Problem 14.5c9

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

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c) Calculate the total number of states in heterotic string theory (bosons plus fermions) at \displaystyle{\alpha' M^2 =4}.

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— This answer is my guess. —

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spacetime bosons:

\displaystyle{NS'+ \otimes NS+}

\displaystyle{\begin{aligned}  \left(  \{ \bar \alpha_{-2}^I, \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  \right)  \otimes  \left(  \{ \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  \right) \end{aligned}}

\displaystyle{\begin{aligned}  I, J, I' &= 2, 3, ..., 9 \\  A, B, C, D &= 1, 2, ..., 32 \\  \end{aligned}}

Number of states:

Let \displaystyle{N(n, k) = {n + k - 1 \choose k - 1}}, the number of ways to put n indistinguishable balls into k boxes.

\displaystyle{\begin{aligned}  &\left( 8+ N(2,8) +8 \times {32 \choose 2} + 32^2 + {32 \choose 4} \right) \times \left( 8^2+8+{8 \choose 3} \right) \\  &= 40996 \times 128  \end{aligned}}

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\displaystyle{R'+ \otimes NS+}

\displaystyle{\begin{aligned}  \left(  |R_\alpha \rangle_L  \right)  \otimes  \left(  \{ \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  \right) \end{aligned}}

Following the same logic:

Postulating a unique vacuum \displaystyle{|0 \rangle}, the creation operators allow us to construct \displaystyle{2^{16}} degenerate Ramond ground states.

Therefore, there are \displaystyle{2^{15}} ground states \displaystyle{|R_\alpha \rangle_L}.

— Me@2018-10-29 03:11:07 PM

\displaystyle{\begin{aligned}  I, J, K &= 2, 3, ..., 9 \\  \end{aligned}}

Number of states:

\displaystyle{\begin{aligned}  &\left( 2^{15} \right) \times \left( 8^2+8+{8 \choose 3} \right) \\  &= 32768 \times 128  \end{aligned}}

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spacetime fermions:

\displaystyle{NS'+ \otimes R-}

\displaystyle{\begin{aligned}  \left(  \{ \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  \right)  \otimes  \left( \alpha_{-1}^{I'} |R_{a} \rangle,~d_{-1}^{I'} |R_{\bar a} \rangle \right)  \\ \end{aligned}}

Number of states:

\displaystyle{\begin{aligned}  &\left( 40996 \right) \times \left( 8^2 + 8^2 \right) \\  &= 40996 \times 128  \end{aligned}}

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\displaystyle{R'+ \otimes R-}

\displaystyle{\begin{aligned}  \left(  |R_\alpha \rangle_L  \right)  \otimes  \left( \alpha_{-1}^{I} |R_{a} \rangle,~d_{-1}^{I} |R_{\bar a} \rangle \right)  \\ \end{aligned}}

Number of states:

\displaystyle{\begin{aligned}  &\left( 2^{15}\right) \times \left( 8^2 + 8^2 \right) \\  &= 32768 \times 128  \end{aligned}}

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— This answer is my guess. —

— Me@2019-01-03 05:26:59 PM

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2019.01.03 Thursday (c) All rights reserved by ACHK