= Solution
For occupations small compared with $S$, use $S^z=S-n$, $S^-\simeq\sqrt{2S}\,a^\dagger$ and $S^+\simeq\sqrt{2S}\,a$. The longitudinal bond contributes $-J\Delta S^2+JS\Delta(n_m+n_{m+1})+O(S^0)$; the transverse pair terms contribute $JS(a_ma_{m+1}+a_m^\dagger a_{m+1}^\dagger)+O(S^0)$. Thus the direct real-space expansion is
$$
H_2=-J\Delta NS^2+JS\sum_m\{2\Delta n_m+a_ma_{m+1}+a_m^\dagger a_{m+1}^\dagger\}.
$$
Take $a_m=N^{-1/2}\sum_k e^{ikm}a_k$, with $k=2\pi j/N$ over one full <Brillouin zone> and the lattice spacing set to one in $k$. Symmetrizing $k$ and $-k$ changes the phase in the pair sum to $\cos k$, giving
$$
H_2=-J\Delta NS^2+JS\sum_k\{2\Delta a_k^\dagger a_k+\cos k(a_ka_{-k}+a_k^\dagger a_{-k}^\dagger)\}.
$$
<normal ordering> the second <Nambu spinor> component then yields
$$
\boxed{H_2=-J\Delta NS(S+1)+JS\sum_k(a_k^\dagger\ \ a_{-k})\begin{pmatrix}\Delta&\cos k\\\cos k&\Delta\end{pmatrix}\binom{a_k}{a_{-k}^\dagger}.}
$$
\b[The printed matrix has a factor-of-two error on its diagonal.] With a full-zone sum its printed $2\Delta$ would produce $4\Delta a_k^\dagger a_k$, and even its vacuum expectation would differ from the direct expansion by $J\Delta NS$. A half-zone convention cannot repair both coefficients: one must then group each $k,-k$ pair and double its off-diagonal pair coefficient as well. The formula above follows from the specified <Hamiltonian> and has the expected gapless isotropic limit; subsequent physical results use this consistent expansion. Terms quartic in <bosons> first enter at order $S^0$.
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