# Problem 14.1a

A First Course in String Theory

14.1 Counting bosonic states

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$n$ is the number of $a$‘s.

$k$ is the number of different $a$‘s.

For $a^{i_1} a^{i_2}$, $n=2$.

(Indices $i_1$ and $i_2$ are not powers. Instead, they are just upper indices for representing different $a$‘s.)

When all $a$‘s commute, $a^1 a^2$ and $a^2 a^1$, for example, represent the same state. So we have to avoid double-counting, except for the same $a$‘s states, such as $a^3 a^3$.

By direct counting, without using the formula, the number of products of the form $a^{i_1} a^{i_2}$ can be built is

$\frac{k(k-1)}{2} + k$
$= \frac{(k + 1)k}{2}$

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

By using the formula, the number of products of the form $a^{i_1} a^{i_2}$ can be built is

$N(2,k)$

$= \frac{(2+k-1)!}{2!(k-1)!}$

$= \frac{(k+1)k}{2}$

— Me@2015-07-26 08:43:22 AM