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I want to apply this transformation to a free-fermion lagrangian: $$ L=\bar{\psi}(\gamma^\mu{\partial_\mu \,- m)\,\psi}$$ $$ \psi ' =\psi\; e^{i \alpha \gamma_5} $$ $$ \bar{\psi}'=\bar{\psi} \;e^{-i \alpha \gamma_5}$$ The lagrangian should be invariant only if the $$ m=0$$ i don't understand why, how this transformation act on the mass term?

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    $\begingroup$ you are getting wrong the transformation of $\bar{\psi}$. There is a $\gamma_0$ factor which changes the sign of the $\alpha$ parameter since it anticommutes with $\gamma_5$. In the end you should get $\bar{\psi} ^\prime = \bar{\psi}e^{i\alpha\gamma_5}$. This means that the mass term goes to $m\bar{\psi} e^{2i\alpha\gamma_5}\psi$, which is not invariant if $m\neq0$ $\endgroup$ – otillaf Jun 18 at 10:50
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The transformation for $\bar \psi$ is not correct.
Since $\bar \psi = \psi^{\dagger} \gamma^{0}$, and the transformation for $\psi$ is $\psi ' = e^{i\alpha\gamma^{5} } \psi$, the correct identity is $$ \bar \psi ' = \psi ^{\dagger} e^{-i\alpha\gamma^{5} } \gamma^{0} = \bar \psi e^{i\alpha\gamma^{5} },$$ where $\{ \gamma^{\mu}, \gamma^{5} \} = 0, \, \mu = \{0 - 3\}$.
It follows that, assuming $\partial_{x}\alpha = 0$, $$L ' =\bar{\psi} ' (\gamma^\mu{\partial_\mu \,- m)\,\psi ' } =\bar{\psi}(\gamma^\mu{\partial_\mu \,- me^{2i\alpha\gamma^{5} })\,\psi}$$ and that $$ L ' = L \leftrightarrow m = 0 $$

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