Problem 13.5b

A First Course in String Theory


13.6 Unoriented closed strings

This problem is the closed string version of Problem 12.12. The closed string \displaystyle{X^{\mu} (\tau, \sigma)} with \displaystyle{\sigma \in [0, 2 \pi]} and fixed \displaystyle{\tau} is a parameterized closed curve in spacetime. The orientation of a string is the direction of the increasing \displaystyle{\sigma} on this curve.

Introduce an orientation reversing twist operator \displaystyle{\Omega} such that

\displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} = X^I (\tau, 2 \pi - \sigma)

Moreover, declare that

\displaystyle{\Omega x_0^- \Omega^{-1} = x_0^-}

\displaystyle{\Omega p^+ \Omega^{-1} = p^+}

(b) Used the closed string oscillator expansion (13.24) to calculate

\displaystyle{\Omega x_0^I \Omega^{-1}}

\displaystyle{\Omega \alpha_0^I \Omega^{-1}}

\displaystyle{\Omega \alpha_n^I \Omega^{-1}}

\displaystyle{\Omega \bar \alpha_n^I \Omega^{-1}}


Equation (13.24):

\displaystyle{X^{\mu} (\tau, \sigma) = x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma})}


\displaystyle{\begin{aligned}   X^{\mu} (\tau, \sigma) &= x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma}) \\   X^I (\tau, 2 \pi - \sigma)  &= x_0^I + \sqrt{2 \alpha'} \alpha_0^I \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} \left( \alpha_n^I e^{- in\sigma} + \bar \alpha_n^I e^{i n \sigma)} \right) \\   \end{aligned}}


\displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} = X^I (\tau, 2 \pi - \sigma)


By comparing \displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} with \displaystyle{X^I (\tau, 2 \pi - \sigma)}, we have:

\displaystyle{\begin{aligned}   \Omega x_0^I \Omega^{-1} &= x_0^I \\  \Omega \alpha_0^I \Omega^{-1} &= \alpha_0^I \\  \Omega \alpha_n^I \Omega^{-1} &= \bar \alpha_n^I \\  \Omega \bar \alpha_n^I \Omega^{-1} &= \alpha_n^I \\   \end{aligned}}

— Me@2019-11-24 04:33:23 PM



2019.11.24 Sunday (c) All rights reserved by ACHK