 Solution

Use the standard canonical anticommutation relations for the two distinct fermionic orbitals. In particular $\{a,b^{†}\}=\{a^{†},b\}=0$, in addition to the relations explicitly listed in the paper; these mixed relations are part of the usual two-orbital fermionic Fock space interpretation. Normalize the original vacuum state by $⟨0|0⟩=1$ and fix the orderingThe canonical anticommutation relations give $a^2=(a^{†}{)}^2=0$ and $a{a}^{†}=1-a^{†}a$. Consequently the number operator is an orthogonal projector:The same holds for $N_b$. Hence each eigenvalue obeys $n^2=n$ and is either $0$ or $1$, expressing the Pauli exclusion principle. Also $[N_a,N_b]=0$, so the two number operators have a simultaneous orthonormal basis of Fock states:Each number operator has eigenvalues $0,1$, each with state degeneracy two. For $N_a$, the zero eigenspace is spanned by $|00⟩,|01⟩$ and the unit eigenspace by $|10⟩,|11⟩$; for $N_b$, these are respectively $|00⟩,|10⟩$ and $|01⟩,|11⟩$. Any normalized linear combination within one such eigenspace is also an eigenstate of that number operator.

The ordering convention will matter below: $b^{†}{a}^{†}|0⟩=-|11⟩$ and