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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 $(2.3.120)$?

ADDENDUM

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 $n^{-27/4}$.

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1 Answer 1

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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.)

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  • $\begingroup$ 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 $\endgroup$
    – Sanjana
    Commented Mar 12 at 19:10
  • $\begingroup$ The lectures indeed give more details. $\endgroup$
    – Qmechanic
    Commented Mar 13 at 8:56

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