If we have a complex scalar ϕ we know that the gauge-invariant interaction with A is given by AμJμ, where J is the Noether current of the U(1) symmetry of the Lagrangian Jμ∼ϕ†∂μϕ−ϕ∂μϕ†
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 Jμ∼ϕ∂μϕ
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 Aμ 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 Aμ 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 Aμ and we assume the field describes particles with spin 1 (avoiding problems with temporal modes) the condition ∂μAμ=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: ∫Aμϕ∂μϕd4x=12∫∂μ(Aμϕ2)d4x
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