# Problem 14.5c6

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 $\displaystyle{\alpha' M_L^2 = 4k}$ 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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— This answer is my guess. —

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The left NS’+ sector:

\displaystyle{\begin{aligned} \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L \end{aligned}}

The left R’+ sector has no massless states.

The right-moving NS+ states:

\displaystyle{\begin{aligned} \alpha'M^2=0, ~~~&N^\perp = \frac{1}{2}: &b_{-1/2}^I~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\ \end{aligned}}

The R- states (that used as right-moving states):

\displaystyle{\begin{aligned} \alpha'M^2=0,~~~&N^\perp = 0:~~~~&|R_a \rangle \\ \end{aligned}}

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Since R’+ has no massless states:

spacetime bosons:

$NS'+ \otimes NS+$

\displaystyle{\begin{aligned} \left( \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} |NS' \rangle_L \right) \otimes \left( b_{-1/2}^I~|NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}

\displaystyle{\begin{aligned} = \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} b_{-1/2}^I |NS' \rangle_L \otimes \left( |NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}

spacetime fermions:

$NS'+ \otimes R-$

\displaystyle{\begin{aligned} \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} |NS' \rangle_L \otimes |R_a \rangle \\ \end{aligned}}

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— This answer is my guess. —

— Me@2018-12-18 07:46:15 PM

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