The stringy scaling loop expansion and stringy scaling violation

In contrast to the finite number of coupling vertices in field theory, there are infinite $n$-point coupling vertices with arbitrary $n$ in string theory due to the infinite number of particles in the string spectrum. This makes the calculation of $n$-point string scattering amplitudes ($SSA$) with $n\geq 5$ much more complicated. Indeed, as was shown by the present authors recently that [1, 2, 3, 4] only $4$-point $SSA$ can be expressed in terms of finite number of terms of Lauricella $SSA$ ($LSSA$). Higher $n$-point $SSA$ with $n\geq 5$ contain infinite number of terms of $LSSA$ [5].

On the other hand, for the case of $4$-point $SSA$, it has long been known that all $4$-point $\left\langle V_{1}V_{2}V_{3}V_{4}\right\rangle$ hard $SSA$ ($E\rightarrow\infty$, fixed $\phi$) of different string states at each fixed mass level of $V_{j}$ ($j=1,2,3,4$) vertex share the same functional form [6, 7, 8, 9]. See the reviews [10, 11]. That is, all $4$-point hard $SSA$ ($HSSA$) at each fixed mass level are proportional to each other with constant ratios [12, 13, 14, 15] (independent of the scattering angle $\phi$, or the deficit of the kinematics variable dim$\mathcal{M}=1$). Moreover, the present authors discovered recently that the reduction of both the number of kinematics variable dependence on the ratios and the number of independent $HSSA$ for the $4$-point $HSSA$ can be generalized to arbitrary $n$-point $HSSA$ with $n\geq 5$ [17, 16]. As an example, all $6$-point $HSSA$ at each fixed mass level are related to each other and the ratios are [17]

$$\frac{\mathcal{T}^{\left(\left\{p_{1},p_{2},p_{3}\right\},m,q\right)}}{\mathcal{T}^{\left(\left\{0,0,0\right\},0,0\right)}}=\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M_{2}}\right)^{2m+q}\left(\cos\theta_{1}\right)^{p_{1}}\left(\sin\theta_{1}\cos\theta_{2}\right)^{p_{2}}\left(\sin\theta_{1}\sin\theta_{2}\right)^{p_{3}}$$

(1.1)

where the number of kinematics variables reduced from $8$ to $2$, $\theta_{1}$ and $\theta_{2}$, and dim$\mathcal{M}=8-2=6$.

These stringy scaling behaviors are reminiscent of Bjorken scaling [18] and the Callan-Gross relation [19] in deep inelastic scattering of electron and proton in the quark-parton model of $QCD$ where, to the leading order in energy, the two structure functions $W_{1}(Q^{2},\nu)$ and $W_{2}(Q^{2},\nu)$ scale, and become not functions of $2$ kinematics variables $Q^{2}$ and $\nu$ independently but only of their ratio $Q^{2}/\nu$. The number of independent kinematics variables thus reduces from $2$ to $1$, or the deficit dim$\mathcal{M}=2-1=1$. That is, the structure functions scale as [18]

$$MW_{1}(Q^{2},\nu)\rightarrow F_{1}(x),\text{ \ \ }\nu W_{2}(Q^{2},\nu)\rightarrow F_{2}(x)\text{ (dim}\mathcal{M}=1\text{)}$$

(1.2)

where $x$ is the Bjorken variable and $M$ is the proton mass. Moreover, due to the spin-$\frac{1}{2}$ assumption of quark, Callan and Gross derived the relation [19]

$$2xF_{1}(x)=F_{2}(x).$$

(1.3)

One easily sees that Eq.(1.1) is the stringy generalization of $QCD$ scaling in Eq.(1.2) and Eq.(1.3). The next interesting issue is then to understand the possible next to leading order stringy scaling violation, similar to the $QCD$ corrections of Bjorken scaling or Bjorken scaling violation through GLAP equation [20, 21] or current algebra.

To compare and make an anology between the stringy scaling and the Bjorken scaling, we give a table for the two behaviors:

Bosonic string	QCD
$SL(K+3,C)$	$SU(3)$
UV soft (exponential fall-off)	Asymptotic freedom
Nambu-Goto string model	Quark-parton model
Stringy scaling	Bjorken scaling
Stringy scaling loop expansion (stringy scaling violation)	GLAP Eq. (Bjorken scaling violation).

Note that it was shown recently that all $n$-point SSA ($n\geq 4$) of the open bosonic string theory can be expressed in terms of the Lauricella functions and form representation of the exact $SL(K+3,C)$ symmetry group [4]. On the other hand, it is interesting to see that while the stringy scaling behavior was recognized only very recently, historically, the Bjorken scaling was proposed before the invention of the idea of parton model, and the discovery of asymptotic freedom was also motivated by the proposal of Bjorken scaling.

To uncover the issue once and for all, in this paper we propose a systematic approximation scheme to calculate general $n$-point $HSSA$ order by order. We will show that the stringy scaling loop expansion scheme we proposed corresponds to finite number of vacuum diagram terms (even for $n\geq 5$) at each order of scattering energy due to a vacuum diagram contraint and a topological graph constraint. Comparing to the traditional effective action calculation for each loop diagram with infinite number of external legs in field theory, finite number of vacuum diagrams without external legs are much more easier to deal with.

