C, T, P transformation mistakes in ``Peskin&Schroeder's QFT''?

Question

I suppose the right way to do C (charge), T (time reversal), P(parity) transformation on the state $\hat{O}| v \rangle$ with operators $\hat{O}$ is that:

$$ C(\hat{O}| v \rangle)=(C\hat{O}C^{-1})(C| v \rangle)\\ P(\hat{O}| v \rangle)=(P\hat{O}P^{-1})(P| v \rangle)\\ T(\hat{O}| v \rangle)=(T\hat{O}T^{-1})(T| v \rangle) $$

Thus to understand how an operator $\hat{O}$ transforms under C,P,T, we care about the following form $$ \hat{O} \to (C\hat{O}C^{-1})\\ \hat{O} \to (P\hat{O}P^{-1})\\ \hat{O} \to (T\hat{O}T^{-1}) $$

Here $\hat{O}=\hat{O}(\hat{\Phi},\hat{\Psi},a,a^\dagger)$ contains possible field operators ($\hat{\Phi},\hat{\Psi}$), or $a,a^\dagger$ etc.

To understand how a state $|v \rangle$ transforms, we care about $$ | v \rangle\to C| v \rangle\\ | v \rangle \to P| v \rangle\\ | v \rangle\to T| v \rangle $$

However, in Peskin and Schroeder QFT book, throughout Chap 3, the transformation is done on the fermion field $\hat{\Psi}$(operator in the QFT) : \begin{align} \hat{\Psi} &\to (C\hat{\Psi}C)? \tag{Eq.3.145}\\ \hat{\Psi} &\to (P\hat{\Psi}P)? \tag{Eq.3.128}\\ \hat{\Psi} &\to (T\hat{\Psi}T)? \tag{Eq.3.139} \end{align}

I suppose one should take one side as inverse operator ($(C\hat{\Psi}C^{-1}),(P\hat{\Psi}P^{-1}),(T\hat{\Psi}T^{-1})$). What have been written there in Peskin and Schroeder QFT Chap 3 is incorrect, especailly because $T \neq T^{-1}$, and $T^2 \neq 1$ in general. ($T^2=-1$ for spin-1/2 fermion)

Am I right?(P&S incorrect here) Or am I wrong on this point? (Why is that correct? I suppose S. Weinberg and M. Srednicki and A Zee use the way I described.)

Answer 1

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 representation, but in the Dirac, Weyl or any other representation where only $\gamma_2$ is imaginary, the choice $T=T^{-1}=i\gamma^{1}\gamma^{3}=T^{\dagger}$ servers our purpose". With the parity and charge conjugation it's the same, being unitary operators. Whatever the representation you use, the end result should be the same. So neither you or P&S are wrong.

Answer 2

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. $$ C(\hat{O}| v \rangle)=(C\hat{O}C^{-1})(C| v \rangle)\\ P(\hat{O}| v \rangle)=(P\hat{O}P^{-1})(P| v \rangle)\\ T(\hat{O}| v \rangle)=(T\hat{O}T^{-1})(T| v \rangle) $$

P&S is wrong there (replacing one side by the inverse operator). But the result of transformation should be correct still.