= Solution
Put $n_n=c_n^\dagger c_n$ and $P_n=\prod_{m<n}(1-2n_m)$. The <Jordan–Wigner transformation> is $S_n^+=P_nc_n$, $S_n^-=P_nc_n^\dagger$ and $S_n^z=1/2-n_n$. Since $\sigma_n^x=P_n(c_n+c_n^\dagger)$, $P_{n+1}=P_n(1-2n_n)$ and $(c_n+c_n^\dagger)(1-2n_n)=c_n^\dagger-c_n$, adjacent bonds become
$$
\sigma_n^x\sigma_{n+1}^x=(c_n^\dagger-c_n)(c_{n+1}+c_{n+1}^\dagger).
$$
Thus, apart from the end bond,
$$
H=-J\sum_n(c_n^\dagger-c_n)(c_{n+1}+c_{n+1}^\dagger)+2Jg\sum_n n_n-JgN.
$$
The end bond contains the global <fermion parity> and sets the sector-dependent periodic or antiperiodic fermion modes. It contributes an order-one boundary term, which is negligible for the thermodynamic energy density; it is not identically zero for the finite spin chain.
Choose the <discrete Fourier transform> convention $c_n=N^{-1/2}\sum_ke^{-ikn}c_k$. Hopping gives $-2J\cos k\,c_k^\dagger c_k$. Opposite-momentum pairing gives $iJ\sin k(c_k^\dagger c_{-k}^\dagger+c_kc_{-k})$, using the <canonical anticommutation relations> to antisymmetrize the coefficient. Therefore
$$
H=2J\sum_k(g-\cos k)c_k^\dagger c_k+iJ\sum_k\sin k(c_k^\dagger c_{-k}^\dagger+c_kc_{-k})-JgN.
$$
For the <Ising-chain Nambu spinor> $\Psi_k=(c_k,-ic_{-k}^\dagger)^T$, expansion of
$$
\boxed{H=J\sum_k\Psi_k^\dagger\begin{pmatrix}g-\cos k&-\sin k\\-\sin k&-(g-\cos k)\end{pmatrix}\Psi_k}
$$
recovers every term: the diagonal contributes $J(g-\cos k)(n_k+n_{-k}-1)$, and the two off-diagonal entries supply the pairing. Since $\sum_k\cos k=0$, the constant is $-JgN$. Reversing the Fourier sign changes the pairing convention; the specified sign makes the displayed matrix agree directly.
Back to article page
