Problem 14.5b

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

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

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— This answer is my guess. —

The naive mass formula in the left R’ sector:

\displaystyle{ \begin{aligned}  \alpha' M_L^2 &= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\  &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\  \end{aligned}}

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\displaystyle{ \begin{aligned}  \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A  &= \frac{1}{2} \sum_{n = -1, -2, ...} n \lambda_{-n}^A \lambda_n^A + \frac{1}{2} \sum_{n = 1, 2, ...} n \lambda_{-n}^A \lambda_n^A  \\  \end{aligned}}

p.316 Equation (14.51):

\displaystyle{ \begin{aligned}  \frac{1}{2} \sum_{n = -1. -2, ...} n \lambda_{-n}^A \lambda_n^A &= \frac{1}{2} \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{1}{24} (D - 2) \\  \end{aligned}}

\displaystyle{ \begin{aligned}  \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A  &= \left[ \frac{1}{2} \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{1}{24} \left[(D - 2)\right]_A \right] + \frac{1}{2} \sum_{n = 1, 2, ...} n \lambda_{-n}^A \lambda_n^A \\  &= \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{32}{24} \\  &= \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{4}{3} \\ \\  \end{aligned}}

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\displaystyle{ \begin{aligned}  \alpha' M_L^2  &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\  &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \left[ \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{4}{3} \right] \\  &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\  \end{aligned}}

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\displaystyle{ \begin{aligned}  \alpha' M_L^2  &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\  \end{aligned}}

— This answer is my guess. —

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2018.10.24 Wednesday (c) All rights reserved by ACHK