I am trying to use the Lagrangian of QED (without kinetic terms for photons) to prove that the axial current of QED satisfies $\partial_\mu j^\mu_5 = 2im\bar\psi\gamma^5\psi,$ where $j^\mu_5 = \bar\psi\gamma^\mu\gamma^5\psi.$ Now, I have used the chiral transformation $\psi \to e^{i\alpha(x)\gamma^5}\psi$ and $\bar \psi \to \bar\psi e^{-i\alpha(x)\gamma^5}$. Working through the calculations, I found that the lagrangian changes to $$\mathcal L - i\alpha(x) (\bar\psi\gamma^5(i\gamma^\mu \partial_\mu - m -e\gamma^\mu A_\mu)\psi) +\alpha(x)(\partial_\mu\bar\psi\gamma^\mu\gamma^5\psi).$$ At this point, I cannot figure out how to get rid of the terms involving $A_\mu$ and the partial derivatives. If you can provide any help, it would be greatly appreciated.


1 Answer 1


The transformations are not what you have. The field $\bar \psi$ is defined by $\bar \psi = \psi^\dagger \gamma_0$, so they should be $$ \psi\to e^{i\gamma^5 \alpha}\psi, \quad \bar\psi \to \bar\psi e^{i\gamma^5 \alpha}. $$ This means that $m\bar\psi \psi$ is not invariant under the axial transformation

  • $\begingroup$ That would seem to make sense, thanks for the clarification. $\endgroup$ Feb 27, 2023 at 20:41

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