# Problem 14.5a4

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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$\displaystyle{\alpha' M_L^2}$

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

$\displaystyle{= \frac{1}{2} \left[ \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{24} \left[ (D - 2) \right]_{I= 2, 3, ..., 9} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

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$\displaystyle{\alpha' M_L^2 = \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

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Equation (14.34):

$\displaystyle{\frac{1}{2} \sum_{r=- \frac{1}{2}, -\frac{3}{2}} r b_{-r}^I b_{r}^I = \frac{1}{2} \sum_{r=\frac{1}{2}, \frac{3}{2}} r b_{-r}^I b_{r}^I - \frac{1}{48} (D - 2)}$

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$\displaystyle{\alpha' M_L^2}$

$\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \frac{1}{2} \sum_{r = - \frac{1}{2}, - \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \left[ \frac{1}{2} \sum_{r=\frac{1}{2}, \frac{3}{2}} r \lambda_{-r}^A \lambda_{r}^A - \frac{1}{48} \left[D - 2\right]_A \right]}$

$\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A - \frac{32}{48}}$

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\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + \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}}

— This answer is my guess. —

— Me@2018-10-10 05:38:08 PM

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