In addition, we give the vacuum diagram representation and its Feynman rules for each term in the expansion of the $HSSA$. In general, there can be many vacuum diagrams, connected and disconnected, corresponds to one term in the expansion. In particular, we match coefficient of each term with sum of the inverse symmetry factors [22] corresponding to all diagrams of the term. As an application to extending our previous calculation of $n$-point leading order stringy scaling behavior, we explicitly calculate some examples of $4$-point next to leading order stringy scaling violation terms.

This paper is organized as following. In the next section, we begin with the stringy scaling loop expansion of the $4$-point $HSSA$. We will calculate in details the functional form and coefficient of each term in the expansion. Moreover, we give Feynman rules of vacuum diagram representation for each term in the expansion. In section III and IV, we generalize the calculation to the $5$-point and general $n$-point $HSSA$ respectively. In section V, we demonstrate explicitly how to draw all the vacuum diagram representation, connected and disconnected, for each term of the expansion. In particular, we will sum over the inverse symmetry factors of all diagrams of the term to consistently match with the coefficient of the term. In section VI, we use the results of section II to calculate some examples of $4$-point next to leading order stringy scaling violation terms.. A brief conclusion is given in section VII.

It can be exprecitly demonstrated that [5] the $t-u$ channel of all $4$-point $SSA$ with four arbitrary tensor states can be written as the following integral form (after $SL(2,R)$ fixing) [5]

$$\mathcal{T}(\Lambda)=\int_{1}^{\infty}dx\mbox{ }u(x)e^{-\Lambda f(x)},$$

(2.1)

where $\Lambda\equiv-(1,2)=-k_{1}\cdot k_{2}$ . The explicit calculation of $4$-point $SSA$ with four arbitrary tensor states can be found in [5]. As a simple example, the $HSSA$ of three tachyons and one high energy state $v_{2}$ [8, 9]

$$\left|N,2m,q\right\rangle=\left(\alpha_{-1}^{T}\right)^{N-2m-2q}\left(\alpha_{-1}^{L}\right)^{2m}\left(\alpha_{-2}^{L}\right)^{q}\left|0;k\right\rangle$$

(2.2)

with $e^{P}=\frac{1}{M_{2}}(E_{2},\mathrm{k}_{2},\vec{0})=\frac{k_{2}}{M_{2}}$ the momentum polarization, $e^{L}=\frac{1}{M_{2}}(\mathrm{k}_{2},E_{2},\vec{0})$ the longitudinal polarization and the transverse polarization $e^{T}=(0,0,1)$ can be written as [8, 9]

	$\displaystyle\mathcal{T}^{(N,2m,q)}$	$\displaystyle=\int_{1}^{\infty}dxx^{(1,2)}(1-x)^{(2,3)}\left[\frac{e^{T}\cdot k_{1}}{x}-\frac{e^{T}\cdot k_{3}}{1-x}\right]^{N-2m-2q}$
		$\displaystyle\cdot\left[\frac{e^{P}\cdot k_{1}}{x}-\frac{e^{P}\cdot k_{3}}{1-x}\right]^{2m}\left[-\frac{e^{P}\cdot k_{1}}{x^{2}}-\frac{e^{P}\cdot k_{3}}{(1-x)^{2}}\right]^{q},$		(2.3)

which can then be put into the form in Eq.(2.1) with

	$\displaystyle\Lambda$	$\displaystyle\equiv-(1,2)\rightarrow\frac{s}{2}\rightarrow 2E^{2},$		(2.4)
	$\displaystyle\tau$	$\displaystyle\equiv-\frac{(2,3)}{(1,2)}\rightarrow-\frac{t}{s}\rightarrow\sin^{2}\frac{\phi}{2},$		(2.5)
	$\displaystyle f(x)$	$\displaystyle\equiv\ln x-\tau\ln(1-x),$		(2.6)
	$\displaystyle u(x)$	$\displaystyle\equiv\left[\frac{(1,2)}{M_{2}}\right]^{2m+q}(1-x)^{-N+2m+2q}(f^{\prime})^{2m}(f^{\prime\prime})^{q}(-e^{T}\cdot k_{3})^{N-2m-2q}.$		(2.7)

In general for four arbitrary string states, we can expand the amplitude in Eq.(2.1) around the saddle point for large $\Lambda$ to obtain

