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A partial answer, is that supposing the gamma matrices, block-diagonal , as $\begin{pmatrix}A&\\&\epsilon A\end{pmatrix}, \begin{pmatrix}&A\\\epsilon A&\end{pmatrix}$, where $A$ is hermitian or anti-hermitian, and $\epsilon =\pm1$, give constraints on $A$ and $\epsilon$ due to $(\gamma^0)^2= \mathbb Id_4, (\gamma^i)^2= - \mathbb Id_4$. For ...

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How about just testing the two different cases? I.e. if $\mu\not=0$ then the LHS becomes $$(\gamma^\mu)^\dagger= (\gamma^i)^\dagger= -\gamma^i$$ while the RHS becomes $$(\gamma^\mu)^\dagger=\gamma^0\gamma^i\gamma^0 = -\gamma^0\gamma^0\gamma^i=-\gamma^i~~~~~~~~ (\text{OK}).$$ For $\mu=0$, the case ...

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Yes, you can show this using only the fact that the Clifford Algebra has a unique representation up to similarity transformation in any dimension. This is shown in the first few pages of http://arxiv.org/pdf/hep-th/9811101.pdf Then you observe that if $\gamma^\mu$ obeys the clifford algebra, then so does $-(\gamma^\mu)^T$. $\mathcal{C}$ is then defined as ...

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I think it's a matter of choice. If you look through several books you'll see all the possible combination $C\Psi(x)C$, $C\Psi(x)C^{-1}$, $C\Psi(x)C^{\dagger}$ (and the same for $P$ and $T$). I think it all comes down to the representation you are using. Like it is said in the book of Sterman (page 524) :"The precise nature of $T$ depends on the ...

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generally under symmetry transformation $S$, $$O \to S O S^{-1}$$ if $S O S^{-1}=O$ then $O$ is invariant under the symmetry transformation $S$, so $S$ commutes with $O$: $$[S,O]=0$$ This is correct as you said.  C(\hat{O}| v \rangle)=(C\hat{O}C^{-1})(C| v \rangle)\\ P(\hat{O}| v \rangle)=(P\hat{O}P^{-1})(P| v \rangle)\\ T(\hat{O}| v ...

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