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I'm revising some string basics, and have come across the following problem. For closed strings one introduces the worldsheet parity operator

$$\Pi : \sigma \mapsto \ell-\sigma$$

where $\ell$ is the string length. In order to relate the left-movers and right-movers by parity in the Fourier expansion one assumes that $X$ and $P$ are parity even.

But why is this reasonable? I don't understand what forces $X$ and $P$ to have a specific behaviour under parity. Indeed surely a generic closed string will not have $X(\sigma)=X(\ell-\sigma)$!

Is there something I'm missing here? Thanks in advance for any helpful thoughts.

Edit

I wonder whether the notes I'm reading are only dealing with unoriented string theory. Perhaps that would be the reason for assuming the evenness of $X$ and $P$. Could someone possibly confirm that I've got the correct meaning of unoriented?

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    $\begingroup$ Yes, it concerns unoriented theories. Without orientation, "increasing" $\sigma$ should be equivalent to "decreasing" $\sigma$, so a new symmetry $\sigma \to l-\sigma, \tau'=\tau$ is needed, generated by the worldsheet parity operator. In consistent string theories, only parity-even states are kept (Ref : Polchinski, Vol $1$, page $29$). $\endgroup$ – Trimok Dec 11 '13 at 19:58
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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 $A_{\rm forward}$ for the closed string to be in a particular generic profile $X(\sigma)$ is the same as the complex amplitude $A_{\rm backward}$ that the string goes along the same path in the opposite direction, $X(\sigma)\to X(\ell-\sigma)$.

It means that in the theory constrained by $\Pi=1$ above, we only allow states composed of $$|\psi\rangle_{\rm forward} + |\psi\rangle_{\rm backward}$$ where the two pieces only differ by the direction of the string (whether the $\sigma$ is increasing or decreasing along a particular part of the string). The relative sign between the two terms is the eigenvalue of $\Pi$ when the ket vectors are properly normalized.

The "forward plus backward" symmetrization obtained for a given closed string's profile is the same as the symmetrization of the opposite, backward profile, and that's why it doesn't matter what the orientation is and why the closed strings are unorientable even though their shape in space is arbitrary.

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