Supercritical linear dilaton capping state using Interpolating functions

Ritabrata Bhattacharya∗

∗Chennai Mathematical Institute,

H1, SIPCOT IT Park, Siruseri, Kelambakkam 603103, India

††footnotetext: electronic address<EMAIL_ADDRESS>
Abstract

We consider time dependent backgrounds with a time-like linear dilaton which
leads to a weakly coupled theory at assymptotic future known as the
Supercritical Linear Dilaton (SCLD) phase. Even after projecting out the
tachyon there are naive divergences in the partition function due to the
spectrum of relevant deformations. Although the partition function seems
divergent, the background is actually well defined in the far future due to
vanishing backreactions, provided we start from an appropriate initial state
called the “capping” state. We use the technique of interpolating functions to
construct this state by embedding the SCLD phase in a strong coupling
completion of string theory. Although we are unable construct a specific
example, we do provide the general strategy in cases where S-duality may
provide the completion.

###### Contents

  1. 1 Introduction
  2. 2 Known results and conventions to be used
    1. 2.1 Linear dilaton backgrounds
    2. 2.2 The one loop partition function
    3. 2.3 Modular Invariance
    4. 2.4 Mass renormalization in string pertubation theory
  3. 3 Two point amplitude at one loop with $Q\neq 0$
    1. 3.1 The real part
    2. 3.2 The imaginary part
  4. 4 Interpolating functions due to Sen
    1. 4.1 Example of the heterotic/type I duality
    2. 4.2 Application to the linear dilaton backgrounds
  5. 5 Discussions
  6. A Relevant correlators for $Q\neq 0$
  7. B Two point graviton one loop amplitude with $Q=0$

##  1 Introduction

Our universe as we know is an expanding one. So to describe a consistent
quantum theory of gravity at early times i.e. near the big bang singularity
one requires the solutions describing quantum gravity in time dependent
backgrounds. String theory in Linear Dilaton backgrounds provide a way to
tackle this problem in a somewhat systematic manner. In [17],[18],[19],[20] it
was noticed that time-like linear dilaton solutions of supercritical string
theories are strongly coupled at assymptotically early times and hence
difficult to solve. Although at assymptotically late times one has much more
control since there, the theories become weakly coupled and one can solve them
perturbatively. In this regard we should mention that we take the following
conventions for the dilaton $\Phi$ and the string coupling $g_{st}$,

$\Phi=-QX^{0}\ ,\quad g_{st}=e^{\langle\Phi\rangle}=e^{-QX^{0}}\ ,\ \
\text{with}\ \ Q>0\ .$

In [1], the authors found that in such cases the spectrum of relevant
deformations of the “matter CFT” has modes which grow in time although their
backreaction on the geometry is small at assymptotically late times due to the
small string coupling. The condition for the backreaction being small is
stated below but one should note that these are not the usual tachyons we are
familiar with. These modes were termed “Pseudo-tachyons” in the literature.
Tachyonic solutions also grow in time but they have large back reactions on
the geometry causing instabilities in the background. If we consider the
deformations,

$\int d^{2}z\ \sum_{i}\mu_{i}\mathcal{O}_{i}e^{\kappa_{i}X^{0}}$

with $\mathcal{O}_{i}$ having dimension $\Delta_{i}$. This is marginal if,

$\Delta_{i}+\frac{1}{4}\kappa_{i}(\kappa_{i}+2Q)=2\ \Rightarrow\
\kappa_{i}=-Q\pm\sqrt{Q^{2}-4(\Delta_{i}-2)}\ .$

For $2<\Delta_{i}<2+Q^{2}/4$ we see that the properly normalised relevant mode
goes like,

$\tilde{\lambda}=\frac{\lambda}{g_{st}}\sim e^{\pm
QX^{0}\sqrt{1-4(\Delta_{i}-2)/Q^{2}}}\ .$

So we see that one of the modes grow in time and these are the ones identified
as Pseudo-tachyons.

The condition that the pseudo-tachyons have small backreaction and hence
provide a well defined i.e. stable background, is that we begin with an
appropriate initial state. This is sometimes called a “capping” state. As per
current understanding, unambiguous description of such states is still an open
problem. Although some work has been done in this regard, where smooth
evolution from “nothing” was considered [24]. Basically the authors studied
the effect of closed string tachyon condensation on the spacetime with a
space-like singularity at the level of general relativity.

Another way to identify the “capping” state is to embed the Supercritical
Linear Dilaton(SCLD) phase in a strong coupling completion of string theory.
The goal of the current paper is to take this route for identifying the
capping state. First let us look at the obstacles we face in achieving this
goal, then we mention cases where there is a systematic way to circumvent
these problems to obtain a reasonably accurate result.

#### Obstacles

  1. 1.

The most obvious obstactle is that we do not know whether there exists a
consistent strong coupling completion for supercritical string theories.

  2. 2.

Even if we are able find certain examples where such completion is possible,
computing physical quantites relevant for the target spacetime is considerably
difficult since spacetime supersymmetry is broken due to the linear dilaton
which couples to the world-sheet curvature.
For example suppose we want to compute the renormalised mass of different
operators at a given mass level. In presence of supersymmetry it is sufficient
to compute the renormalised mass of one component of the supermultiplet or
superfield, the rest are fixed by supersymmetry. This will obviously not be
true if supersymmetry is broken.

  3. 3.

Full non-perturbative treatment of string theory is yet to be as well
understood as perturbative string theory which is why even if we resolve the
first obstacle, the resolution may be true only in some special cases.

We argue that if we have two supercritical theories which are S-dual to each
other describing the weak and strong coupling regime then we can attempt to
construct the capping state within reasonable accuracy using the technique of
interpolating functions. Although we fail to find a specific example and
produce numbers, it may help to know the strategy of constructing the capping
state in case we suceed in finding an example.

The technique of interpolating function involves computing the renormalized
mass function $F(g)$ as a perturbative expansion in string coupling. If we
have two theories at the weak and strong coupling end which are S-dual to each
other, then we have two functions $F_{W}$ and $F_{S}$ which have perturbative
expansions in the weak and strong coupling respectively at the two ends. We
will closely follow the approach described in [2] to find an interpolating
function $F_{m,n}(g)$ which matches $F_{W}$ upto $m$-th power in the weak
coupling expansion and $F_{S}$ upto the $n$-th power in the strong coupling
expansion. This interpolating function is supposed to give the renormalized
mass of the theory for the whole range of coupling to a reasonable accuracy.
Other approaches are also available in the literatre [25] and the Padé
approximant approach.

The rest of the paper is organised as follows. In Section 2 we discuss briefly
the known results and the conventions we use for linear dilaton backgrounds
and mass renormalization in string perturbation theory. We will also derive
the partition function for SCLD backgrounds and show that modular invarince
dictates that only certain spacetime dimensions are allowed, which is well
known. In Section 3 we compute the two point amplitude of a pseudo-tachyonic
mode on the torus, which gives the renormalized mass at one loop for this
mode. In Section 4 we describe strategy for determining the “capping” state
with the help of interpolating functions. And finally we conclude with some
discussions in Section 5.

##  2 Known results and conventions to be used

###  2.1 Linear dilaton backgrounds

In this section we give a brief review of known results in the case when a
dilaton which is linear in the string fields, couples to the world-sheet
curvature [5]. So,

$\phi=V_{\mu}X^{\mu}\ ,$ (1)

and the world-sheet Eucledian action for the strings is given by,

$S_{P}=\frac{1}{4\pi\alpha^{\prime}}\int
d^{2}x\sqrt{\gamma}\gamma^{ab}\partial_{a}X^{\mu}\partial_{b}X_{\mu}\ +\
\frac{1}{4\pi}\int d^{2}x\sqrt{\gamma}\phi R$ (2)

Varrying this action w.r.t the world-sheet metric $\gamma$ we obtain the
energy momentum tensor for this theory. We give the result here for the
holomorphic and anti-holomorphic part of the E.M. tensor,

$\displaystyle T(z)$ $\displaystyle=$
$\displaystyle-\frac{1}{\alpha^{\prime}}:\partial X^{\mu}\partial X_{\mu}:\ +\
V_{\mu}\partial^{2}X^{\mu}$ $\displaystyle\bar{T}(\bar{z})$ $\displaystyle=$
$\displaystyle-\frac{1}{\alpha^{\prime}}:\bar{\partial}X^{\mu}\bar{\partial}X_{\mu}:\
+\ V_{\mu}\bar{\partial}^{2}X^{\mu}\ .$ (3)

With this energy momentum tensor we want to find the modified central charge
as well as the modifed conformal weights of the vertex operators. Now we want
to work with time-like dilaton so we choose $V_{\mu}\equiv(Q,\vec{0})$ and
hence we end up with (space-time metric is the minkowski metric)

$T(z)=-\frac{1}{\alpha^{\prime}}:\partial X^{\mu}\partial X_{\mu}:\ -\
Q\partial^{2}X^{0}$ (4)

and similarly for $\bar{T}(\bar{z})$.

We want to compute the the $T(z)T(0)$ as well as the $T(z)e^{ik.X}$ OPE’s. We
know that now $\partial X^{0}$ is no longer a primary while $e^{-ik^{0}X^{0}}$
is still a primary. Looking only at the leading order divergences, we get the
following results111Here $D$ is the dimensionality of the target space-time.,

$\displaystyle T(z)T(0)$ $\displaystyle\sim$
$\displaystyle\frac{D-6\alpha^{\prime}Q^{2}}{2z^{4}}\ +\ ....$ (5)
$\displaystyle T(z)e^{ik.X}(0)$ $\displaystyle\sim$
$\displaystyle\left(\frac{\alpha^{\prime}k^{2}}{4}-\frac{iQk^{0}\alpha^{\prime}}{2}\right)\frac{e^{ik.X}}{z^{2}}\
+\ ....$ (6)

Thus we have $c=\bar{c}=D-6\alpha^{\prime}Q^{2}$. And the conformal dimension
of the vertex operator,

$h_{v}=\frac{\alpha^{\prime}k^{2}}{4}-\frac{iQk^{0}\alpha^{\prime}}{2}=\frac{\alpha^{\prime}}{4}\left[\vec{k}^{2}-(k^{0})^{2}-2ik^{0}Q)\right]$

###  2.2 The one loop partition function

Before computing the partition function for such theories we include world-
sheet supersymmetry so that in the case of $Q=0$ we have a GSO projected
spectrum of critical superstrings in 10 dimension where the tachyon is
projected out.

