Problem 14.5b2

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

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

We have 32 zero modes $\displaystyle{\lambda_0^A}$ and 16 linear combinations behave as creation operators.

As usual half of the ground states have $\displaystyle{(-1)^{F_L} = +1}$ and the other half have $\displaystyle{(-1)^{F_L} = -1}$ .

Let $\displaystyle{|R_\alpha \rangle_L}$ denote ground states with $\displaystyle{(-1)^{F_L} = +1}$ .

How many ground states $\displaystyle{|R_\alpha \rangle_L}$ are there?

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— This answer is my guess. —

$\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}$

p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states. Postulating a unique vacuum $\displaystyle{|0 \rangle}$ , the creation operators allow us to construct $\displaystyle{16 = 2^4}$ degenerate Ramond ground states.”

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 $\displaystyle{|R_\alpha \rangle_L}$ .

— This answer is my guess. —

— Me@2018-10-29 03:11:07 PM

Physics Town

continues

Problem 14.5b2

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