Problem 14.5c2

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

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?

~~~

— This answer is my guess. —

The left NS’+ sector:

$\displaystyle{\begin{aligned} \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&|NS' \rangle_L, \\ \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L, \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, ... \} \\ & & \{ ..., \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L \\ \end{aligned}}$

The left R’+ sector:

$\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_\alpha \rangle_L \\ \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&\alpha_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\ \end{aligned}}$

The right-moving NS+ states:

NS+ equations of (14.38):

$\displaystyle{\begin{aligned} \alpha'M^2=0, ~~~&N^\perp = \frac{1}{2}: &b_{-1/2}^I~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\ \alpha'M^2=1, ~~~&N^\perp = \frac{3}{2}: &\{ \alpha_{-1}^I b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^I, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \}~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\\ \alpha'M^2=2, ~~~&N^\perp = \frac{5}{2}: &\{\alpha_{-2}^I b_{\frac{-1}{2}}^J, \alpha_{-1}^I \alpha_{-1}^J b_{\frac{-1}{2}}^K, \alpha_{-1}^I b_{\frac{-3}{2}}^J, \alpha_{-1}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M, ...\}~& \\ &&\{ ..., b_{\frac{-5}{2}}^I, b_{\frac{-3}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M b_{\frac{-1}{2}}^N \}~&|NS \rangle \otimes |p^+, \vec p_T \rangle \end{aligned}}$

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

Mass levels of R- and R+ (Equations 14.54):

$\displaystyle{\begin{aligned} \alpha'M^2=0,~~~&N^\perp = 0:~~~~&|R_a \rangle~~&||~~|R_{\bar a} \rangle \\ \alpha'M^2=1,~~~&N^\perp = 1:~~~~&\alpha_{-1}^I |R_{a} \rangle,~d_{-1}^I |R_{\bar a} \rangle ~~&||~~ ... \\ \alpha'M^2=2,~~~&N^\perp = 2:~~~~&\{ \alpha_{-2}^I,~\alpha_{-1}^I \alpha_{-1}^J,~d^I_{-1} d^J_{-1} \} |R_{a} \rangle,~&|| \\ &&\{\alpha_{-1}^I d_{-1}^J,~d_{-2}^I \} |R_{\bar a} \rangle~~&||~~ ... \\ \end{aligned}}$

There are no tachyonic states in heterotic string theory, since neither of the right-moving parts (NS+ and R-) has states with $\displaystyle{\begin{aligned} \alpha' M^2 < 0\end{aligned}}$ .

— This answer is my guess. —

— Me@2018-11-22 12:00:30 PM

Physics Town

continues

Problem 14.5c2

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