We use the known results for fermionic traces [6],

$\displaystyle Z^{\alpha}_{\
\beta}=\text{Tr}_{\alpha}\left[q^{H}\text{exp}(i\pi\beta
F)\right]=\frac{1}{\eta(\tau)}\vartheta\begin{bmatrix}\alpha/2\\\
\beta/2\end{bmatrix}(0|\tau)\ .$ (7)

Here $F$ is the world-sheet fermion number, $\alpha$ and $\beta$ take integer
values modulo 2. For $\alpha$ even we have trace over the NS and for odd we
have trace over R states.

The total central charge due to the bc and $\beta\gamma$ ghosts is given by,
$c_{g}=-26+11=-15$ and the matter CFT has central charge
$c_{m}=D-6\alpha^{\prime}Q^{2}+\frac{D}{2}$. So for the central charge for the
matter + ghost CFT to vanish we need,

$\frac{3D}{2}-6\alpha^{\prime}Q^{2}=15\ \ \Rightarrow\
D-4\alpha^{\prime}Q^{2}=10\ .$ (8)

For specificity we assume $D/2$ to be odd and we will GSO project onto even
world-sheet fermion number to remove the tachyon.

Now we have the following well results on the torus,

$\langle
c(w_{1})b(w_{2})\bar{c}(\bar{w}_{3})\bar{b}(\bar{w}_{4})\rangle_{T^{2}}=|\eta(\tau)|^{4}\
.$ (9)

The scalar partition function ($q=e^{2\pi i\tau}\ \Rightarrow\
q\bar{q}=e^{-4\pi\tau_{2}}$),

$\displaystyle Z_{s}(\tau)$ $\displaystyle=$
$\displaystyle(q\bar{q})^{-\frac{D-6\alpha^{\prime}Q^{2}}{24}}\text{Tr}\left(q^{L_{0}}\bar{q}^{\bar{L}_{0}}\right)$
(10) $\displaystyle Z_{s}(\tau)$ $\displaystyle=$ $\displaystyle
V_{D}(q\bar{q})^{-\frac{D-6\alpha^{\prime}Q^{2}}{24}}\int\frac{d^{D-1}k}{(2\pi)^{D-1}}\frac{dk^{0}}{2\pi}\text{exp}\left[-\pi\alpha^{\prime}\tau_{2}\big{(}\vec{k}^{2}-(k^{0})^{2}-2ik^{0}Q\big{)}\right]$
$\displaystyle\hskip 170.71652pt\times\prod_{\mu,n}\sum_{N_{\mu
n},\bar{N}_{\mu n}}q^{nN_{\mu n}}\bar{q}^{n\bar{N}_{\mu n}}$ $\displaystyle
Z_{s}(\tau)$ $\displaystyle=$ $\displaystyle
iV_{D}(q\bar{q})^{-\frac{D}{24}}(q\bar{q})^{\frac{\alpha^{\prime}Q^{2}}{4}}\int\frac{d^{D-1}k}{(2\pi)^{D-1}}\frac{d\omega}{2\pi}\text{exp}\left[-\pi\alpha^{\prime}\tau_{2}\big{(}\vec{k}^{2}+\omega^{2}+2\omega
Q\big{)}\right]$ $\displaystyle\hskip
170.71652pt\times\prod_{n}\frac{1}{(1-q^{n})^{D}(1-\bar{q}^{n})^{D}}$
$\displaystyle Z_{s}(\tau)$ $\displaystyle=$ $\displaystyle
iV_{D}|\eta(\tau)|^{-2D}\int\frac{d^{D-1}k}{(2\pi)^{D-1}}\frac{d\omega}{2\pi}\text{exp}\left[-\pi\alpha^{\prime}\tau_{2}\big{(}\vec{k}^{2}+\omega^{2}+2\omega
Q+Q^{2}\big{)}\right]$ (11)

Now just redifing the variable $\omega\rightarrow\omega+Q$ and integrating
over the momenta variables we get,

$Z_{s}(\tau)=iV_{D}\left((4\pi^{2}\alpha^{\prime}\tau_{2})^{-\frac{1}{2}}|\eta(\tau)|^{-2}\right)^{D}=iV_{D}(Z_{X}(\tau))^{D}$
(12)

From the fermionic traces we obtain after projecting onto even world-sheet
fermion number we get,

$Z_{\psi}^{+}=\frac{1}{2}\left(Z^{0}_{\ 0}(\tau)^{\frac{D}{2}-1}-Z^{0}_{\
1}(\tau)^{\frac{D}{2}-1}-Z^{1}_{\ 0}(\tau)^{\frac{D}{2}-1}-Z^{1}_{\
1}(\tau)^{\frac{D}{2}-1}\right)\ .$ (13)

Now we know that $Z^{1}_{\ 1}(\tau)=0$ so,

$Z_{\psi}^{+}=\frac{1}{2}\left(Z^{0}_{\ 0}(\tau)^{\frac{D}{2}-1}-Z^{0}_{\
1}(\tau)^{\frac{D}{2}-1}-Z^{1}_{\ 0}(\tau)^{\frac{D}{2}-1}\right)\ .$ (14)

Thus the full partition function is given by222Here we consider the case of
Type IIB theories although generalization to other cases is simple.,

$Z_{LD}=\int\frac{d\tau
d\bar{\tau}}{4\tau_{2}}iV_{D}\left((4\pi^{2}\alpha^{\prime}\tau_{2})^{-\frac{1}{2}}|\eta(\tau)|^{-2}\right)^{D}|\eta(\tau)|^{4}Z_{\psi}^{+}(\tau)Z_{\bar{\psi}}^{+}(\bar{\tau})$
$\Rightarrow Z_{LD}=iV_{D}\int\frac{d\tau
d\bar{\tau}}{16\pi^{2}\alpha^{\prime}\tau_{2}^{2}}\left((4\pi^{2}\alpha^{\prime}\tau_{2})^{-\frac{1}{2}}|\eta(\tau)|^{-2}\right)^{D-2}Z_{\psi}^{+}(\tau)Z_{\bar{\psi}}^{+}(\bar{\tau})$
(15)

###  2.3 Modular Invariance

While testing the modular invariance of the above partition function one
immediately notices that for arbitrary values of $D>10$ we do not have modular
invariance i.e. even though the measure $d\tau d\bar{\tau}/\tau_{2}^{2}$ and
$Z_{X}(\tau)$ are modular invariant by themselves
$Z^{+}_{\psi}Z^{+}_{\bar{\psi}}$ break this invarince. But one can show
explicitly that for $D=10+16n$ with integer values of $n$,333In [1] this
result was stated and also this result was argued by K. Narayan and
collaborators in an unpublished notes. $Z^{+}_{\psi}Z^{+}_{\bar{\psi}}$ is
modular invariant owing to the following transformation laws,

$\displaystyle Z^{+}_{\psi}(-1/\tau)=Z^{+}_{\psi}(\tau)$ ; $\displaystyle
Z^{+}_{\psi}(\tau+1)=\exp\left(\frac{2\pi(2n+1)i}{3}\right)Z^{+}_{\psi}(\tau)$
(16) $\displaystyle
Z^{+}_{\bar{\psi}}(-1/\bar{\tau})=Z^{+}_{\bar{\psi}}(\bar{\tau})$ and
$\displaystyle
Z^{+}_{\bar{\psi}}(\bar{\tau}+1)=\exp\left(-\frac{2\pi(2n+1)i}{3}\right)Z^{+}_{\bar{\psi}}(\bar{\tau})$
(17)

Hence for the rest of this paper we take $D=10+16n$. Thus we have including
(8),

$4\alpha^{\prime}Q^{2}=16n\ ,\quad\Rightarrow\quad
Q=2\sqrt{\frac{n}{\alpha^{\prime}}}\ .$ (18)

So we see that due to the modular invarince we have,

$Z_{LD}=iV_{D}\int_{\mathbb{F}_{0}}\frac{d\tau
d\bar{\tau}}{16\pi^{2}\alpha^{\prime}\tau_{2}^{2}}\left(Z_{X}(\tau)\right)^{8+16n}Z_{\psi}^{+}(\tau)Z_{\bar{\psi}}^{+}(\bar{\tau})$
(19)

where, $\mathbb{F}_{0}$ denotes the fundamental domain in the complex $\tau$
upper half plane. Notice that only for $n=0$ i.e. the critical case $Z_{LD}$
vanishes by the use of Riemann Identity. For $n\neq 0$ the partition function
is non vanishing which implies that linear dilaton backgrounds break spacetime
supersymmetry.

