= Solution
Apply a <discrete Fourier transform>, with lattice spacing one:
$$
a_n=\frac1{\sqrt N}\sum_k a_ke^{ikn},\qquad k=\frac{2\pi m}{N}\pmod{2\pi}.
$$
The <wavevectors> lie in a <Brillouin zone>, which we choose as $[-\pi,\pi)$. Pairing $k$ with $-k$ makes the quadratic <Hamiltonian>
$$
H_2=-NJS^2+\sum_k\left[Aa_k^\dagger a_k+\frac{B_k}{2}(a_ka_{-k}+a_k^\dagger a_{-k}^\dagger)\right],\qquad A=2JS,\quad B_k=2JS\cos k.
$$
A <bosonic Bogoliubov diagonalization> uses $a_k=\cosh\theta_k\,\alpha_k-\sinh\theta_k\,\alpha_{-k}^\dagger$. The <canonical commutation relations> hold because $\cosh^2\theta_k-\sinh^2\theta_k=1$. The anomalous terms vanish when $\tanh2\theta_k=B_k/A=\cos k$, giving
$$
H_2=E_0+\sum_k\omega_k\alpha_k^\dagger\alpha_k,\qquad E_0=-NJS^2+\frac12\sum_k(\omega_k-A),\qquad \boxed{\omega_k=\sqrt{A^2-B_k^2}=2JS|\sin k|}.
$$
Near $k=0$ and $k=\pi$, the <dispersion relation> is linear: $\omega_k\sim2JS|k-k_0|$. These are the low-energy <antiferromagnetic spin waves>; the two zeroes are related by the two-sublattice description. The transformation is singular at the exact zero modes, which require an infrared regulator or separate treatment of the collective rotation.
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