# 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: $\displaystyle{\alpha' M_L^2}$ $\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$ $\displaystyle{= \frac{-1}{8} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

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

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Equation (13.116): $\displaystyle{\sum_{k \in \mathbf{Z}^+_{\text{odd}}} k = \frac{1}{12}}$

. \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}}

. \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 \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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