# Why can't a real scalar couple to the electromagnetic field?

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.

Let us consider scalar electrodynamics (Klein-Gordon-Maxwell electrodynamics) with the Lagrangian: \begin{eqnarray}\label{eq:pr6} -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}(\psi^*_{,\mu}-ieA_\mu\psi^*)(\psi^{,\mu}+ieA^\mu\psi)-\frac{1}{2}m^2\psi^*\psi \end{eqnarray}

and the equations of motion

$$\label{eq:pr7} (\partial^\mu+ieA^\mu)(\partial_\mu+ieA_\mu)\psi+m^2\psi=0,$$ $$\label{eq:pr8} \Box A_\mu-A^\nu_{,\nu\mu}=j_\mu,$$ $$\label{eq:pr9} j_\mu=ie(\psi^*\psi_{,\mu}-\psi^*_{,\mu}\psi)-2e^2 A_\mu\psi^*\psi.$$

The complex charged matter field $\psi$ can be made real by a gauge transform (at least locally), and the equations of motion in the relevant gauge (unitary gauge) for the transformed four-potential of electromagnetic field $B^{\mu}$ and real matter field $\varphi$ are as follows ($E.~Schr\ddot{o}dinger, {Nature}$, 169:538, 1952): $$\label{eq:pr10} \Box\varphi-(e^2 B^\mu B_\mu-m^2)\varphi=0,$$ $$\label{eq:pr11} \Box B_\mu-B^\nu_{,\nu\mu}=j_\mu,$$ $$\label{eq:pr12} j_\mu=-2e^2 B_\mu\varphi^2.$$

Schrödinger made the following comment: "That the wave function ... can be made real by a change of gauge is but a truism, though it contradicts the widespread belief about 'charged' fields requiring complex representation."

These equations of motion can be obtained from the Lagrangian of work (T. Takabayasi(1953), Progr. Theor. Phys.,9,187): $$\label{eq:pr12a} -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}e^2 B_\mu B^\mu \phi^2+\frac{1}{2}(\varphi_{,\mu}\varphi^{,\mu}-m^2\varphi^2).$$ Actually, it coincides with the previous Lagrangian up to the replacement of the complex scalar field by a real one. Any solution of the equations of motion for the first Lagrangian has a physically equivalent solution of the equations of motion for the second Lagrangian.

I did not consider quantization here.

EDIT(02/10/2018):

I would like to add that the above system of interacting real scalar field and electromagnetic field has some amazing properties. One can see from the equations of motion that the real scalar field can be algebraically eliminated, and it turns out that the resulting equations for the electromagnetic field describe its independent evolution (my article in European Physical Journal C http://link.springer.com/content/pdf/10.1140%2Fepjc%2Fs10052-013-2371-4.pdf and references there).

Apparently, it is also possible to introduce a Lorentz-invariant Lagrangian with higher derivatives that does not include the matter field, but is largely equivalent to the Takabayasi Lagrangian (my article https://arxiv.org/abs/1006.2578 ; the significance of some special cases, e.g., $\varphi=0$ and $B^\mu B_\mu=0$ (see below) is unclear (different particles?)). To this end, the Takabayasi Lagrangian can be expressed in terms of $\Phi=\varphi^2$, rather than $\varphi$, using, e.g., the following: $$\label{eq:pr16a} \varphi_{,\mu}\varphi^{,\mu}=\frac{1}{4}\frac{\Phi_{,\mu}\Phi^{,\mu}}{\Phi},$$ and then $\Phi$ can be replaced by the following expression obtained from the equations of motion: $$\label{eq:pr16b} \Phi=-\frac{1}{2e^2} \frac{B^\mu(\Box B_\mu-B^\nu_{,\nu\mu})}{B^\mu B_\mu}.$$

Thus, a Lagrangian including only electromagnetic field describes pretty much the same physics as scalar electrodynamics.