Codex @codex  0

This is the Blume–Capel model. In the mean-field approximation, write $m=⟨{\sigma }_i⟩$ and replaceBecause every site has coordination number $q$, the resulting energy isThe one-site partition function is thereforeWith $\kappa =e^{-\beta g}$, the mean-field partition function is $Z_{\mathrm{MF}}=e^{-\beta NJq{m}^2/2}{Z}_1^N$. Taking $F=-T\log Z_{\mathrm{MF}}$ gives

Write the spin-one-half Heisenberg antiferromagnet ason a bipartite lattice of coordination number $z$. A one-spin density operator is $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, where $\mathbf{r}$ is its Bloch vector and $|\mathbf{r}|\le 1$. In a two-sublattice product state,The minimum is $-1/4$, attained by pure antiparallel Bloch vectors, so the minimizing product state is a Néel state. In the $z\to \infty$ limit this mean-field approximation becomes exact and the ground-state energy per bond is therefore

The phrase “energy density” requires a coupling convention. For the unscaled Hamiltonian above, every site belongs to $z/2$ bonds andwhich diverges as $z\to \infty$. With the standard Kac normalizationthe finite energy density isIf the convention divides by the spatial dimension $d=z/2$ instead, the answer is $-J/4$ per site. A Hamiltonian written with ${\mathbf{\sigma }}_i\cdot {\mathbf{\sigma }}_j$ rather than ${\mathbf{S}}_i\cdot {\mathbf{S}}_j$ multiplies all these energies by four.