Quick Calculation 14.6

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.

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\alpha' M^2 = 2~~~\to~~~N^\perp = 3

The first possibility group includes the cases of one a_3^\dagger. Since there are 24 species, this group has 24 possibilities.

The second possibility group includes the cases of one a_1^\dagger with one a_2^\dagger. (Their order is not important because interchanging the order would create the same state anyway.) Since there are 24 species for each of a_1^\dagger and a_2^\dagger, this group has 24 \times 24 possibilities.

The third possibility group includes the cases of three a_1^\dagger‘s. We need to further divide this group into 3 sub-groups.

For the first sub-group, three a_1^\dagger‘s are all different:

a_{1A}^\dagger a_{1B}^\dagger a_{1C}^\dagger

Since there are 24 species for each a_1^\dagger, there should be 24 \times 23 \times 22 possibilities.

However, since their order is not important, there are repetitions among those possibilities. To eliminate repetitions, we divide the number with 3!. So, this subgroup has

\frac{24 \times 23 \times 22}{6}

possibilities.

For the second sub-group, two of the three a_1^\dagger‘s are identical:

a_{1A}^\dagger a_{1B}^\dagger a_{1B}^\dagger

There should be 24 \times 24 possibilities.Me@2018-04-20 12:18:38 PM

There should be 24 \times 23 possibilities.

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For the third sub-group, all three are identical. In other words, there are 24 possibilities.

— Me@2015-06-04 10:19:30 PM

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2015.06.04 Thursday (c) All rights reserved by ACHK