Quick Calculation 14.6

A First Course in String Theory

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.

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

— Me@2015-06-04 10:19:30 PM

Physics Town

continues

Quick Calculation 14.6

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