Quick Calculation 13.1

A First Course in String Theory

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Verify that

\displaystyle{\left[ \bar L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_m},

\displaystyle{\left[ L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \alpha^I_m}.

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Equation (13.37):

\displaystyle{\bar L_m^\perp = \frac{1}{2} \sum_{p \in \mathbf{Z}} \bar \alpha_p^J \bar \alpha_{n-p}^J},\displaystyle{~~~L_m^\perp = \frac{1}{2} \sum_{p \in \mathbf{Z}} \alpha_p^J \alpha_{n-p}^J}.

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\displaystyle{ \begin{aligned} \left[ \bar L_m^\perp, x_0^I \right] &= \frac{1}{2} \sum_{p \in \mathbb{Z}} \left[ \bar \alpha^J_p \bar \alpha^J_{m-p}, x_0^I \right] \\  &= \frac{1}{2} \sum_{p \in \mathbb{Z}} \bar \alpha^I_p \left[ \bar \alpha^J_{m-p}, x_0^I \right] + \frac{1}{2} \sum_{p \in \mathbb{Z}} \left[ \bar \alpha^J_p, x_0^I \right] \bar \alpha^I_{m-p} \\  &= \frac{1}{2} \bar \alpha^J_m \left[ \bar \alpha^J_{0}, x_0^I \right] + \frac{1}{2} \left[ \bar \alpha^J_0, x_0^I \right] \bar \alpha^J_{m} \\  \end{aligned} }

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By Equation (13.33):

\displaystyle{ \begin{aligned}  \left[ \bar L_m^\perp, x_0^I \right]  &= - \frac{1}{2} \bar \alpha^J_m \left[ i \sqrt{\frac{\alpha'}{2}} \eta^{IJ} \right] - \frac{1}{2} \left[ i \sqrt{\frac{\alpha'}{2}} \eta^{IJ} \right] \bar \alpha^J_{m} \\  &= - i \sqrt{\frac{\alpha'}{2}} \eta^{IJ} \bar \alpha^I_m \\  \end{aligned} }

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Since I and J are transverse coordinate indices, neither of them can be zero.

\displaystyle{ \begin{aligned}  \left[ \bar L_m^\perp, x_0^I \right]  &=  - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_m \\  \end{aligned} }

— Me@2019-12-25 10:56:15 AM

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2019.12.25 Wednesday (c) All rights reserved by ACHK