# Problem 14.5c3

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

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At any mass level $\displaystyle{\alpha' M^2 = 4k}$ of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = k}$ with the right-moving NS+ states with $\displaystyle{\alpha' M_R^2 = k}$.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = k}$ with the right-moving R- states with $\displaystyle{\alpha' M_R^2 = k}$.

c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

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Open String: \displaystyle{ \begin{aligned} N^\perp &= \sum_{p=1}^{\infty} \alpha_{-p}^I \alpha_p^I \\ L_n^{\perp} &= \frac{1}{2} \sum_{-\infty}^{\infty} \alpha_{n-p}^I \alpha_{p}^I~~~(n \ne 0) \\ L_0^{\perp} &= \alpha' p^I p^I + N^\perp \end{aligned} }

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Closed String: \displaystyle{ \begin{aligned} N^\perp &= \sum_{p=1}^{\infty} \alpha_{-p}^I \alpha_p^I \\ \bar N^\perp &= \sum_{p=1}^{\infty} \bar \alpha_{-p}^I \bar \alpha_p^I \\ L_0^\perp &= \frac{\alpha'}{4} p^I p^I + N^\perp \\ \bar L_0^\perp &= \frac{\alpha'}{4} p^I p^I + \bar N^\perp \end{aligned} }

— A First Course in String Theory

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We have the following well-known bosonic string mass formulae $\displaystyle{\alpha' = \frac{1}{2}}$:

open string: $\displaystyle{\frac{1}{2} M^2 = N - a}$

closed string: $\displaystyle{\frac{1}{8} M^2 = N_L - a}$ $\displaystyle{\frac{1}{8} M^2 = N_R - a}$

p.55

— Solutions to K. Becker, M. Becker, J. Schwarz String Theory And M-theory

— Mikhail Goykhman

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What is the meaning of $\displaystyle{\alpha'}$?

How come $\displaystyle{\alpha' = \frac{1}{2}}$?

— Me@2018-12-07 10:43:10 PM

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