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For finite-range interactions and bounded wall couplings, only sites within an interaction range of the wall have altered local environments. In a regular container of linear size $L$ and dimension $d$, the surface area scales as $A\sim L^{d-1}$ while volume scales as $V\sim L^d$. A boundary contribution of order $A$ therefore gives an order-$1/L$ correction to free-energy density. Away from a critical phenomenon, a regular surface contribution often has the form $F=Vf+A\gamma +\cdots$, with $\gamma$ a wall Helmholtz free energy per area. Critical finite-size rounding, edges and long-range interactions require separate care; vanishing relative boundary size does not guarantee a simple surface expansion in every regime.

An exact check uses the zero-field nearest-neighbour Ising model on a chain of $N\ge 3$ sites with fixed coupling $J>0$ and $K=J/(k_{B}T)$. The open chain has $N-1$ bonds, while the periodic ring has $N$. Their finite partition functions and dimensionless Helmholtz free energies per spin $g_N=F_N/(N{k}_{B}T)$ areFor the open chain, sum the last Ising spin successively: each removed bond contributes $2\cosh K$, leaving a factor of two. For the ring, the transfer matrix for the one-dimensional Ising model has eigenvalues $2\cosh K$ and $2\sinh K$, and its $N$th power's trace gives the displayed sum. The two ends are a constant-size boundary in one dimension, so the open-chain total correction is $k_{B}T\log \cosh K$, and its per-spin correction is exactly proportional to $1/N$.

Both sets of finite results approach the same $g_{\infty }$ at the fixed plotted $K=0.8$. Multiplying the correction by $N$ exposes the open-chain boundary term, while the ring correction vanishes exponentially at fixed finite $K$. This model's thermodynamic limit remains a real analytic function at every positive temperature: having a thermodynamic limit does not require a phase transition.

At zero field and with spin inversion symmetry in its boundary conditions, a finite Ising model pairs every configuration with its spin-reversed configuration. Their equal Boltzmann weights force the expected value of its spin magnetization $m=N^{-1}{\sum }_i{s}_i$ to vanish. Below a model's ordering critical temperature, an arbitrarily small positive field can instead select the positive spontaneous magnetization if size is taken to infinity first:The distinction is the order of limits and the stated symmetric boundary conditions; plus boundary conditions already break spin inversion symmetry. A zero-field symmetric mixture of the two ordered states can retain zero mean spin magnetization even in the thermodynamic limit. Also, $⟨|m|⟩>0$ in a finite system is not by itself proof of spontaneous symmetry breaking.^[ref]

A standard sufficient tempering condition is this tail bound beyond a fixed radius, with $ϵ>0$ in spatial dimension $d$. It makes the distant interaction tail integrable; finite-range pair potentials satisfy it. Together with a stable pair potential and suitable growing containers, such assumptions support the conventional thermodynamic limit. A Curie–Weiss model with its size-dependent coupling uses a different prescription.

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.

Consider the original two-Electron example obtained by taking $N=2$ in the Laughlin wavefunction, with no additional polynomial for the centre of mass. Introduce $Z=(z_1+z_2)/2$, relative complex coordinate $w=z_1-z_2$, separation $r=|w|$, and dimensionless separation $u=r/{\ell }_B$. Since $|z_1{|}^2+|z_2{|}^2=2|Z{|}^2+|w{|}^2/2$, the probability density factorizes:Integrating out $Z$ leaves an isotropic relative probability density. Define $D_m(u)$ per unit dimensionless area and $p_m(u)$ per unit dimensionless separation:The physical areal probability density is $D_m(r/{\ell }_B)/{\ell }_B^2$. The probability of a separation between $r$ and $r+dr$ is $p_m(r/{\ell }_B)dr/{\ell }_B$. The extra $u$ comes from the Jacobian determinant in polar coordinates: a thin annulus has area $2\pi rdr$. Omitting that factor changes the question being answered.

Both wavefunction normalization and the peak shift can be checked exactly. The substitution $t=u^2/4$ gives the Gamma function integralFor $m>0$, logarithmic differentiation givesConsequently the areal maximum lies at $u=2\sqrt{m}$, while the most probable separation lies farther out, at $u=\sqrt{4m+2}$. For $m=1$ these are $2$ and $\sqrt{6}\simeq 2.449$; for $m=3$, $\sqrt{12}\simeq 3.464$ and $\sqrt{14}\simeq 3.742$. The same substitution gives $⟨u^2⟩=4(m+1)$, so the root mean square separations are $\sqrt{8}$ and $4$ respectively. The distinction between a probability density maximum and a radial maximum therefore has a directly calculable geometric origin.

This two-Electron model explains exchange and pair avoidance. It does not establish a nonzero many-particle spectral gap, an extended quantum Hall plateau, or the topological properties of a fluid in the thermodynamic limit. Those require additional many-Electron reasoning and experimental tests.