Problem 14.5a2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

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\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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— This answer is my guess. —

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Equation at Problem 14.5:

\displaystyle{\alpha' M_L^2}

\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{-1}{8} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}
\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right]}

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Equation (13.116):

\displaystyle{\sum_{k \in \mathbf{Z}^+_{\text{odd}}} k = \frac{1}{12}}

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$latex \displaystyle{\begin{aligned} &\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A \\

&= \sum_{r = \frac{1}{2}, \frac{3}{2}, …} r \left[ 2 \lambda_{-r}^A \lambda_r^A – 1 \right] \\ &= – \sum_{r = \frac{1}{2}, \frac{3}{2}, …} r + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, …} r \lambda_{-r}^A \lambda_r^A \\

&= – \frac{1}{2} \sum_{r = 1, 3, …} r + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, …} r \lambda_{-r}^A \lambda_r^A \\ &= – \frac{1}{24} + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, …} r \lambda_{-r}^A \lambda_r^A \end{aligned}}$

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$latex \displaystyle{
\begin{aligned}

\alpha’ M_L^2

&= \frac{-7}{48}
+ \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I
+ \sum_{r = \frac{1}{2}, \frac{3}{2}, …} r \lambda_{-r}^A \lambda_r^A \\

\end{aligned}}$

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If we define N^\perp in the way similar to equation (14.37), we have

$latex \displaystyle{
\begin{aligned}

\alpha’ M_L^2

&= \frac{-7}{48} + N^\perp \\

\end{aligned}}$

— This answer is my guess. —

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— Me@2018-09-01 06:05:29 AM

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2018.09.01 Saturday (c) All rights reserved by ACHK