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Let $n_n=c_n^{†}{c}_n$ and $P_n={\prod }_{m<n}(1-2n_m)$. The Jordan–Wigner transformation gives $2S_n^x=P_n(c_n+c_n^{†})$ and $P_{n+1}=P_n(1-2n_n)$. The canonical anticommutation relations implyConsequently the bondwise cancellation of a Jordan-Wigner string yieldsSubstitution therefore givesThe finite-ring parity qualification of an Ising quasiparticle gap matters here. For an exact periodic spin chain, the closing bond retains total fermion parity. The parity-dependent boundary condition of the Jordan-Wigner chain is $c_{N+1}=-P{c}_1$, with $P=(-1{)}^{N_f}$. Fixing that parity determines the allowed Fourier grid, and the resulting many-body states must still have the chosen parity. Dropping this boundary issue preserves the thermodynamic energy density, but can lose a degeneracy or the correct finite-ring excitation gap. ^[ref]

For a bulk bond, $K_{n+1}=K_n(1-2n_n)$ and $c_n(1-2n_n)=-c_n$. Reordering the remaining different-site operators then givesThe periodic spin boundary condition requires separate attention. Set $P={\prod }_{n=1}^N(1-2n_n)=(-1{)}^{N_f}$. The end bond becomes $S_N^{+}{S}_1^{-}=-P{c}_1^{†}{c}_N$, with the conjugate relation for its adjoint. Consequently the exact finite-ring Hamiltonian isfermion parity is conserved, so diagonalize in a sector $P=\pm 1$. There the end bond can be written like every other bond provided $c_{N+1}=-P{c}_1$. This parity-dependent boundary condition of the Jordan-Wigner chain is antiperiodic for even fermion number and periodic for odd fermion number. Simply assuming periodic fermions in both sectors changes the finite-size spectrum.