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Work with $\eta =\mathrm{diag}(1,-1,-1,-1)$ and $/\partial ={\gamma }^{\mu }{\partial }_{\mu }$. Multiplying the free Dirac equation by the operator with the opposite mass sign givesThe ordinary derivatives commute, so only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ contributes. Requiring this operator to factor the Klein-Gordon equation, $({\partial }^{\mu }{\partial }_{\mu }+m^2)\psi =0$, for arbitrary free momenta givesThus the Dirac matrices represent the corresponding Clifford algebra. The inference uses factorization for the general free equation; the fact that one specially chosen wavefunction happens to solve both equations would not alone determine every matrix anticommutator.

Let $U(x)=e^{ie\alpha (x)}$, ${\psi }^{\prime }=U\psi$, and $A_{\mu }^{\prime }=A_{\mu }+{\partial }_{\mu }\alpha$. Direct differentiation in the gauge covariant derivative givesThe gradient of the local phase is exactly cancelled by the potential transformation. Since $U$ is scalar in spinor space and commutes with the Dirac matrices,Consequently a solution is mapped into a solution: the minimally coupled Dirac equation is covariant under the local U(1) gauge symmetry. The operator and wavefunction both transform; ordinary differentiation of the transformed wavefunction alone would leave an extra phase-gradient term.

For a free particle the Dirac matrices are constant and the momenta commute. Squaring the Hamiltonian cancels all mixed terms by their anticommutators:Applying another time derivative to the Dirac equation therefore givesfor every component. This is the Klein-Gordon equation. Its componentwise satisfaction is a necessary consequence of the first-order Dirac equation, not a converse selecting all spinor solutions of the squared equation.