# New bosonic hard string scattering amplitudes and extended Gross conjecture
in superstring theory

Sheng-Hong Lai<EMAIL_ADDRESS>Department of Electrophysics, National
Chiao-Tung University, Hsinchu, Taiwan, R.O.C.  Jen-Chi Lee
<EMAIL_ADDRESS>Department of Electrophysics, National Chiao-Tung
University, Hsinchu, Taiwan, R.O.C.  Yi Yang<EMAIL_ADDRESS>Department of Electrophysics, National Chiao-Tung University, Hsinchu, Taiwan,
R.O.C.

###### Abstract

We discover new hard string scattering amplitudes (HSSA) with polarizations
orthogonal to the scattering plane in the NS sector of $10D$ open superstring
theory. The corresponding HSSA in the bosonic string theory are of subleading
order in energy. In addition, all HSSA are found to be proportional to each
other and the ratios among these HSSA are calculated to justify Gross
conjecture in 1988 on high energy symmetry of string theory. Moreover,
together with these new HSSA, the total number of HSSA calculated in the first
massive level of NS sector are found to agree with those of hard polarized
fermion string scattering amplitudes (PFSSA) in the R sector calculated
recently.

##  I Introduction

In contrast to high energy behaviour of quantum field theories, e.g. the well-
known asymptotic freedom of QCD discovered in early 70s PGW , many important
issues of high energy behaviour of string theory GM  remained to be studied.
Historically, it was first conjectured in 1988 by Gross Gross  in his pioneer
works on this subject that there exist infinite linear relations GrossManes
among hard string scattering amplitudes (HSSA) of different string states in
the hard scattering limit. Moreover, the conjecture states the linear
relations are so powerful that all the HSSA are equal, up to a numerical
(angle-independent) constant. This conjecture was later corrected and proved
by Taiwan group using the decoupling of zero norm states ZNS1 ; 2D  in
ChanLee1 ; ChanLee2 ; CHL ; PRL ; CHLTY ; susy . For more details, see the
recent review review ; over .

In addition to the bosonic string theory, there has been attempts to extend
Gross conjecture (GC) to the NS sector of superstring theory susy ; RRsusy .
Recently, the present authors calculated a class of polarized fermion string
scattering amplitudes (PFSSA) PFSSA  at arbitrary mass levels which involve
the leading Regge trajectory fermion string states of the R sector, and
successfully extended GC to the fermionic sector. Since it is a nontrivial
task to construct the general massive fermion string vertex operators vertex ;
RRR ; m1 ; m2 ; m3 ; m4 , the proof of extended GC for the case of superstring
theory remains an open question.

One important issue of extended GC for superstring is to compare the HSSA of
NS and R sectors at each fixed mass level of the theory. Since so far the
complete hard PFSSA is available only for the first massive level (no non-
leading Regge trajectory fermion string states) in the R sector PFSSA , as a
first step, in this paper we will calculate the complete HSSA for the first
massive level of the NS sector. On the other hand, we will also calculate and
count the number of HSSA for both NS and R sectors and check whether they
match to each other or not.

Since Gross conjecture was shown to be valid for both GSO even and odd states
in the NS sector susy , we are going to ignore the GSO projection in this
paper. To be more specific, the two HSSA we want to compare are

$\left\langle V^{(-1)}Tensor^{(-1)}\phi^{(0)}\phi^{(0)}\right\rangle$ (1.1)

for the NS sector and

$\left\langle\psi^{(-\frac{1}{2})}\chi^{(-\frac{1}{2})}\phi^{(-1)}\phi^{(0)}\right\rangle$
(1.2)

for the R sector. In Eq.(1.1) the first vertex $V^{(-1)}$ is chosen to be a
massless vector, the second vertex $Tensor^{(-1)}$ is one of the two massive
tensor states and the third and the fourth vertices are tachyons $\phi^{(0)}$.
The total ghost charges sum up to $-2$. In Eq.(1.2) the first vertex
$\psi^{(-\frac{1}{2})}$ is chosen to be a massless spinor, the second vertex
$\chi^{(-\frac{1}{2})}$ is the $10D$ massive Majorana spinor and the third and
the fourth vertices are tachyons $\phi^{(-1)}$ and $\phi^{(0)}$ in $(-1)$ and
$(0)$ ghost pictures respectively. Again, the total ghost charges sum up to
$-2$.

The two first vertices $\\{V^{(-1)}$ $,\psi^{(-\frac{1}{2})}\\}$ in Eq.(1.1)
and Eq.(1.2) form a massless $10D$ SUSY multiplet. Similarly,
$\\{Tensor^{(-1)},\chi^{(-\frac{1}{2})}\\}$ form a massive SUSY multiplet RRR
in the spectrum. The number of HSSA in Eq.(1.2) was calculated to be $16$
recently PFSSA . On the other hand, the HSSA of Eq.(1.1) with polarizations on
the scattering plane was calculated in susy  and was listed in Eq.(3.35). We
will use slightly different approach to obtain the result in this paper. Since
the number of bosonic HSSA calculated in susy  was less than $16$, we expect
the existence of HSSA with polarizations orthogonal to the scattering plane.
These bosonic HSSA were mainly due to the ”worldsheet fermion exchange” and
were first noted in susy . Motivated by the recent result of PFSSA PFSSA , we
will calculate in this paper the complete bosonic HSSA including all HSSA with
polarizations orthogonal to the scattering plane.

Recently there has been a new perspective to demonstrate the Gross conjecture
regarding the high energy symmetry of string theory, see the recent review
paper in LSSA . The authors in LSSA  constructed the exact string scattering
amplitudes (SSA) of three tachyons and one arbitrary string state, or the
Lauricella SSA (LSSA), in the $26D$ open bosonic string theory. These LSSA
form an infinite dimensional representation of the $SL(K+3,\mathbb{C})$ group.
Moreover, they show that the $SL(K+3,\mathbb{C})$ group can be used to solve
all the LSSA through infinite recurrence relations and express them in terms
of $1$ amplitude. This seems to agree with the result obtained in Mizera  that
there are $((n-3)!)^{2}=1$ ”linearly-independent integrals” for the case of
$n=4$. Moreover, as an application in the hard scattering limit, the LSSA can
be used to rederive infinite linear relations (with constant coefficients
independent of the scattering angle), which directly prove Gross conjecture.

