eigenfunctions to satisfy [@Gutierrez_2017]. For a massless particle ($m=0$), the Dirac-Pauli equation takes the simplified form $$\label{eige_1}
\left[ \left(\hat{p}^{\mu} - \frac{e}{c} A^{\mu} \right) \left(\hat{p}_{\mu} - \frac{e}{c} A_{\mu} \right) -\frac{i e \hbar}{4c} \left[ \gamma^{\mu}, \gamma^{\nu} \right]_{-} F_{\mu \nu}  \right] \psi_{P} = 0 \, ,$$ with $$\label{eige_2}
F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} \,$$ the electromagnetic tensor. Solutions $\psi_{D}$ to the Dirac equation are then built from solutions $\psi_{P}$ of this auxiliary second-order equation by applying the Dirac operator: $$\label{eige_0}
\psi_{D} = \gamma^{\mu}\left( \hat{p}_{\mu} - \frac{e}{c}A_{\mu} \right) \psi_{P} \, .$$ For the physical scenario presented above, the relativistic invariants $$\begin{aligned}
\label{eige_3}
F_{\mu \nu} F^{\mu \nu} = \mathbf{B} \cdot \mathbf{B} - \mathbf{E} \cdot \mathbf{E} \, , \\
\epsilon_{\mu \nu \rho \eta} F^{\mu \nu} F^{\rho \eta} = 2 \mathbf{E} \cdot \mathbf{B} \, ,\end{aligned}$$ will allow us to reduce the solutions of the system to only two cases, which correspond to the possible phases of light encountered in the quantum optical system. Given that when $F_{\mu \nu}F^{\mu \nu}$ is positive (negative) the existence of a reference frame where the electric (magnetic) field vanishes is guaranteed, the exact solution of the Dirac equation can be obtained in this privileged frame. Afterwards, by applying a 