 Solution

Because the present ${\gamma }_0$ is anti-Hermitian, choose the Dirac adjoint with Hermitian intertwining matrix $A=i{\gamma }_0$:This convention makes the Dirac scalar bilinear $\bar{\psi }\psi$ Hermitian. For real $m$, a Lorentz invariant mass term isTo check infinitesimal Lorentz invariance, put $K={\sigma }_3/2$. Then $K^{†}=K$ and $K^{†}A+AK=\{K,A\}=0$. Using a,Equivalently, $\delta ({\psi }_{+}^{\ast }{\psi }_{-})=\theta {\psi }_{+}^{\ast }{\psi }_{-}/2-\theta {\psi }_{+}^{\ast }{\psi }_{-}/2=0$, and similarly for its complex conjugate. Exponentiation gives $S^{†}AS=A$ and therefore ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$.

At fixed coordinates ${\mathcal{L}}_m$ has only the orbital variation of a Lorentz scalar; the spacetime measure is invariant under the connected Lorentz group, so the integrated action is invariant. The mass term pairs the two opposite chiralities; retaining only one Weyl spinor removes this Dirac mass. An equivalent spinor phase convention gives a real multiple of ${\psi }_{+}^{\ast }{\psi }_{-}+{\psi }_{-}^{\ast }{\psi }_{+}$ instead.