Action of a Dirac fermion

One can take the coupling $bar{psi} gamma^mu psi A_mu$ with a prefactor $e$ that determines the strength of the coupling.

This works since $bar{psi}gamma^mu psi A_mu$ is a scalar with respect to spinors($bar{psi}gamma^mu psi$ is scalar, $A_mu$ is a scalar) and vectors($bar{psi}gamma^mu psi$ is a vector, $A_mu$ is a vector, their contraction is a scalar).

Formally, one includes this term in the quantum electrodynamics lagrangian through the covariant derivative $D_mu = partial_mu +ieA_mu$ that replaces the $partial_mu$ in the Dirac lagrangian. Since the new field $A_mu$ is a new degree of freedom, one has to include all possible terms containing it. However, one can argue that they can all be reduced to $-frac{1}{4}F_{munu} F^{munu}$ with the $F^{munu}$ that one knows from classical electromagnetism(without covariant derivatives in it). One gets $$mathcal{L}=bar{psi}left( igamma^mu partial_mu -mright)psi to bar{psi}left( igamma^mu D_mu-mright)psi -frac{1}{4}F_{munu} F^{munu} = bar{psi}left( igamma^mu partial_mu -mright)psi -ebar{psi}gamma^mu psi A_mu -frac{1}{4}F_{munu} F^{munu}. $$