###  2.4 Mass renormalization in string pertubation theory

We know that one has to start with the BV master action when attempting an
off-shell formulation of string theory [7],[8],[9],[10],[11],[12],[15]. It is
then clear that if one is able to solve the resulting equation of motion
pertubatively, then there exists a well defined procedure to compute several
physical quantities such as mass renormalization [13],[14], vacuum shift [16]
etc. order by order in perturbation theory. From [14] we learn that using the
1PI effective action to some given order in pertubation theory, we have to sum
over all the 1PR diagrams for the two point function to get the renormalised
mass at that order in the pertubation expansion. The full propagator is given,
in accordance with the above statement by,

$\Pi=\Delta+\Delta\hat{\mathcal{F}}\Delta+\Delta\hat{\mathcal{F}}\Delta\hat{\mathcal{F}}\Delta+....=\Delta(1-\hat{\mathcal{F}}\Delta)^{-1}=(1-\Delta\hat{\mathcal{F}})^{-1}\Delta\
.$ (20)

Here $\hat{\mathcal{F}}$ denotes the contribution from the Riemann surfaces
that are 1PI in the external momenta. And,

$\Delta=\frac{1}{2(L_{0}+\bar{L}_{0})}\delta_{L_{0},\bar{L}_{0}}$

gives the tree level propagator. If $\mathcal{F}$ is the full off-shell two
point funtion then,

$\Pi=\Delta+\Delta\mathcal{F}\Delta.$ (21)

Now suppose that we project on to a given mass level $m$, if the full off-
shell two point function restricted to this level be denoted by $F_{T}$ then
we have,

$\tilde{F}_{T}=F_{T}(1+(k^{2}+m^{2})^{-1}\mathcal{I}F_{T})^{-1}$ (22)

with,

$\mathcal{I}=\begin{pmatrix}&I&\\\ I&&\\\ &&I\end{pmatrix}\ .$

The propagator restricted to the modified physical sector is given by,

$\mathcal{V}=U(k)(k^{2}+m^{2}-\tilde{F}_{d}(k))^{-1}U(k)^{\dagger},\
\text{with}\ \tilde{F}(k)=U(k)\tilde{F}_{d}(k)U(k)^{\dagger}\ \text{and}\
\tilde{F}_{\alpha\beta}(k)=\langle\bar{\alpha}|\tilde{F}_{T}|\bar{\beta}\rangle_{p}.$

So we see that $\tilde{F}(k)$ is the renormalized mass matrix for the physical
sector. On diagonalizing we will get the renormalized masses for each physical
state. As an exercise we work out the mass renormalization of the massless
states like the graviton, in critical superstring theory i.e. $Q=0$ case in
Appendix B. For the critical case, the two point graviton amplitude should
vanish since we expect general co-ordinate invarince in the quantum theory of
superstrings as well. So the massless states remain massless under quantum
corrections.

##  3 Two point amplitude at one loop with $Q\neq 0$

In case of $Q\neq 0$ i.e. the linear dilaton background with world sheet
supersymmetry, instead of (3) we get the EM tensor and the world-sheet
supercurrent given by,

$\displaystyle T_{B}(z)$ $\displaystyle=$
$\displaystyle-\frac{1}{\alpha^{\prime}}:\partial X^{\mu}\partial X_{\mu}:\ +\
V_{\mu}\partial^{2}X^{\mu}\ -\ \frac{1}{2}\psi^{\mu}\partial\psi_{\mu}$ (23)
$\displaystyle T_{F}(z)$ $\displaystyle=$ $\displaystyle
i(2/\alpha^{\prime})^{1/2}\psi^{\mu}\partial X_{\mu}\ -\
i(2\alpha^{\prime})^{1/2}V_{\mu}\partial\psi^{\mu}$ (24)

With these results we can straight forwardly compute the BRST current,

$Q_{B}=\frac{1}{2\pi i}\oint(dzj_{B}(z)\ -\ d\bar{z}\bar{j}_{B}(\bar{z}))$
(25)

where,

$j_{B}=cT_{B}+\gamma T_{F}+bc\partial c+\frac{3}{4}\partial
c\beta\gamma+\frac{1}{4}c\partial(\beta\gamma)-c\beta\partial\gamma-b\gamma^{2}$
(26)

and the same expression with antiholomorphic fields for the antiholomorphic
part.

Let us first find the equation of motion for the first excited level with this
BRST current. The on shell condition for an operator $\Phi$, as always is
given by, $Q_{B}\Phi=0$. Since the tachyon has been projected out the first
excited level gives the lowest energy i.e. the ground state. The vertex
operator for the NS-NS state in the -1 picture number is given by,

$\bar{c}cV_{-1}(k,z,\bar{z})=\bar{c}c\
e^{-(\phi+\bar{\phi})}\zeta_{\mu\nu}\psi^{\mu}\bar{\psi}^{\nu}e^{ik.X}(z,\bar{z})$
(27)

Until the next subsection, we concentrate on the holomorphic part only since
the anti-holomorphic part works out in a similar fashion. So applying the BRST
charge we get,

$\displaystyle Q_{B}(c\ e^{-\phi}\zeta_{\mu\nu}\psi^{\mu}e^{ik.X}(0,0))$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi i}\oint dz\ j_{B}(z)c\
e^{-\phi}\zeta_{\mu\nu}\psi^{\mu}e^{ik.X}(0,0)$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi i}\oint dz\ \frac{(\partial
c)c(0)e^{-\phi}\zeta_{\mu\nu}\psi^{\mu}e^{ik.X}(0,0)}{z}\left(\frac{\alpha^{\prime}k^{2}}{4}+\frac{iV.k\alpha^{\prime}}{2}+\frac{1}{2}\right)$
$\displaystyle+$ $\displaystyle\frac{1}{2\pi i}\oint dz\
\left(\frac{\alpha^{\prime}}{2}\right)^{1/2}\frac{\eta(0)c(0)e^{ik.X}(0,0)}{z}\zeta_{\mu\nu}(k^{\mu}+2iV^{\mu})$
$\displaystyle+$ $\displaystyle\frac{1}{2\pi i}\oint dz\
\left(-\frac{1}{2}\right)\frac{(\partial
c)c(0)e^{-\phi}\zeta_{\mu\nu}\psi^{\mu}e^{ik.X}(0,0)}{z}\ +\ ....\ .$ (29)

The $....$ denote terms which are regular in $z$ and hence they vanish under
the closed contour integral. Just to be precise, the first two terms are the
contributions due to $cT_{B}$ and $\gamma T_{F}$ while the third term is the
single pole contribution due to the ghost part. Adding up all the contribution
we get the on-shell conditions,

$k^{2}+2iV.k=0\ ,\quad\text{and}\quad\zeta_{\mu\nu}(k^{\mu}+2iV^{\mu})=0\ .$
(30)

Notice that if we take $V_{\mu}\rightarrow 0$ we get back the expected on-
shell conditions for the massless states in critical superstring theory. For
time like $V$ we have,

$k^{2}-2iQk^{0}=0\ \Rightarrow\ k^{2}=2iQk^{0}\
,\quad\text{and}\quad\zeta_{\mu\nu}k^{\mu}=2iQ\zeta_{0\nu}\ .$ (31)

From the above equation (31) we see that,

$k^{2}-2iQk^{0}=0\ \Rightarrow\ -(k^{0}+iQ)^{2}+|\vec{k}|^{2}=Q^{2}\
\Rightarrow\ -M^{2}=Q^{2}$ (32)

i.e. the tree level mass squared go as $-Q^{2}$.

In what follows we take $Q$ to be large. We will work with the convention of
$\alpha^{\prime}=2$. Thus (18) becomes

$Q=\sqrt{2n}\quad\Rightarrow\ \text{$n\rightarrow\infty$ as
$Q\rightarrow\infty$.}$ (33)

While computing the two point amplitude on the torus we will focus on the
terms that have the highest power of $n$ or equivalently $Q$. We will first
determine the 0-picture vertex operators and determine first the real part and
then the imaginary part of the two-point amplitude.

From the -1 picture vertex in (27) and the PCO,

$\chi(z)=e^{\phi}T_{F}(z)+c\partial\xi(z)+\frac{1}{4}\partial b\eta
e^{2\phi}(z)+\frac{1}{4}b(2\partial\eta e^{2\phi}+\eta\partial(e^{2\phi}))(z)\
,$ (34)

we get for the holomorphic part,

$\displaystyle\chi(z)cV_{-1}(k,0,0)$ $\displaystyle=$ $\displaystyle
zT_{F}(z)c\zeta_{\mu}\psi^{\mu}e^{ik.X}(0,0)\ +\
z\partial\xi(z)e^{-\phi}(0)c\zeta_{\mu}\psi^{\mu}e^{ik.X}(0,0)$ (35)
$\displaystyle-$
$\displaystyle\frac{1}{4}z^{2}.\frac{1}{z^{2}}\eta(z)e^{\phi}(0)c\zeta_{\mu}\psi^{\mu}e^{ik.X}(0,0)\
+\
\frac{1}{4}2z.\frac{1}{z}\eta(z)e^{\phi}(0)c\zeta_{\mu}\psi^{\mu}e^{ik.X}(0,0)$
$\displaystyle+$
$\displaystyle\frac{1}{2}z^{2}.\frac{1}{z}\partial\eta(z)e^{\phi}(0)c\zeta_{\mu}\psi^{\mu}e^{ik.X}(0,0)\
.$

For the coefficient of $z$ we need the single pole contribution for a finite
answer in the $z\rightarrow 0$ limit. While for the $z^{0}$ coefficient we
need the constant regular piece. Thus in the $z\rightarrow 0$ limit we have,

$V_{0}(k,0,0)=\zeta_{\mu}\left(i\partial
X^{\mu}-\psi^{\mu}(k.\psi)\right)e^{ik.X}(0,0)+\frac{1}{4}\eta
e^{\phi}(0)\zeta_{\mu}\psi^{\mu}e^{ik.X}(0,0)\ .$

The last term above does not conserve the picture number inside the two point
one loop amplitude and hence we can drop it. We see that we reach at the same
result as in the $Q=0$ case, but now with on shell conditons (31). And
similarly for the anti-holomorphic part.

Now we can compute the two-point amplitide. As in the case with $Q=0$, we have
the full amplitude given by,

$\mathcal{M}_{g}=g_{s}^{2}\int_{\mathbb{F}_{0}}\frac{d\tau
d\bar{\tau}}{4\tau_{2}}\int d^{2}z\ \langle
b\bar{b}\bar{c}c(0)V_{0}(k_{1},0,0)V_{0}(k_{2},z,\bar{z})\rangle\ .$ (36)

###  3.1 The real part

Let us first turn our attention to the real part of the correlator.

