Past exam of the advanced Quantum Condensed Matter Physics course of the University of Cambridge / 2012 / Paper 1 / 1 / Solution

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.