Quick Calculation 13.2

A First Course in String Theory

Verify that

$\displaystyle{\left[ \bar L_0^\perp, X^I (\tau, \sigma) \right] = - \frac{i}{2} \left( {\dot{X}}^I + {X^I}' \right)}$ ,

$\displaystyle{\left[ L_0^\perp, X^I (\tau, \sigma) \right] = - \frac{i}{2} \left( {\dot{X}}^I - {X^I}' \right)}$ ,

~~~

Equation (13.24):

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

Equation (13.39):

$\displaystyle{{\dot{X}}^- + {X^-}' = \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \bar \alpha_n^- e^{-in (\tau + \sigma)}}$

$\displaystyle{{\dot{X}}^- - {X^-}' = \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \alpha_n^- e^{-in (\tau - \sigma)}}$

Equation (13.51):

$\displaystyle{\left[\bar L_m^\perp, \bar \alpha_n^J \right] = - n \bar{\alpha}_{m+n}^J}$

$\displaystyle{\left[L_m^\perp, \alpha_n^J \right] = - n \alpha_{m+n}^J}$

Equation (13.52):

$\displaystyle{\left[L_m^\perp, \bar \alpha_n^J \right] = 0}$

$\displaystyle{\left[\bar L_m^\perp, \alpha_n^J \right] = 0}$

Equation (13.53):

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

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

$\displaystyle{\left[ \bar L_0^\perp, X^I (\tau, \sigma) \right]}$ ,

$\displaystyle{= \left[ \bar L_0^\perp, x_0^I + \sqrt{2 \alpha'} \alpha_0^I \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^I e^{i n \sigma} + \bar \alpha_n^I e^{-in \sigma}) \right]}$

$\displaystyle{= - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_0 + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} e^{-in \sigma} (-n \bar \alpha_n^I)}$