$\displaystyle\left[\int d^{2}z\
\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\left\langle\left\\{\partial
X^{\mu}(0)\bar{\partial}X^{\nu}(0)-\psi^{\mu}(0)(k_{1}.\psi(0))\bar{\psi}^{\nu}(0)(k_{1}.\bar{\psi}(0))\right\\}\right.\right.$
$\displaystyle\left.\left.\times\\{\partial
X^{\rho}(z)\bar{\partial}X^{\sigma}(\bar{z})-\psi^{\rho}(z)(k_{2}.\psi(z))\bar{\psi}^{\sigma}(\bar{z})(k_{2}.\bar{\psi}(\bar{z}))\\}e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\right\rangle\right.$
$\displaystyle\left.-\int d^{2}z\
\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\left\langle\left\\{\psi^{\mu}(0)(k_{1}.\psi(0))\bar{\partial}X^{\nu}(0)+\partial
X^{\mu}(0)\bar{\psi}^{\nu}(0)(k_{1}.\bar{\psi}(0))\right\\}\right.\right.$
$\displaystyle\left.\left.\times\\{\psi^{\rho}(z)(k_{2}.\psi(z))\bar{\partial}X^{\sigma}(\bar{z})+\partial
X^{\rho}(z)\bar{\psi}^{\sigma}(\bar{z})(k_{2}.\bar{\psi}(\bar{z}))\\}e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\right\rangle\right]\
.$ (37)

We will look at the different terms that contribute to this correlator and
focus our attention to the terms that have the highest power of $n$.

We will be using the result,

$\displaystyle\langle e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle$
$\displaystyle=$ $\displaystyle
iV_{D}(Z_{X}(\tau))^{D}(2\pi)^{D}\delta^{D}(k_{1}+k_{2})\left|\vartheta_{1}(z,\tau)\exp\left(-\frac{\pi(\text{Im}\
z)^{2}}{\tau_{2}}\right)\right|^{\alpha^{\prime}k_{1}.k_{2}}$
$\displaystyle\times\prod_{i=1}^{2}\left|\frac{2\pi}{\vartheta_{1}^{\prime}(0,\tau)}\right|^{-\frac{\alpha^{\prime}k_{i}^{2}}{2}}\
.$ (38)

Note: For linear dilaton backgrounds with charge $Q$, the correlation function
of tachyon vertices yields, $\delta^{D}(k_{1}+k_{2}+2Q(1-h))$ with $h$ being
the genus of the Riemann surface we consider [4]. In our case we have the
torus with $h=1$. Hence the factor of $\delta^{D}(k_{1}+k_{2})$, which forces
$k_{2}=-k_{1}$.

Thus applying (31), (33) and $\alpha^{\prime}=2$ we have,

$\displaystyle\langle e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle$
$\displaystyle=$ $\displaystyle
i(2\pi)^{D}\delta^{D}(k_{1}+k_{2})\mathcal{C}_{X}^{T^{2}}\left|\frac{2\pi\vartheta_{1}(z,\tau)}{\vartheta^{\prime}_{1}(0,\tau)}\exp\left(-\frac{\pi(\text{Im}\
z)^{2}}{\tau_{2}}\right)\right|^{-4iQk_{1}^{0}}.$ (39) $\displaystyle=$
$\displaystyle
i(2\pi)^{D}\delta^{D}(k_{1}+k_{2})\mathcal{C}_{X}^{T^{2}}\left|\frac{2\pi\vartheta_{1}(z,\tau)}{\vartheta^{\prime}_{1}(0,\tau)}\exp\left(-\frac{\pi(\text{Im}\
z)^{2}}{\tau_{2}}\right)\right|^{8n}$

With $\mathcal{C}_{X}^{T^{2}}=V_{D}(Z_{X}(\tau))^{D}$. Since we are computing
everything in a Lorentz’s invariant manner we can go to the rest frame of the
pseudo-tachyonic mode for which,

$(k_{1}^{0})^{2}=M^{2}=-Q^{2}\quad\Rightarrow\quad k_{1}^{0}=iQ=i\sqrt{2n}$

In the rest of this section we will directly state the results for the
different correlators we need. The details on how to obtain these results are
provided in the Appendix A.

  * •

First let us focus on the piece,

$\left\langle\partial X^{\mu}(0)\partial
X^{\rho}(z)\bar{\partial}X^{\nu}(0)\bar{\partial}X^{\sigma}(\bar{z})e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\right\rangle.$
(40)
$\displaystyle=\left[\delta^{\mu\rho}\delta^{\nu\sigma}\left(\frac{(\vartheta_{11}\partial^{2}_{z}\vartheta_{11}-\partial_{z}\vartheta_{11}\partial_{z}\vartheta_{11})(z)}{\vartheta_{11}(z)^{2}}+\frac{1}{4\pi\tau_{2}}\right)\left(\frac{(\vartheta_{11}\bar{\partial}^{2}_{z}\vartheta_{11}-\bar{\partial}_{z}\vartheta_{11}\bar{\partial}_{z}\vartheta_{11})(\bar{z})}{\vartheta_{11}(\bar{z})^{2}}+\frac{1}{4\pi\tau_{2}}\right)\right.$
$\displaystyle\left.+\
\frac{\pi^{2}}{\tau_{2}^{2}}\delta^{\mu\sigma}\delta^{\rho\nu}+\big{(}\delta^{\mu\rho}k_{1}^{\sigma}k_{2}^{\nu}(..)+k_{1}^{\sigma}k_{2}^{\mu}\delta^{\rho\nu}(..)\big{)}+k_{1}^{\sigma}k_{1}^{\rho}k_{2}^{\mu}k_{2}^{\nu}(..)+....\right]\left\langle\prod_{i}e^{ik_{i}.X}\right\rangle.$
(41)

When we now contract the indeces with
$\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}$ and use (30) along with
$k_{1}=-k_{2}$, we see that the first two terms are $\sim O(1)$, the third
term which is quadratic in the momenta contribute at $\sim O(2n)$ and the term
which is quartic in momenta contributes at $\sim O(4n^{2})$. So we focus on
the last piece and it is given by,

$\displaystyle
64n^{2}\zeta^{(1)}_{00}\zeta^{(2)}_{00}\left[\left|\left(\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}\right)^{2}-\frac{4\pi^{2}}{\tau_{2}^{2}}(\text{Im}(-z))^{2}\right|^{2}+\frac{16\pi^{2}}{\tau_{2}^{2}}\left|\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}\text{Im}(-z)\right|^{2}\right.$
$\displaystyle\left.+\frac{8\pi}{\tau_{2}}\text{Im}\left\\{\frac{\vartheta^{\prime}_{11}(z)(\bar{\vartheta}^{\prime}_{11}(\bar{z}))^{2}}{\vartheta_{11}(z)(\bar{\vartheta}_{11}(\bar{z}))^{2}}\text{Im}(-z)\right\\}\right]\left\langle\prod_{i=1,2}e^{ik_{i}.X}\right\rangle.$
(42)

This contribution is positive as we can see from the above expression.

  * •

Next we consider the cross terms of the type,

$\displaystyle\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\left\langle\partial
X^{\mu}(0)\bar{\partial}X^{\nu}(0)\psi^{\rho}(z)(k_{2}.\psi(z))\bar{\psi}^{\sigma}(\bar{z})(k_{2}.\bar{\psi}(\bar{z}))e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\right\rangle$
$\displaystyle=-\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}k_{2}^{\mu}k_{2}^{\nu}k_{2}^{\rho}k_{2}^{\sigma}\left(\left|\frac{\vartheta_{11}^{\prime}(z)}{\vartheta_{11}(z)}\right|^{2}+\frac{4\pi^{2}}{\tau_{2}^{2}}(\text{Im}(-z))^{2}+\frac{4\pi}{\tau_{2}}\text{Im}\left\\{\frac{\vartheta_{11}^{\prime}(z)}{\vartheta_{11}(z)}\text{Im}(-z)\right\\}\right)\
$ $\displaystyle=\
-64n^{2}\zeta^{(1)}_{00}\zeta^{(2)}_{00}\left(\left|\frac{\vartheta_{11}^{\prime}(z)}{\vartheta_{11}(z)}\right|^{2}+\frac{4\pi^{2}}{\tau_{2}^{2}}(\text{Im}(-z))^{2}+\frac{4\pi}{\tau_{2}}\text{Im}\left\\{\frac{\vartheta_{11}^{\prime}(z)}{\vartheta_{11}(z)}\text{Im}(-z)\right\\}\right)\
.$ (43)

Hence we see that these terms go as $\sim\ O(Q^{4})\ \sim\ O(4n^{2})$. And all
the cross terms in (37) come with a minus sign hence these terms also give
positive contributions to the amplitude.

