quantum electrodynamics - Why can't a real scalar couple to the electromagnetic field?

If we have a complex scalar $\phi$ we know that the gauge-invariant interaction with $A$ is given by $A^\mu J_\mu$, where $J$ is the Noether current of the $U(1)$ symmetry of the Lagrangian $$ J_\mu\sim \phi^\dagger\partial_\mu\phi-\phi\partial_\mu \phi^\dagger \tag{1} $$

If we have a real scalar instead, $J=0$ and the field doesn't couple to $A$: real particles aren't charged.

My question is: what if we took $$ J_\mu\sim \phi\partial_\mu\phi \tag{2} $$ to couple $\phi$ to $A$? The interaction is Lorentz invariant and renormalisable, but it is not gauge-invariant, which is probably a bad thing. At what point would this theory break down?

This is probably very naïve, but I think that the Feynman rules for this theory are straightforward. So I guess the theory makes sense at least perturbatively. The theory is probably flawed at a more fundamental level, but I cant't seem to find where (some kind of gauge anomaly perhaps?)

Answer

The fact that the theory is not gauge invariant implies that all degrees of freedom of $A_\mu$ must have physical meaning: This is not the theory of photons where only transverse degrees of freedom make sense. This way you must tackle some non-trivial issue like the negative norm associated with temporal modes. This could be avoided by adding a mass to $A_\mu$ and giving spin $1$ (instead of helicity) to the associated particles. However, once again, this is not the EM field.

ADDENDUM. Actually if we add a mass to $A_\mu$ and we assume the field describes particles with spin $1$ (avoiding problems with temporal modes) the condition $\partial_\mu A^\mu =0$ has to be added just to remove a degree of freedom (or is even automatic if using Proca action as observed by AccidentalFourierTransform). This has the devastating consequence that the interaction Lagrangian becomes a boundary term, i.e., it vanishes: $$\int A_\mu \phi \partial^\mu \phi\,\mathrm d^4x = \frac{1}{2}\int \partial^\mu \left(A_\mu \phi^2\right)\,\mathrm d^4x$$ I think we have a hopeless theory from each viewpoint.