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It has been pointed out in a previous post that the convention used for the gauge covariant derivative tends to be different in nearly every textbook. As was pointed out in this excellent answer, these differences fundamentally boil down to the choice of the metric signature and the sign convention for the elementary charge $e$.

I am a bit confused about the interpretation of the sign of $e$ in this context. To illustrate, suppose I am considering a simple complex scalar theory such as scalar QED

$$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-(D^\mu \phi)^*D_\mu\phi-m^2|\phi^2|$$

with metric signature $(-,+,+,+)$ and gauge covariant derivative $D_\mu=\partial_\mu-ieA_\mu$. This system describes a complex scalar field coupled to electromagnetism.

If I derive the (classical) equations of motion for this system, there will be terms proportional to $e$. What is the interpretation of plugging in $e>0$ versus $e<0$ into these equations of motion? Does picking $e>0$ describe the evolution of a negatively charged field, while $e<0$ a positive one? How can one see this?

Fundamentally, I guess my confusion is this: In classical E&M I know that picking $q>0$ or $q<0$ will lead to different motion for positive and negative charges (for example via the Lorentz force law). But for the above theory, there is no clear distinction between a positive and negative point charge, only a continuous scalar field coupled to the gauge field. What then is the interpretation of picking a value of $e$ in the equations of motion?

Qmechanic
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Superbee
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    Are you asking how the system's time-evolution would change in the presence of a background EM field if we flip the sign of the charge without also flipping the sign of the background field? (If we flip both, then nothing observable is changed, just like in the Lorentz force equation). – Chiral Anomaly Oct 26 '20 at 22:10
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    e is a positive constant. It makes no sense to give it a minus sign. What changes sign is $\partial_0$. – my2cts Oct 26 '20 at 23:27
  • @ChiralAnomaly I understand that $e$ determines the coupling strength between the scalar and gauge field, and in this context it makes sense that $e$ should only be a positive constant. But my question is about point 2) from this answer. It looks like there are two conventions: picking $e>0$ or $e<0$. But it seems to me like each would lead to a different equations of motion! So I guess my question is that I don't fully understand the meaning of "$q=−|e|$" and the choice of "$e = −|e| < 0$" or "$e = |e| > 0$" in the linked answer... – Superbee Oct 27 '20 at 00:15
  • @my2cts The constant $e$ is just a parameter in the model, and either sign is a perfectly legitimate choice. – Chiral Anomaly Oct 27 '20 at 03:11
  • @ChiralAnomaly This is a bad idea that only creates confusion. – my2cts Oct 27 '20 at 08:13

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Fundamentally, I guess my confusion is this: In classical E&M I know that picking q>0 or q<0 will lead to different motion for positive and negative charges (for example via the Lorentz force law).

This is not true. It's certainly true that in a given electromagnetic field, the motion of a positively charged particle will be different than the motion of a negatively charged particle. However, if we flipped the sign of every charge in the universe, then the motion of charged particles would be unaffected.

Ultimately this is a reflection of the fact that (as far as we know) all electromagnetic fields are ultimately generated by charged particles. If you flip the charge of the particle, you flip the direction of the fields which it generates. As a result, the Lorentz force law is unaffected.

Of course, if it turns out that there is some primordial electromagnetic field which is fixed by the boundary conditions of the universe and is not generated by charged matter, then this would no longer be true. But if that's the case, then Lorentz invariance is out the window as well, and fundamental physics would suddenly become much more exciting.

Anyway, in QED (scalar or otherwise) Lorentz invariance is assumed from the start, so the aforementioned excitement doesn't come into play. Any physically observable quantities (e.g. cross sections, etc) will be the same whether $e>0$ or $e<0$. The only thing that matters is that the electron and positron have opposite charges, which is guaranteed by the fact that the coupling term $ieA_\mu$ is purely imaginary (and therefore flips sign under conjugation).

J. Murray
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