Two-mode fermionic oscillator for the $N=1$ rest algebra

The occupation formulas give $a_1^{\ast }\mathrm{\mid }0,0⟩=\mathrm{\mid }1,0⟩$, $a_1^{\ast }\mathrm{\mid }0,1⟩=\mathrm{\mid }1,1⟩$, $a_2^{\ast }\mathrm{\mid }0,0⟩=\mathrm{\mid }0,1⟩$, and $a_2^{\ast }\mathrm{\mid }1,0⟩=-\mathrm{\mid }1,1⟩$, with creation zero when its mode is occupied. Thus $a_1$ removes the first occupied mode without sign and $a_2$ removes the second with the sign $(-1{)}^{n_1}$. If $i=j$, applying $a_i{a}_i^{\ast }+a_i^{\ast }{a}_i$ to a basis vector gives that vector: when mode $i$ is empty only the first term returns it, and when occupied only the second does. If $i\mathrm{\ne }j$, the two orders remove and insert distinct modes with opposite signs from the displayed $(-1{)}^{n_1}$ rule, so they cancel. Therefore $\{a_i,a_j^{\ast }\}={\delta }_{ij}I$. Two annihilations or two creations also anticommute, since exchanging their order changes the same occupation sign and repeated creation or annihilation is zero.

Scaling the CAR by $\sqrt{2M}$ gives $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2M{\delta }_{\alpha \beta }I$. By [F1], the rest right-hand side is $-2{\sigma }_{\alpha \dot{\beta }}^0{P}_0=2M{\delta }_{\alpha \beta }I$. The vanishing same-chirality anticommutators follow from the two-annihilation and two-creation relations in step 1.1, and all central charges were set to zero. Thus every stated rest anticommutator holds.

Each annihilation or creation changes $n_1+n_2$ by one, so every $Q_{\alpha }$ and ${\bar{Q}}_{\dot{\alpha }}$ reverses parity. The four occupation vectors have parities $+,-,-,+$, giving two states in each sector. The common domain is all of finite-dimensional $\mathcal{H}$, so all products and adjoints used above are defined.