# 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.

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

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