In the next section, we will first calculate all physical states which include
zero norm states (ZNS) of the first mass level in the NS sector of old
covariant quantization string spectrum. In particular, we will pay attention
to ZNS and the enlarged gauge symmetries induced by them. Surprisingly, in
contrast to the open bosonic string theory ZNS1 , there is $\mathit{no}$
”inter-particle gauge transformation” for the two positive-norm states at this
mass level. Since at each fixed mass level, it is believed that string states
of both NS sector and R sector form a large gauge multiplet, this result seems
to tell us that for the case of superstring, one cannot ignore the effect of R
sector when considering high energy symmetry of superstring theory.

##  II ZNS and enlarged gauge transformations in the NS sector

In this section, we will first calculate all physical states for the first
massive level in the NS sector (0 ghost picture). This will include all ZNS
and the corresponding enlarged gauge transformations induced by them.
Surprisingly, in contrast to the open bosonic string theory ZNS1 , there is
$\mathit{no}$ inter-particle enlarged gauge transformation for the positive-
norm anti-symmetric spin three state and symmetric spin two state at this mass
level. The calculation is lengthy but elementary and we will mainly list the
results.

In the old covariant first quantized (OCFQ) spectrum of 10D open superstring
theory, the solutions of physical state conditions include positive-norm
states and two types of ZNS. In the NS sector, the ZNS are GSW

Type I $\displaystyle:G_{\frac{-1}{2}}\left|\chi\right\rangle,$ (2.1) where
$\displaystyle\text{ \
}G_{\frac{1}{2}}\left|\chi\right\rangle=G_{\frac{3}{2}}\left|\chi\right\rangle=L_{0}\left|\chi\right\rangle=0;$
(2.2) Type II
$\displaystyle:\left(G_{\frac{-3}{2}}+2G_{\frac{-1}{2}}L_{-1}\right)\left|\chi^{\prime}\right\rangle,$
(2.3) where $\displaystyle\text{ \ \
}G_{\frac{1}{2}}\left|\chi^{\prime}\right\rangle=G_{\frac{3}{2}}\left|\chi^{\prime}\right\rangle=\left(L_{0}+1\right)\left|\chi^{\prime}\right\rangle=0.$
(2.4)

In this paper, we will use the notation of reference GSW . Note that while
type I ZNS are zero-norm at any spacetime dimension, type II ZNS are zero-norm
only at $D=10$.

The most general positive-norm states for the first massive level in the NS
sector can be calculated to be

$\left(\epsilon_{\mu\nu\rho}b_{\frac{-1}{2}}^{\mu}b_{\frac{-1}{2}}^{\nu}b_{\frac{-1}{2}}^{\rho}+\epsilon_{\mu\nu}b_{\frac{-1}{2}}^{\mu}\alpha_{-1}^{\nu}+\epsilon_{\mu}b_{\frac{-3}{2}}^{\mu}\right)\left|0,k\right\rangle$
(2.5)

with the on-shell conditions

$3\epsilon_{\left[\mu\nu\rho\right]}k^{\rho}-\epsilon_{\left[\mu\nu\right]}=0\text{,
}\left(\epsilon_{\left(\mu\nu\right)}k^{\mu}+\epsilon_{\nu}\right)=0.$ (2.6)

The above on-shell conditions imply that we can divide Eq.(2.5) into two
states, the anti-symmetric spin three state which is described by the doublet
$\\{\epsilon_{\mu\nu\rho},\epsilon_{\left[\mu\nu\right]}\\}$

$\displaystyle\left(\epsilon_{\mu\nu\rho}b_{\frac{-1}{2}}^{\mu}b_{\frac{-1}{2}}^{\nu}b_{\frac{-1}{2}}^{\rho}+\epsilon_{[\mu\nu]}b_{\frac{-1}{2}}^{\mu}\alpha_{-1}^{\nu}\right)\left|0,k\right\rangle,$
(2.7) $\displaystyle\text{ \ with
}3\epsilon_{\left[\mu\nu\rho\right]}k^{\rho}-\epsilon_{\left[\mu\nu\right]}=0;$
(2.8)

and the symmetric spin two state which is described by the doublet
$\\{\epsilon_{(\mu\nu)},\epsilon_{\mu}\\}$

$\displaystyle\left(\epsilon_{(\mu\nu)}b_{\frac{-1}{2}}^{\mu}\alpha_{-1}^{\nu}+\epsilon_{\mu}b_{\frac{-3}{2}}^{\mu}\right)\left|0,k\right\rangle,$
(2.9) $\displaystyle\text{ \ with
}\epsilon_{\left(\mu\nu\right)}k^{\mu}+\epsilon_{\nu}=0.$ (2.10)

We are now ready to calculate the ZNS. The most general form of type I ZNS
turns out to be

$\displaystyle\left|ZNS\right\rangle$
$\displaystyle=\left[\left(k_{[\mu}\theta_{\nu]}+k_{(\mu}\theta_{\nu)}-2\theta_{\left[\mu\nu\right]}\right)b_{\frac{-1}{2}}^{\mu}\alpha_{-1}^{\nu}\right.$
$\displaystyle+\left.k_{[\rho}\theta_{\mu\nu]}b_{\frac{-1}{2}}^{\rho}b_{\frac{-1}{2}}^{\mu}b_{\frac{-1}{2}}^{\nu}+\theta_{\mu}b_{\frac{-3}{2}}^{\mu}\right]\left|0,k\right\rangle$
(2.11)

with the on-shell conditions

$2k^{\mu}\theta_{\left[\mu\nu\right]}+\theta_{\nu}=0,(\text{automatically
}k\cdot\theta=0).$ (2.12)

For the rest of this paper, we define
$e^{P}=\frac{1}{M_{2}}(E_{2},\mathrm{k}_{2},0)=\frac{k_{2}}{M_{2}}$ the
momentum polarization, $e^{L}=\frac{1}{M_{2}}(\mathrm{k}_{2},E_{2},0)$ the
longitudinal polarization and $e^{T}=(0,0,1)$ the transverse polarization on
the scattering plane. In addition, we define
$e^{T_{I}}=\\{e^{T},e^{T_{3}},e^{T_{4}},....,e^{T_{9}})=\\{e^{T},e^{T_{i}}\\}$
to represent all transverse directions including directions $e^{T_{i}}$
orthogonal to the scattering plane.