  * •

The term with four holomorphic and four anti-holomorphic fermions is given by,

$\displaystyle
k_{1\alpha}k_{2\beta}k_{1\gamma}k_{2\delta}\langle\psi^{\mu}\psi^{\alpha}(0)\psi^{\rho}\psi^{\beta}(z)\bar{\psi}^{\nu}\bar{\psi}^{\gamma}(0)\bar{\psi}^{\sigma}\bar{\psi}^{\delta}(z)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle$
(44)

Now for the fermionic correlator we have,

$k_{1\alpha}k_{2\beta}\langle\psi^{\mu}\psi^{\alpha}(0)\psi^{\rho}\psi^{\beta}(z)\rangle\
=\
k_{1\alpha}k_{2\beta}\left(A_{z}\delta^{\mu\alpha}\delta^{\rho\beta}+B_{z}\delta^{\mu\rho}\delta^{\alpha\beta}+C_{z}\delta^{\mu\beta}\delta^{\alpha\rho}\right)\
.$ (45)

The coefficients can be computed easily and are given by [2], [3],

$\displaystyle A_{z}$ $\displaystyle=$
$\displaystyle\langle\psi^{1}(0)\psi^{1}(0)\psi^{2}(z)\psi^{2}(z)\rangle\ =\
1\ ,$ (46) $\displaystyle B_{z}$ $\displaystyle=$
$\displaystyle\langle\psi^{1}(0)\psi^{2}(0)\psi^{1}(z)\psi^{2}(z)\rangle$
$\displaystyle=$
$\displaystyle-\frac{1}{8(\eta(\tau)^{4})^{2n+1}}\left(\frac{\vartheta_{11}^{\prime}(0)}{\vartheta_{11}(z)}\right)^{2}\sum_{\nu}\delta_{\nu}\left(\vartheta_{\nu}(z)^{2}+2\vartheta_{\nu}(z)\vartheta_{\nu}(-z)+\vartheta_{\nu}(-z)^{2}\right)\big{(}\vartheta_{\nu}(0)^{2}\big{)}^{4n+1}\
,$ $\displaystyle C_{z}$ $\displaystyle=$
$\displaystyle\langle\psi^{1}(0)\psi^{2}(0)\psi^{2}(z)\psi^{1}(z)\rangle\ =\
-B_{z}\ .$ (48)

So the full correlator on using (18) behaves like,

$\displaystyle\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\big{(}A_{z}k_{1}^{\mu}k_{2}^{\rho}-B_{z}(k_{1})^{2}\delta^{\mu\rho}+C_{z}k_{1}^{\rho}k_{2}^{\mu}\big{)}\big{(}A_{\bar{z}}k_{1}^{\nu}k_{2}^{\sigma}-B_{\bar{z}}(k_{1})^{2}\delta^{\nu\sigma}+C_{\bar{z}}k_{1}^{\sigma}k_{2}^{\nu}\big{)}$
$\displaystyle=16n^{2}\left(4\zeta^{(1)}_{00}\zeta^{(2)}_{00}|A_{z}+C_{z}|^{2}+\zeta^{(1)}_{\mu\nu}\zeta^{(2)\mu\nu}|B_{z}|^{2}+4Re[B_{z}(A_{z}+C_{z})]\zeta^{(1)}_{0\mu}\zeta^{(2)\mu}_{\
0}\right)$ (49)

This contribution as we can see gives positive contribution as well and
appears in (37) with a $`+$’ sign.

  * •

Finally from the last two lines of (37) also, we have negative contributions
but since they appear with a $`-$’ sign in the equation. So their full
contribution to the amplitude is also positive when we keep the highest order
terms $n$ i.e. prortional $4n^{2}$ like in the above cases. The explicit
details are given in Appendix A.

###  3.2 The imaginary part

Let us now look at the contribution from the imaginary part which is given by,

$\left[\int d^{2}z\
\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\left\langle\left\\{\psi^{\mu}(0)(k_{1}.\psi(0))\bar{\partial}X^{\nu}(0)+\partial
X^{\mu}(0)\bar{\psi}^{\nu}(0)(k_{1}.\bar{\psi}(0))\right\\}\right.\right.$
(51) $\left.\left.\times\left\\{\partial
X^{\rho}(z)\bar{\partial}X^{\sigma}(\bar{z})-\psi^{\rho}(z)(k_{2}.\psi(z))\bar{\psi}^{\sigma}(\bar{z})(k_{2}.\bar{\psi}(\bar{z}))\right\\}e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\right\rangle\right.$
$\left.+\int d^{2}z\
\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\langle(\mu\leftrightarrow\rho,\
\nu\leftrightarrow\sigma,\ k_{1}\leftrightarrow k_{2},\
z,\bar{z}\leftrightarrow
0,0)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle\right]\ .$

From the first two lines of (51) we have four terms. We first look at,

$\displaystyle\langle\bar{\partial}X^{\nu}(0)\partial
X^{\rho}(z)\bar{\partial}X^{\sigma}(\bar{z})\psi^{\mu}(0)(k_{1}.\psi(0))e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle$
(52) $\displaystyle+\ \langle\partial X^{\mu}(0)\partial
X^{\rho}(z)\bar{\partial}X^{\sigma}(\bar{z})\bar{\psi}^{\mu}(0)(k_{1}.\bar{\psi}(0))e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle.$

Now using (66), (45) and the onshell condition (31) one can easily check the
following facts

  * •

The first term has two contributions, one at $\sim O(2n)$ which we will
neglect with respect to the contribution at $\sim O(4n^{2})$,

$-64n^{2}\zeta^{(1)}_{00}\zeta^{(2)}_{00}\left(-i\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}+\frac{2\pi}{\tau_{2}}\text{Im}(z)\right)\left(-i\frac{\bar{\vartheta}^{\prime}_{11}(\bar{z})}{\bar{\vartheta}_{11}(\bar{z})}-\frac{2\pi}{\tau_{2}}\text{Im}(z)\right)\left(i\frac{\bar{\vartheta}^{\prime}_{11}(\bar{z})}{\bar{\vartheta}_{11}(\bar{z})}+\frac{2\pi}{\tau_{2}}\text{Im}(z)\right)$
$\Rightarrow
64n^{2}\zeta^{(1)}_{00}\zeta^{(2)}_{00}\left(\left|\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}\right|^{2}+\frac{4\pi^{2}}{\tau_{2}^{2}}(\text{Im}(z))^{2}+\frac{4\pi}{\tau_{2}}\text{Im}\left\\{\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}\text{Im}(z)\right\\}\right)\left(i\frac{\bar{\vartheta}^{\prime}_{11}(\bar{z})}{\bar{\vartheta}_{11}(\bar{z})}+\frac{2\pi}{\tau_{2}}\text{Im}(z)\right)$
(53)

Same argument are true for the second term and the contribution at $\sim
Q^{4}$ or $4n^{2}$ is given by,

$-64n^{2}\zeta^{(1)}_{00}\zeta^{(2)}_{00}\left(-i\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}+\frac{2\pi}{\tau_{2}}\text{Im}(z)\right)\left(-i\frac{\bar{\vartheta}^{\prime}_{11}(\bar{z})}{\bar{\vartheta}_{11}(\bar{z})}-\frac{2\pi}{\tau_{2}}\text{Im}(z)\right)\left(i\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}-\frac{2\pi}{\tau_{2}}\text{Im}(z)\right).$

Thus adding them up we get,

$i64n^{2}\zeta^{(1)}_{00}\zeta^{(2)}_{00}\left(\left|\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}\right|^{2}+\frac{4\pi^{2}}{\tau_{2}^{2}}(\text{Im}(z))^{2}+\frac{4\pi}{\tau_{2}}\text{Im}\left\\{\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}\text{Im}(z)\right\\}\right)\left(\frac{\bar{\vartheta}^{\prime}_{11}(\bar{z})}{\bar{\vartheta}_{11}(\bar{z})}+\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}\right).$
(54)

  * •

And one can check that from the last line in (51) these contribution which is
conjugate to the one described above comes out to be

$-i64n^{2}\zeta^{(1)}_{00}\zeta^{(2)}_{00}\left(\left|\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}\right|^{2}+\frac{4\pi^{2}}{\tau_{2}^{2}}(\text{Im}(z))^{2}+\frac{4\pi}{\tau_{2}}\text{Im}\left\\{\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}\text{Im}(z)\right\\}\right)\left(\frac{\bar{\vartheta}^{\prime}_{11}(\bar{z})}{\bar{\vartheta}_{11}(\bar{z})}+\frac{\vartheta^{\prime}_{11}(z)}{\vartheta_{11}(z)}\right).$

So when we add this piece to (54) the total contribution vanishes.

  * •

One can check that similar arguments hold for the other type of term as well
i.e. terms with one bosonic field and rest all fermionic fields. When we take
the contribution from all possible terms of this type the total contribution
vanishes.

One can choose a particular co-ordinate patch i.e. fix some gauge to carry out
the integral over the moduli space, but we will not do it in this note. There
is one point we should mention; for the bosonic piece we see explicitly that
there are no other divergences other than $z\rightarrow 0$ i.e. they are free
from spurious divergences. But for a function like $B_{z}$ or $C_{z}$ since we
do not completely know zeros and poles for the spin structure sum it is hard
to conclude the absence of spurious poles. This can be an interesting
direction to check that the perturbative amplitude is indeed completely well
defined.

Now from [14] we have to get the renomalized mass one has to solve the
following equation iteratively,

$E^{2}=m^{2}-\tilde{F}(E)\ ,$ (55)

to a given order in perturbation theory. Here $m^{2}$ is the tree level mass
squared and $\tilde{F}(E)$ is the quantum correction to the propagator. in the
current case,

$m^{2}=-Q^{2}\ ,\quad\text{and}\quad\text{Im}(\tilde{F}(E))=0\
,\quad\tilde{F}(E)>0\ .$

from the one loop correction. Also since the imaginary part is zero at the
order we considered, we have zero decay width so the modes are stable. Hence
atleast to this order we can confirm the following statement unambiguously.

  * •

The Pseudotachyon modes remain pseudotachyonic under quantum corrections and
the system stabilizes towards condensation of such modes.

##  4 Interpolating functions due to Sen

We now attempt to construct an interpolating function which gives the
renormalized mass for all coupling and matches the pertubative expansion to a
given order at both the weak and strong coupling end if such expansions indeed
exists. We will follow the strategy described in the seminal work [2] for this
purpose. We will first briefly discuss the know example where this technique
succeeds to within $10\%$ accuracy, and then try to implement it for the
Linear Dilaton backgrounds.

