= Solution
Adopt the <Minkowski metric> $\eta=\operatorname{diag}(-1,+1)$ and the <Clifford algebra> convention $\{\gamma_\mu,\gamma_\nu\}=2\eta_{\mu\nu}I$. The source's <Pauli matrices> give
$$
\gamma_0=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad
\gamma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
They 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 boost>
$$
\begin{pmatrix}t'\\x'\end{pmatrix}
=\Lambda(\theta)\begin{pmatrix}t\\x\end{pmatrix},\qquad
\Lambda(\theta)=\begin{pmatrix}\cosh\theta&-\sinh\theta\\-\sinh\theta&\cosh\theta\end{pmatrix}.
$$
The corresponding <Spinor representation of the Lorentz group> is
$$
S(\theta)=\exp\left(\frac\theta2\gamma^0\gamma^1\right)
=\exp\left(\frac\theta2\sigma_3\right)
=\operatorname{diag}(e^{\theta/2},e^{-\theta/2}).
$$
For example $S^{-1}\gamma^0S=\gamma^0\cosh\theta-\gamma^1\sinh\theta$ and $S^{-1}\gamma^1S=\gamma^1\cosh\theta-\gamma^0\sinh\theta$, establishing <Lorentz covariance> by $S^{-1}\gamma^\mu S=\Lambda^\mu{}_{\nu}\gamma^\nu$.
Thus the <Dirac spinor> transforms as $\psi'(x')=S(\theta)\psi(x)$, with infinitesimal law
$$
\boxed{\delta\psi=\frac\theta2\sigma_3\psi,\qquad
\delta\psi_+=\frac\theta2\psi_+,\qquad
\delta\psi_-=-\frac\theta2\psi_-.}
$$
These 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>.
Back to article page
