Quick Calculation 14.6b

A First Course in String Theory

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Construct explicitly all the states with \alpha' M^2=2 and count them, verifying that there are indeed a total of 3200 states. You may find the counting formula in Problem 12.11 useful.

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Let N(n, k) = {n + k - 1 \choose k - 1}, the number of ways to put n indistinguishable balls into k boxes.

p.318 “For open bosonic strings \alpha' M^2 = N^\perp - 1, …”

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When \alpha' M^2 = 2, N^\perp = 3, the cases are:

1. three a_1^\dagger‘s:

N(3,24) = 2600

2. one a_2^\dagger and one a_1^\dagger:

24 \times 24 = 576

3. one a_3^\dagger:

24

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Total number of possible states for N^\perp = 3 is 3200.

— Me@2018-04-20 02:45:35 PM

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2018.04.20 Friday (c) All rights reserved by ACHK