###  4.1 Example of the heterotic/type I duality

In case of the Heterotic and type I, S-dual theories the couplings at the two
ends are related by,

$g_{H}g_{I}=2^{5}\pi^{7}\ .$

Thus we can parametrize them as follows,

$g_{H}=(2\pi)^{7/2}g,\quad g_{I}=2^{3/2}\pi^{7/2}g^{-1},$

so that the two formulae meet at $g=1$. The renormalized mass function of the
second excited state444In our convention we always state the tachyons to be
the ground state or zero excitation state, even though in superstring theories
they are projected out by GSO projection.  (i.e. the first massive level) was
considered. We define, the renormalized mass function $F(g)$ by,

$M(g)=2^{15/8}\pi^{7/8}F(g)$

For weak coupling end we have the Heterotic theory whereas for the strong
coupling end we have the type I theory with,

$F^{W}_{2}(g)=g^{1/4}\left(1+K_{w}g^{2}+O(g^{4})\right),\ \
F^{S}_{1}(g)=g^{3/4}\left(1+K_{s}g^{-1}+O(g^{-2})\right);\ K_{w}\simeq 0.23,\
K_{s}\simeq 0.351$ (56)

where one loop or the subleading corrections at both ends where considered.

We now look at the interpolating function which has to agree with the coupling
expansions to order $m$ in the weak coupling and order $n$ in strong coupling
end.

$F_{m,n}(g)=g^{1/4}\left[1+a_{1}g+...+a_{m}g^{m}+b_{n}g^{m+1}+b_{n-1}g^{m+2}+...+b_{1}g^{m+n}+g^{m+n+1}\right]^{\frac{1}{2(m+n+1)}}$
(57)

In the present example it is known that for the Heterotic theory being a
closed superstring theory has the expansion in $g^{2}$ so $a_{m}=0$ for $m$
odd. The interpolating function which is within $10\%$ accuracy of the actual
function was calculated to be,

$F_{3,1}=g^{1/4}\left(1+10K_{w}g^{2}+10K_{s}g^{4}+g^{5}\right)^{1/10}$

###  4.2 Application to the linear dilaton backgrounds

In our case we would like to apply the same technique to the first excited
state of strings which are the Pseudotachyonic modes. It is fairly obvious
that now all the coefficients for the interpolating function may depend on the
background charge $Q$.

Note: It should be understood that this technique is useful only in the cases
where it is possible to determine the physical quantities of the theory at
both ends perturbatively, with the use of S-duality. In cases where this is
not true, this process cannot yield sensible results and one indeed has to
carry out a full non-perturbative computation.

So we will begin with a theory where the weak coupling ($g_{w}$) regime is
S-dual to another theory at the strong coupling ($g_{s}$) regime. Hence we
will have,

$g_{w}g_{s}=\mathcal{C}_{0}\ ,\quad\text{where $\mathcal{C}_{0}$ is a
constant.}$ (58)

We do the following parametrization,

$g_{w}=c_{0}g,\ \text{and}\ g_{s}=(\mathcal{C}_{0}/c_{0})g^{-1}$

such that the formulae meet at $g=1$ i.e. both the strong and weak couplings
become of the same order ($\sim\ O(1)$) while satisfying (58). This parameter
now can be treated as,

$g\equiv g_{st}=e^{-QX^{0}}\ \ \text{i.e. the coupling varrying with time.}$

As in the earlier example we define the renormalized mass function via,

$M(g)=\text{(const.)}\times F(g)\ .$

Let us first try to see the leading order coefficient of the interpolating
function for the case at hand. From (32) we have,

$k^{2}-2iQk^{0}=0\ \Rightarrow\ M^{2}=-Q^{2}$ (59)

and the leading order coupling behavior at the strong and weak coupling end
are assumed to be,

$F_{W}(g)\sim g^{\kappa},\quad F_{S}(g)\sim g^{\delta}\ .$ (60)

Note that in case of the example in previous subsection $\kappa=1/4$ and
$\delta=3/4$. Here we don’t fix the values since they are theory dependent.

With these details, now we can venture to make an ansatz for the interpolating
function that agrees the expansions at both ends upto some given order.

$\displaystyle F_{m,n}(g)$ $\displaystyle=$
$\displaystyle-Q^{2}g^{\kappa}\left[1\ +\ a_{1}(Q)g\ +\ a_{2}(Q)g^{2}\ +\
...+\ a_{m}(Q)g^{m}\ +\ b_{n}(Q)g^{m+1}\ +\ ...\right.$ $\displaystyle\left.+\
b_{1}(Q)g^{m+n}\ +\ g^{m+n+1}\right]^{\frac{\delta-\kappa}{m+n+1}}.$ (61)

We saw that the weak coupling expansion is in powers of $g^{2}$ since it is a
cosed superstring perturbation theory. So $a_{m}(Q)=0$ for odd $m$. And the
coefficient $a_{2}(Q)\sim O(Q^{2})$ in the large $Q$ limit. Similarly at the
strong coupling end if the theory is an open superstring theory then
$b_{1}(Q)$ starts at $\sim O(Q^{2})$ otherwise, $b_{2}(Q)$ starts at $\sim
O(Q^{2})$ and $b_{n}(Q)=0$ for odd $n$. The above interpolating function
matches the leading order expansion at both ends by construction555In the
large $Q$ limit we saw that the corrections go as $\sim\ Q^{4}$. So if an
overall factor of $-Q^{2}$ is pulled out we are left with a correction factor
of $Q^{2}$ within the parenthesis..

Expanding around $g=0$ $\displaystyle:$ $\displaystyle
F_{m,n}(g)\approx-Q^{2}g^{\kappa}\left(1+\frac{a_{2}(Q)(\delta-\kappa)}{m+n+1}g^{2}+O(g^{4})\right)$
(62) Expanding around $g=\infty$ $\displaystyle:$ $\displaystyle
F_{m,n}(g)\approx-Q^{2}g^{\delta}\left(1+\frac{b_{1}(Q)(\delta-\kappa)}{m+n+1}g^{-1}+O(g^{-2})\right)$
(63)

For the one loop corrected mass at both ends one can compute the interpolating
function $F_{3,1}$ (like in the previous example) we have,

$F_{3,1}(g)=-Q^{2}g^{\kappa}\left(1\ +\ a_{2}Q^{2}g^{2}\ +\ b_{1}Q^{2}g^{4}\
+\ g^{5}\right)^{(\delta-\kappa)/5}.$ (64)

We have written,

$a_{2}(Q)\equiv a_{2}Q^{2},\quad b_{1}(Q)\equiv b_{1}Q^{2},\ \text{with}\
a_{2}>0,\ b_{1}>0$ in accordance with their large $Q$ expansion.

Now that we have a function which gives the renormalized mass for the lowest
energy state we can take the limit $g\rightarrow 1$ i.e. at time $X^{0}=0$ to
find the lowest energy state to which the universe tunnels to from early time
strongly coupled region in case of the linear dilaton (time dependent)
background. Hence after tunneling we reach a capping state configuration with
the lowest energy state given by,

$M_{cs}=-Q^{2}\left[2\ +\ Q^{2}(a_{2}\ +\
b_{1})\right]^{(\delta-\kappa)/5}\times(\text{const.}).$ (65)

The over all constant is also theory specific like $\kappa$ and $\delta$. The
accuracy of the result obviously depends on the accuracy of the interpolating
function and it increases or stabilizes as we go to higher order in
pertubation theory at both ends.

##  5 Discussions

In this note we have presented a strategy of constructing the so called
“capping” state in SCLD theories by embedding the SCLD phase in a strong
coupling completion. Although we are unable to produce exact numbers by
working with some specific theories, we do give consistency arguments to
support that the strategy employed may give sensible results. Let us list the
results we obtained for summarising this work,

  * •

We show for the two point function on the torus, that in the large $Q$ limit
the imaginary part vanishes to the highest order in $Q$ implyng a vanishing
decay width to this order. From the real part we get a positive contribution
so that the energy of the Pseudo-tachyonic mode lowers under the one loop
correction of the mass.

  * •

We find the interpolating function which is supposed to match at both the weak
and strong coupling end in case of theories which are S-dual to each other.

An interesting future direction is to actually compute the fermionic
correlation function using some mathematical tools to figure out whether there
are any spurious poles for this amplitude. Another open problem is to find
specific theories satisfying the relevant conditions so that this procedure
can be checked explicitly. But these are currently at the level of
speculation. If we do infact find specific theories then we can produce
numbers which will render a concrete example where this strategy can be tested
for determining the initial capping state within some reasonable accuracy.

## Acknowledgement

I would like to thank Ashoke Sen, Harold Erbin and K. Narayan for their useful
comments on the earlier versions of the draft which improved my understanding
significantly.

##  Appendix A Relevant correlators for $Q\neq 0$

  1. 1.

Correlation function due to the bososnic piece is given by
($\alpha^{\prime}=2$)666Since, $\vartheta_{11}(-z)=-\vartheta_{11}(z)$, hence
$\left(i\partial_{z}\text{ln}[\vartheta_{11}(-z)]+\frac{2\pi}{\tau_{2}}\text{Im}(-z)\right)=\left(i\partial_{z}\text{ln}[\vartheta_{11}(z)]-\frac{2\pi}{\tau_{2}}\text{Im}(z)\right)$.,

$\left\langle\partial X^{\mu}(0)\partial
X^{\rho}(z)\bar{\partial}X^{\nu}(0)\bar{\partial}X^{\sigma}(\bar{z})e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\right\rangle$
$\displaystyle=\
\delta^{\mu\rho}\left(\frac{(\vartheta_{11}\partial^{2}_{z}\vartheta_{11}-\partial_{z}\vartheta_{11}\partial_{z}\vartheta_{11})(z)}{\vartheta_{11}(z)^{2}}+\frac{1}{4\pi\tau_{2}}\right)\langle\bar{\partial}X^{\nu}(0)\bar{\partial}X^{\sigma}(z)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle$
$\displaystyle+k_{2}^{\mu}\left(i\partial_{z}\text{ln}[\vartheta_{11}(z)]-\frac{2\pi}{\tau_{2}}\text{Im}(z)\right)\langle\partial
X^{\rho}(z)\bar{\partial}X^{\nu}(0)\bar{\partial}X^{\sigma}(z)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle$
$\displaystyle+\frac{\pi}{\tau_{2}}\delta^{\mu\sigma}\langle\partial
X^{\rho}(z)\bar{\partial}X^{\nu}(0)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle\
.$ (66)

If we further evaluate the correlators for each term it is fairly obvious as
to what will be the expansion in powers of $\alpha^{\prime}/2$ and $Q$ once we
apply (31),

$\displaystyle\sim\ \left\\{O\left(\frac{\alpha^{\prime
2}}{4}\right)+O\left(\frac{\alpha^{\prime
3}Q^{2}}{8}\right)+O\left(\frac{\alpha^{\prime 4}Q^{4}}{16}\right)\right\\}$
(67)

  2. 2.