	$\displaystyle\mathcal{T}(\Lambda)$	$\displaystyle=\int_{1}^{\infty}dx\text{ }u\left(x\right)e^{-\Lambda f\left(x\right)}$
		$\displaystyle=\int_{1}^{\infty}dx\left(\sum_{p=0}\frac{u_{0}^{\left(p\right)}}{p!}\left(x-x_{0}\right)^{p}\right)e^{-\Lambda f_{0}-\frac{1}{2}\Lambda f_{0}^{\prime\prime}\left(x-x_{0}\right)^{2}-\Lambda\sum_{j=3}\frac{f_{0}^{\left(j\right)}}{j!}\left(x-x_{0}\right)^{j}}$
		$\displaystyle=e^{-\Lambda f_{0}}\int_{1}^{\infty}dx\left(\sum_{p=0}\sum_{q=0}\frac{\left(-\Lambda\right)^{q}u_{0}^{\left(p\right)}}{i!p!}\left(x-x_{0}\right)^{p}\left[\sum_{j=3}\frac{f_{0}^{\left(j\right)}}{j!}\left(x-x_{0}\right)^{j}\right]^{q}\right)e^{-\frac{1}{2}\Lambda f_{0}^{\prime\prime}\left(x-x_{0}\right)^{2}}.$		(2.8)

Let’s rewrite the bracket term in the last line of the above equation as

	$\displaystyle\left[\sum_{j=3}\frac{f_{0}^{\left(j\right)}}{j!}\left(x-x_{0}\right)^{j}\right]^{q}$	$\displaystyle=\left[\sum_{n_{1}=3}a_{n_{1}}z^{n_{1}}\right]\left[\sum_{n_{2}=3}a_{n_{2}}z^{n_{2}}\right]\cdots\left[\sum_{n_{q}=3}a_{n_{q}}z^{n_{q}}\right]$
		$\displaystyle=\sum_{n_{1},\cdots n_{q}=3}a_{n_{1}}\cdots a_{n_{q}}z^{\sum_{r=1}^{q}n_{r}}=\sum_{n_{1},\cdots n_{q}=0}a_{n_{1}+3}\cdots a_{n_{q}+3}z^{\sum_{r=1}^{q}(n_{r}+3)}.$		(2.9)

Inserting Eq.(2.9) into Eq.(2.8), and using the Gaussian integral

$$\int_{-\infty}^{\infty}dzz^{2n}e^{-\frac{1}{2}z^{2}}=\sqrt{2\pi}\frac{(2n)!}{2^{n}n!}$$

(2.10)

to perform the integration, we obtain

	$\displaystyle\mathcal{T}(\Lambda)$	$\displaystyle=\sqrt{\frac{2\pi}{\Lambda f_{0}^{\prime\prime}}}e^{-\Lambda f_{0}}\sum_{q=0}\sum_{n_{1},\cdots n_{q}=0}\sum_{p=0}\frac{(-)^{q}(2M+2q)!}{q!2^{M+q}(M+q)!}$
		$\displaystyle\cdot\frac{u_{0}^{(p)}f_{0}^{(n_{1}+3)}\cdots f_{0}^{(n_{q}+3)}}{p!(n_{1}+3)!\cdots(n_{q}+3)!(f_{0}^{\prime\prime})^{M+q}}\frac{1}{\Lambda^{M}}$
		$\displaystyle\equiv\sqrt{\frac{2\pi}{\Lambda f_{0}^{\prime\prime}}}e^{-\Lambda f_{0}}\left[\mathcal{A}(\Lambda^{0})+\frac{1}{\Lambda}\mathcal{A}(\Lambda^{-1})+\frac{1}{\Lambda^{2}}\mathcal{A}(\Lambda^{-2})+O\left(\frac{1}{\Lambda^{M}}\right)\right]$		(2.11)

where

$$2M=p+\sum_{r=1}^{q}(n_{r}+1)\geq 0.$$

(2.12)

In Eq.(2.12), $M$, $p$, $q$ and $n_{r}$ are nonnegative integers. It is important to note that for a given inverse energy order $\frac{1}{\Lambda^{M}}$, there are only finite number of terms in Eq.(2.11) due to the condition in Eq.(2.12). We can now explicitly calculate $\mathcal{A}(\Lambda)$ in Eq.(2.11) order by order.

For the leading order $M=0$, we have $p=0$, $q=0$ and there is no $n_{r}$. The amplitude is

$$\mathcal{A}(\Lambda^{0})=u_{0}.$$

(2.13)

For the next to leading order $M=1$, there are $4$ terms:

	$\displaystyle\mathcal{A}_{1}(\Lambda^{-1})$	$\displaystyle=-\frac{u_{0}f_{0}^{(4)}}{8(f_{0}^{\prime\prime})^{2}},\text{\ }(p=0,q=1,n_{1}=1)$		(2.14)
	$\displaystyle\mathcal{A}_{2}(\Lambda^{-1})$	$\displaystyle=\frac{5u_{0}(f_{0}^{(3)})^{2}}{24(f_{0}^{\prime\prime})^{3}},\text{ \ }(p=0,q=2,n_{1}=n_{2}=0)$		(2.15)
	$\displaystyle\mathcal{A}_{3}(\Lambda^{-1})$	$\displaystyle=-\frac{u_{0}^{\prime}f_{0}^{(3)}}{2(f_{0}^{\prime\prime})^{2}},\text{\ }(p=1,q=1,n_{1}=0)$		(2.16)
	$\displaystyle\mathcal{A}_{4}(\Lambda^{-1})$	$\displaystyle=\frac{u_{0}^{\prime\prime}}{2f_{0}^{\prime\prime}}.\text{\ }(p=2,q=0)$		(2.17)

