Problem 14.5d2

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

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:

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

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

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

$\displaystyle{N^\perp:}$

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

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

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

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

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

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

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

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

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— This answer is my guess. —

— Me@2019-01-20 09:09:37 PM

Physics Town

continues

Problem 14.5d2

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