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?

~~~

— 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

.

.

2018.10.29 Monday (c) All rights reserved by ACHK