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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.

The Holstein–Primakoff transformation converts a spin algebra into bosonic operators while retaining the finite physical spin Hilbert space. With $[a_n,a_m^{†}]={\delta }_{nm}$, writeThe Holstein–Primakoff occupation constraint is $0\le n_n\le 2S$. Acting on $|n⟩$, the raising matrix element is $\sqrt{n(2S-n+1)}$ and the lowering element is $\sqrt{(n+1)(2S-n)}$. Their squared difference is $2(S-n)$, reproducing $[S^{+},S^{-}]=2S^z$. Operator ordering inside the square roots is essential. For low boson occupation relative to $S$, expand them to obtain $S^{+}\simeq \sqrt{2S}a$ and $S^{-}\simeq \sqrt{2S}{a}^{†}$. Keeping quadratic terms gives the linear spin-wave approximation; higher terms describe magnon interactions. Large $S$ helps control the expansion, but a low-dimensional infrared divergence can invalidate it even at large finite $S$.

For the ferromagnet choose $H_F=-J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, with unit lattice spacing and $\hslash =1$. Expand around the all-up state. A longitudinal bond contributes $-J{S}^2+JS(n_n+n_{n+1})$, and the transverse exchange contributes $-JS(a_n^{†}{a}_{n+1}+a_{n+1}^{†}{a}_n)$. After a Fourier transform,These are ferromagnetic magnons. The polarized vacuum is an exact ground state of the full ferromagnetic Hamiltonian: every bond is in its maximum total-spin sector, minimizing that bond energy. It therefore has no quantum zero-point reduction of its magnetization. The one-magnon dispersion is also exact, although interacting many-magnon states need not be free bosons. Rotating the polarization generates the maximum-total-spin ground-state multiplet. Selecting a definite direction in a thermodynamic symmetry-breaking description breaks $SU(2)$ to $U(1)$; the two transverse spin components form a canonical pair, producing a single quadratic type-B Goldstone boson. A small aligning field gaps it by the Zeeman energy.

At nonzero temperature the one-dimensional ferromagnet is different: its thermal magnon number contains $\int dk[e^{\beta JS{k}^2}-1{]}^{-1}$, with infrared behaviour $\int dkT/k^2$. The divergent magnetization reduction signals the absence of finite-temperature long-range order in an infinite short-range isotropic chain, consistent with the Mermin-Wagner theorem. It does not disallow the polarized zero-temperature ground states just constructed.^[ref]

For the antiferromagnet use $H_{AF}=+J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, still $J>0$, on an even bipartite chain. The classical reference state alternates up and down. Rotate one sublattice through $\pi$ so that both local quantization axes point along their reference spins. In unrotated physical components, to leading order $S_A^z=S-a^{†}a$, $S_B^z=-S+b^{†}b$, $S_A^{+}\simeq \sqrt{2S}a$ and $S_B^{+}\simeq \sqrt{2S}{b}^{†}$. Choosing Fourier phases using the actual site positions givesThe reduced Brillouin zone has $M=N/2$ momenta and length $\pi$. The pair creation and annihilation terms express the fact that an antiparallel reference product state is not an eigenstate. A bosonic Bogoliubov transformation with $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, $\tanh 2{\theta }_k=\cos k$ diagonalizes each two-mode block. With the sign convention $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$, the result isThus the harmonic antiferromagnetic modes have linear, rather than quadratic, long-wave dispersion. In an unfolded Brillouin-zone description the same result is represented by zeros at both zero and the antiferromagnetic wavevector; these are not extra independent degrees of freedom beyond the two-sublattice accounting.

The Bogoliubov vacuum has no $\alpha$ or $\beta$ excitations but contains original spin bosons, with $v_k^2=\frac{1}{2}(|\sin k{|}^{-1}-1)$. Its staggered magnetization is $m_s=S-M^{-1}{\sum }_{k\in \mathrm{RBZ}}{v}_k^2$. This quantum depletion of Néel order diverges logarithmically at small $k$ in one dimension. Linear spin-wave theory therefore does not establish a stable ordered antiferromagnetic ground state for this chain. Its nominal gapless dispersion is a result of expanding around an invalid ordered reference, not a general proof of the exact excitation spectrum. For the nearest-neighbour isotropic chain, integer-spin systems have a Haldane gap, whereas half-integer-spin systems are gapless; the spin-one-half chain in particular has fractional spin excitations rather than a simple dilute gas of ordered-state magnons. This distinction requires physics beyond the harmonic expansion, including the topological term in the continuum theory.^[ref]

Near zero the Langevin function has expansionFor positive $x$, $\mathcal{L}(x)$ rises toward one and its ratio $\mathcal{L}(x)/x$ decreases from $1/3$ to zero. Therefore a line $\tau x$ has only the zero intersection for $\tau \ge 1/3$, and also a nonzero positive intersection for $0<\tau <1/3$. In a fixed signed component there is a corresponding negative solution; in the isotropic vector problem the nonzero solutions have all spin-space orientations.

The mean-field transition temperature isAt the critical temperature the line is tangent at the origin. Just below it, $S\simeq \sqrt{(5/3)(1-T/T_c^{\mathrm{MF}})}$, the usual mean-field square-root onset. The ordered minimum is stable below this temperature while $S=0$ becomes unstable.

This is a mean-field prediction, not a genuine finite-temperature ordering transition for the ideal two-dimensional short-range isotropic model. Long-wavelength spin fluctuations invalidate the predicted nonzero thermodynamic magnetization at every $T>0$. The relevant obstruction is the Mermin-Wagner theorem for continuous spin symmetry; anisotropy, long-range interactions or a finite sample change the assumptions. Its original paper establishes the absence of finite-temperature magnetic ordering for one- and two-dimensional finite-range isotropic Heisenberg models.^[ref]