One obvious solution of Eq.(2.12) is to choose

$\theta_{\nu}^{\left(1\right)}=0\text{ and
}\theta_{\left[\mu\nu\right]}^{\left(1\right)}=\frac{1}{2}\left(e_{\mu}^{I}e_{\nu}^{J}-e_{\nu}^{I}e_{\mu}^{J}\right)\text{,
}I,J\neq P\text{,}$ (2.13)

and one obtains the anti-symmetric spin two ZNS

$\left|ZNS\right\rangle=\left(k_{[\rho}\theta_{\mu\nu]}b_{\frac{-1}{2}}^{\rho}b_{\frac{-1}{2}}^{\mu}b_{\frac{-1}{2}}^{\nu}-2\theta_{\left[\mu\nu\right]}b_{\frac{-1}{2}}^{\mu}\alpha_{-1}^{\nu}\right)\left|0,k\right\rangle,$
(2.14)

with $k^{\mu}\theta_{\left[\mu\nu\right]}=0$. It can be shown that the ZNS in
Eq.(2.14) generates an enlarged on-shell gauge transformation

$\epsilon_{\left[\mu\nu\rho\right]}^{\prime}=\left(\epsilon_{\left[\mu\nu\rho\right]}+k_{[\mu}\theta_{\nu\rho]}\right),\epsilon_{\left[\mu\nu\right]}^{\prime}=\epsilon_{\left[\mu\nu\right]}-\theta_{\left[\mu\nu\right]}$
(2.15)

for the anti-symmetric spin three state in Eq.(2.7).

The second solution of Eq.(2.12) is to choose

$\displaystyle\theta_{\nu}^{\left(2\right)}$
$\displaystyle=-2k^{\mu}\theta_{\left[\mu\nu\right]}=\frac{-1}{M}k^{\mu}\left(e_{\mu}^{P}e_{\nu}^{K}-e_{\nu}^{P}e_{\mu}^{K}\right)=e_{\nu}^{K},$
(2.16) $\displaystyle\theta_{\left[\mu\nu\right]}^{\left(2\right)}$
$\displaystyle=\frac{1}{2M}\left(e_{\mu}^{P}e_{\nu}^{K}-e_{\nu}^{P}e_{\mu}^{K}\right)\text{,
}K\neq P.$ (2.17)

For this choice, Eq.(2.11) got simplified and one obtains a vector ZNS

$\left|ZNS\right\rangle=\left(k_{(\mu}\theta_{\nu)}b_{\frac{-1}{2}}^{\mu}\alpha_{-1}^{\nu}+\theta_{\mu}b_{\frac{-3}{2}}^{\mu}\right)\left|0,k\right\rangle,$
(2.18)

with $k\cdot\theta=0$. It can be shown that the ZNS in Eq.(2.18) generates an
enlarged on-shell gauge transformation

$\epsilon_{\left(\mu\nu\right)}^{\prime}=\epsilon_{\left(\mu\nu\right)}+k_{(\mu}\theta_{\nu)},\epsilon_{\mu}^{\prime}=\left[\epsilon_{\mu}+\theta_{\mu}\right]$
(2.19)

for the symmetric spin two state in Eq.(2.9). To obtain the vector ZNS state
in Eq.(2.18), one can alternatively propose

$\left|\chi\right\rangle=\left[\frac{1}{2M^{2}}\left(k_{\mu}\theta_{\nu}-k_{\nu}\theta_{\mu}\right)b_{\frac{-1}{2}}^{\mu}b_{\frac{-1}{2}}^{\nu}+\theta_{\mu}\alpha_{-1}^{\mu}\right]\left|0,k\right\rangle,$
(2.20)

with $k\cdot\theta=0$ and put it in Eq.(2.1).

The only type II ZNS is a scalar and can be calculated to be

$\left|ZNS\right\rangle=\left[3k_{\mu}b_{\frac{-3}{2}}^{\mu}+\left(\eta_{\mu\nu}+2k_{\mu}k_{\nu}\right)b_{\frac{-1}{2}}^{\mu}\alpha_{-1}^{\nu}\right]\left|0,k\right\rangle.$
(2.21)

It can be shown that the ZNS in Eq.(2.21) generates an enlarged on-shell gauge
transformation

$\epsilon_{\left(\mu\nu\right)}^{\prime}=\epsilon_{\left(\mu\nu\right)}+\eta_{\mu\nu}+2k_{\mu}k_{\nu},\epsilon_{\mu}^{\prime}=\epsilon_{\mu}+3k_{\mu}$
(2.22)

for the symmetric spin two state in Eq.(2.9).

We note that the number of ZNS in Eq.(2.14), Eq.(2.18) and Eq.(2.21) can be
counted to be $36+9+1=46$. On the other hand, the number of ”spurious states”
in the $-1$ ghost picture were calculated RRR  to be $36+9+1=46,$ which is
consistent with the results calculated in the $0$ ghost picture in this paper.

Surprisingly, in contrast to the open bosonic string theory ZNS1 , there is
$\mathit{no}$ ”inter-particle gauge transformation” for the positive-norm
anti-symmetric spin three state and symmetric spin two state at this mass
level.

Here we briefly review the results of $26D$ open bosonic string theory at mass
level $M^{2}=4$ ZNS1 . At this mass level, there are two positive-norm states,
the symmetric spin three state and the anti-symmetric spin two state. In
addition there are three type I ZNS, the symmetric spin two, a vector and a
scalar, and one type II vector ZNS. One can do a linear combination of the two
degenerate vector ZNS to obtain the so-called $D_{2}$ vector ZNS ZNS1

