 Solution

The canonical anticommutation relations make $(a^{†}{b}^{†}{)}^2=0$, so the proposed exponential function terminates exactly:Use the inverse Bogoliubov transformation $c=ua-v{b}^{†}$, $d=v{a}^{†}+ub$, and the actions $a|11⟩=b^{†}|0⟩$ and $b|11⟩=-a^{†}|0⟩$. The two vacuum state conditions becomeBoth therefore require $B=v/u$. Normalizing this quantum state gives $1=A^2(1+B^2)$. Choosing positive $A$, which fixes an irrelevant overall global phase, yields $A=u$ because $u^2+v^2=1$. Thus the two-mode fermionic paired vacuum isThe other sign for $A$ describes the same physical quantum state. The positive relative coefficient is fixed by the paper's pairing order and the convention $|11⟩=a^{†}{b}^{†}|0⟩$.

As a direct eigenvector check, $\Delta v=(E-\epsilon )u$ and $\Delta u=(E+\epsilon )v$ implyIt is therefore the unique ground state, as also follows from positive quasiparticle excitation energy $E$. It has definite even fermion parity but is not an eigenstate of the original total number operator: the empty-pair and occupied-pair probabilities are $u^2$ and $v^2$. In particular $⟨N_a⟩=⟨N_b⟩=v^2$. The pair amplitude is $⟨ab⟩=-uv$, consistently with $ab|11⟩=-|0⟩$.

For the limiting check $\Delta \to 0^{+}$ at fixed positive $\epsilon$, $A\to 1$, $B\to 0$, so $|\tilde{0}⟩\to |0⟩$. For $\epsilon \to 0^{+}$ at fixed positive $\Delta$, $A\to 1/\sqrt{2}$, $B\to 1$, giving equal amplitudes for the empty and occupied pair. These checks also fix which branch of the Bogoliubov transformation produces the ground state.