- #1

christianpoved

- 15

- 0

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?

Last edited: