 Solution

The inverse Bogoliubov transformation gives $c=ua-v{b}^{†}$ and $d=v{a}^{†}+ub$. Because $(a^{†}{b}^{†}{)}^2=0$ by the canonical anticommutation relations, the exponential truncates:The fermionic signs are fixed byIt follows thatBoth annihilation conditions hold precisely when $B=v/u$. Normalizing the paired fermionic vacuum, using the orthogonality of its two Fock states, gives $A^2(1+B^2)=1$. Choose the irrelevant overall phase so that $A>0$. The result isEquivalently,Its positive pair coefficient follows from the specified order $ab+b^{†}{a}^{†}$ in the Hamiltonian operator. The state has even fermion parity and is a quantum superposition of zero and two original particles. In the limit $\Delta \to 0^{+}$, $A\to 1$ and $B\to 0$, so it reduces to the original vacuum state, as expected for $\epsilon >0$.