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?)

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    $\begingroup$ 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 transversal degrees of freedom make sense. $\endgroup$ Commented Apr 24, 2016 at 14:46
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    $\begingroup$ @ValterMoretti and therefore the time component $A^0$ would be a real field and we would have physical negative norm sates in scattering amplitudes. I cant believe it was so simple :-) thank you! (you should post that as an answer) $\endgroup$ Commented Apr 24, 2016 at 14:48
  • $\begingroup$ I do not know if it is the only reason... $\endgroup$ Commented Apr 24, 2016 at 14:49
  • $\begingroup$ @ValterMoretti I guess Ill wait in case someone has something else to say, but to me your comment pretty much answers my question. $\endgroup$ Commented Apr 24, 2016 at 14:52
  • $\begingroup$ An additional comment: the simplest way to make a gauge invariant theory is to have fields that transform as a non-trivial linear representation of the gauge group. The EM group $U(1)$ has no non-trivial one-dimensional real representations, so single real scalars can't couple in this way. There are trickier ways to couple a real scalar to a gauge field (Higgs boson, dilaton, axion, etc.) but their interactions are not what you might naively expect. $\endgroup$ Commented Apr 25, 2016 at 19:45

2 Answers 2


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.

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    $\begingroup$ The theory may be hopeless in high energy physics, but boundary terms may have physical consequences in condensed matter physics. $\endgroup$ Commented Apr 25, 2016 at 18:16
  • $\begingroup$ You are right, unfortunately I am not an expert on that subject... $\endgroup$ Commented Apr 25, 2016 at 19:31
  • $\begingroup$ @ValterMoretti: So do you think the system with real fields described in my answer is problematic? And another question. Are the conclusions of your new work arxiv.org/abs/1611.09029 applicable to this system? I am trying to understand, not to criticize. $\endgroup$
    – akhmeteli
    Commented Dec 1, 2016 at 4:37
  • $\begingroup$ Yes, because it breaks gauge invariance which is fundamental in EM theory as it corresponds to the fact that photons have two internal degrees of freedom only. Regarding your last question, my recent paper with one of my PhD students M. Oppio concerns more relativistic quantum mechanics than QFT, so it is difficult to say anything without developing further the theory. $\endgroup$ Commented Dec 1, 2016 at 16:10
  • $\begingroup$ @ValterMoretti: I don't think it breaks gauge invariance any more than any other choice of gauge condition. In this case, the unitary gauge condition is used (the matter field is real). $\endgroup$
    – akhmeteli
    Commented Dec 1, 2016 at 18:30

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

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

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): \begin{equation}\label{eq:pr10} \Box\varphi-(e^2 B^\mu B_\mu-m^2)\varphi=0, \end{equation} \begin{equation}\label{eq:pr11} \Box B_\mu-B^\nu_{,\nu\mu}=j_\mu, \end{equation} \begin{equation}\label{eq:pr12} j_\mu=-2e^2 B_\mu\varphi^2. \end{equation}

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): \begin{equation}\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). \end{equation} 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.


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: \begin{equation}\label{eq:pr16a} \varphi_{,\mu}\varphi^{,\mu}=\frac{1}{4}\frac{\Phi_{,\mu}\Phi^{,\mu}}{\Phi}, \end{equation} and then $\Phi$ can be replaced by the following expression obtained from the equations of motion: \begin{equation}\label{eq:pr16b} \Phi=-\frac{1}{2e^2} \frac{B^\mu(\Box B_\mu-B^\nu_{,\nu\mu})}{B^\mu B_\mu}. \end{equation}

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

  • $\begingroup$ Sorry for taking so long to comment, I've been out for some time. Thank you very much for your answer :-) If I got you right, you are saying that the Lagrangian you wrote couples the $B$ field to a real $\phi$, right? In that case, should we say that $\phi$ is charged or uncharged? $\endgroup$ Commented Aug 9, 2016 at 13:40
  • $\begingroup$ @AccidentalFourierTransform: I would say in this situation it describes a charged field. $\endgroup$
    – akhmeteli
    Commented Aug 10, 2016 at 1:09

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