For the next next to leading order $M=2$, there are $12$ terms:

	$\displaystyle\mathcal{A}_{1}(\Lambda^{-2})$	$\displaystyle=-\frac{u_{0}f_{0}^{(6)}}{48(f_{0}^{\prime\prime})^{3}},\text{\ }(p=0,q=1,n_{1}=3)$		(2.18)
	$\displaystyle\mathcal{A}_{2}(\Lambda^{-2})$	$\displaystyle=\frac{7u_{0}f_{0}^{(3)}f_{0}^{(5)}}{48(f_{0}^{\prime\prime})^{4}},\text{ \ }(p=0,q=2,(n_{1},n_{2})=(2,0)\text{ or }(0,2))$		(2.19)
	$\displaystyle\mathcal{A}_{3}(\Lambda^{-2})$	$\displaystyle=\frac{35u_{0}(f_{0}^{(4)})^{2}}{384(f_{0}^{\prime\prime})^{4}},\text{\ }(p=0,q=2,n_{1}=n_{2}=1)$		(2.20)
	$\displaystyle\mathcal{A}_{4}(\Lambda^{-2})$	$\displaystyle=-\frac{35u_{0}f_{0}^{(4)}(f_{0}^{(3)})^{2}}{64(f_{0}^{\prime\prime})^{5}},\text{\ }(p=0,q=3,(n_{1},n_{2},n_{3})=(1,0,0)\text{ or permutation})$		(2.21)
	$\displaystyle\mathcal{A}_{5}(\Lambda^{-2})$	$\displaystyle=\frac{385u_{0}(f_{0}^{(3)})^{4}}{1152(f_{0}^{\prime\prime})^{6}},\text{\ }(p=0,q=4,n_{1}=n_{2}=n_{3}=n_{4}=0)$		(2.22)
	$\displaystyle\mathcal{A}_{6}(\Lambda^{-2})$	$\displaystyle=-\frac{u_{0}^{{}^{\prime}}f_{0}^{(5)}}{8(f_{0}^{\prime\prime})^{3}},\text{\ }(p=1,q=1,n_{1}=2)$		(2.23)
	$\displaystyle\mathcal{A}_{7}(\Lambda^{-2})$	$\displaystyle=\frac{35u_{0}^{{}^{\prime}}f_{0}^{(3)}f_{0}^{(4)}}{48(f_{0}^{\prime\prime})^{4}},\text{\ }(p=1,q=2,(n_{1},n_{2})=(1,0)\text{ or }(0,1))$		(2.24)
	$\displaystyle\mathcal{A}_{8}(\Lambda^{-2})$	$\displaystyle=-\frac{35u_{0}^{{}^{\prime}}(f_{0}^{(3)})^{2}}{48(f_{0}^{\prime\prime})^{5}},\text{\ }(p=1,q=3,n_{1}=n_{2}=n_{3}=0)$		(2.25)
	$\displaystyle\mathcal{A}_{9}(\Lambda^{-2})$	$\displaystyle=-\frac{5u_{0}^{{}^{\prime\prime}}f_{0}^{(4)}}{16(f_{0}^{\prime\prime})^{3}},\text{\ }(p=2,q=1,n_{1}=1)$		(2.26)
	$\displaystyle\mathcal{A}_{10}(\Lambda^{-2})$	$\displaystyle=\frac{35u_{0}^{\prime\prime}(f_{0}^{(3)})^{2}}{48(f_{0}^{\prime\prime})^{4}},\text{\ }(p=2,q=2,n_{1}=n_{2}=0)$		(2.27)
	$\displaystyle\mathcal{A}_{11}(\Lambda^{-2})$	$\displaystyle=-\frac{5u_{0}^{(3)}f_{0}^{(3)}}{12(f_{0}^{\prime\prime})^{3}},\text{\ }(p=3,q=1,n_{1}=0)$		(2.28)
	$\displaystyle\mathcal{A}_{12}(\Lambda^{-2})$	$\displaystyle=\frac{u_{0}^{(4)}}{8(f_{0}^{\prime\prime})^{2}}\text{\ }.(p=4,q=0,\text{ No }n_{r})$		(2.29)

To study the general higher order amplitudes, we note that a typical term at each order $\Lambda^{-M}$ in the expansion of Eq.(2.11) can be written as

$$\mathcal{A}(\Lambda^{-M})\sim\frac{\left(2P\right)!}{P!2^{P}}\frac{1}{m!}\left[\underset{n\geq 3}{\prod}\frac{1}{\left(-n!\right)^{V\left(n\right)}V\left(n\right)!}\right]\cdot\frac{u_{0}^{\left(m\right)}\underset{n\geq 3}{\prod}\left(f_{0}^{\left(n\right)}\right)^{V\left(n\right)}}{\left(f_{0}^{\left(2\right)}\right)^{P}}.$$

(2.30)

The rules (corresponding to symmetry factors of Feynman rules in field theory, see section V for more details) to assign constant factors in the bracket of Eq.(2.30) are

