Can a Path Integral Formulation for Photons Start from a Massless Premise?

[gerald V]

TL;DR Summary

Can one do first quantization of electromagnetism starting from a photon action?

I am aware that one usually starts from the Maxwell equations and then derives the masslessness of a photon. But can one do it the other way round? The action of photon would be $S = \hbar \int \nu (1 - \dot{x}^2) \mbox{d}t$, where $\nu$ is the frequency acting as a Lagrange multiplier, forcing the velocity squared to be unity and the action to be null.

Does it make sense in principle to use this action for a path integral formulation?

If yes, how to deal with the factor $\nu$? Can one assume it to be constant if the photon is free?

Can one add to the action a hypothetical „potential“ making the frequency vary, for example to let the photon couple to some electromagnetic current? How then to deal with the frequency inside the path integral?

Thank you very much in advance

 

[anuttarasammyak]

Light path has zero world interval s or proper time so we cannot take s as parameter for integral. We should find other parameter than s. Fermat's principle or geodesic of light would suggest you a hint to your question.

 

[throw]

The action for a relativistic particle is
$S = - m \int ds = - m \int \sqrt{-\dot{x}^2} d \tau$
This clearly assumes $m \neq 0$, however we must also recognize that the system is a constrained system since $p_{\mu}$ satisfies $p^2 = m^2$, so one should really first reformulate the problem as a constrained system. On doing this one can show the action can be reformulated as
$S = \frac{1}{2} \int e (e^{-2} \dot{x}^2 + m^2) d \tau$
where $e$ can be interpreted as a metric. This action reproduces the original action on using the equation of motion for $e$, and it also encodes the $p^2 = m^2$ constraint directly in the action rather than as a constraint. This form of the action admits a massless limit $m \to 0$. Quantizing the action in this form results in the Klein-Gordon equation applied to a quantum wave function, which all free particle wave equations must satisfy. Why such a particle is a photon as opposed to simply a scalar, starting from the classical picture, requires justification.

 

[dextercioby]

There is no particle action for a massless vector theory. One can only build that for a massive scalar (einbein formulation, see the post by @throw ), or a massive spin 1/2 particle (the Brink-Howe-DiVecchia action for a fermionic elementary particle).