Can we understand from basic QED, why is the photon electrically neutral?

Question

The QED Lagrangian $$\mathcal{L} =-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}+\bar\Psi(i\gamma^\alpha D_\alpha-m)\Psi,$$ where $D_\alpha=\partial_\alpha-ieA_\alpha$, is invariant under $$\Psi\to \Psi'(x)=e^{-ie\theta(x)}\Psi, \quad A_\alpha\to A_\alpha+\partial_\alpha\theta(x).$$ Do these transformation properties tell us why the photon must be electrically neutral?

Answer 1

Actually I could give a similar answer as given by "Silly Goose" that the gauge transformations of QED form an abelian group, therefore the structure constants of its Lie-algebra are zero, which would lead to a linear expression of the field tensor $F^{\mu\nu}$ (consequently no self-coupling).

But the charge-less character of the photon can already be seen on classical level: The Maxwell-equations are linear equations.

Actually there exists are classical nonlinear theory which does not share this property: Einstein's gravitation. Einstein's field equations are nonlinear and this is due to the fact that curvature couples to energy. But as the gravitational field also contains energy, it couples to itself.

Therefore upon quantization a quantum of the gravitational field, also called graviton, would couple to itself, which photons don't do. Gravitons carry its "charge" around: energy.

Recap: Photons are charge-less since the equations which govern them are linear. There is no self-coupling.

Answer 2

From the wikipedia page for Photon. "The quanta of an Abelian gauge field must be massless, uncharged bosons..." The electromagnetic field is an abelian gauge field. The quanta of the electromagnetic field are called photons. Thus, photons are massless and uncharged bosons.

For a derivation on why a gauge field being Abelian leads to the quanta being uncharged, see the answer to this stack post.