= Solution
With the specified <Fourier transform> convention, <normal ordering> gives
$$
\sum_n(c_n^\dagger c_{n+1}-c_nc_{n+1}^\dagger)=2\sum_k\cos k\,c_k^\dagger c_k,
$$
where the possible constant $\sum_ke^{ik}$ vanishes on the complete lattice momentum grid. Antisymmetry of opposite-momentum pairs similarly gives
$$
\sum_nc_n^\dagger c_{n+1}^\dagger=-i\sum_k\sin k\,c_k^\dagger c_{-k}^\dagger,\qquad
\sum_nc_nc_{n+1}=i\sum_k\sin k\,c_kc_{-k}.
$$
Hence
$$
H=-JgN+2J\sum_k(g-\cos k)c_k^\dagger c_k
+iJ\sum_k\sin k\,(c_k^\dagger c_{-k}^\dagger+c_kc_{-k}).
$$
Use the <Ising-chain Nambu spinor> $\Psi_k=(c_k,-ic_{-k}^\dagger)^T$. Expanding its two diagonal terms uses $c_{-k}c_{-k}^\dagger=1-c_{-k}^\dagger c_{-k}$, retaining the constant. The result is
$$
\boxed{H=J\sum_k\Psi_k^\dagger\left[(g-\cos k)\sigma_z-\sin k\,\sigma_x\right]\Psi_k.}
$$
This is a full-zone sum. Its opposite-momentum components are redundant; treating each matrix eigenvalue as an independent physical excitation would miss the factor of two in the quasiparticle energy. Self-paired momenta have zero anomalous term and are ordinary number levels.
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