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What you're missing is that the condition $$\Pi |\psi\rangle = |\psi\rangle$$ does not imply $X(\sigma)=X(\ell-\sigma)$ – a condition which would force the closed string to go back and forth along the same path and effective become an open string. Instead, the condition above implies (is equivalent to) a much weaker condition that the complex amplitude ...

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Sorry I found David Z' answer a bit confused just when discussing the crucial point. Since the two functions ψ(x) and ψ(−x) satisfy the same equation, you should get the same solutions for them, except for an overall multiplicative constant; in other words, ψ(x)=aψ(−x) Normalizing ψ requires that |a|=1, which leaves two possibilities: a=+1 ...

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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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