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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.

In the chirality basis of a, the two rows of the Dirac Hamiltonian equation giveThe Dirac mass term thus couples the two opposite chirality components. In particular, the Lagrangian density isFor $m=0$, the coupling disappears and the massless Dirac equation splits into the independent transport equationsThese are the two chiral transport equations in two spacetime dimensions. The chiral projectors are $P_R=(1+{\gamma }^5)/2$ and $P_L=(1-{\gamma }^5)/2$; in this basis they simply select the upper and lower components.