ distinguishable sites and Hamiltonian levels $E_{\pm }=\pm \mu B$, where $\mu >0$ is the magnitude of the field-axis magnetic moment and $B>0$ is a uniform externally controlled magnetic field. Assume positive temperature, no interactions between sites, and no appreciable lattice or load entropy. The two levels differ by $2\mu B$. With inverse temperature $\beta$, the canonical ensemble gives the one-site and total partition functionsHere the minus subscript denotes the lower energy $-\mu B$. Factorization of the partition function requires additive energy and independent sites; there is no particle-exchange factor $N!$ for these fixed distinguishable sites.

Differentiate the partition function, holding $B$ fixed when differentiating with respect to $\beta$. The internal energy, Helmholtz free energy, total field-axis magnetic moment $m$, and magnetic entropy areThe extensive moment $m$ is distinct from magnetization, the moment per unit volume. Direct substitution of $p_{\pm }$ into the Gibbs entropy gives the same $s(x)$ independently of the free-energy derivative.

At fixed $B$, the magnetic heat capacity isStrict decrease makes $s$ invertible on $x>0$. Constant magnetic entropy therefore fixes $x$, not $B$ or $T$ separately:The partition function may be used to label these equilibrium states even though the isolated cooling leg has no external heat bath. Actual thermalization of the lattice and load additionally requires a sufficiently fast spin-lattice relaxation; perfectly uncoupled spins by themselves cannot refrigerate an attached object.

The limiting cases are instructive. For $x\ll 1$, $s(x)=\ln 2-x^2/2+O(x^4)$, $m\simeq N{\mu }^{2}B/(k_{B}T)$, and $C_{B,\mathrm{mag}}\simeq N{k}_B{x}^2$. The moment response reproduces Curie's law. For $x\gg 1$, $s(x)\simeq (1+2x)e^{-2x}$ and the upper-state occupation is exponentially small. Taking $B\to 0$ at fixed $T>0$ gives $s\to \ln 2$; taking $T\to 0$ at fixed $B>0$ gives $s\to 0$. The origin has a path-dependent limit. Extrapolating a line of constant $x$ to $B=T=0$ does not construct a usable zero-temperature state or prove that a real material retains independent degenerate spins there.

A zero-energy mode lies in the kernel of the quadratic elastic filament operator. For pure bending, zero curvature makes the displacement affine. Free-free endpoints therefore leave translation and tilt modes, while slope-constrained, force-free endpoints leave only translation. Both clamped-clamped and hinged-hinged endpoints remove these modes. An unrestricted modal amplitude has constant Boltzmann distribution weight, so no normalizable canonical ensemble or finite displacement variance exists. Fixing the rigid degrees of freedom allows the equipartition theorem on the positive complement.

At fixed odd $N$, the zero-temperature canonical ensemble equally weights the $2N$ ground states of an antiferromagnetic Ising ring. The single frustrated bond in an antiferromagnetic Ising ring lies on a chosen $n$-bond path with probability $n/N$. The Ising spin product is $(-1{)}^n$ if the path excludes that bond and $-(-1{)}^n$ if it includes it, proving the displayed two-point correlation function. The loss of staggered alignment upon averaging persists at zero temperature and is caused by geometric frustration, rather than a positive density of thermal defects. The expression remains symmetric under $n\mapsto N-n$ for odd $N$.