	$\displaystyle u_{0}^{\left(m\right)}$	$\displaystyle\Rightarrow\frac{1}{m!},$		(2.31)
	$\displaystyle\underset{n\geq 3}{\prod}\left(f_{0}^{\left(n\right)}\right)^{V\left(n\right)}$	$\displaystyle\Rightarrow\underset{n\geq 3}{\prod}\frac{1}{\left(-n!\right)^{V\left(n\right)}V\left(n\right)!},$		(2.32)
	$\displaystyle\left(f_{0}^{\left(2\right)}\right)^{P}$	$\displaystyle\Rightarrow\frac{\left(2P\right)!}{P!2^{P}}=\left(2P-1\right)!!.$		(2.33)

Note that the factor in Eq.(2.33) can be interpreted as the coefficient of $x_{2}^{P}$ term in the expansion of the incomplete Bell polynomials $B_{n,k}$ $(x_{1},x_{2},\cdots,x_{n-k+1})$ with $n=2P$ and $k=P$ since there are $P$ propagators each with $2$ end points. We have verified coefficients of all terms in Eq.(2.14) to Eq.(2.17) calculated previously in $\mathcal{A}_{j}(\Lambda^{-1})$ and all terms in Eq.(2.18) to Eq.(2.29) calculated in $\mathcal{A}_{j}(\Lambda^{-2})$ by using Eq.(2.30).

It is remarkable that each typical term in Eq.(2.30) corresponds to (at least) one vacuum Feynman diagram (no external legs). Here we list the rules regarding the expansion and the construction of a vacuum diagram corresponds to the typical term in Eq.(2.30):

• •

  $V\left(n\right)$ $n$-vertex $\sim\left(f_{0}^{\left(n\right)}\right)^{V\left(n\right)}$ for $n\geq 3$,

• •

  $P$ propagators $\sim\left(f_{0}^{(2)}\right)^{P}$,

• •

  a loop with $m$ legs $\sim$ $u_{0}^{\left(m\right)}$ ( if $m=0$, $u_{0}$ will be treated as a disconnedted loop),

• •

  $M=$ # of loops $-$ # of the connected components $\geq 1$,

• •

  Note that some terms in Eq.(2.30) can correspond to more than one diagram. However, for each order of $M$, there are only finite number of terms (diagrams) in the stringy scaling loop expansion scheme.

The constraints for the parameters are

	$\displaystyle P-\sum_{n=3}V\left(n\right)$	$\displaystyle=M\text{,}$		(2.34)
	$\displaystyle m+\sum_{n=3}nV\left(n\right)$	$\displaystyle=2P\Rightarrow\text{vacuum diagram.}$		(2.35)

Note that Eq.(2.34) can be read from Eq.(2.11), and Eq.(2.35) is equivalent to Eq.(2.12). On the other hand, Eq.(2.34) means that $M$ is the difference between the number of $f_{0}^{\left(n\right)}$ in the numerator and the number of $f_{0}^{(2)}$ in the denominator, and Eq.(2.35) means that the number of differentiations of $f$ in the numerator equals to the number of differentiations in the denominator. We will see that Eq.(2.34) and Eq.(2.35) give a vacuum diagram representation for each term in Eq.(2.30). While Eq.(2.35) gives the vacuum diagram condition, topologically, Eq.(2.34) follows from the Euler characteristics $\chi(\mathbb{M})$ with dim$\mathbb{M}=1$

$$\chi(\mathbb{M})=\sum_{n=3}V\left(n\right)-P=-M$$

(2.36)

where the number of the $n$-vertex is $V\left(n\right)$, the number of edges is $P$ and the number of faces of the $1D$ graph manifold $\mathbb{M}$ is zero. Indeed, for this case, the Euler characteristics can also be written as

$$\chi(\mathbb{M})=b_{0}-b_{1}=-M$$

(2.37)

where $b_{j}$ is the $j$th Betti number of $\mathbb{M}$. Here $b_{0}$ counts the number of the connected components of the diagram and $b_{1}$ counts the total number of loops of the diagram.

Eliminating $P$ from the above constraints Eq.(2.34) and Eq.(2.35), we obtain the following equation

$$\sum_{n=3}^{2l}\left(n-2\right)V\left(n\right)=2M-m\geq 0.$$

(2.38)

For a given integer $M\geq 1$,

$$m=0,1,\cdots,2M.$$

(2.39)

One can solve all non-negative integer solutions for $V\left(n\right)$ with $n\geq 3$ in Eq.(2.38).

For the $\Lambda^{-1}$ order, i.e. $M=1$, we get

V(3)0210V(4)1000⇒4​ terms\begin{tabular}[c]{|c|c|c|c|c|}\hline\cr$m$&$0$&$0$&$1$&$2$\\ \hline\cr$V\left(3\right)$&$0$&$2$&$1$&$0$\\ \hline\cr$V\left(4\right)$&$1$&$0$&$0$&$0$\\ \hline\cr\end{tabular}\Rightarrow 4\text{ terms}

		m	0	0	1	2		(2.40)

as expected from the previous calculation.