We now look at the fermionic correlator to determine $B_{z}$, given by,

$\displaystyle B_{z}$ $\displaystyle=$
$\displaystyle\langle\psi^{1}(0)\psi^{2}(0)\psi^{1}(z)\psi^{2}(z)\rangle$ (68)
$\displaystyle=$
$\displaystyle-\frac{1}{4}\left(\langle\chi^{1}(0)\tilde{\chi}^{1}(z)\chi^{2}(0)\tilde{\chi}^{2}(z)\rangle\
+\
\langle\tilde{\chi}^{1}(0)\chi^{1}(z)\chi^{2}(0)\tilde{\chi}^{2}(z)\rangle\right.$
$\displaystyle+\
\left.\langle\chi^{1}(0)\tilde{\chi}^{1}(z)\tilde{\chi}^{2}(0)\chi^{2}(z)\rangle\
+\
\langle\tilde{\chi}^{1}(0)\chi^{1}(z)\tilde{\chi}^{2}(0)\chi^{2}(z)\rangle\right)\
.$

Although the expression for $\langle\chi(0)\tilde{\chi}(z)\rangle,$ and
$\langle\tilde{\chi}(0)\chi(z)\rangle$ obeys (86), but now taking in
contribution from all the fermions and superconformal ghosts we get (with
$D=10+16n$),

$B_{z}=-\frac{1}{8(\eta(\tau)^{4})^{2n+1}}\left(\frac{\vartheta_{11}^{\prime}(0)}{\vartheta_{11}(z)}\right)^{2}\sum_{\nu}\delta_{\nu}\left(\vartheta_{\nu}(z)^{2}+2\vartheta_{\nu}(z)\vartheta_{\nu}(-z)+\vartheta_{\nu}(-z)^{2}\right)\big{(}\vartheta_{\nu}(0)^{2}\big{)}^{4n+1}.$
(69)

  3. 3.

Let us now consider the last two lines of (37). There are two types of terms
that we need to consider if we just want to check their expansion in powers of
$\alpha^{\prime}$ and $Q$.
Firstly we have,

$k_{1\alpha}k_{2\beta}\langle\psi^{\mu}\psi^{\alpha}(0)\psi^{\rho}\psi^{\beta}(z)\bar{\partial}X^{\nu}(0)\bar{\partial}X^{\sigma}(z)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})$
(70)

and,

$k_{1\alpha}k_{2\beta}\langle\partial X^{\mu}(0)\partial
X^{\rho}(z)\bar{\psi}^{\nu}\bar{\psi}^{\alpha}(0)\bar{\psi}^{\sigma}\bar{\psi}^{\beta}(z)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\
.$ (71)

Using (45), (66) and of course (31) we get the behavior of these type of terms
in the large $Q$ limt to be negative and proportional to $4n^{2}$. And they
occur in the amplitide with a ‘$-$’ sign, so contribution to the amplitude is
positive.

Lastly we have the terms of type,

$\displaystyle k_{1\alpha}k_{2\beta}\left\langle\partial
X^{\rho}(z)\bar{\partial}X^{\nu}(0)\psi^{\mu}\psi^{\alpha}(0)\bar{\psi}^{\sigma}\bar{\psi}^{\beta}(\bar{z})e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\right\rangle$
(72) $\displaystyle\text{and}\quad k_{1\alpha}k_{2\beta}\left\langle\partial
X^{\mu}(0)\bar{\partial}X^{\sigma}(\bar{z})\psi^{\rho}\psi^{\beta}(z)\bar{\psi}^{\nu}\bar{\psi}^{\alpha}(0)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\right\rangle$
(73)

Using the same set of equations one can again show that the same argument
holds.

But now we can see clearly that the piece which proprotional to $2n$ is
completely regular in $z$ and $\bar{z}$ while the term proportional to
$4n^{2}$ has a single pole for both $z$ and $\bar{z}$. Hence when we carry out
the integration over $z$ plane the first term vanishes while for the second
term we can pick up the residue at $z=\bar{z}=0$. So these type of terms are
free from spurious poles and are proportional to $4n^{2}$.

##  Appendix B Two point graviton one loop amplitude with $Q=0$

For this computation we need a -1 picture unintegrated vertex whose location
on the torus can be fixed (say (0,0)) by the CKV of the torus and a 0 picture
integrated vertex operator and a PCO. The following definitions will be useful
in our case,

-1 picture vertex : $\displaystyle V_{-1}(k,z,\bar{z})=i\ \bar{c}ce^{-(\phi+\bar{\phi})}\zeta_{\mu\nu}\psi^{\mu}(z)\bar{\psi}^{\nu}(\bar{z})e^{ik.X}(z,\bar{z})$ (74) World-sheet supercurrent : $\displaystyle T_{F}(z)=i(2/\alpha^{\prime})^{\frac{1}{2}}\psi^{\rho}\partial X_{\rho}\ ,\ \bar{T}_{F}(\bar{z})=i(2/\alpha^{\prime})^{\frac{1}{2}}\bar{\psi}^{\rho}\bar{\partial}X_{\rho}$ (75) PCO : $\displaystyle\chi=\oint dz\ e^{\phi}T_{F}(z)+...\ ,\quad\bar{\chi}=\oint d\bar{z}\ e^{\bar{\phi}}\bar{T}_{F}(\bar{z})+...$ (76)

The $...$ denotes involving $\eta,\ \partial\xi$ etc. We will not be needing
those for our computation.

When $\chi\bar{\chi}$ hits $V_{-1}(k_{1},0,0)$ is converted to a 0 picture
unintegrated vertex given by,

$\bar{c}cV_{0}(k_{1},0,0)=\left(\frac{2}{\alpha^{\prime}}\right)\bar{c}c\zeta^{(1)}_{\mu\nu}\left\\{\partial
X^{\mu}+i\left(\frac{\alpha^{\prime}}{2}\right)\psi^{\mu}(k_{1}.\psi)\right\\}\left\\{\bar{\partial}X^{\nu}+i\left(\frac{\alpha^{\prime}}{2}\right)\bar{\psi}^{\nu}(k_{1}.\bar{\psi})\right\\}e^{ik_{1}.X}(0,0)$
(77)

while the 0 picture integrated vertex is given by,

$U_{0}(k_{2})=\left(\frac{2}{\alpha^{\prime}}\right)\int d^{2}z\
\zeta^{(2)}_{\mu\nu}\\{\partial
X^{\mu}+i\psi^{\mu}(k_{2}.\psi)\\}\\{\bar{\partial}X^{\nu}+i\bar{\psi}^{\nu}(k_{2}.\bar{\psi})\\}e^{ik_{2}.X}(z,\bar{z})$
(78)

The correlation function that we will be computing is,777We will be using the
convention $\alpha^{\prime}=2$ for the rest of this appendix.

$\int_{\mathbb{F}_{0}}\frac{d\tau d\bar{\tau}}{4\tau_{2}}\langle
b\bar{b}(0)\bar{c}c(0)V_{0}(k_{1},0,0)U_{0}(k_{2})\rangle\ .$

One can immediately see that there are four possible terms for each of the the
real and the imaginary parts. We will look at a few of these terms and guess
the result for the rest from the simarities. The correlation function due to
the $bc$ ghost zero mode has the well known behavior,

$\langle b(0)\bar{b}(0)\bar{c}(0)c(0)\rangle_{bc}\ =\ |\eta(\tau)|^{4}\ .$
(79)

Let us now first look at the imaginary part of the matter correlation
function,

$\int d^{2}z\
\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\langle\\{\psi^{\mu}(0)(k_{1}.\psi(0))\bar{\partial}X^{\nu}(0)+\partial
X^{\mu}(0)\bar{\psi}^{\nu}(0)(k_{1}.\bar{\psi}(0))\\}$ (80) $\times\\{\partial
X^{\rho}(z)\bar{\partial}X^{\sigma}(\bar{z})-\psi^{\rho}(z)(k_{2}.\psi(z))\bar{\psi}^{\sigma}(\bar{z})(k_{2}.\bar{\psi}(\bar{z}))\\}e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle$
$+\int d^{2}z\
\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\langle(\mu\leftrightarrow\rho,\
\nu\leftrightarrow\sigma,\ k_{1}\leftrightarrow k_{2},\
z,\bar{z}\leftrightarrow 0,0)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle\
.$

Once we compute the first term the second one should work out in a similar
manner with the relevant exchanging of indices. The tensor structure for this
part is given by,

$\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\left(\delta^{\mu\alpha}k_{1\alpha}\langle\bar{\partial}X^{\nu}\mathcal{K}^{\rho\sigma}e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle+k_{1\alpha}\delta^{\nu\alpha}\langle\partial
X^{\mu}\mathcal{K}^{\rho\sigma}e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle\right)\
,$ (81)

where,

$\mathcal{K}^{\rho\sigma}=\partial
X^{\rho}(z)\bar{\partial}X^{\sigma}(\bar{z})-A\
k_{2\alpha}k_{2\beta}\delta^{\rho\alpha}\delta^{\sigma\beta}\ .$

So we see that both contributions are proportional to,

$\text{either}\ k_{1}^{\mu}\zeta^{(1)}_{\mu\nu}\ \text{or}\
k_{1}^{\nu}\zeta^{(1)}_{\mu\nu}$

Thus for the graviton on-shell condition these contributions vanish. Similarly
the second term in (80) also vanishes which suggests that the imaginary part
of the amplitude is zero.

