= 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 <2/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,
$$
\mathbf S_j\cdot\mathbf S_k
=S^2-S(n_j+n_k)+S(a_j^\dagger a_k+a_k^\dagger a_j)+O(S^0).
$$
On a periodic <cubic lattice> with $N$ sites and spacing $a$, there are $3N$ undirected bonds. Consequently,
$$
H_{\rm LSW}=-NS^2(3J+B)
+JS\sum_{\langle j,k\rangle}(n_j+n_k-a_j^\dagger a_k-a_k^\dagger a_j)
+BS\sum_jn_j.
$$
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:
$$
\boxed{H_{\rm LSW}=E_{\rm GS}+\sum_{\mathbf k}\varepsilon_{\mathbf k}a_{\mathbf k}^\dagger a_{\mathbf k},\qquad
\varepsilon_{\mathbf k}=S\left[B+2J\sum_{\alpha=1}^3(1-\cos k_\alpha a)\right],\quad
E_{\rm GS}=-NS^2(3J+B).}
$$
There are no anomalous $aa$ or $a^\dagger a^\dagger$ terms, so no <Bogoliubov transformation> or extra zero-point correction is needed. At long wavelengths,
$$
\boxed{\varepsilon_{\mathbf k}=BS+JSa^2|\mathbf k|^2+O(k^4).}
$$
At $B=0$ this is the quadratic ferromagnetic <Goldstone mode>; for $B>0$ it is gapped by $BS$. Energies are written with $\hbar=1$; an excitation frequency is $\omega_{\mathbf k}=\varepsilon_{\mathbf k}/\hbar$. 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>.
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