Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 75 / 1 / b

Put $n_n=c_n^{†}{c}_n$ and $P_n={\prod }_{m<n}(1-2n_m)$. The Jordan–Wigner transformation is $S_n^{+}=P_n{c}_n$, $S_n^{-}=P_n{c}_n^{†}$ and $S_n^z=1/2-n_n$. Since ${\sigma }_n^x=P_n(c_n+c_n^{†})$, $P_{n+1}=P_n(1-2n_n)$ and $(c_n+c_n^{†})(1-2n_n)=c_n^{†}-c_n$, adjacent bonds becomeThus, apart from the end bond,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 $c_n=N^{-1/2}{\sum }_k{e}^{-ikn}{c}_k$. Hopping gives $-2J\cos k{c}_k^{†}{c}_k$. Opposite-momentum pairing gives $iJ\sin k(c_k^{†}{c}_{-k}^{†}+c_k{c}_{-k})$, using the canonical anticommutation relations to antisymmetrize the coefficient. ThereforeFor the Ising-chain Nambu spinor ${\Psi }_k=(c_k,-i{c}_{-k}^{†}{)}^T$, expansion ofrecovers 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.