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) :
$$
\hat{\Psi} \to (C\hat{\Psi}C)? (Eq.3.145)\\
\hat{\Psi} \to (P\hat{\Psi}P)? (Eq.3.128)\\
\hat{\Psi} \to (T\hat{\Psi}T)? (Eq.3.139)
$$

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.)

This post imported from StackExchange Physics at 2014-06-04 11:39 (UCT), posted by SE-user Idear