$\displaystyle|D_{2}\rangle$
$\displaystyle=\left[\left(\frac{1}{2}k_{\mu}k_{\nu}\theta_{\lambda}^{2}+2\eta_{\mu\nu}\theta_{\lambda}^{2}\right)\alpha_{-1}^{\mu}\alpha_{-1}^{\nu}\alpha_{-1}^{\lambda}\right.$
$\displaystyle+\left.9k_{\mu}\theta_{\nu}^{2}\alpha_{-2}^{[\mu}\alpha_{-1}^{\nu]}-6\theta_{\mu}^{2}\alpha_{-3}^{\mu}\right]\left|0,k\right\rangle,$
(2.23)

with $k\cdot\theta^{2}=0$. The above $D_{2}$ ZNS generates an ”inter-particle
gauge transformation” for the positive-norm symmetric spin three state and the
anti-symmetric spin two state. Thus these two states form a large gauge
multiplet. The other three ZNS generate gauge transformations for the
symmetric spin three state. An important consequence of this inter-particle
symmetry is the linear relation which relates hard string scattering
amplitudes of the two positive-norm states. Indeed, all HSSA are proportional
to each others with the ratios ChanLee1 ; ChanLee2

$\displaystyle\mathcal{T}_{TTT}:\mathcal{T}_{LLT}:\mathcal{T}_{(LT)}:\mathcal{T}_{[LT]}$
$\displaystyle=$
$\displaystyle\alpha_{-1}^{T}\alpha_{-1}^{T}\alpha_{-1}^{T}:\alpha_{-1}^{L}\alpha_{-1}^{L}\alpha_{-1}^{T}:\alpha_{-1}^{(L}\alpha_{-2}^{T)}:\alpha_{-1}^{[L}\alpha_{-2}^{T]}$
$\displaystyle=$ $\displaystyle 8:1:-1:-1.$ (2.24)

Although there is no ”inter-particle gauge transformation” at mass level
$M^{2}=2$ in the NS sector of the spectrum of the superstring, in the next
section we will see that the $16$ HSSA originated from the two positive-norm
states are still related and are indeed again proportional to each others.
Presumably, this is due to the spacetime SUSY and the spacetime massive
fermion string scattering amplitudes of the fermionic sector of the theory
PFSSA .

##  III Hard string scattering amplitudes in the NS sector

In this section we will calculate all HSSA in Eq.(1.1) including some
$\mathit{new}$ ones which were not considered in susy . We will demonstrate
the linear relations among the HSSA between the two positive-norm states, and
thus extend and justify the original Gross conjecture to the case of
superstring. Similar inter-particle symmetry was discovered ChanLee1 ;
ChanLee2  in the 26D bosonic string theory.

For the case of the bosonic string, string scattering amplitudes with
polarizations orthogonal to the scattering plane are of subleading order in
energy in the hard scattering limit. However, it was noted that susy  for the
HSSA of the NS sector of superstring, there existed leading order HSSA with
polarizations orthogonal to the scattering plane. This was due to the
”worldsheet fermion exchange” in the correlation functions and was argued to
be related to the massive spacetime fermion HSSA of the theory.

In the subsection A of this section, we will first calculate ratios among HSSA
with polarizations on the scattering plane by using the hard Ward identities
(or decoupling of ZNS in the hard scattering limit), and then by using
explicit HSSA calculations. In subsections B and C, we will calculate two
types of new HSSA with polarizations orthogonal to the scattering plane.

###  III.1 Hard SSA with polarizations on the scattering plane

We first do the hard Ward identity (HWI) calculation for polarizations on the
scattering plane. In the hard scattering limit, one identifies $e^{L}=e^{P}$
ChanLee1 ; ChanLee2 . In the following calculation, each scattering amplitude
has been assigned a relative energy power. For each longitudinal $L$
component, the energy order is $E^{2}$ while for each transverse $T$
component, the energy order is $E.$ This is due to the definitions of $e^{L}$
and $e^{T}$ in the paragraph below Eq.(2.12) where $e^{L}$ got one energy
power more than that of $e^{T}.$

In the HWI calculation, we put the ZNS in the second vertex and suppress the
indices for the other vertices. The ZNS in Eq.(2.14) gives the HWI

$\left[\overset{\left(3\rightarrow
1\right)}{\overbrace{\alpha_{-1}^{L}b_{\frac{-1}{2}}^{T}-\alpha_{-1}^{T}b_{\frac{-1}{2}}^{L}}}\right]\left|0,k\right\rangle=0.$
(3.25)

For the choice of $\theta^{\left(1\right)}=e^{L}$, the ZNS in Eq.(2.18) gives
the HWI

$\left[\overset{\left(2\right)}{\overbrace{b_{\frac{-3}{2}}^{L}}}+\frac{1}{M}\overset{\left(4\rightarrow
2\right)}{\overbrace{\alpha_{-1}^{L}b_{\frac{-1}{2}}^{L}}}\right]\left|0,k\right\rangle.$
(3.26)

For the choice of $\theta^{\left(1\right)}=e^{T}$, the ZNS in Eq.(2.18) gives
the HWI

$\left[\overset{1}{\overbrace{b_{\frac{-3}{2}}^{T}}}+\frac{1}{2M}\overset{\left(3\rightarrow
1\right)}{\overbrace{\left(\alpha_{-1}^{L}b_{\frac{-1}{2}}^{T}+\alpha_{-1}^{T}b_{\frac{-1}{2}}^{L}\right)}}\right]\left|0,k\right\rangle=0.$
(3.27)

The ZNS in Eq.(2.21) gives the HWI

$\left(\frac{3}{M}b_{\frac{-3}{2}}^{L}+\eta_{\mu\nu}\alpha_{-1}^{\mu}b_{\frac{-1}{2}}^{\nu}+\frac{2}{M^{2}}\alpha_{-1}^{L}b_{\frac{-1}{2}}^{L}\right)\left|0,k\right\rangle=0$
(3.28)

where

$\eta_{\mu\nu}=-e_{\mu}^{P}e_{\nu}^{P}+e_{\mu}^{L}e_{\nu}^{L}+e_{\mu}^{T}e_{\nu}^{T}+\underset{\alpha_{-1}^{T_{i}}b_{\frac{1}{2}}^{T_{i}}\rightarrow\text{not
contribute}}{\underbrace{\overset{7}{\underset{i=1}{\sum}}e_{\mu}^{T_{i}}e_{\nu}^{T_{i}}}.}$
(3.29)

In sum, ZNS in Eq.(2.21) gives the HWI

$\left[3M\overset{\left(2\right)}{\overbrace{b_{\frac{-1}{2}}^{L}}}+M^{2}\overset{\left(2\right)}{\overbrace{\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T}}}+2\overset{\left(4\rightarrow
2\right)}{\overbrace{\alpha_{-1}^{L}b_{\frac{-1}{2}}^{L}}}\right]\left|0,k\right\rangle=0.$
(3.30)

