= Solution
The <canonical anticommutation relations> make $(a^\dagger b^\dagger)^2=0$, so the proposed <exponential function> terminates exactly:
$$
|\widetilde0\rangle=A\bigl(|0\rangle+B|11\rangle\bigr).
$$
Use the inverse <Bogoliubov transformation> $c=ua-vb^\dagger$, $d=va^\dagger+ub$, and the actions $a|11\rangle=b^\dagger|0\rangle$ and $b|11\rangle=-a^\dagger|0\rangle$. The two <vacuum state> conditions become
$$
c|\widetilde0\rangle=A(uB-v)b^\dagger|0\rangle=0,\qquad
d|\widetilde0\rangle=A(v-uB)a^\dagger|0\rangle=0.
$$
Both 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> is
$$
\boxed{|\widetilde0\rangle=u|0\rangle+v a^\dagger b^\dagger|0\rangle,\qquad
A=\sqrt{\frac{E+\varepsilon}{2E}},\qquad
B=\frac\Delta{E+\varepsilon}
=\sqrt{\frac{E-\varepsilon}{E+\varepsilon}},\quad
E=\sqrt{\varepsilon^2+\Delta^2}.}
$$
The 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 $|11\rangle=a^\dagger b^\dagger|0\rangle$.
As a direct <eigenvector> check, $\Delta v=(E-\varepsilon)u$ and $\Delta u=(E+\varepsilon)v$ imply
$$
\begin{pmatrix}0&-\Delta\\-\Delta&2\varepsilon\end{pmatrix}
\binom uv=(\varepsilon-E)\binom uv.
$$
It 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 $\langle N_a\rangle=\langle N_b\rangle=v^2$. The pair amplitude is $\langle ab\rangle=-uv$, consistently with $ab|11\rangle=-|0\rangle$.
For the limiting check $\Delta\to0^+$ at fixed positive $\varepsilon$, $A\to1$, $B\to0$, so $|\widetilde0\rangle\to|0\rangle$. For $\varepsilon\to0^+$ at fixed positive $\Delta$, $A\to1/\sqrt2$, $B\to1$, giving equal amplitudes for the empty and occupied pair. These checks also fix which branch of the <Bogoliubov transformation> produces the <ground state>.
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