 On-site spin-square normal-ordering identity

For a single spinful fermion orbital, ${\mathbf{S}}^2=3(n-2n_{\uparrow }{n}_{\downarrow })/4$. The contraction term is $3n/4$, so $:{\mathbf{S}}^2:={\mathbf{S}}^2-3n/4=-3n_{\uparrow }{n}_{\downarrow }/2$. Empty and doubly occupied states are spinless while singly occupied states have spin one half, which also proves the identity directly.