The naive energy order of $2\alpha_{-1}^{L}b_{\frac{-1}{2}}^{L}$ term in
Eq.(3.30) is $E^{4}$, the HWI forces its energy order to drop to $E^{2}$.
Similar calculations were performed in Eq.(3.25),Eq.(3.26) and Eq.(3.27). We
note that the energy order $E^{3}$ term in Eq.(3.27) gives

$\alpha_{-1}^{L}b_{\frac{-1}{2}}^{T}+\alpha_{-1}^{T}b_{\frac{-1}{2}}^{L}=0,$
(3.31)

which together with Eq.(3.25) force the following two $E^{3}$ order terms
vanish

$\overset{\left(3\right)}{\overbrace{\alpha_{-1}^{L}b_{\frac{-1}{2}}^{T}}}=0=\overset{\left(3\right)}{\overbrace{\alpha_{-1}^{T}b_{\frac{-1}{2}}^{L}}}.$
(3.32)

Thus the real leading order terms are those of energy order $E^{2}$ in
Eq.(3.26) and Eq.(3.30)

$\displaystyle\alpha_{-1}^{L}b_{\frac{-1}{2}}^{L}+Mb_{\frac{-3}{2}}^{L}$
$\displaystyle=0,$ (3.33) $\displaystyle
3Mb_{\frac{-1}{2}}^{L}+M^{2}\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T}+2\alpha_{-1}^{L}b_{\frac{-1}{2}}^{L}$
$\displaystyle=0,$ (3.34)

which can be solved to give the ratios

$b_{\frac{-3}{2}}^{L}:\alpha_{-1}^{L}b_{\frac{-1}{2}}^{L}:\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T}=\frac{-1}{M}:\frac{1}{M^{2}}:1.$
(3.35)

The result in Eq.(3.35) is consistent with the calculation obtained in susy
by a slightly different approach.

The ratios in Eq.(3.35) can be rederived by calculating a set of sample HSSA.
We first calculate

$\displaystyle\left\langle
V^{(-1)}Tensor^{(-1)}\phi^{(0)}\phi^{(0)}\right\rangle={\displaystyle\int}dz_{1}dz_{2}dz_{3}dz_{4}\frac{\left|z_{13}z_{14}z_{34}\right|}{dz_{1}dz_{3}dz_{4}}$
$\displaystyle\cdot\left\langle\begin{array}[c]{c}\left(\epsilon_{1}^{\mu_{1}}\psi_{1\mu_{1}}e^{-\phi_{1}}e^{ik_{1}X_{1}}\right)\left(\epsilon_{2}^{\mu_{2}}\psi_{2\mu_{2}}e^{-\phi_{2}}i\epsilon_{\mu}\partial
X_{2}^{\mu}e^{ik_{2}X_{2}}\right)\\\
\times\left(k_{3}^{\mu_{3}}\psi_{3\mu_{3}}e^{ik_{3}X_{3}}\right)\left(k_{4}^{\mu_{4}}\psi_{4\mu_{4}}e^{ik_{4}X_{4}}\right)\end{array}\right\rangle$
(3.38)

where the first vertex $V^{(-1)}$ is chosen to be a vector, the second vertex
is the $\alpha_{-1}^{T}b_{-\frac{1}{2}}^{T}\left|0;k\right\rangle$ tensor
state and the third and the fourth vertices are tachyons $\phi^{(0)}$. The
total ghost charges sum up to $-2$. We first put $\epsilon=e^{T}$ to get

$\displaystyle\left\langle
V^{(-1)}Tensor^{(-1)}\phi^{(0)}\phi^{(0)}\right\rangle$
$\displaystyle=\epsilon_{1}^{\mu_{1}}\epsilon_{2}^{\mu_{2}}k_{3}^{\mu_{3}}k_{4}^{\mu_{4}}{\displaystyle\int}dz_{2}\left[\frac{\eta_{\mu_{1}\mu_{2}}}{z_{12}}\frac{\eta_{\mu_{3}\mu_{4}}}{z_{34}}-\frac{\eta_{\mu_{1}\mu_{3}}}{z_{13}}\frac{\eta_{\mu_{2}\mu_{4}}}{z_{24}}+\frac{\eta_{\mu_{1}\mu_{4}}}{z_{14}}\frac{\eta_{\mu_{2}\mu_{3}}}{z_{23}}\right]$
$\displaystyle\times\left|z_{12}\right|^{k_{1}\cdot
k_{2}-1}\left|z_{13}\right|^{k_{1}\cdot
k_{3}+1}\left|z_{14}\right|^{k_{1}\cdot
k_{4}+1}\left|z_{23}\right|^{k_{2}\cdot k_{3}}\left|z_{24}\right|^{k_{2}\cdot
k_{4}}\left|z_{34}\right|^{k_{3}\cdot
k_{4}+1}\left(\frac{k_{3}^{T}}{z_{23}}+\frac{k_{4}^{T}}{z_{24}}\right).$
(3.39)

The next step is to fix the gauge $z_{1}=0,z_{3}=1,z_{4}\rightarrow\infty$
$(0<z_{2}<1)$ and use $k_{1}\cdot k_{2}=-\frac{s}{2}+1,k_{2}\cdot
k_{3}=-\frac{t}{2}+\frac{1}{2}$ to perform the integration to obtain

$\displaystyle\left\langle
V^{(-1)}Tensor^{(-1)}\phi^{(0)}\phi^{(0)}\right\rangle=B\left(\frac{-s}{2},\frac{-t}{2}+\frac{1}{2}\right)$
$\displaystyle\cdot\left(-k_{3}^{T}\right)\left[\begin{array}[c]{c}\left(\epsilon_{1}\cdot\epsilon_{2}\right)\left(k_{3}\cdot
k_{4}\right)\\\
-\left(k_{3}\cdot\epsilon_{1}\right)\left(k_{4}\cdot\epsilon_{2}\right)\frac{s}{u+1}\\\
+\left(k_{3}\cdot\epsilon_{2}\right)\left(k_{4}\cdot\epsilon_{1}\right)\frac{s}{t+1}\end{array}\right].$
(3.43)

where
$B\left(\frac{-s}{2},\frac{-t}{2}+\frac{1}{2}\right)=\frac{\Gamma\left(\frac{-s}{2}\right)\Gamma\left(\frac{-t}{2}+\frac{1}{2}\right)}{\Gamma\left(\frac{u}{2}+\frac{1}{2}\right)}$
is the Beta function.

