The Lagrangian [tex]\mathcal L =\psi^{\dagger}\gamma^0 \gamma^\mu (1-\gamma^5)\partial_\mu \psi [/tex] should violate parity, but I'm getting that it doesn't.(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \psi(x) [/tex] changes to [tex]\gamma^0 \psi( Px) [/tex] where Px=(t,-x) and x=(t,x).

[tex] \gamma^j [/tex] goes to [tex] - \gamma^j [/tex], while [tex] \gamma^0 [/tex] stays the same. [tex] \gamma^5 [/tex] goes to [tex] -\gamma^5 [/tex].

[tex] \partial_j [/tex] goes to [tex] -\partial_j [/tex] while [tex] \partial_0 [/tex] stays the same.

So [tex]P^{-1}\mathcal L P=(\psi^{\dagger}(Px)\gamma^0)\gamma^0 \gamma^\mu (1+\gamma^5)\partial_\mu (\gamma^0 \psi(Px)) [/tex] (the contraction between [tex] \gamma^\mu [/tex] and [tex] \partial_\mu [/tex] stays the same under parity).

So: [tex]P^{-1}\mathcal L P=(\psi^{\dagger}(Px) \gamma^0 \gamma^\mu (1-\gamma^5)\partial_{P\mu} \psi(Px) [/tex]. Here [tex]\partial_{P\mu}=\frac{\partial}{\partial (Px)^\mu} [/tex]

Therefore doesn't [tex]P^{-1}\mathcal L(x) P= \mathcal L(Px)[/tex]? Which means the Lagrangian doesn't change under parity?

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# How to check if Lagrangian is parity invariant?

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