Problem 14.5d2

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

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d) Write a generating function \displaystyle{f_L(x) = \sum_{r} a(r) x^r} for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where \displaystyle{a(r)} counts the number of states with \displaystyle{\alpha' M_L^2 = r}.

Use \displaystyle{f_L(x)} and an algebraic manipulator to find the total number of states in heterotic string theory at \displaystyle{\alpha' M_L^2 = 8}.

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— This answer is my guess. —

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The left R’+ sector:

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\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}}

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

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\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

\displaystyle{N^\perp:}

\displaystyle{\begin{aligned}  \left(  1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ...  \right)^8  \left(  1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ...  \right)^8  ...  \left( 1 + \lambda_{-1} x^{1} \right)^{32}  \left( 1 + \lambda_{-2} x^{2} \right)^{32}  ... \\  \end{aligned}}

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However, there are \displaystyle{2^{15}} ground states \displaystyle{|R_\alpha\rangle_L} and \displaystyle{2^{15}} ground states \displaystyle{|R_\alpha \rangle_R}:

\displaystyle{\begin{aligned}  (2^{15} + 2^{15})   \left[ \left(  1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ...  \right)^8  ...  \left( 1 + \lambda_{-1} x^{1} \right)^{32}   ... \right] \\   \end{aligned}}

\displaystyle{\begin{aligned}  2^{16} \prod_{r=1}^\infty  \frac{1}{(1 - x^r)^8}  (1 + x^{r})^{32} \\  \end{aligned}}

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“Keep only states with \displaystyle{(-1)^{F_L} = +1}; this defines the left R’+ sector.”

\displaystyle{\begin{aligned}  \frac{2^{16}}{2} \prod_{r=1}^\infty  \frac{(1 + x^{r})^{32}}{(1 - x^r)^8}   \\  \end{aligned}}

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\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

\displaystyle{\begin{aligned}  &f_{L, R'+}(x) \\ &= a_{R'+} (r) x^r \\ &= 2^{15} x \prod_{r=1}^\infty \frac{(1 + x^{r})^{32}}{(1 - x^r)^8} \\  & \\  &= 32768 \, x+1310720 \, x^{2}+27131904 \, x^{3}+387973120 \, x^{4}+4312727552 \, x^{5}+39739981824 \, x^{6} + ...   \end{aligned}}

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— This answer is my guess. —

— Me@2019-01-20 09:09:37 PM

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2019.01.20 Sunday (c) All rights reserved by ACHK