# Problem 14.5a4

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

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

The GSO projection here keeps the states with $\displaystyle{(-1)^{F_L} = + 1}$; this defines the left NS’+ sector.

Write explicitly and count the states we keep for the three lowest mass levels, indicating the corresponding values of $\displaystyle{\alpha' M_L^2}$. [This is a long list.]

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p.314 “Let us declare that number to be minus one, thus making the ground states fermionic:”

Equation (14.39): $\displaystyle{(-1)^F |NS \rangle \otimes |p^+, \overrightarrow{p}_T \rangle = - |NS \rangle \otimes |p^+, \overrightarrow{p}_T \rangle}$

Equation (14.40): $\displaystyle{(-1)^F |\lambda \rangle = -(-1)^{\sum_{r,J} \rho_{r,J}} |\lambda \rangle}$

p.315 “So all the states with integer $\displaystyle{N^{\perp}}$ have $\displaystyle{(-1)^F = -1}$; they are fermionic states.”

However, in this problem:

“The left NS’ sector is built with oscillators $\displaystyle{\bar \alpha_{-n}^I}$ and $\displaystyle{\lambda_{-r}^A}$ acting on the vacuum $\displaystyle{|NS' \rangle_L}$, declared to have $\displaystyle{(-1)^{F_L} = + 1}$:” $\displaystyle{(-1)^{F_L} |NS' \rangle_L = + |NS' \rangle_L}$

So all the states with integer $\displaystyle{N^{\perp}}$ have $\displaystyle{(-1)^F = +1}$.

— This answer is my guess. — \displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

If we define $N^\perp$ in the way similar to equation (14.37), we have \displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}} \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}}

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} \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&1 \\ \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&8 + \frac{32 \times 31}{2} = 504 \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\left( \frac{8 \times 8}{2} + \frac{8}{2} \right) = 36, 8 \times \left( \frac{32 \times 31}{2} \right) = 3968, 32 \times 32 = 1024, {32 \choose 4} = 35960 \\ \end{aligned}} \displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&N(2,8) = 36, 8 \times {32 \choose 2} = 3968, 32 \times 32 = 1024, {32 \choose 4} = 35960 \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~& 36 + 3968 + 1024 + 35960 = 40988 \\ \end{aligned}}

— This answer is my guess. —

— Me@2018-10-14 03:25:08 PM

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