= Solution
Use the <Discrete Fourier transform>[unitary discrete Fourier transform]
$$
a_j=\frac1{\sqrt L}\sum_ke^{ikj}a_k,\qquad k=\frac{2\pi n}{L},\quad n=0,\ldots,L-1.
$$
The transformed <Linear operators>[operators] obey $[a_k,a_q^\dagger]=\delta_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. Thus
$$
H_{\mathrm{LSW}}=\sum_k\left[\alpha_k a_k^\dagger a_k+\frac{\Delta_k}{2}(a_ka_{-k}+a_k^\dagger a_{-k}^\dagger)\right],
$$
where
$$
\alpha_k=S(H_f+2J-J\cos k),\qquad \Delta_k=JS\cos k.
$$
The factor one half in the full <Sum>[momentum sum] avoids counting an <Bosonic Bogoliubov diagonalization>[anomalous pair] twice. For $H_f>0$, $\alpha_k>|\Delta_k|$ for every allowed <momentum>. Set
$$
a_k=\cosh\theta_k\,b_k-\sinh\theta_k\,b_{-k}^\dagger,\qquad
\tanh2\theta_k=\frac{\Delta_k}{\alpha_k}.
$$
Real even <angles> ensure consistency between opposite <momenta>, and $\cosh^2\theta_k-\sinh^2\theta_k=1$ preserves the <canonical commutation relations>. The anomalous coefficient becomes $\Delta_k\cosh2\theta_k-\alpha_k\sinh2\theta_k=0$. <Normal ordering> then gives
$$
\boxed{H_{\mathrm{LSW}}=\sum_k\epsilon(k)b_k^\dagger b_k+\frac12\sum_k[\epsilon(k)-\alpha_k],}
$$
with
$$
\boxed{\epsilon(k)=\sqrt{\alpha_k^2-\Delta_k^2}=S\sqrt{(H_f+2J)[H_f+2J(1-\cos k)]}.}
$$
The <field-polarized planar ferromagnetic spin wave> has <energy gap> $\epsilon(0)=S\sqrt{H_f(H_f+2J)}$. The <momenta> run over the whole <Brillouin zone> once, including the self-inverse <momenta> $0$ and, for even $L$, $\pi$; these use <Single-mode squeeze operator>[single-mode squeezing] rather than introducing an extra independent partner.
\b[The zero-field case has a <zero-mode obstruction to bosonic Bogoliubov diagonalization>.] Its nonzero-momentum modes have $\epsilon(k)=2\sqrt2JS|\sin(k/2)|$, but $\alpha_0=\Delta_0=JS$, so no finite <Bogoliubov transformation>[squeezing angle] exists for the uniform mode. With $Q_0=(a_0+a_0^\dagger)/\sqrt2$, its <Hamiltonian> is $JSQ_0^2-JS/2$, not an identically zero oscillator. For example, its <matrix element> between the <Fock vacuum>[vacuum] and the two-boson <Fock state>[number state] is $JS/\sqrt2$, whereas $\epsilon(0)b_0^\dagger b_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero <Magnetic field>[field]; at zero <Magnetic field>[field] treat the collective orientation separately or introduce a <Magnetic field>[field] regulator before taking a limit.
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