Finally in the hard scattering limit we have

$\displaystyle k_{1}$
$\displaystyle=\left(E,-E,0\right),k_{2}=\left(E,E,0\right),$ $\displaystyle
k_{3}$ $\displaystyle=\left(-E,-E\cos\theta,-E\sin\theta\right),$
$\displaystyle k_{4}$ $\displaystyle=\left(-E,E\cos\theta,E\sin\theta\right)$
(3.44)

and

$\displaystyle\epsilon_{1}$
$\displaystyle=\epsilon_{2}=e^{T}=\left(0,0,1\right),s=4E^{2},$
$\displaystyle\frac{s}{u}$
$\displaystyle\rightarrow\frac{-1}{\cos^{2}\frac{\theta}{2}},\frac{s}{t}\rightarrow\frac{-1}{\sin^{2}\frac{\theta}{2}},k_{3}\cdot
k_{4}=\frac{-s}{2}-1.$ (3.45)

After a bit calculation, one ends up with

$\left\langle
V^{(-1)}Tensor_{(\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T})}^{(-1)}\phi^{(0)}\phi^{(0)}\right\rangle=-2k_{3}^{T}E^{2}B$
(3.46)

where $B$ is the hard scattering limit of the Beta function.

Similar calculations can be performed if one replaces the second vertex
$\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T}$ by
$\alpha_{-1}^{L}b_{\frac{-1}{2}}^{L}$and
$b_{-\frac{3}{2}}^{T}\left|0;k\right\rangle$ respectively. The results are

$\displaystyle\left\langle
V^{(-1)}Tensor_{(\alpha_{-1}^{L}b_{\frac{-1}{2}}^{L})}^{(-1)}\phi^{(0)}\phi^{(0)}\right\rangle$
$\displaystyle=\frac{-2k_{3}^{T}E^{2}B}{M^{2}},$ (3.47)
$\displaystyle\left\langle
V^{(-1)}Tensor_{(b_{-\frac{3}{2}}^{T})}^{(-1)}\phi^{(0)}\phi^{(0)}\right\rangle$
$\displaystyle=\frac{2k_{3}^{T}E^{2}B}{M}.$ (3.48)

These results justify the ratios calculated in Eq.(3.35). They also agree with
the results obtained by the saddle-point method susy .

###  III.2 New hard SSA I

In addition to the three HSSA calculated in Eq.(3.46), Eq.(3.47) and
Eq.(3.48), it was noted that susy  there existed HSSA with polarizations
orthogonal to the scattering plane. In particular, one can replace
$b_{\frac{-1}{2}}^{T}$ operator in the vertex by
$b_{\frac{-1}{2}}^{T_{i}}$where $T_{i},i=3,4,5,\cdots,9$ represents directions
orthogonal to the scattering plane. Note that the corresponding string
scattering amplitudes are of subleading order in energy for the case of
bosonic string theory. We first propose the following HSSA (we use the obvious
brief notation)

$\displaystyle\left\langle
b_{\frac{-1}{2}}^{T_{i}},\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T_{j}},\phi_{3},\phi_{4}\right\rangle=B\left(\frac{-s}{2},\frac{-t}{2}+\frac{1}{2}\right)$
$\displaystyle\cdot\left(-k_{3}^{T}\right)\left[\begin{array}[c]{c}\left(e^{T_{i}}\cdot
e^{T_{j}}\right)\left(k_{3}\cdot k_{4}\right)\\\ -\left(k_{3}\cdot
e^{T_{i}}\right)\left(k_{4}\cdot e^{T_{j}}\right)\frac{s}{u+1}\\\
+\left(k_{3}\cdot e^{T_{j}}\right)\left(k_{4}\cdot
e^{T_{i}}\right)\frac{s}{t+1}\end{array}\right].$ (3.52)

The calculation of the above amplitude is similar to Eq.(3.43).

Finally in the hard scattering limit, one ends up with

$\left\langle
b_{\frac{-1}{2}}^{T_{i}},\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T_{j}},\phi_{3},\phi_{4}\right\rangle=\delta_{ij}2k_{3}^{T}E^{2}B.$
(3.53)

Eq.(3.53) contributes $7$ more HSSA in the NS sector at this mass level.

###  III.3 New hard SSA II

So far all HSSA calculated in this section are from the positive-norm
symmetric spin two state. One expects HSSA contributed from the anti-symmetric
spin three state. In this subsection we propose the following HSSA

$\left\langle
b_{\frac{-1}{2}}^{T_{K}},b_{\frac{-1}{2}}^{L}b_{\frac{-1}{2}}^{T_{I}}b_{\frac{-1}{2}}^{T_{J}},\phi_{3},\phi_{4}\right\rangle,$
(3.54)

which was not considered in susy . The calculation of this HSSA is lengthy but
straightforward. We begin with the standard calculation of string scattering
amplitude

