Magnitude 4-Vector Lorenz Gauge: Klein-Gordon Eq.

[Thomas Rigby]

The Klein-Gordon equation is based on the relation
(E-eΦ)²-(pc-eA)²=m²^2c², which is the magnitude of the difference between the momentum four-vector and the four-potential.
Since the magnitude of the momentum four-vector is given by
E²-p²c²=m²c⁴, does it follow that the magnitude of the four-potential is zero? If so, does this follow from choosing the Lorenz gauge?

 

[vanhees71]

No! You mix up canonical momentum and mechanical momentum. The KG equation reads (with natural units, $\hbar=c=1$)
$$(\mathrm{i} \partial_{\mu}-q A_{\mu})(\mathrm{i} \partial^{\mu} - q A^{\mu}) \Phi=m^2 \Phi.$$
In the first-quantization formalism (which is not really applicable here, but let's give a simple though heuristic argument to clarify the issue) $\mathrm{i} \partial_{\mu}$ represents the canonical energy-momentum vector. So you cannot conclude $E^2-p^2=m^2$ from the equation. Also this is the on-shell condition for (asymptotic) free particles, but it's not valid for interacting particles.

 

[Thomas Rigby]

Thank you for the answer