 Solution

Use the canonical anticommutation relations $\{c_n,c_m^{†}\}={\delta }_{nm}$ and $\{c_n,c_m\}=0$. In particular, $n_m=c_m^{†}{c}_m$ is a projector, so $(1-2n_m{)}^2=1$. Define $K_n={\prod }_{m<n}(1-2n_m)$. Then $K_n^{†}=K_n$, $K_n^2=1$, and $K_n$ commutes with $c_n$ and $c_n^{†}$. Taking the adjoint of the given raising operator gives $S_n^{-}=c_n^{†}{K}_n$. HenceAlso $[n_n,c_n]=-c_n$ and $[n_n,c_n^{†}]=c_n^{†}$ imply $[S_n^z,S_n^{\pm }]=\pm S_n^{\pm }$. For different sites, exchanging two fermion operators contributes one minus sign and commuting an operator through the other site's parity string contributes another; their product is positive. Thus distinct spin algebras commute. The string supplies precisely the signs required for the Jordan–Wigner transformation. The fermion vacuum corresponds to spin up, while one occupied site corresponds to spin down; this convention fixes subsequent signs.