 Solution

Quantize along the selected field direction by the right-handed relabeling ${\tilde{S}}^z=S^x$, ${\tilde{S}}^x=S^y$, ${\tilde{S}}^y=S^z$. Then the exact Holstein–Primakoff transformation in c gives $S_j^x=S-n_j$. Keep exchange terms of order $S^2$ and $S$, with $J$ and the printed $B$ held fixed. For a nearest-neighbor bond,On a periodic cubic lattice with $N$ sites and spacing $a$, there are $3N$ undirected bonds. Consequently,Use the unitary lattice Fourier transform $a_j=N^{-1/2}{\sum }_{\mathbf{k}}{e}^{i\mathbf{k}\cdot {\mathbf{r}}_j}{a}_{\mathbf{k}}$. Each positive lattice direction contributes $2(1-\cos k_{\alpha }a)$, so the Hamiltonian operator is already diagonal in momentum space:There are no anomalous $aa$ or $a^{†}{a}^{†}$ terms, so no Bogoliubov transformation or extra zero-point correction is needed. At long wavelengths,At $B=0$ this is the quadratic ferromagnetic Goldstone mode; for $B>0$ it is gapped by $BS$. Energies are written with $\hslash =1$; an excitation frequency is ${\omega }_{\mathbf{k}}={\epsilon }_{\mathbf{k}}/\hslash$. In this isotropic ferromagnet the fully aligned vacuum and its one-magnon dispersion are also exact; higher-magnon interactions are among the terms discarded from the linear spin-wave Hamiltonian operator.