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

PhD, 3.8.1

財政自由 1.3.1


The secret to creativity is knowing how to hide your sources.

— Not Einstein


In 1924, while working as a Reader (Professor without a chair) at the Physics Department of the University of Dhaka, Bose wrote a paper deriving Planck’s quantum radiation law without any reference to classical physics by using a novel way of counting states with identical particles. This paper was seminal in creating the very important field of quantum statistics. Though not accepted at once for publication, he sent the article directly to Albert Einstein in Germany. Einstein, recognising the importance of the paper, translated it into German himself and submitted it on Bose’s behalf to the prestigious Zeitschrift für Physik. As a result of this recognition, Bose was able to work for two years in European X-ray and crystallography laboratories, during which he worked with Louis de Broglie, Marie Curie, and Einstein.

— Wikipedia on Satyendra Nath Bose






















— Me@2019-10-29 10:20:33 PM


Bose adapted this lecture into a short article called Planck’s Law and the Hypothesis of Light Quanta and submitted it to the Philosophical Magazine. However, the referee’s report was negative, and the paper was rejected. Undaunted, he sent the manuscript to Albert Einstein requesting publication in the Zeitschrift für Physik. Einstein immediately agreed, personally translated the article from English into German (Bose had earlier translated Einstein’s article on the theory of General Relativity from German to English), and saw to it that it was published. Bose’s theory achieved respect when Einstein sent his own paper in support of Bose’s to Zeitschrift für Physik, asking that they be published together. The paper came out in 1924.

— Wikipedia on Bose–Einstein statistics



2019.10.29 Tuesday (c) All rights reserved by ACHK