= Solution
Variation of the <action> with respect to the independent <Dirac adjoint> gives $(i\gamma^0\partial_t+i\gamma^1\partial_x-m)\psi=0$. The <gamma matrices> chosen in <3/a> make the <Dirac equation>
$$
\boxed{\begin{pmatrix}
-m&i(\partial_t-\partial_x)\\
i(\partial_t+\partial_x)&-m
\end{pmatrix}\begin{pmatrix}\psi_R\\\psi_L\end{pmatrix}=0.}
$$
Multiplication by $\gamma^0$ gives the equivalent <Dirac Hamiltonian> form
$$
i\partial_t\psi=(-i\sigma_3\partial_x+m\sigma_1)\psi.
$$
For a <plane wave> $\psi=u\,e^{-iEt+ipx}$, where $p$ is the signed one-dimensional <momentum>, the equation is
$$
\begin{pmatrix}E-p&-m\\-m&E+p\end{pmatrix}u=0.
$$
A nonzero <spinor> exists precisely when the <determinant> vanishes:
$$
\boxed{E^2-p^2-m^2=0,\qquad E=\pm\sqrt{p^2+m^2}.}
$$
For $m\ne0$, one possible unnormalized <spinor> solution is $u=(E+p,m)^{\mathsf T}$. Alternatively, $(p\sigma_3+m\sigma_1)^2=(p^2+m^2)1$ follows from the <Pauli matrix multiplication law>, proving the same <relativistic energy-momentum relation>. Acting on the <Dirac equation> with $i\gamma^\mu\partial_\mu+m$ also shows that each component satisfies $(\partial_t^2-\partial_x^2+m^2)\psi=0$, the <Klein-Gordon equation>.
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