Would it be consistent with QED to have leptons of different charges?

Question

A recent question, Equality of electric charges of all leptons, made me wonder about a specific aspect of why exactly the charges of the different (free) fundamental particles are all the same. Specifically, why are the charges of the different (charged) leptons all equal?

I understand this is a requirement in any treatment that includes the weak force, since the existence of $W^-\to e^-\bar{\nu}_e$, $W^-\to \mu^-\bar{\nu}_\mu$ and $W^-\to \tau^-\bar{\nu}_\tau$ vertices requires all the charges present to be equal. These vertices enable decays of the form $\mu^-\to e^-\bar{\nu}_e\nu_\mu$, which can only conserve charge if there's a single lepton charge unit.

If you take away this interaction, though, and you simply consider a multi-species QED theory, this mechanism goes away, and it would seem that you could in principle have different charges for the different species. Is this consistent with QED? Or is there something else going on, within QED, which requires the charges to be equal?

Answer 1

Multispecies spinor QED is described by the following Lagrangian:

\begin{equation} {\cal L} = \sum_i\bar{\psi} _i i \left( \partial _\mu + i e _i A _\mu \right) \gamma ^\mu \psi _i - m \bar{\psi} _i \psi _i - \frac{1}{4} F _{ \mu \nu } F ^{ \mu \nu } \end{equation} If $ e _i $ is the same for every particle then we have a flavor symmetry under, $ \psi _i \rightarrow e ^{ i T _a \theta ^a } \psi _i $. If it isn't then we don't. With this alone there is no apriori reason to believe such a symmetry exists.

In the Standard Model we see a lot of similar flavor symmetries. We don't currently have any good justification for these symmetries, and since we don't know of any reason for their existence, there is a lot of reserach that is done to try to explain them.