# Free fermion Lagrangian invariance under chiral symmetry

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?

• 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 $m\neq0$ – otillaf Jun 18 '19 at 10:50

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 $$L ' = L \leftrightarrow m = 0$$