For the $\Lambda^{-2}$ order, i.e.$M=2$, we get

$$\begin{tabular}[c]{|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline\cr$m$&$0$&$0$&$0$&$0$&$0$&$1$&$1$&$1$&$2$&$2$&$3$&$4$\\ \hline\cr$V\left(3\right)$&$0$&$4$&$1$&$2$&$0$&$3$&$0$&$1$&$2$&$0$&$1$&$0$\\ \hline\cr$V\left(4\right)$&$0$&$0$&$0$&$1$&$2$&$0$&$0$&$1$&$0$&$1$&$0$&$0$\\ \hline\cr$V\left(5\right)$&$0$&$0$&$1$&$0$&$0$&$0$&$1$&$0$&$0$&$0$&$0$&$0$\\ \hline\cr$V\left(6\right)$&$1$&$0$&$0$&$0$&$0$&$0$&$0$&$0$&$0$&$0$&$0$&$0$\\ \hline\cr\end{tabular}\Rightarrow 12\text{ terms}$$

(2.41)

as expected from the previous calculation.

For the higher order amplitudes, the total number of terms are

$$\begin{tabular}[c]{|c|c|c|c|c|c|c|c|c|c|c|}\hline\cr$M$&$1$&$2$&$3$&$4$&$5$&$6$&$7$&$8$&$9$&$\cdots$\\ \hline\cr$\#$ of terms&$4$&$12$&$30$&$67$&$139$&$272$&$508$&$915$&$1597$&$\cdots$\\ \hline\cr\end{tabular}.$$

(2.42)

On the other hand, for a given $M$, we can count the number of terms for each $m$

$$\begin{tabular}[c]{|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline\cr$m$&$0$&$1$&$2$&$3$&$4$&$5$&$6$&$7$&$8$&$9$&$10$&total\\ \hline\cr$M=1$&$2$&$1$&$1$&&&&&&&&&$4$\\ \hline\cr$M=2$&$5$&$3$&$2$&$1$&$1$&&&&&&&$12$\\ \hline\cr$M=3$&$11$&$7$&$5$&$3$&$2$&$1$&$1$&&&&&$30$\\ \hline\cr$M=4$&$22$&$15$&$11$&$7$&$5$&$3$&$2$&$1$&$1$&&&$67$\\ \hline\cr$M=5$&$42$&$30$&$22$&$15$&$11$&$7$&$5$&$3$&$2$&$1$&$1$&$139$\\ \hline\cr\end{tabular}\ \ \ .$$

(2.43)

We observe from the above table that the distribution on $m$ for a given $M$

$$1,1,2,3,5,7,11,15,22,30,42,\cdots$$

(2.44)

can be generated by the generating function

	$\displaystyle\prod\limits_{n=1}^{\infty}\left(1-q^{n}\right)^{-1}$	$\displaystyle=1+q+2q^{2}+3q^{3}+5q^{4}+7q^{5}+11q^{6}+15q^{7}$
		$\displaystyle+22q^{8}+30q^{9}+42q^{10}+56q^{11}+77q^{12}+\cdots$
		$\displaystyle=\sum_{n=0}^{\infty}P(n)q^{n}$		(2.45)

which is the inversed Dedekind eta function. It corresponds to the scalar partition function on the torus containing the information of the number of states at each energy level or character of a conformal family. $P(n)$ in Eq.(2.45) is the number of ways of writing $n$ as a sum of positive integer. From Eq.(2.38), we easily see that the numer of terms $N_{(M,m)}$ for given $M$ and $m$ presented in Eq.(2.43) is

$$N_{(M,m)}=P(2M-m).$$

(2.46)

The $5$-point $SSA$ can be written in the following integral form (after $SL(2,R)$ fixing)

$$\mathcal{T}(\Lambda)=\int dx_{2}dx_{3}\text{ }u\left(x_{2},x_{3}\right)e^{-\Lambda f\left(x_{2},x_{3}\right)},\text{ }\Lambda=-k_{1}\cdot k_{2},$$

(3.1)

where

$$f\left(x_{2},x_{3}\right)=-\frac{k_{1}\cdot k_{2}}{\Lambda}\ln x_{2}-\frac{k_{1}\cdot k_{3}}{\Lambda}\ln x_{3}-\frac{k_{2}\cdot k_{3}}{\Lambda}\ln\left(x_{3}-x_{2}\right)-\frac{k_{2}\cdot k_{4}}{\Lambda}\ln\left(1-x_{2}\right)-\frac{k_{3}\cdot k_{4}}{\Lambda}\ln\left(1-x_{3}\right).$$

(3.2)

Since we are going to use the Gaussian approximation and perform the integration of Eq.(3.1) by Eq.(2.10), for the time being, we will ignore the range of integration in Eq.(3.1).

