# 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

.

.