 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 a make the Dirac equationMultiplication by ${\gamma }^0$ gives the equivalent Dirac Hamiltonian formFor a plane wave $\psi =u{e}^{-iEt+ipx}$, where $p$ is the signed one-dimensional momentum, the equation isA nonzero spinor exists precisely when the determinant vanishes:For $m\ne 0$, 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.