 Solution

Let $N_c=c^{†}c$, $N_d=d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Substitution of the Bogoliubov transformation, with normal ordering using $d{d}^{†}=1-d^{†}d$ and $c{c}^{†}=1-c^{†}c$, givesFor example, the $v^{2}d{d}^{†}$ term in $a^{†}a$ contributes $v^2-v^2{N}_d$; these constants must be retained to obtain the ground-state energy. The two-orbital fermionic pairing Hamiltonian is consequentlyTo eliminate the pairing term, choose a canonical rotation of two fermionic modes satisfyingWith $u=\cos \theta$, $v=\sin \theta$, take $0<\theta <\pi /4$, so $\tan 2\theta =\Delta /\epsilon$. Setting $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$ givesThis positive choice fixes a convenient convention for the vacuum state in part (d). It yieldsBoth number operators independently have eigenvalues $0,1$. Hence the energy eigenvalues and state degeneracies areThere is one fermionic Bogoliubov quasiparticle of energy $E$ per occupied transformed orbital. The two Nambu spinor matrix eigenvalues $\pm E$ should not be mistaken for two independently occupiable levels of opposite energy; the normal ordering constant and the two positive quasiparticle excitation costs give the four physical energy eigenvalues above.

An independent diagonalization of a matrix checks both the pairing sign and the additive constant. In the ordered orthonormal basis $|0⟩,|11⟩$, the even-fermion parity block isbecause $ab|11⟩=-|0⟩$ and $b^{†}{a}^{†}|0⟩=-|11⟩$. Its characteristic polynomial is $x^2-2\epsilon x-{\Delta }^2$, giving $x=\epsilon \pm E$. The odd-fermion parity Fock states $a^{†}|0⟩,b^{†}|0⟩$ are unaffected by pairing and both have energy $\epsilon$.

Equivalently, with the Nambu spinor $\Psi =(a,b^{†}{)}^{\mathsf{T}}$,Replacing $\Psi$ by $R(c,d^{†}{)}^{\mathsf{T}}$ gives $E{c}^{†}c-Ed{d}^{†}+\epsilon =E(N_c+N_d)+\epsilon -E$, reproducing the result by normal ordering.