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

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.)

• "I suppose S. Weinberg and M. Srednicki and A Zee use the way I described." Did you check that? – Prahar Nov 24 '13 at 4:12
• yes, they use the one I believe to be right. I am sure. – wonderich Nov 24 '13 at 4:13
• I guess the reason that both the ways of transformation are valid is the following. $C,P,T$ are discrete operators. Moreover $C^2$~$I$,$P^2$~$I$,$T^2$~$I$. So even if (e.g.) you use $C^{-1}$ instead of $C$ to transform a state (and hence $COC$ to transform an operator) the result would differ only by a phase and hence physically be the same. – user10001 Nov 24 '13 at 6:07
• To user10001: Thanks. But I could not see why $T^2=I$ in general. Does $T^2=1$ or $T^2=-1$ does not matter at all? – wonderich Nov 26 '13 at 3:47

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