Canonical rotation of two fermionic modes ($a=uc+v{d}^{†},b=-v{c}^{†}+ud,u^2+v^2=1$)

A real two-mode Bogoliubov transformation preserving the full canonical anticommutation relations. The self anticommutators become $\{a,a^{†}\}=\{b,b^{†}\}=u^2+v^2$, while $\{a,b\}=-uv+uv=0$ and every other cross anticommutator vanishes. Thus its matrix is an orthogonal matrix. For the two-orbital fermionic pairing Hamiltonian, let $E=\sqrt{{\epsilon }^2+{\Delta }^2}$, take $u=\sqrt{(E+\epsilon )/(2E)}$ and $v=\sqrt{(E-\epsilon )/(2E)}$, and obtain $H=E(c^{†}c+d^{†}d)+\epsilon -E$. Inverting gives $c=ua-v{b}^{†}$, $d=v{a}^{†}+ub$.

The normalized vacuum state annihilated by $c=ua-v{b}^{†}$ and $d=v{a}^{†}+ub$, for positive $u,v$ with $u^2+v^2=1$. Since $(a^{†}{b}^{†}{)}^2=0$, it can also be written $u\exp [(v/u)a^{†}{b}^{†}]|0⟩$. Acting with either annihilation operator makes the two terms cancel. Its original occupation numbers each have expected value $v^2$ and its pair amplitude is $⟨ab⟩=-uv$. It has even fermion parity although the original total number operator is not sharp. For the two-orbital fermionic pairing Hamiltonian, it is the unique ground state of energy $\epsilon -\sqrt{{\epsilon }^2+{\Delta }^2}$.