 Solution

With $\eta =\mathrm{diag}(1,-1)$, the Clifford algebra gives $({\gamma }^0{)}^2=1$, $({\gamma }^1{)}^2=-1$, and ${\gamma }^0{\gamma }^1=-{\gamma }^1{\gamma }^0$. Henceso $\{{\gamma }^5,{\gamma }^0\}=\{{\gamma }^5,{\gamma }^1\}=0$. Its square isChoose the following gamma matrices, expressed using the Pauli matrices:These satisfy $({\gamma }^0{)}^{†}={\gamma }^0$, $({\gamma }^1{)}^{†}=-{\gamma }^1$ and $({\gamma }^5{)}^{†}={\gamma }^5$. The chirality eigenvalues are therefore $\boxed{+1,-1}$, and we order the spinor as $\psi =({\psi }_R,{\psi }_L{)}^{\mathsf{T}}$ throughout. Its Dirac adjoint is $\bar{\psi }=({\psi }_L^{†},{\psi }_R^{†})$. No additional factor of $i$ multiplies ${\gamma }^0{\gamma }^1$ in this spacetime dimension. This realizes a Dirac field in two spacetime dimensions.