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A Heisenberg ferromagnet has negative exchange coupling in $H=J{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$, so neighboring spins energetically prefer parallel alignment.

A magnon is one quantum of a spin wave, a collective oscillation of magnetic order. Exchange-coupled spins do not fluctuate independently: a disturbance spreads through the ordered system as a mode with a definite wavevector and energy.

In a simple collinear Heisenberg ferromagnet, a single magnon reduces the total spin projection along the ordered axis by one unit of $\hslash$ while distributing the disturbance across many sites. At low excitation density, magnon modes can be treated approximately as bosonic oscillators, motivating the Holstein–Primakoff transformation and linear spin-wave approximation. Their population contributes to the reduction of magnetization with increasing temperature. Strong interactions, noncollinear order and large excitation densities can require corrections to the simple independent-magnon picture.

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.

Use dimensionless spin angular momentum operators, so their commutators have no additional $\hslash$. A fully polarized state $|F⟩={⨂}_n|S,S{⟩}_n$ is an exact ground state of the Heisenberg ferromagnet, with $E_0=-NJ{S}^2$. On each bond the largest eigenvalue of ${\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$ is $S^2$, obtained in the two-spin total-spin-$2S$ sector. Hence the bondwise lower bound on the Hamiltonian is attained by this state.

The ferromagnetic ground-state multiplet has total spin $NS$. Applying the global lowering operator $(S_{\mathrm{tot}}^{-}{)}^r$, $r=0,\dots ,2NS$, generates its $2NS+1$ magnetic states; their linear combinations are also ground states. Global spin rotations generate polarized ground states pointing in arbitrary directions and span the same multiplet. Uniform reorientation therefore costs no energy. Slowly varying orientations give gapless ferromagnetic magnons, with quadratic rather than linear long-wavelength dispersion. In the ordered thermodynamic ground state, the two broken spin generators are canonically paired because $⟨[S_{\mathrm{tot}}^x,S_{\mathrm{tot}}^y]⟩=i⟨S_{\mathrm{tot}}^z⟩\ne 0$, producing one type-B Goldstone boson. For a finite chain the zero-wavevector motion remains inside the exact ground multiplet; nonzero wavevectors have a finite-size spacing that vanishes as $N^{-2}$. This is a ground-state statement, not a claim of finite-temperature long-range order in one dimension.

Letwhere the two signs describe the Heisenberg antiferromagnet and Heisenberg ferromagnet. A standard Lieb-Robinson bound is obtained by iterating the Heisenberg picture equation and bounding nested commutators by operator norms. Its constants depend on the interaction only through quantities such asChanging $J$ to $-J$ changes neither the supports nor the norms $\Vert \Phi (X)\Vert$. Every term in the nested-commutator estimate acquires at most an irrelevant sign before its absolute value is taken. Therefore both chains obey exactly the same estimatewith the same $C,\mu$, and Lieb-Robinson velocity $v_{\mathrm{LR}}$. No unitary equivalence of the two Hamiltonians is required.

The large effective field primarily represents exchange interactions, which arise from Coulomb interactions, electronic hopping and the Pauli exclusion principle. It is not usually the much smaller classical magnetic dipole field of neighbouring ions. Direct exchange involves overlapping magnetic-ion orbitals. Superexchange instead uses virtual charge transfer through a ligand while real charge transport remains gapped, allowing substantial magnetic order in an electrical insulator.

Use dimensionless spin operators ${\mathbf{S}}_i$ with eigenvalue $S=1/2$, and count each bond once. One consistent Heisenberg model convention isThen $J_{ij}<0$ favours a Heisenberg ferromagnet, while $J_{ij}>0$ favours a Heisenberg antiferromagnet. Replacing neighbouring spins by their mean values gives the effective fieldFor $z$ equivalent ferromagnetic neighbours with $J_{ij}=-J_F<0$ and uniform order, $M=n{g}_J{\mu }_B⟨S_z⟩$, soThe last equality is the same molecular-field result derived above, not an exact transition temperature for every lattice.

Microscopically, a half-filled orbital pair with virtual hopping $t_{ij}$ and a large on-site repulsion $U$ permits an energy lowering of the singlet but not the parallel-spin triplet, yielding antiferromagnetic $J\simeq 4|t_{ij}{|}^2/U$ in this convention. Ligand-mediated hopping is the analogous mechanism in many ionic insulators. Different orbital occupancies and nearly orthogonal ligand paths can suppress this antiferromagnetic term and permit ferromagnetic exchange to dominate. Thus the sign depends on electronic structure and bonding geometry, not simply whether the sample conducts.

For antiferromagnetic coupling on a bipartite lattice, the field on each sublattice is proportional to the opposite sublattice magnetization, with opposite alignment preferred. Two coupled self-consistency equations then permit $M_A=-M_B\ne 0$ and vanishing total magnetization: the order parameter is staggered. The single positive-$\lambda$ uniform equation describes ferromagnetism and must be extended to these sublattices to describe an antiferromagnet.

The negative sign of the exchange term and $J>0$ make this a Heisenberg ferromagnet. Each classical bond minimizes its energy when its two fixed-length spins are parallel. On the connected cubic lattice all ground states are therefore uniform:using length $S$ for the classical large-spin convention. Any two orientations are related by a common spatial rotation of every spin. The global spin-rotation symmetry is $SO(3)$, and a chosen direction retains its $SO(2)$ subgroup of rotations about that axis.

Spontaneous symmetry breaking means selecting one member of this continuous set in the thermodynamic limit, for example by taking volume to infinity before sending an aligning field to zero. The zero-field Hamiltonian operator remains rotationally invariant even though the selected state is not. There are gapless transverse spin waves. Although two rotation generators are broken, an isotropic ferromagnet has one quadratic magnon branch, whose two transverse spin components form a conjugate dynamical pair. It is not necessary to invent two independent linear-dispersion branches from the number of broken generators. The long-wavelength dispersion is derived in d.