As a simple example, for the $5$-point $HSSA$ with $4$ tachyons and $1$ high energy state at mass level $M^{2}=2(N-1)$

$$\left|p_{1},p_{2};2m,2q\right\rangle=\left(\alpha_{-1}^{T_{1}}\right)^{N+p_{1}}\left(\alpha_{-1}^{T_{2}}\right)^{p_{2}}\left(\alpha_{-1}^{L}\right)^{2m}\left(\alpha_{-2}^{L}\right)^{q}\left|0;k\right\rangle$$

(3.3)

where $p_{1}+p_{2}=-2(m+q)$ with two transverse directions $T_{1}$ and $T_{2}$, $u\left(x_{2},x_{3}\right)$ can be calculated to be

$$u\left(x_{2},x_{3}\right)=\left(k^{T_{1}}\right)^{N+p_{1}}\left(k^{T_{2}}\right)^{p_{2}}\cdots\left(k^{T_{r}}\right)^{p_{r}}\left(k^{L}\right)^{2m}\left(k^{\prime L}\right)^{q},$$

(3.4)

where we have defined

$$k=-\frac{k_{i}}{x_{1}-x_{2}}-\frac{k_{i}}{x_{3}-x_{2}}-\frac{k_{i}}{x_{4}-x_{2}}.$$

(3.5)

We perform Taylor expansions on the saddle point $\left(x_{20},x_{30}\right)$ of both the $u$ and $f$  functions to obtain

	$\displaystyle\mathcal{T}(\Lambda)$	$\displaystyle=\int dx_{2}dx_{3}\text{ }\left[u\left(x_{20},x_{30}\right)+\cdots\right]$
		$\displaystyle\cdot e^{-\Lambda\left[f\left(x_{20},x_{30}\right)+\frac{1}{2}\frac{\partial^{2}f\left(x_{2},x_{3}\right)}{\partial x_{2}^{2}}\left(x_{2}-x_{20}\right)^{2}+\frac{\partial^{2}f\left(x_{2},x_{3}\right)}{\partial x_{2}\partial x_{3}}\left(x_{2}-x_{20}\right)\left(x_{3}-x_{30}\right)+\frac{\partial^{2}f\left(x_{2},x_{3}\right)}{\partial x_{3}^{2}}\left(x_{3}-x_{30}\right)^{2}+\cdots\right]}$		(3.6)

where $\left(x_{20},x_{30}\right)$ satisfies

$$\frac{\partial f\left(x_{20},x_{30}\right)}{\partial x_{2}}=0,\text{ \ \ }\frac{\partial f\left(x_{20},x_{30}\right)}{\partial x_{3}}=0.$$

(3.7)

We observe that in the Taylor expansion of the function $f$ , there are crossing terms such as $\frac{\partial^{2}f\left(x_{2},x_{3}\right)}{\partial x_{2}\partial x_{3}}$ which involves $(x_{2}-x_{20})(x_{3}-x_{30})$. These crossing terms will result in an infinite number of terms at each order $\Lambda$ of expansion of $\mathcal{T}(\Lambda)$ in the limit as $\Lambda\rightarrow\infty$ . Therefore, we need to do a change of variables here to eliminate these crossing terms and obtain

$$\mathcal{T}(\Lambda)=\int dx_{2}^{\prime}dx_{3}^{\prime}\text{ }\left[u\left(x_{20}^{\prime},x_{30}^{\prime}\right)+\cdots\right]e^{-\Lambda\left[f\left(x_{20}^{\prime},x_{30}^{\prime}\right)+\frac{1}{2}\frac{\partial^{2}f\left(x_{2}^{\prime},x_{3}^{\prime}\right)}{\partial x_{2}^{\prime 2}}\left(x_{2}^{\prime}-x_{20}^{\prime}\right)^{2}+\frac{1}{2}\frac{\partial^{2}f\left(x_{2}^{\prime},x_{3}^{\prime}\right)}{\partial x_{3}^{\prime 2}}\left(x_{3}^{\prime}-x_{30}^{\prime}\right)^{2}+\cdots\right]}$$

(3.8)

where $\left(x_{20}^{\prime},x_{30}^{\prime}\right)$ satisfies

$$\frac{\partial f\left(x_{20}^{\prime},x_{30}^{\prime}\right)}{\partial x_{2}^{\prime}}=0,\text{ \ \ }\frac{\partial f\left(x_{20}^{\prime},x_{30}^{\prime}\right)}{\partial x_{3}^{\prime}}=0.$$

(3.9)

Let’s define the coefficients in the Taylor expansion of $f$ and $u$ at $\left(x_{20}^{\prime},x_{30}^{\prime}\right)$ as follows:

	$\displaystyle u\left(x_{20}^{\prime},x_{30}^{\prime}\right)$	$\displaystyle=u_{0},$		(3.10)
	$\displaystyle\frac{\partial^{m_{2}+m_{3}}u\left(x_{20}^{\prime},x_{30}^{\prime}\right)}{\partial\left(x_{2}^{\prime}\right)^{m_{2}}\partial\left(x_{3}^{\prime}\right)^{m_{3}}}$	$\displaystyle=u_{0}^{\left(m_{2},m_{3}\right)},$		(3.11)
	$\displaystyle\frac{\partial^{n_{2}+n_{3}}f\left(x_{20}^{\prime},x_{30}^{\prime}\right)}{\partial\left(x_{2}^{\prime}\right)^{n_{2}}\partial\left(x_{3}^{\prime}\right)^{n_{3}}}$	$\displaystyle=f_{0}^{\left(n_{2},n_{3}\right)}.$		(3.12)

