# How to get the factor of $n^{-27/4}$ in number of open string states from the calculation in GSW's book?

In section 2.3.5 of Green, Schwarz, Witten's book on string theory (volume-1) pp. 116-118, the objective is to calculate an Asymptotic Formula for Level Densities $$d_n$$ for open bosonic string theory. $$d_n$$ is related to the classical partition function $$f(w)$$ via the generating function $$G(w)=\sum_{n=0}^\infty d_n w^n$$ and $$G(w)=f(w)^{-24}.$$ Two estimates are found out in the book in equations $$(2.3.112)$$ and $$(2.3.117)$$ respectively which we state here for convenience

1. $$f(w) \sim \exp\left(-\frac{\pi^2}{6(1-w)}\right).\tag{2.3.112}$$
2. $$f(w) \sim (1-w)^{-1/2}\exp\left(-\frac{\pi^2}{6(1-w)}\right).\tag{2.3.117}$$

Using this and the integral formula for Laurent series coefficients $$d_n=\frac{1}{2 \pi i}\oint dw \frac{G(w)}{w^{n+1}} \tag{2.3.118}$$ one can find out $$d_n$$ after making a saddle point approximation ($$w \to 1$$ as $$n \to \infty$$) as explained in the book. The final result in eqn. $$(2.3.120)$$ is

$$d_n \sim n^{-27/4} \exp(4 \pi \sqrt{n}).\tag{2.3.120}$$

I am able to find out the exponential piece which comes directly from the exponential piece in $$f(w) \sim \exp\left(-\frac{\pi^2}{6(1-w)}\right)$$, but not the extra $$n^{-27/4}$$ factor.

I thought that maybe then the more accurate approximation $$(2.3.117)$$ is to be used. But that will change the saddle points and what I am finding is a factor of $$n^{-6}$$. Given that the approximation $$(2.3.117)$$ is not even mentioned in a newer book ("Becker-Becker-Schwarz" pg. 52 subsection "The number of states") by one of the authors seems to imply that the first approximation $$(2.3.112)$$ is used to obtain the correct $$n$$ factor. But how? So in a nutshell my question is: Which approximation formula is being used in the texts and how does it give a factor of $$n^{-27/4}$$ in $$(2.3.120)$$?

This is how I got the exponential piece: The integrand in $$(2.3.118)$$ can be rewritten using $$(2.3.112)$$ as $$\frac{G(w)}{w^{n+1}}=\frac{1}{w^{n+1}}f(w)^{-24}\sim \frac{1}{w^{n+1}}\exp\left(\frac{4 \pi^2}{1-w}\right)=\exp \left(-(n+1) \ln(w) \right)\exp\left(\frac{4 \pi^2}{1-w}\right)$$ $$\Rightarrow \frac{G(w)}{w^{n+1}}=\exp\left(\frac{4 \pi^2}{(1-w)}-(n+1) \ln(w)\ \right)$$ Now as $$w \sim 1$$, using Taylor expansion of $$\ln(1-w)$$ we can write $$-\ln(w)=-\ln(1-(1-w))\sim (1-w)$$ Therefore putting $$-\ln(w)$$ instead of $$(1-w)$$ in the first term of the argument of the exponential in the equation for $$G(w)/w^{n+1}$$, we get $$\frac{G(w)}{w^{n+1}}=\exp\left(-\frac{4 \pi^2}{\ln(w)}-(n+1) \ln(w)\ \right) \tag{2.3.119}$$ From this the saddle point of the integrand is easily evaluated by differentiating w.r.t $$w$$ and setting the result to zero which gives $$\ln(w) \sim -2 \pi/\sqrt{n}$$ as the saddle point, substituting which back to $$(2.3.119)$$ readily gives the exponential part of $$(2.3.120)$$. Doing a similar analysis using the approximation $$(2.3.117)$$ instead yields $$n^{-6}$$ instead of the expected $$n^{-27/4}$$.

TL;DR: The extra factor $$n^{-3/4}$$ comes from the Hessian/second-order derivative (i.e. the Gaussian integral) in the WKB/saddle point approximation.

In more detail,

\begin{align} d_n~\stackrel{(2.3.118)}{=}&\oint \frac{dw}{2 \pi i} \frac{f(w)^{-24}}{w^{n+1}} ~\stackrel{(2.3.117)}{=}~\oint \frac{dw}{2 \pi i} e^{-S(w)}\cr ~\stackrel{\rm WKB}{\sim}~&\frac{e^{-S(w_0)}}{i\sqrt{2\pi S^{\prime\prime}(w_0)}}~=~\frac{n^{-6}e^{4\pi\sqrt{n}}}{\sqrt{2n^{3/2}}}\cr ~=~~~&\frac{1}{\sqrt{2}}n^{-27/4}e^{4\pi\sqrt{n}} \quad{\rm for}\quad n~\to~\infty ,\end{align} where \begin{align} S(w)~\stackrel{(2.3.117)}{=}& -12\ln(1-w) -\frac{4\pi^2}{1-w}\cr & +(n+1)\ln(\underbrace{w}_{=1-(1-w)}),\cr 0~\approx~S^{\prime}(w)~=~~& \frac{12}{1-w} -\frac{4\pi^2}{(1-w)^2} +\frac{n+1}{w}\cr &\quad\Rightarrow\quad 1-w_0~\sim~2\pi n^{-1/2} \quad{\rm for}\quad n~\to~\infty,\cr S^{\prime\prime}(w)~=~~& \frac{12}{(1-w)^2} -\frac{8\pi^2}{(1-w)^3} -\frac{n+1}{w^2},\end{align} so that \begin{align} S(w_0)~\sim~&6\ln n -2\pi \sqrt{n} - n 2\pi n^{-1/2}~=~ 6\ln n -4\pi \sqrt{n},\cr S^{\prime\prime}(w_0)~\sim~&-\frac{n^{3/2}}{\pi} \quad{\rm for}\quad n~\to~\infty. \end{align}

References:

1. M.B. Green, J.H. Schwarz and E. Witten, Superstring theory, Vol. 1, 1986; subsection 2.3.5.

2. J. Bedford, arXiv:1107.3967; p. 65 eq. (A.22). (Hat tip: Sanjana.)

• Thanks. It seems that we have to use $(2.3.117)$ indeed and $(1-w) \sim -\ln(w)$ is to be used in the $(1-w)^{12}$ factor also. Got some help from Green's own lectures pg 61 Commented Mar 12 at 19:10
• The lectures indeed give more details. Commented Mar 13 at 8:56