After replacing the derivative in the Klein-Gordon equation with the minimal coupling prescription i.e. $\partial^\mu \to D^\mu \equiv \partial^\mu+ieA^\mu$. The equation becomes:
\begin{equation*}
((\partial_\mu +ieA_\mu)(\partial^\mu+ieA^\mu)+m^2)\phi(x)=0
\end{equation*}
\begin{equation*}
(\partial_\mu\partial^\mu +m^2 +ie(\partial_\mu A^\mu +A_\mu \partial ^\mu)-e^2A_\mu A^\mu)\phi(x)=0
\end{equation*}
In the book "Introduction to Particle and Astroparticle Physics" by Alessandro De Angelis, the book makes the following substitution in the above equation calling it potential:
\begin{equation}
V(x)=ie(\partial_\mu A^\mu +A_\mu \partial ^\mu)
\end{equation}
which gives us the final equation. (neglecting $e^2$ term which is of second order):
\begin{equation*}
(\partial_\mu\partial^\mu +m^2 +V(x))\phi(x)=0
\end{equation*}
My question is what is the significance of V(x)? As in what is this potential? Further if we assume the lorenz guage i.e. $\partial_\mu A^\mu =0$, how does that change the description of this potential?
Best Answer
Well, I don't have an explanation as I don't dispose of the mentioned book. Probably it is the best to continue to read to find out why this notation was chosen. But I would like to point a couple of details which might help.
One should understand $V(x)$ as a differential operator. Therefore one should better write:
$$V(x)\phi = ie (\partial_\mu (A^\mu \phi) + A_\mu \partial^\mu\phi)$$
which can be changed to:
$$V(x)\phi = ie (\phi \partial_\mu A^\mu + 2 A^\mu \partial_\mu \phi)\stackrel{Lorenz\, gauge}{=} 2ieA^\mu \partial_\mu \phi $$
I even see applications of this result. Let's assume that the potential is only electrostatic and constant: $A^\mu = (W,0,0,0)$. So we get the following Klein-Gordon equation:
$$(\Box +m^2)\phi = -2ie W\frac{\partial \phi}{\partial t}$$
So in this case $V(x)\phi = 2ie W \partial_t \phi$.
BTW this equation has a simple solutions if the particle described by the field is at rest (momentum is zero):
$$\phi_\pm \approx e^{i (\pm m -eW)t }$$
One can now think about what these 2 solutions mean, in particular because one of them is of negative energy (better assume $eW <<m$). But this goes far beyond the question.
So the main message here is that one has to consider $V(x)$ as differential operator and applications exist.