13.1 Commutation relations for oscillators

A First Course in String Theory

(a) Use the lower-sign version of equation (13.28) and the appropriate mode expansion to verify explicitly that the unbarred commutation relations of (13.29) emerge.

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Eq. (13.28):

$[(\dot X^I - X^{I'}) (\tau, \sigma), (\dot X^J - X^{J'})(\tau, \sigma')] = - 4 \pi \alpha' i \eta^{IJ} \frac{d}{d \sigma} \delta(\sigma - \sigma')$

Eq. (13.29):

$\left[ \alpha_m^I, \alpha_n^J \right] = m \delta_{m+n, 0} \eta^{IJ}$

Eq. (13.24):

$\displaystyle{X^\mu (\tau, \sigma) = x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \neq 0} \frac{e^{-i n \tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma})}$

$\displaystyle{\dot{X}^\mu(\tau, \sigma) = \sqrt{2 \alpha'} \alpha_0^\mu + \sqrt{\frac{\alpha'}{2}} \sum_{n \neq 0} e^{-i n \tau} \left( \alpha_n^\mu e^{i n \sigma} + \bar{\alpha}_n^\mu e^{-i n \sigma} \right)}$

$\displaystyle{X^{\mu'}(\tau, \sigma) = \sqrt{\frac{\alpha'}{2}} \sum_{n \neq 0} e^{-i n \tau} \left( \bar{\alpha}_n^\mu e^{-i n \sigma} - \alpha_n^\mu e^{i n \sigma} \right)}$

$\displaystyle{\begin{aligned} &\int_0^{2 \pi} f(\sigma) d \sigma\frac{d}{d \sigma} \delta(\sigma - \sigma') \\ &= \int_0^{2 \pi} f(\sigma) d \delta(\sigma - \sigma') \\ &= f(2 \pi) \delta(2 \pi - \sigma') - f(0) \delta(0 - \sigma') - \int_0^{2 \pi} \delta(\sigma - \sigma') \frac{df(\sigma)}{d \sigma} d \sigma \\ \end{aligned}}$

Eq. (13.26):

$\displaystyle{\begin{aligned} \left[(\dot X^I - X^{I'}) (\tau, \sigma), (\dot X^J - X^{J'})(\tau, \sigma')\right] &= - 4 \pi \alpha' i \eta^{IJ} \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \left[ \sqrt{2 \alpha'} \sum_{m \in \mathbb{Z}} \alpha_m^I e^{-i m (\tau - \sigma)}, \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \alpha_n^J e^{-i n (\tau - \sigma')} \right] &= - 4 \pi \alpha' i \eta^{IJ} \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \sum_{m \in \mathbb{Z}} \sum_{n \in \mathbb{Z}} e^{-i m (\tau - \sigma)} e^{-i n (\tau - \sigma')} \left[ \alpha_m^I , \alpha_n^J \right] &= - 2 \pi i \eta^{IJ} \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{1}{2 \pi} \int_0^{2 \pi} d \sigma e^{iq\sigma} \sum_{m \in \mathbb{Z}} \sum_{n \in \mathbb{Z}} e^{-i (m+n) \tau} e^{i m \sigma} e^{i n \sigma'} \left[ \alpha_m^I , \alpha_n^J \right] &= - 2 \pi i \eta^{IJ} \frac{1}{2 \pi} \int_0^{2 \pi} d \sigma e^{iq\sigma} \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \sum_{n \in \mathbb{Z}} e^{-i (-q+n) \tau} e^{i n \sigma'} \left[ \alpha_{-q}^I , \alpha_n^J \right] &= - i \eta^{IJ} \left[ \delta(2 \pi - \sigma') - \delta(0 - \sigma') - iq \int_0^{2 \pi} \delta (\sigma - \sigma') e^{i q \sigma} d\sigma \right] \\ \sum_{n \in \mathbb{Z}} e^{-i (-q+n) \tau} e^{i n \sigma'} \left[ \alpha_{-q}^I , \alpha_n^J \right]&= - i \eta^{IJ} \left[ \delta(2 \pi - \sigma') - \delta(0 - \sigma') - iq e^{i q \sigma'} \right] \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{1}{2 \pi} \int_0^{2 \pi} d \sigma' e^{ip\sigma'} \sum_{n \in \mathbb{Z}} e^{-i (-q+n) \tau} e^{i n \sigma'} \left[ \alpha_{-q}^I , \alpha_n^J \right]&= - i \eta^{IJ} \frac{1}{2 \pi} \int_0^{2 \pi} d \sigma' e^{ip\sigma'} \left[ \delta(2 \pi - \sigma') - \delta(0 - \sigma') - iq e^{i q \sigma'} \right] \\ \end{aligned}}$

$\displaystyle{\begin{aligned} e^{i (p+q) \tau} \left[ \alpha_{-q}^I , \alpha_{-p}^J \right]&= - \eta^{IJ} \frac{1}{2 \pi} q \int_0^{2 \pi} d \sigma' e^{ip\sigma'} e^{i q \sigma'} \\ e^{i (p+q) \tau} \left[ \alpha_{-q}^I , \alpha_{-p}^J \right]&= - \eta^{IJ} q \delta_{p+q,0} \\ \\ \\ e^{i (m+n) \tau} \left[ \alpha_{-n}^I , \alpha_{-m}^J \right]&= - n \delta_{m+n,0} \eta^{IJ} \\ \end{aligned}}$

— Me@2025-04-17 12:26:46 PM

Physics Town

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13.1 Commutation relations for oscillators

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