I have a slightly technical combinatorics question. Consider the degeneracy $D_n$ of bound states of $n$ D0 branes and one D4 brane. This is given in Polchinski by (13.6.24),

\begin{align} \sum_{n=0}^{\infty}q^n D_n=2^8\prod_{k=1}^{\infty}\left(\frac{1+q^k}{1-q^k}\right)^{8}. \end{align}

I was able to verify this up to $n=3$, basically you have to count all ways to form bound states of the D0 branes and then bound them to the D4 brane. However, I wasn't able to verify it in general, since brute forcing it is a bit of a mess. Is there some clever mathematical formalism that allows one to deal with combinatorical problems like this?


1 Answer 1


This formula is actually pretty simple to understand.

First, the $2^8$ is the number of possible $D4$ states. Then for each (indistinguishable) $D0$, they can be in either a fermionic or bosonic state, of which there are $8$ each.

Next, the coefficient of $q^n$ in $(1+q)^8$ is the number of ways for $n$ independent $D0$ branes to fit in $8$ fermionic states.

The coefficient of $q^n$ in $(1-q)^{-8}$ is the number of ways for $n$ independent $D0$ branes to fit in $8$ bosonic states.

We multiply these two to allow $D0$ branes to occupy either bosonic or fermionic states.

By taking the products over $q^k$, we allow $D0$ branes to first form $k$-tuply bound states which occupy a single $D0$ state.

  • $\begingroup$ Thanks @Ryan! This is very helpful. One minor correction: I think you meant (1-q)^(-8) for the bosonic counting instead of (1-q)^(8). $\endgroup$
    – Matthew
    Nov 22, 2013 at 4:57
  • 1
    $\begingroup$ Good catch! Yes I do. $\endgroup$ Nov 22, 2013 at 20:21

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