 Solution

Adopt the Minkowski metric $\eta =\mathrm{diag}(-1,+1)$ and the Clifford algebra convention $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\eta }_{\mu \nu }I$. The source's Pauli matrices giveThey satisfy ${\gamma }_0^2=-I$, ${\gamma }_1^2=I$ and ${\gamma }_0{\gamma }_1=-{\gamma }_1{\gamma }_0=-{\sigma }_3$, which verifies the required Clifford algebra. Raising the indices gives ${\gamma }^0=-{\gamma }_0$ and ${\gamma }^1={\gamma }_1$.

Fix the sign of the rapidity $\theta$ by taking the Lorentz boostThe corresponding Spinor representation of the Lorentz group isFor example $S^{-1}{\gamma }^{0}S={\gamma }^0\cosh \theta -{\gamma }^1\sinh \theta$ and $S^{-1}{\gamma }^{1}S={\gamma }^1\cosh \theta -{\gamma }^0\sinh \theta$, establishing Lorentz covariance by $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$.

Thus the Dirac spinor transforms as ${\psi }^{\prime }(x^{\prime })=S(\theta )\psi (x)$, with infinitesimal lawThese variations compare the fields at corresponding spacetime points. At a fixed coordinate one adds the usual orbital term $-\delta x^{\mu }{\partial }_{\mu }\psi$. Reversing the rapidity convention reverses both boost signs consistently. Infinitesimal transformations describe the proper orthochronous Lorentz group.