28 | October | 2024 | Physics Town

A First Course in String Theory

Prove (13.109).

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

~~~

Eq. (13.107):

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

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

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

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

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

— Me@2024-10-28 01:34:14 PM

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