Past exam of the condensed matter physics major option course of the University of Oxford / 2015 / 4 / b

For the classical XY ferromagnet choose every spin parallel within the $xy$ plane. Each nearest-neighbor and next-nearest-neighbor in-plane dot product is $S^2$. The printed sum counts one bond of each range per spin, soThere is no additional factor of two from counting both directions: the Hamiltonian operator already lists only $n\to n+1$ and $n\to n+2$. Alignment along $z$ would give zero XY exchange energy and is not the in-plane ordered state used for these expressions.

For collinear antiferromagnetic order in the $xy$ plane, take ${\mathbf{S}}_n=(-1{)}^{n}S\hat{\mathbf{x}}$. Nearest neighbors are antiparallel, giving $-S^2$, whereas second neighbors lie on the same sublattice and remain parallel, giving $+S^2$. ThusThe result follows from substituting into the specified XY Hamiltonian operator, regardless of whether that trial state is the global minimum for the chosen exchanges. Comparing trial-state energies does not eliminate the possibility of a lower helical state.