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In even dimensions all the representations of the gamma matrices are equivalent, in particular $\gamma^\mu$ and $-\gamma^\mu$ are equivalent. Usually the Dirac Lagrangian is \begin{equation} \psi^\dagger \gamma^0 (i \gamma^\mu \partial_\mu - m) \psi. \end{equation} It follows that if this is written considering the $\gamma^\mu$ representation, in terms of the equivalent $-\gamma^\mu$ it reads \begin{equation} \psi^\dagger \gamma^0 (i \gamma^\mu \partial_\mu + m) \psi. \end{equation} Is the sign of the mass in the Dirac action irrelevant?

Weyl
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    Does this answer your question? Has the relative sign in the Dirac equation any meaning? – Connor Behan Mar 14 '23 at 22:21
  • The most general mass term of a fermion is $m\bar{\psi} e^{\theta i\gamma_5} \psi$. The two cases you mentioned are the special cases of $\theta$ being $\pi$ or $0$. This $\theta$ angle can be rotated away by an axial phase transformation at the Lagrangian level. It's a common practice when you are dealing with the phases in the quark mass/mixing CKM matrix and the QCD strong CP phase $\Theta$. See details here: https://physics.stackexchange.com/questions/611255/is-there-a-square-root-of-klein-gordon-equation-which-is-different-from-dirac?rq=1 – MadMax Mar 16 '23 at 14:53
  • @MadMax thank you for your comment. Maybe I should have not closed the question, since, more precisely, what I have in mind is the following: if the sign of the mass is irrelevant, does this means that even in the massive case the transformation $\psi \rightarrow \exp[(i \gamma^5 \pi /2] \psi$ is a discrete symmetry of the Dirac action? For this value of the chiral angle, should I not expect a mass term in axial Ward Identity? And what about the anomaly for this case? – Weyl Mar 16 '23 at 16:19
  • @Weyl, no, transformation $\psi \rightarrow \exp[(i \gamma^5 \pi /2] \psi$ is not a symmetry of the Dirac action for massive case. Breaking of axial symmetry is a fact regardless of the sign/phase of the non-zero mass term. BTW, the chiral/axial/ABJ anomaly is a different story, its existence is independent of mass being zero or not. – MadMax Mar 16 '23 at 16:43
  • @MadMax My doubt is: staying at the classical level, if $\psi^\dagger \gamma^0 (i \gamma^\mu \partial_\mu - m) \psi$ is equivalent to $\psi^\dagger \gamma^0 (i \gamma^\mu \partial_\mu + m) \psi$ (since the Dirac action is written independently from the gammas representation chosen), why is it not true that $\psi \rightarrow \exp[i \gamma^5 \pi/2] \psi$ is a symmetry? – Weyl Mar 16 '23 at 18:23
  • @Weyl, yes you can choose whatever conventions you like. But you have to stay with ONE convention when you are talking about symmetry properties. Or you can argue that the combination of $\gamma^\mu \rightarrow -\gamma^\mu$ and $\psi \rightarrow \exp[i \gamma^5 \pi/2] \psi$ is a discrete symmetry, called "Weyl" symmetry ;) – MadMax Mar 16 '23 at 19:39

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