Quick Calculation 13.5

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A First Course in String Theory

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Show that [ \bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] = \frac{q}{2} \bar{\alpha}_{- \frac{q}{2}} and explain why \bar{N}^\perp is properly called a number operator.

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\bar N^\perp = \sum_{p=1}^{\infty} \bar \alpha^i_{-p} \bar \alpha^i_{~p} + \sum_{k \in \mathbb{Z}^+_\text{odd}} \bar \alpha_{- \frac{k}{2}} \bar \alpha_{\frac{k}{2}}

Eq. (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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[AB,C]=A[B,C]+[A,C]B

\begin{aligned} &[ \bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] \\ &= \sum_{p=1}^{\infty} [ \bar \alpha^i_{-p} \bar \alpha^i_{~p}, \bar{\alpha}_{- \frac{q}{2}} ] + \sum_{k \in \mathbb{Z}^+_\text{odd}} [ \bar \alpha_{- \frac{k}{2}} \bar \alpha_{\frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \end{aligned}

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If q=0,

[\bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] = 0

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If q>0,

\begin{aligned} &[ \bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] \\ &= \sum_{k \in \mathbb{Z}^+_\text{odd}}\left( \bar \alpha_{- \frac{k}{2}} [ \bar \alpha_{\frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] + [ \bar \alpha_{- \frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \bar \alpha_{\frac{k}{2}} \right) \\ &= \sum_{k \in \mathbb{Z}^+_\text{odd}}\left( \bar \alpha_{- \frac{k}{2}} [ \bar \alpha_{\frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \right) + 0 \\ &= \bar \alpha_{- \frac{q}{2}} [ \bar \alpha_{\frac{q}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \\ &= \bar \alpha_{- \frac{q}{2}} \frac{q}{2} \\ \end{aligned}

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If q<0,

\begin{aligned} &[ \bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] \\ &= \sum_{k \in \mathbb{Z}^+_\text{odd}}\left( \bar \alpha_{- \frac{k}{2}} [ \bar \alpha_{\frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] + [ \bar \alpha_{- \frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \bar \alpha_{\frac{k}{2}} \right) \\ &= 0 + \sum_{k \in \mathbb{Z}^+_\text{odd}} + [ \bar \alpha_{- \frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \bar \alpha_{\frac{k}{2}} \\ &= \frac{q}{2} \bar \alpha_{\frac{-q}{2}} \\ \end{aligned}

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\bar{N}^\perp should not be called a number operator. Instead, [ \bar{N}^\perp, \cdot ] should be.

It's through the commutation relation [ \bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] = \frac{q}{2} \bar{\alpha}_{- \frac{q}{2}} that we see \bar{N}^\perp functioning as a number operator, not by \bar{N}^\perp acting directly on \bar{\alpha}_{- \frac{q}{2}} operators.

— Me@2025-02-04 11:56:57 AM

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2025.02.04 Tuesday (c) All rights reserved by ACHK