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For $H=-2J\sum S_i^z{S}_{i+1}^z-g{\mu }_{B}B\sum S_i^z$, with dimensionless $S_i^z=\pm 1/2$, define $K=\beta J/2$ and $b=\beta g{\mu }_{B}B/2$. The transfer matrix for the one-dimensional Ising model has largest eigenvalue $e^K[\cosh b+\sqrt{{\sinh }^{2}b+e^{-4K}}]$. Differentiating its logarithm gives the displayed moment per site. In particular the zero-field susceptibility per site is $(g{\mu }_B{)}^2{e}^{\beta J}/(4k_{B}T)$. Factors of two differ from a model whose spin variables themselves take values $\pm 1$.

For a finite-range isotropic Heisenberg model, the Mermin-Wagner theorem excludes spontaneous magnetic order in one dimension at every nonzero temperature. Long-wavelength rotations cost very little energy: a smooth orientational fluctuation has $\delta F=({\rho }_s/2)\int dx({\partial }_x\theta {)}^2$, so its thermal variance contains $(k_{\mathrm{B}}T/{\rho }_s)\int dq/q^2$, which diverges in the infrared. In a Heisenberg ferromagnet, the same obstruction appears as a divergent thermal occupation of quadratic long-wave spin waves. This explains why a finite magnetization cannot be supported, although correlations can be strong over a long finite distance. The rigorous statement requires short-range interactions and the stated continuous symmetry; interchain coupling and anisotropy change the problem. Symmetry forcing the mean to zero in a finite chain alone would not prove absence of thermodynamic order. ^[ref]

For the Ising model, use dimensionless $s_i=2S_i^z=\pm 1$, $h=g{\mu }_{\mathrm{B}}B$, $K=\beta J/2$ and $b=\beta h/2$. Its pair Boltzmann factor isMultiplying N such factors around the periodic ring counts each bond once and each field term twice with weight one half. Summing the closed spin-index sequence therefore gives the transfer matrix for the one-dimensional Ising modelThe two eigenvalues areAt finite temperature every entry of P is positive, so the Perron–Frobenius theorem gives ${\lambda }_{+}>|{\lambda }_{-}|$. ConsequentlyThe exact finite-N correction to the extensive expression is $-k_{\mathrm{B}}T\ln [1+({\lambda }_{-}/{\lambda }_{+}{)}^N]$. No assumption that ${\lambda }_{-}$ is positive is needed.

Differentiate at fixed temperature. For $D=\sqrt{{\sinh }^{2}b+e^{-2\beta J}}$, ${\partial }_b\ln (\cosh b+D)=\sinh b/D$. The Ising-chain magnetization with spin-half normalization gives total magnetic momentThis is the thermodynamic-limit result; the moment per site is M/N. For every finite $T>0$, ${\lim }_{B\to 0}M/N=0$, even with the thermodynamic limit taken first. The zero-field magnetic susceptibility per site is $(g{\mu }_{\mathrm{B}}{)}^2{e}^{\beta J}/(4k_{\mathrm{B}}T)$, large but finite.

The Ising-chain domain-wall energy convention explains the result for $J>0$: changing a parallel bond into an antiparallel bond costs J in this spin-half normalization. Domain walls, occurring in pairs on a ring, acquire a nonzero density at any nonzero temperature. Indeed the zero-field correlation length is $\xi =-a/\ln [\tanh (\beta J/2)]\sim (a/2)e^{\beta J}$ at low temperature. The Ising symmetry is discrete, so this is a separate one-dimensional domain-wall mechanism rather than an application of the continuous-symmetry theorem. At zero temperature a ferromagnetic chain can select a fully polarized state; the $B\to 0$ and $T\to 0$ limits do not commute.

Assume short-range ferromagnetic exchange $J>0$ and zero external field. In the one-dimensional Ising model, a domain wall costs only $2J$, independent of the lengths of the domains it separates. There are order-$N$ positions for one wall, so its positional entropy offsets any finite energy cost at nonzero temperature as the system grows. Domain walls consequently have a positive density at every $T>0$, destroying a uniform spontaneous spin magnetization. Periodic boundaries require an even number of walls, but that constraint does not remove their nonzero thermodynamic-limit density.

The One-dimensional Ising correlation function makes this argument explicit. A bond is broken with probability $p_w=e^{-2{\beta }_{\mathrm{th}}J}/(1+e^{-2{\beta }_{\mathrm{th}}J})$, and the infinite-chain correlation isThe dominant eigenvalue of the transfer matrix for the one-dimensional Ising model is $2\cosh {\beta }_{\mathrm{th}}J$, so the free energy per spin $-k_{B}T\log [2\cosh {\beta }_{\mathrm{th}}J]$ is analytic for $T>0$. There is no finite-temperature transition in one dimension; only the zero-temperature limit orders.

In two dimensions a reversed cluster is enclosed by a Peierls contour. A contour of length $\ell$ costs $2J\ell$, while the number of such contours surrounding a fixed site grows at most as a polynomial in $\ell$ times $3^{\ell }$: after the initial edge, a nonbacktracking square-lattice walk has at most three choices per step. With plus boundary conditions, flipping the interior of a specified contour yields the Boltzmann suppression $e^{-2{\beta }_{\mathrm{th}}J\ell }$. ThereforeFor sufficiently low temperature this bound is below $1/2$, giving a positive spontaneous magnetization. This is the energy-entropy mechanism of the Peierls argument.

At sufficiently high temperature, the graphical expansion bounds correlations by sums of connecting paths weighted by $(\tanh {\beta }_{\mathrm{th}}J{)}^{\ell }$. The same nonbacktracking count converges geometrically when $3\tanh {\beta }_{\mathrm{th}}J<1$, giving exponential decay and a disordered phase. Since an ordered low-temperature phase and a disordered high-temperature phase both occur, the two-dimensional model has a finite-temperature phase transition. The argument establishes its existence and the dimensional distinction, not the exact value of its critical temperature.

Expanding the trace in the spin basis givesso the transfer matrix for the one-dimensional Ising model reproduces the partition function.

In the ordered basis $s=+1,-1$,Its eigenvalues areSince ${\lambda }_{+}>|{\lambda }_{-}|$ at every positive temperature, the thermodynamic limit gives the Helmholtz free energy per spinThis is an analytic function of $T$ and $B$ for $T>0$, so the one-dimensional short-range Ising model has no finite-temperature phase transition.

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.