Hey! Maybe this is a "piece of cake" question, but here is the thing, i have the Maxwell equations in the Lorenz gauge are
\begin{array}{c}
\partial_{\mu}\partial^{\mu}A^{\nu}=\mu_{0}j^{\nu}
\end{array}
In vacuum this gets reduced into
\begin{array}{c}
\partial_{\mu}\partial^{\mu}A^{\nu}=0
\end{array}
Also the Klein-Gordon equation says that
\begin{array}{c}
\left(\square-\frac{m^{2}c^{2}}{\hbar^{2}}\right)\psi=0
\end{array}
I guess that for a massless particle this is just
\begin{array}{c}
\square\psi=0
\end{array}
This leads to my question, is there any relation between the wavefunction of the photon and the four-potential?
\begin{array}{c}
\partial_{\mu}\partial^{\mu}A^{\nu}=\mu_{0}j^{\nu}
\end{array}
In vacuum this gets reduced into
\begin{array}{c}
\partial_{\mu}\partial^{\mu}A^{\nu}=0
\end{array}
Also the Klein-Gordon equation says that
\begin{array}{c}
\left(\square-\frac{m^{2}c^{2}}{\hbar^{2}}\right)\psi=0
\end{array}
I guess that for a massless particle this is just
\begin{array}{c}
\square\psi=0
\end{array}
This leads to my question, is there any relation between the wavefunction of the photon and the four-potential?