Quick Calculation 13.4

A First Course in String Theory

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Prove (13.109).

\left[ \bar \alpha_{\frac{m}{2}}, \bar \alpha_{\frac{n}{2}} \right] = \frac{m}{2} \delta_{m+n, 0}

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Eq. (13.107):

\begin{aligned}   \sum_{m',n'\in \mathbb{Z}_\text{odd}} e^{-i\frac{m'}{2}(\tau+\sigma)} e^{-i\frac{n'}{2}(\tau+\sigma')} \left[ \bar \alpha_{\frac{m'}{2}}, \bar \alpha_{\frac{n'}{2}} \right]   &= 2 \pi i \frac{d}{d \sigma} \delta(\sigma - \sigma') \\   \end{aligned}

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\begin{aligned}   \frac{1}{2\pi} \int_0^{2 \pi} d \sigma e^{i \frac{m}{2} \sigma}\sum_{m',n'\in \mathbb{Z}_\text{odd}} \cdots   &= \int_0^{2 \pi} d \sigma e^{i \frac{m}{2} \sigma} i \frac{d}{d \sigma} \delta(\sigma - \sigma') \\     \end{aligned}

\begin{aligned}     \frac{1}{2\pi} \int_0^{2 \pi} d \sigma \sum_{m',n'\in \mathbb{Z}_\text{odd}} e^{-i\frac{m'}{2}\tau} e^{i\frac{(m-m')}{2}\sigma} \cdots   &= \int_0^{2 \pi} d \sigma e^{i \frac{m}{2} \sigma} i \frac{d}{d \sigma} \delta(\sigma - \sigma') \\     \sum_{m',n'\in \mathbb{Z}_\text{odd}} e^{-i\frac{m'}{2}\tau} \delta_{m,m'} \cdots  &= i \int_0^{2 \pi} e^{i \frac{m}{2} \sigma} d \delta(\sigma - \sigma') \\   \end{aligned}

\begin{aligned}     \sum_{n'\in \mathbb{Z}_\text{odd}} e^{-i\frac{m}{2}\tau}  \cdots  &= i \left(  \delta(2\pi - \sigma') - \delta(- \sigma')  - \int_0^{2 \pi} \delta(\sigma - \sigma') d e^{i \frac{m}{2} \sigma} \right) \\    \end{aligned}

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\begin{aligned}     e^{-i\frac{m}{2}\tau}  e^{-i\frac{n}{2}\tau} \left[ \cdots \right]   &= i \frac{1}{2\pi} \left( - e^{i\frac{n}{2} 2 \pi} - e^{i\frac{n}{2} 0}  - \int_0^{2 \pi} e^{i\frac{n}{2}\sigma} d e^{i \frac{m}{2} \sigma} \right) \\        e^{-i\frac{m+n}{2}\tau} \left[ \bar \alpha_{\frac{m}{2}}, \bar \alpha_{\frac{n}{2}} \right]   &= i \frac{1}{2\pi} \left(  - i\frac{m}{2} \int_0^{2 \pi} e^{i\frac{m+n}{2}\sigma} d \sigma  \right) \\    e^{-i\frac{m+n}{2}\tau} \left[ \bar \alpha_{\frac{m}{2}}, \bar \alpha_{\frac{n}{2}} \right]   &= i \frac{1}{2\pi} \left(   - i\frac{m}{2} 2\pi \delta_{m+n,0} \right) \\    e^{-i\frac{m+n}{2}\tau} \left[ \bar \alpha_{\frac{m}{2}}, \bar \alpha_{\frac{n}{2}} \right]   &= \frac{m}{2} \delta_{m+n,0}  \\    \end{aligned}

— Me@2024-10-28 01:34:14 PM

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2024.10.28 Monday (c) All rights reserved by ACHK