$\displaystyle\left\langle
V^{(-1)}Tensor_{bbb}^{(-1)}\phi^{(0)}\phi^{(0)}\right\rangle={\displaystyle\int}dz_{1}dz_{2}dz_{3}dz_{4}\frac{\left|z_{13}z_{14}z_{34}\right|}{dz_{1}dz_{3}dz_{4}}\left\langle\begin{array}[c]{c}\left(\epsilon_{\mu_{1}}^{T_{K}}\psi_{1}^{\mu_{1}}e^{-\phi_{1}}e^{ik_{1}X_{1}}\right)\left(\epsilon_{\mu_{2}}^{L}\epsilon_{\mu
3}^{T_{I}}\epsilon_{\mu
4}^{T_{J}}\psi_{2}^{\mu_{2}}\psi_{2}^{\mu_{3}}\psi_{2}^{\mu_{4}}e^{-\phi_{1}}e^{ik_{2}X_{2}}\right)\\\
\times\left(k_{\mu_{5}}^{3}\psi_{3}^{\mu_{5}}e^{ik_{3}X_{3}}\right)\left(k_{\mu_{6}}^{4}\psi_{4}^{\mu_{6}}e^{ik_{4}X_{4}}\right)\end{array}\right\rangle$
(3.57)
$\displaystyle=\epsilon_{\mu_{1}}^{T_{K}}\epsilon_{\mu_{2}}^{L}\epsilon_{\mu
3}^{T_{I}}\epsilon_{\mu
4}^{T_{J}}k_{\mu_{5}}^{3}k_{\mu_{6}}^{4}{\displaystyle\int}dz_{2}\left|z_{13}z_{14}z_{34}\right|\left\langle\psi_{1}^{\mu_{1}}\psi_{2}^{\mu_{2}}\psi_{2}^{\mu_{3}}\psi_{2}^{\mu_{4}}\psi_{3}^{\mu_{5}}\psi_{4}^{\mu_{6}}\right\rangle\left\langle
e^{-\phi_{1}}e^{-\phi_{2}}\right\rangle\left\langle
e^{ik_{1}X_{1}}e^{ik_{2}X_{2}}e^{ik_{3}X_{3}}e^{ik_{4}X_{4}}\right\rangle.$
(3.58)

The next step is to fix the gauge $z_{1}=0,z_{3}=1,z_{4}\rightarrow\infty$
$(0<z_{2}<1)$ to get

$\displaystyle\left\langle
V^{(-1)}Tensor_{bbb}^{(-1)}\phi^{(0)}\phi^{(0)}\right\rangle$ $\displaystyle=$
$\displaystyle-\epsilon_{\mu_{1}}^{T_{K}}\epsilon_{\mu_{2}}^{L}\epsilon_{\mu
3}^{T_{I}}\epsilon_{\mu
4}^{T_{J}}k_{\mu_{5}}^{3}k_{\mu_{6}}^{4}{\displaystyle\int}dz_{2}\left(z_{2}\right)^{k_{1}\cdot
k_{2}-2}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}-1}$
$\displaystyle\cdot\left(\begin{array}[c]{c}\eta^{\mu_{1}\mu_{2}}\eta^{\mu_{4}\mu_{5}}\eta^{\mu_{3}\mu_{6}}-\eta^{\mu_{1}\mu_{2}}\eta^{\mu_{3}\mu_{5}}\eta^{\mu_{4}\mu_{6}}\\\
\eta^{\mu_{1}\mu_{3}}\eta^{\mu_{2}\mu_{5}}\eta^{\mu_{4}\mu_{6}}-\eta^{\mu_{1}\mu_{3}}\eta^{\mu_{4}\mu_{5}}\eta^{\mu_{2}\mu_{6}}\\\
\eta^{\mu_{1}\mu_{4}}\eta^{\mu_{3}\mu_{5}}\eta^{\mu_{2}\mu_{6}}-\eta^{\mu_{1}\mu_{4}}\eta^{\mu_{2}\mu_{5}}\eta^{\mu_{3}\mu_{6}}\end{array}\right).$
(3.62)

We can now use $k_{1}\cdot k_{2}=-\frac{s}{2}+1,k_{2}\cdot
k_{3}=-\frac{t}{2}+\frac{1}{2}$ and perform the integration to get

$\displaystyle\left\langle
V^{(-1)}Tensor_{bbb}^{(-1)}\phi^{(0)}\phi^{(0)}\right\rangle$ $\displaystyle=$
$\displaystyle-\epsilon_{\mu_{1}}^{T_{K}}\epsilon_{\mu_{2}}^{L}\epsilon_{\mu
3}^{T_{I}}\epsilon_{\mu
4}^{T_{J}}k_{\mu_{5}}^{3}k_{\mu_{6}}^{4}{\displaystyle\int}dz_{2}\left(z_{2}\right)^{-\frac{s}{2}-1}\left(1-z_{2}\right)^{-\frac{t}{2}-\frac{1}{2}}$
$\displaystyle\cdot\left(\begin{array}[c]{c}\eta^{\mu_{1}\mu_{3}}\eta^{\mu_{2}\mu_{5}}\eta^{\mu_{4}\mu_{6}}-\eta^{\mu_{1}\mu_{3}}\eta^{\mu_{4}\mu_{5}}\eta^{\mu_{2}\mu_{6}}\\\
\eta^{\mu_{1}\mu_{4}}\eta^{\mu_{3}\mu_{5}}\eta^{\mu_{2}\mu_{6}}-\eta^{\mu_{1}\mu_{4}}\eta^{\mu_{2}\mu_{5}}\eta^{\mu_{3}\mu_{6}}\end{array}\right).$
(3.65) $\displaystyle=$
$\displaystyle-\frac{\Gamma\left(\frac{-s}{2}\right)\Gamma\left(\frac{-t}{2}+\frac{1}{2}\right)}{\Gamma\left(\frac{u}{2}+\frac{1}{2}\right)}$
$\displaystyle\cdot\left(\begin{array}[c]{c}\delta_{IK}\frac{1}{M}\left(\frac{-t}{2}+\frac{1}{2}\right)k_{4}^{T_{J}}-\delta_{IK}\frac{1}{M}\left(\frac{-u}{2}+\frac{1}{2}\right)k_{3}^{T_{J}}\\\
\delta_{JK}\frac{1}{M}\left(\frac{-u}{2}+\frac{1}{2}\right)k_{3}^{T_{I}}-\delta_{JK}\frac{1}{M}\left(\frac{-t}{2}+\frac{1}{2}\right)k_{4}^{T_{I}}\end{array}\right).$
(3.68)

In the hard scattering limit, the HSSA reduces to

$\displaystyle\left\langle
b_{\frac{-1}{2}}^{T_{K}},b_{\frac{-1}{2}}^{L}b_{\frac{-1}{2}}^{T_{I}}b_{\frac{-1}{2}}^{T_{J}},\phi_{3},\phi_{4}\right\rangle$
$\displaystyle=$
$\displaystyle-\left[\begin{array}[c]{c}\delta_{IK}\frac{1}{M}\left(\frac{-t}{2}\right)k_{4}^{T_{J}}-\delta_{IK}\frac{1}{M}\left(\frac{-u}{2}\right)k_{3}^{T_{J}}\\\
\delta_{JK}\frac{1}{M}\left(\frac{-u}{2}\right)k_{3}^{T_{I}}-\delta_{JK}\frac{1}{M}\left(\frac{-t}{2}\right)k_{4}^{T_{I}}\end{array}\right]B.$
(3.71)