Now we turn our attention to the real part of the amplitude. This part has
contributions from the following terms,

$\int d^{2}z\ \zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\langle\\{\partial
X^{\mu}(0)\bar{\partial}X^{\nu}(0)-\psi^{\mu}(0)(k_{1}.\psi(0))\bar{\psi}^{\nu}(0)(k_{1}.\bar{\psi}(0))\\}$
$\times\\{\partial
X^{\rho}(z)\bar{\partial}X^{\sigma}(\bar{z})-\psi^{\rho}(z)(k_{2}.\psi(z))\bar{\psi}^{\sigma}(\bar{z})(k_{2}.\bar{\psi}(\bar{z}))\\}e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle$
$-\int d^{2}z\
\zeta^{(1)}_{\mu\nu}\zeta^{(2)}_{\rho\sigma}\langle\\{\psi^{\mu}(0)(k_{1}.\psi(0))\bar{\partial}X^{\nu}(0)+\partial
X^{\mu}(0)\bar{\psi}^{\nu}(0)(k_{1}.\bar{\psi}(0))\\}$
$\times\\{\psi^{\rho}(z)(k_{2}.\psi(z))\bar{\partial}X^{\sigma}(\bar{z})+\partial
X^{\rho}(z)\bar{\psi}^{\sigma}(\bar{z})(k_{2}.\bar{\psi}(\bar{z}))\\}e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle\
.$ (82)

For the second piece we see that only terms with four fermions all of which
are either left or right moving, contribute. Other terms vanish on-shell by
previous arguments. So we need to compute the correlation function,

$\langle\psi^{\mu}(0)\psi^{\alpha}(0)\psi^{\rho}(z)\psi^{\beta}(z)\rangle\equiv\
A\delta^{\mu\alpha}\delta^{\rho\beta}\ +\
B\delta^{\mu\rho}\delta^{\alpha\beta}\ +\
C\delta^{\mu\beta}\delta^{\alpha\rho}$ (83)

The first term in the above equation vanishes due to previous arguments.
Although the second and third term cannot be determined by these arguments.
The correlation function for such tensor structures receive contributions from
terms of the form,

$\langle\psi^{i}(0)\psi^{j}(0)\psi^{i}(z)\psi^{j}(z)\rangle\ \text{or}\
\langle\psi^{i}(0)\psi^{j}(0)\psi^{j}(z)\psi^{i}(z)\rangle\quad\text{with}\ \
i\neq j\ .$ (84)

Let us look at the cases where $i$, $j$ takes values 1, 2, 3, 4. Here we keep
in mind that $\psi^{0}$ is wick rotataed to $\psi^{4}$ at the cost of picking
up a factor of $i$, and also we combine the fermions into complex pairs via,

$\chi_{k}=\frac{1}{\sqrt{2}}\left(\psi^{k}+i\psi^{k+4}\right),\quad\tilde{\chi}^{k}=\frac{1}{\sqrt{2}}\left(\psi^{k}-i\psi^{k+4}\right),\quad
1\leq k\leq 4\ .$

As a concrete example let us look at the following possible contribution888One
can easily check that for, $i=k$ and $j=k+4$ the total contribution always
vanishes in this case.,

$\displaystyle\langle\psi^{1}(0)\psi^{2}(0)\psi^{1}(z)\psi^{2}(z)\rangle$
$\displaystyle=$
$\displaystyle-\frac{1}{4}\left(\langle\chi^{1}(0)\tilde{\chi}^{1}(z)\chi^{2}(0)\tilde{\chi}^{2}(z)\rangle\
+\
\langle\tilde{\chi}^{1}(0)\chi^{1}(z)\chi^{2}(0)\tilde{\chi}^{2}(z)\rangle\right.$
$\displaystyle+\
\left.\langle\chi^{1}(0)\tilde{\chi}^{1}(z)\tilde{\chi}^{2}(0)\chi^{2}(z)\rangle\
+\
\langle\tilde{\chi}^{1}(0)\chi^{1}(z)\tilde{\chi}^{2}(0)\chi^{2}(z)\rangle\right)\
.$ (85)

Using the results of [3], [2] we have for a single complex fermion $\chi^{i}$,

$\langle\chi^{i}(0)\tilde{\chi}^{i}(z)\rangle\ =\
\frac{1}{\eta(\tau)}\frac{\vartheta^{\prime}_{11}(0)}{\vartheta_{11}(z)}\vartheta_{\nu}(z)\
,\quad\langle\tilde{\chi}^{i}(0)\chi^{i}(z)\rangle\ =\
\frac{1}{\eta(\tau)}\frac{\vartheta^{\prime}_{11}(0)}{\vartheta_{11}(z)}\vartheta_{\nu}(-z)\
.$ (86)

If we now take contributions from all the holomorphic fermions and the
superconformal ghosts we end up with,

$\displaystyle\langle\psi^{1}(0)\psi^{2}(0)\psi^{1}(z)\psi^{2}(z)\rangle$
$\displaystyle=$
$\displaystyle-\frac{1}{8\eta(\tau)^{4}}\left(\frac{\vartheta_{11}^{\prime}(0)}{\vartheta_{11}(z)}\right)^{2}\sum_{\nu}\delta_{\nu}\left(\vartheta_{\nu}(z)^{2}\vartheta_{\nu}(0)^{2}+2\vartheta_{\nu}(z)\vartheta_{\nu}(-z)\vartheta_{\nu}(0)^{2}\right.$
$\displaystyle\left.+\vartheta_{\nu}(-z)^{2}\vartheta_{\nu}(0)^{2}\right).$

The factor of half from the GSO projection is accounted for within the overall
factor $1/8$. The r.h.s. of equation (LABEL:eq:frmn) vanishes by the use of
the Riemann Identity,

$\displaystyle\sum_{\nu}\delta_{\nu}\vartheta_{\nu}(z_{1})\vartheta_{\nu}(z_{2})\vartheta_{\nu}(z_{3})\vartheta_{\nu}(z_{4})$
$\displaystyle=$ $\displaystyle
2\vartheta_{11}\left((z_{1}+z_{2}+z_{3}+z_{4})/2\right)\vartheta_{11}\left((z_{1}+z_{2}-z_{3}-z_{4})/2\right)$
$\displaystyle\vartheta_{11}\left((z_{1}-z_{2}-z_{3}+z_{4})/2\right)\vartheta_{11}\left((z_{1}-z_{2}+z_{3}-z_{4})/2\right)\
.$

Thus we see that the only possible term that might contribute to (82) is,

$\langle\partial X^{\mu}(0)\partial
X^{\rho}(z)\bar{\partial}X^{\nu}(0)\bar{\partial}X^{\sigma}(z)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle\
.$ (89)

Following [ref] we find that this is given by,

$\displaystyle\delta^{\mu\rho}\left(\frac{(\vartheta_{11}\partial^{2}_{z}\vartheta_{11}-\partial_{z}\vartheta_{11}\partial_{z}\vartheta_{11})(z)}{\vartheta_{11}(z)^{2}}+\frac{1}{4\pi\tau_{2}}\right)\langle\bar{\partial}X^{\nu}(0)\bar{\partial}X^{\sigma}(z)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle$
$\displaystyle+k_{2}^{\mu}\left(i\partial_{z}\text{ln}[\vartheta_{11}(-z)]+\frac{2\pi}{\tau_{2}}\text{Im}(-z)\right)\langle\partial
X^{\rho}(z)\bar{\partial}X^{\nu}(0)\bar{\partial}X^{\sigma}(z)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle$
$\displaystyle-\frac{\pi}{\tau_{2}}\delta^{\mu\sigma}\langle\partial
X^{\rho}(z)\bar{\partial}X^{\nu}(0)e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle\
.$ (90)

Notice that both the first piece and the second piece have double poles at
$z=0$ owing to the fact that the correlation function due to the exponentials
give,

$\langle
e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle\sim(2\pi)^{10}\delta^{10}(k_{1}+k_{2})\left|\vartheta_{1}(z,\tau)\exp\left(-\frac{\pi(\text{Im}\
z)^{2}}{\tau_{2}}\right)\right|^{\alpha^{\prime}k_{1}.k_{2}}\prod_{i=1}^{2}\left|\frac{2\pi}{\vartheta_{1}^{\prime}(0,\tau)}\right|^{\frac{-\alpha^{\prime}k_{i}^{2}}{2}}$

and due to the delta function we get $k_{1}=-k_{2}$ which implies
$k_{1}.k_{2}=-k_{1}^{2}=-k_{2}^{2}=0$ from the on-shell condition. Thus,

$\langle e^{ik_{1}.X}(0,0)e^{ik_{2}.X}(z,\bar{z})\rangle\sim(2\pi)^{10}\ .$

We have an integration over $z$ since one vertex is an integrated vertex
operator. Near the pole we see the behavior,

$\oint\frac{dz}{z^{2}}(2\pi)^{10}f(k,\zeta)\quad\text{where $f(k,\zeta)$ is a
function of polarization and momenta.}$ (91)

which vanishes from Cauchy’s integral formulae. Where as terms proportional to
$1/4\pi\tau_{2}$ or $\pi^{2}/\tau_{2}^{2}$ vanish since they have no
singularities on the torus described by the $z$-coordinate. Terms which have a
single pole at $z=0$ survive at the end and it is clear from the above
expression that they are proportional to,

$k_{2}^{\mu}\zeta^{(1)}_{\mu\nu},\ k_{1}^{\rho}\zeta^{(2)}_{\rho\sigma},\
\text{or}\ k_{1}^{\rho}\zeta^{(2)}_{\rho\sigma}\zeta^{(1)\mu}_{\ \nu}$

Now from the correlation function of the exponentials we obtain an overall
momentum conserving delta function $\delta^{10}(k_{1}+k_{2})$ so that
$k_{2}=-k_{1}$, and we get zero contributions imposing the on-shell
conditions.

Thus we see that the graviton two point one loop amplitude vanishes on-shell
as we had expected.

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