Consider a theory containing a Dirac field $\psi$ and a scalar field $\varphi$ where the only interaction is given by a Yukawa potential $$ V = -g\bar{\psi}\varphi\psi $$ I know that real scalar fields cannot couple to the electromagnetic field because their Noether current density $$\mathcal{J}^\mu = \varphi\partial^\mu\varphi - \varphi\partial^\mu\varphi$$ under the $\text{U}(1)$ symmetry of the EM field vanishes and therefore there is no coupling (cf. this question). Is the reason that the real scalar field theory is not invariant under $\text{U}(1)$ transformations at all?
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1$\begingroup$ $V$ has to be real. What does this imply for $\varphi$? $\endgroup$– PraharCommented Jun 12 at 20:53
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$\begingroup$ it implies that $\varphi$ must be real. thanks! $\endgroup$– paulinaCommented Jun 12 at 20:57
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The current $\bar\psi \gamma^\mu \psi$ will be equal to zero if $\psi$ is a Majorana fermion. This condition is the fermion analogue of being "real."
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1$\begingroup$ I am looking explicitly for the case where $\psi$ is a Dirac fermion. $\endgroup$– paulinaCommented Jun 12 at 13:37
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1$\begingroup$ Then I do not understand your question. Perhaps you say what you mean by a "similar argument"? What are you trying to show? $\endgroup$ Commented Jun 12 at 13:44
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1$\begingroup$ I just realized that the way I wrote it makes no sense. I will think about it again and then edit the question accordingly. $\endgroup$– paulinaCommented Jun 12 at 13:53
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2$\begingroup$ Wow, reminding me of congressmen talking PASS each other. $\endgroup$– MadMaxCommented Jun 12 at 14:28