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Experimentally, we determine the value of the coupling constant by measuring the scattering cross-section and compare it with the results calculated by the scattering amplitude. However, in the expression of the scattering cross-section, it is proportional to the square of scattering amplitude, which means it is proportional to the square of coupling constants at tree level. So actually it seems that we cannot determine the sign of coupling constants.

However, through some basic assumptions in quantum field theory, like analyticity and unitarity, it seems that it will impose constraints on the sign of coupling constants, see paper http://arxiv.org/abs/hepth/0602178, so I'm wondering if we could determine the sign of coupling constant by experiment? So that we could see if their papers are right or not.

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By properly redefining the gauge field $$ eA_\mu \rightarrow A_\mu, $$ one can reduce the coupling constant to the YM term only, $$ \sim \frac{1}{4e^2} F^{\mu\nu}F_{\mu\nu}. $$

Therefore, the absolute sign of $e$ does not matter, since only $e^2$ matters. (Of course, if you are concerned with the sign in front of $\pm \frac{1}{4e^2}$, it surely matters. )

That said, if there are multiple particles with different charges, the relative sign of the coupling constant does make a difference. However, experimentally, you can't distinguish between $$ +e\text{, for A} \\ -e\text{, for B} $$ and $$ -e\text{, for A} \\ +e\text{, for B.} $$ So you still don't have a clue of the absolute sign. The negative charge for the electron and positive charge for the proton are just man made conventions.

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  • $\begingroup$ I think you mean $\frac{1}{4e^2} F^{\mu\nu} F_{\mu\nu} $ $\endgroup$ – Prahar Jul 13 at 19:34
  • $\begingroup$ @Prahar, good catch! corrected. $\endgroup$ – MadMax Jul 13 at 19:35

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