# Relation between wavefunction of the photon and the Four-potential

## Main Question or Discussion Point

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?

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dextercioby
Homework Helper
1. It's the LORENZ gauge (no 't').
2. The wavefunction of the photon is not a well-defined concept.
3. Photons are described by Fock space states, quantizing the 4-potential leads to fields as operators acting on those states.

vanhees71
Gold Member
2019 Award
A photon has no wave function! You cannot apply the single-particle interpretation from non-relativistic quantum theory to relativistic particles. It's totally impossible to do so for massless particles like the photon. There is not even a position operator for a photon!

From the point of view of quantum-field theory classical electromagnetic fields are approximations to coherent photon states which are a superposition of Fock states adding up components from all photon-number states, i.e., for these states the photon number is indetermined but the phase is pretty well determined (there's a Heisenberg uncertainty relation between the phase of the field and the photon number).

For the same reason to introduce the photon concept by "very dimmed light" is misleading and strictly speaking even wrong! "Very dimmed light" is correctly described within QED by a coherent state with very small average (!!!) photon number or energy. Thus the state mostly consists of the vacuum state.

Only for the last 30 or so years quantum opticians are able to prepare really single-photon states on demand. This is done by creating an entangled photon pair, (e.g., by shooting a laser into a birefrigerent crystal, using parametric downconversion) and then measuring and absorbing one of the entangled photons. Then you are sure to have a true single-photon Fock state ("heralded single-photon preparation").

Another misconception, often found in high-school textbooks, is that the photoelectric effect proves the existence of photons. This goes back to Einstein's famous paper of 1905, which was historically very important to establish the "old quantum mechanics", which finally lead to modern quantum mechanics and after all relativistic quantum field theory and especially QED. Nevertheless it's incorrect, because the photoelectric effect can be explained semiclassically, i.e., by a model where the electric wave is treated as a classical field and only the electrons are described as bound states in the metal. Then the first-order time-dependent perturbation theory, with the electromagnetic wave field as the time-dependent perturbation, leads to the correct formula for the energy bilance of the photoelectric effect, which coincides with Einstein's formula from 1905.

A better experimental argument for the quantization of the electromagnetic field is the Lamb shift or (a bit easier to comprehend) quantum beats:

http://en.wikipedia.org/wiki/Quantum_beats

or better the book cited therein

Marlan Orvil Scully & Muhammad Suhail Zubairy (1997). Quantum optics. Cambridge UK: Cambridge University Press

Whenever you find (semi-)popular descriptions about what a photon might be, be very careful, as most explanations you find (even in textbooks on the university level!) are not up-to-date (to say it friendly) :-(.

Whenever you

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Drakkith
Staff Emeritus