The energy-momentum relation of a free particle is (in SI Units):
$$ m^2c^4 =- c^2 \vec{p}^2 + E^2 $$ Minimal coupling is a way to fix a gauge freedom for the choice of canonical momentum (which I can in special relativity give as $ p_{\mu} = \left( \begin{matrix} \frac{E}{c} \\ p_x \\ p_y \\ p_z \end{matrix} \right)$. One way to write the minimal coupling would be by writing the canonical momentum now as:
$$ p_{\mu} = \left( \begin{matrix} \frac{E - e \Phi}{c} \\ p_x + A_x \\ p_y + A_x \\ p_z + A_z \end{matrix} \right)$$ (Here I emply kind of a relativistic hamiltonian formalism, where the movement of a particle in space time, parameterised by $s$, is given by the hamiltonian $\lambda (m^2 + p^{\mu} p_{\mu})$, with $\lambda$ being an abitrary positive function of s, that describes how fast the path in space time is gone through).
Does this mean that I can generalize the energy-momentum relation to: $$ m^2c^4 = - c^2 (\vec{p} + \vec{A})^2 + (E-e\Phi)^2~? $$
If I solve this for $E$ for $\vec{A} = 0$, it resolves to: $$ E = e \Phi + \sqrt{m^2 c^4 + c^2 p^2} $$ which to me seems plausible, and that's why I'm asking.
Edit: the question is not only wether the modified relation holds, but also wether the reasoning behind is right.
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