In eq. (3.128) of page 66 of "An Introduction to QFT", by Peskin & Schroeder, a step involves: $$P\,\overline{\psi}\left(t,\textbf{x}\right)P^{-1}=P\,\psi^{\dagger}\left(t,\textbf{x}\right)\gamma^0P^{-1}=P\,\psi^{\dagger}\left(t,\textbf{x}\right)P^{-1}\gamma^0\,.$$ Without knowing a priori the form of matrix $P$, except that it reverses the momentum of a particle without flipping its spin, how can I justify that $P$ commutes with $\gamma^0$ as suggested by Peskin & Schroeder?
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1$\begingroup$ We get asked this exact same question at least once a week. $P$ and $\gamma^0$ act on different vector spaces, so they commute. Do we have a canonical question on this? I never know which one to link to... $\endgroup$– AccidentalFourierTransformNov 17, 2021 at 22:05
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$\begingroup$ I was expecting this answer concerning the different vector spaces and I'm glad you gave it, since it pertains to my real doubt: at some point, we conclude that $P$ is of the form $P$ = \gamma^0, which still entails the same commutation relations. However, they now live in the same vector space. Unless I consider that stating $P$ is of the form above is unrelated to the space it acts upon. Am I correct? $\endgroup$– RicardoPNov 18, 2021 at 1:10
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Make both matrices diagonally dominate. Compute the product of the 2 matrices. If all of the vectors of the matrix are linearly independent, the matrices commute.
