Chiral symmetry of the Dirac Lagrangian

Question

I need to show that in the mass to zero limit the lagrangian density: $$\mathcal{L}=\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$ is invariant under the transformations: $$\psi'=e^{i\alpha\gamma^5} \psi$$ $${\psi^{\dagger}}'=\psi^\dagger e^{-i\alpha\gamma^5} $$ I've used the identity: $e^{-i\alpha\gamma^5}\gamma^0=\gamma^0e^{i\alpha\gamma^5}$ to arrive at the lagrangian: $$\mathcal{L'}=\psi^\dagger e^{-i\alpha\gamma^5}\gamma^0(i\gamma^\mu\partial_\mu-m)e^{i\alpha\gamma^5}\psi=\bar{\psi}e^{i\alpha\gamma^5}(i\gamma^\mu\partial_\mu-m)e^{i\alpha\gamma^5}\psi=\bar{\psi}e^{i\alpha\gamma^5}(i\gamma^\mu\partial_\mu) e^{i\alpha\gamma^5}\psi-\bar{\psi} me^{2i\alpha\gamma^5}\psi$$ I see that when the mass goes to zero the term on the right dissapears. I can't see how to get rid of the exponentials on the left term, should I find a similar identity to swap $e^{i\alpha\gamma^5}$ and $\gamma^\mu$?

Answer 1

The already used identity for $\gamma^0$ can be generalized to arbitrary $\mu$, i.e. $\{\gamma^\mu, \gamma^5\} =0$, hence one picks up a sign in $$\gamma^\mu e^{i\alpha \gamma^5} = e^{-i\alpha \gamma^5} \gamma^\mu\ .$$

This means that $$ \bar{\psi}e^{i\alpha\gamma^5}(i\gamma^\mu\partial_\mu) e^{i\alpha\gamma^5}\psi = \bar{\psi}e^{i\alpha\gamma^5}e^{-i\alpha\gamma^5}(i\gamma^\mu\partial_\mu) \psi = \bar{\psi}(i\gamma^\mu\partial_\mu) \psi\ $$ and proves the invariance (i.e. global symmetry and means that the parameter $\alpha$ is a constant not a function $\alpha(x)$) under one-flavor chiral rotations in the chiral limit (i.e. for massless Dirac spinors).