We can then simplify the integral into the following form

$$\mathcal{T}(\Lambda)=\int dx_{2}^{\prime}dx_{3}^{\prime}\text{ }\left[u_{0}+\cdots\right]e^{-\Lambda\left[f_{0}+\frac{1}{2}f_{0}^{\left(2,0\right)}\left(x_{2}^{\prime}-x_{20}^{\prime}\right)^{2}+\frac{1}{2}f_{0}^{\left(0,2\right)}\left(x_{3}^{\prime}-x_{30}^{\prime}\right)^{2}+\cdots\right]}.$$

(3.13)

Expanding the integral up to the second order in $\Lambda$, we obtain the following expression:

$$\mathcal{T}(\Lambda)=\sqrt{\frac{2\pi}{\Lambda f_{0}^{\left(2,0\right)}}}\sqrt{\frac{2\pi}{\Lambda f_{0}^{\left(0,2\right)}}}\left[u_{0}+\frac{1}{\Lambda}\mathcal{B}(\Lambda^{-1})+\frac{1}{\Lambda^{2}}\mathcal{B}\left(\Lambda^{-2}\right)+O\left(\frac{1}{\Lambda^{3}}\right)\right]$$

(3.14)

where

	$\displaystyle\mathcal{B}(\Lambda^{-1})$	$\displaystyle=\frac{3}{8}\frac{u_{0}\left(f_{0}^{\left(2,1\right)}\right)^{2}}{\left(f_{0}^{\left(2,0\right)}\right)^{2}\left(f_{0}^{\left(0,2\right)}\right)}+\frac{3}{8}\frac{u_{0}\left(f_{0}^{\left(1,2\right)}\right)^{2}}{\left(f_{0}^{\left(2,0\right)}\right)\left(f_{0}^{\left(0,2\right)}\right)^{2}}+\frac{5}{24}\frac{u_{0}\left(f_{0}^{\left(3,0\right)}\right)^{2}}{\left(f_{0}^{\left(2,0\right)}\right)^{3}}+\frac{5}{24}\frac{u_{0}\left(f_{0}^{\left(0,3\right)}\right)^{2}}{\left(f_{0}^{\left(0,2\right)}\right)^{3}}$		(3.15)
		$\displaystyle-\frac{1}{4}\frac{u_{0}\left(f_{0}^{\left(2,2\right)}\right)}{\left(f_{0}^{\left(2,0\right)}\right)\left(f_{0}^{\left(0,2\right)}\right)}-\frac{1}{8}\frac{u_{0}\left(f_{0}^{\left(4,0\right)}\right)}{\left(f_{0}^{\left(2,0\right)}\right)^{2}}-\frac{1}{8}\frac{u_{0}\left(f_{0}^{\left(0,4\right)}\right)}{\left(f_{0}^{\left(0,2\right)}\right)^{2}}+\frac{1}{4}\frac{u_{0}\left(f_{0}^{\left(2,1\right)}\right)\left(f_{0}^{\left(0,3\right)}\right)}{\left(f_{0}^{\left(2,0\right)}\right)\left(f_{0}^{\left(0,2\right)}\right)^{2}}\frac{1}{4}\frac{u_{0}\left(f_{0}^{\left(1,2\right)}\right)\left(f_{0}^{\left(3,0\right)}\right)}{\left(f_{0}^{\left(2,0\right)}\right)^{2}\left(f_{0}^{\left(0,2\right)}\right)}$		(3.16)
		$\displaystyle-\frac{1}{2}\frac{u_{0}^{\left(1,0\right)}\left(f_{0}^{\left(1,2\right)}\right)}{\left(f_{0}^{\left(2,0\right)}\right)\left(f_{0}^{\left(0,2\right)}\right)}-\frac{1}{2}\frac{u_{0}^{\left(0,1\right)}\left(f_{0}^{\left(2,1\right)}\right)}{\left(f_{0}^{\left(2,0\right)}\right)\left(f_{0}^{\left(0,2\right)}\right)}-\frac{1}{2}\frac{u_{0}^{\left(1,0\right)}\left(f_{0}^{\left(3,0\right)}\right)}{\left(f_{0}^{\left(2,0\right)}\right)^{2}}-\frac{1}{2}\frac{u_{0}^{\left(0,1\right)}\left(f_{0}^{\left(0,3\right)}\right)}{\left(f_{0}^{\left(0,2\right)}\right)^{2}}$		(3.17)
		$\displaystyle+\frac{1}{2}\frac{u_{0}^{\left(2,0\right)}}{\left(f_{0}^{\left(2,0\right)}\right)}+\frac{1}{2}\frac{u_{0}^{\left(0,2\right)}}{\left(f_{0}^{\left(0,2\right)}\right)}.$		(3.18)

There are $15$ terms in $\mathcal{B}(\Lambda^{-1})$ above and $151$ terms in $\mathcal{B}\left(\Lambda^{-2}\right)$ calculated by direct expansion using Maple, which are consistent with the results we will calculate by hand in the following.