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

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(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 and 16 linear combinations behave as creation operators.

As usual half of the ground states have and the other half have .

Let denote ground states with .

How many ground states are there?

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— This answer is my guess. —

p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states. Postulating a unique vacuum , the creation operators allow us to construct degenerate Ramond ground states.”

Following the same logic:

Postulating a unique vacuum , the creation operators allow us to construct degenerate Ramond ground states.

Therefore, there are ground states .

— This answer is my guess. —

— Me@2018-10-29 03:11:07 PM

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2018.10.29 Monday (c) All rights reserved by ACHK