# Textbook question - number of states

1. Jun 6, 2009

### trinitron

Hi all,

I've been going through Becker and Schwarz this summer and I'm a bit stuck at the moment. I'm not sure if this is better suited for the homework forum, so feel free to move it.

For those that have the book, this is page 52, eq 2.143.

Given the number operator,

$$N = \sum_{i=1}^{D-2}\sum_{n=1}^\infty \alpha^i_{-n}\alpha^i_{n}$$

one can calculate the number of physical states of mass n-1 by calculating the coefficient of w^n-1 in the expansion of trace(w^N). That much I understand; however, there are a couple of things which seem unclear in the calculation of trace(w^N).

The book writes,

$$tr w^N = \prod_{n=1}^\infty\prod_{i=1}^{D-2} tr w^{\alpha_{-n}^i \alpha_{n}^i} = \prod_{n=1}^\infty (1-w^n)^{-24}$$

Thinking about the trace as a sum of eigenvalues, pulling the products out of the trace in the first equality seems to amount to changing the order of a sum and product (sum of eigenvalues and product over {i,n}). More directly, tr(AB)!=tr(A)tr(B). So what makes this equality valid?

I'm also having trouble with the second equality. It seems that the claim is $$tr w^{\alpha_{-n}^i\alpha_n^i} = w^n + w^{2n} + w^{3n} + \dots$$.

However, to me it seems that $$tr w^{\alpha_{-n}^i\alpha_n^i} = \infty w^n + \infty w^{2n} + \infty w^{3n} + \dots$$

Because if I want a state $$|\phi>$$ s.t. $$\alpha_{-n}^i\alpha_n^i |\phi> = m\times n|\phi>$$, I can simply take $$|0,k>$$ and act on it with $$m$$ copies of $$\alpha_{-n}^i$$ and an arbitrary number of $$\alpha_{-k}^j$$ for k!=n.

Any help would be greatly appreciated.