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nrqed

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You know how ##\psi## transforms, right? Write ##\psi_L## in terms of ##\psi##, apply the transformation to ##\psi## and then show that the result can be written as ##e^{i\alpha\gamma^5}\psi_L## .JD_PM said:OK I was at this point

$$\psi_L = \frac 1 4 ( 1 - \gamma^5)e^{i \alpha \gamma_5} (1- \gamma^5) e^{i \alpha \gamma_5} \psi = \frac 1 4 e^{2 i \alpha \gamma_5} \psi + \frac 1 4 \gamma^5 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5} \psi - \frac 1 4 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5} \psi - \frac 1 4 \gamma^5 e^{2 i \alpha \gamma_5} \psi$$

Do you agree so far? If yes, could you please give me a hint on how could I use ##\left[(1- \gamma^5), e^{i\alpha\gamma^5}\right]=0## to show that ##\psi_L\rightarrow e^{i\alpha\gamma^5}\psi_L## ?

Thanks.