Problem 14.5d1.1.2

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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p.322

\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}

\displaystyle{\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2}

\displaystyle{\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2}

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The left NS’+ sector:

\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_{-\frac{1}{2}} x^{\frac{1}{2}} \right)^{32}   \left( 1 + \lambda_{-\frac{3}{2}} x^{\frac{3}{2}} \right)^{32}  ... \\  \end{aligned}}

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

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

\displaystyle{\begin{aligned}  &f_{L, NS'}(x) \\ &= a_{NS'} (r) x^r \\  &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\   \end{aligned}}

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“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}.

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\displaystyle{ \begin{aligned} &f_{L, NS'}(x) \\ \end{aligned}}

\displaystyle{ = \frac{1}{x}+\frac{32}{\sqrt{x}}+504+5248 \, \sqrt{x}+40996 \, x+258624 \, x^{\frac{3}{2}}+1384320 \, x^{2}+6512384 \, x^{\frac{5}{2}} + ...}

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Let

\displaystyle{ \begin{aligned}  &g (\sqrt{x}) \\  &= \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\   &= 1+32 \, \sqrt{x}+504 \, x+5248 \, x^{\frac{3}{2}}+40996 \, x^{2}+258624 \, x^{\frac{5}{2}}+1384320 \, x^{3}+6512384 \, x^{\frac{7}{2}} + ... \\   \end{aligned}}

Then

\displaystyle{ \begin{aligned}  &g (-\sqrt{x}) \\  &= \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\   &= 1 - 32 \sqrt{x} + 504 x - 5248 \, x^{\frac{3}{2}} + 40996 \, x^{2} - 258624 \, x^{\frac{5}{2}}+1384320 \, x^{3} - 6512384 x^{\frac{7}{2}} + ... \\   \\ \end{aligned}}

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\displaystyle{ \begin{aligned}  &f_{L, NS'+}(x) \\  &= \frac{1}{x} + 504 + 40996 x + 1384320 x^{2} + ... \\  &= \frac{1}{2x} \left[ g(\sqrt{x}) + g(-\sqrt{x}) \right] \\   &= \frac{1}{2x} \left[   \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8}   + \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8}   \right] \\   \end{aligned}}

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The left R’+ sector:

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

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— This answer is my guess. —

— Me@2019-01-14 04:28:10 PM

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2019.01.14 Monday (c) All rights reserved by ACHK