There are three cases to further reduce the result:

Case one: for
$\epsilon^{T_{K}}=e^{T},\epsilon^{T_{I}}=e^{T},\epsilon^{T_{J}}=e^{T_{j}},$

$\displaystyle-\left[\frac{1}{M}\left(\frac{-t}{2}\right)k_{4}^{T_{j}}-\frac{1}{M}\left(\frac{-u}{2}\right)k_{3}^{T_{j}}+0-0\right]$
$\displaystyle=\frac{1}{M}\left(\frac{t}{2}\right)\times
0+\frac{1}{M}\left(\frac{-u}{2}\right)\times 0=0.$ (3.72)

Case two: for
$\epsilon^{T_{K}}=e^{T},\epsilon^{T_{I}}=e^{T_{i}},\epsilon^{T_{J}}=e^{T_{j}},$

$-\left[\begin{array}[c]{c}0\times\frac{1}{M}\left(\frac{-t}{2}\right)k_{4}^{T_{j}}-0\times\frac{1}{M}\left(\frac{-u}{2}\right)k_{3}^{T_{j}}\\\
0\times\frac{1}{M}\left(\frac{-u}{2}\right)k_{3}^{T_{i}}-0\times\frac{1}{M}\left(\frac{-t}{2}\right)k_{4}^{T_{i}}\end{array}\right]=0.$
(3.73)

Case three: for
$\epsilon^{T_{K}}=e^{T_{k}},\epsilon^{T_{I}}=e^{T},\epsilon^{T_{J}}=e^{T_{j}},$

$\displaystyle-\left[\begin{array}[c]{c}0\times\frac{1}{M}\left(\frac{-t}{2}\right)k_{4}^{T_{j}}-0\times\frac{1}{M}\left(\frac{-u}{2}\right)k_{3}^{T_{j}}\\\
+\delta_{jk}\frac{1}{M}\left(\frac{-u}{2}\right)k_{3}^{T}-\delta_{jk}\frac{1}{M}\left(\frac{-t}{2}\right)k_{4}^{T}\end{array}\right]$
(3.76)
$\displaystyle=\delta_{jk}\frac{1}{2M}\left(t+u\right)k_{3}^{T}=\frac{-2E^{2}\delta_{jk}k_{3}^{T}}{M}.$
(3.77)

In summary, we get the following ratios in the NS sector

$\displaystyle\left\langle
b_{\frac{-1}{2}}^{T},\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T}\right\rangle:\left\langle
b_{\frac{-1}{2}}^{T},\alpha_{-1}^{L}b_{\frac{-1}{2}}^{L}\right\rangle:\left\langle
b_{\frac{-1}{2}}^{T},b_{\frac{-3}{2}}^{L}\right\rangle:\left\langle
b_{\frac{-1}{2}}^{T_{i}},\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T_{j}}\right\rangle:\left\langle
b_{\frac{-1}{2}}^{T_{k}},b_{\frac{-1}{2}}^{L}b_{\frac{-1}{2}}^{T}b_{\frac{-1}{2}}^{T_{l}}\right\rangle$
$\displaystyle=$
$\displaystyle-2k_{3}^{T}E^{2}B:\frac{-2k_{3}^{T}E^{2}B}{M^{2}}:\frac{2k_{3}^{T}E^{2}B}{M}:\delta_{ij}2k_{3}^{T}E^{2}B:\delta_{lk}\frac{-2k_{3}^{T}E^{2}}{M}B=1:\frac{1}{M^{2}}:-\frac{1}{M}:-\delta_{ij}:\frac{\delta_{lk}}{M}.$
(3.78)

where we have, for simplicity, omitted the last two tachyon vertices in the
notation of each HSSA in Eq.(3.78), which are the superstring version of the
bosonic string ratios in Eq.(2.24). However, there is no inter-particle gauge
transformation for the NS sector of the superstring case as in the analysis of
section II.

At this stage, one might think that there are $1+1+1+7+7=17$ HSSA in Eq.(3.78)
which is inconsistent with the $2^{4}=16$ hard PFSSA calculated in PFSSA . To
resolve the inconsistency, one notes that $b_{\frac{-3}{2}}^{L}$ in the third
amplitude in Eq.(3.78) cannot be alone a (high energy) physical state. So one
needs to do a linear combination of the first three amplitudes in Eq.(3.78) to
transform the HSSA from the operator basis to the physical state basis. The
first physical state amplitude can be obtained by choosing

$\text{\
}\epsilon_{\left(\mu\nu\right)}=e_{\mu}^{T}e_{\nu}^{T},\epsilon_{\nu}=-\epsilon_{\left(\mu\nu\right)}k^{\mu}=0$
(3.79)

in Eq.(2.9) to obtain

$\left(\epsilon_{(\mu\nu)}b_{\frac{-1}{2}}^{\mu}\alpha_{-1}^{\nu}+\epsilon_{\mu}b_{\frac{-3}{2}}^{\mu}\right)\left|0,k\right\rangle\rightarrow\left(b_{\frac{-1}{2}}^{T}\alpha_{-1}^{T}\right)\left|0;k\right\rangle\rightarrow
1.$ (3.80)

The second physical state amplitude can be obtained by choosing

$\displaystyle\epsilon_{\nu}$
$\displaystyle=-\epsilon_{\left(\mu\nu\right)}k^{\mu}=\frac{1}{2}e_{\mu}^{P}e_{\nu}^{L}k^{\mu}=\frac{-M}{2}e_{\nu}^{L},$
$\displaystyle\epsilon_{\left(\mu\nu\right)}$
$\displaystyle=\frac{1}{2}\left(e_{\mu}^{P}e_{\nu}^{L}+e_{\nu}^{P}e_{\mu}^{L}\right)$
(3.81)

in Eq.(2.9) to obtain

$\displaystyle\left(\epsilon_{\left(\mu\nu\right)}b_{\frac{-1}{2}}^{\mu}\alpha_{-1}^{\nu}+\epsilon_{\mu}b_{\frac{-3}{2}}^{\mu}\right)\left|0;k\right\rangle$
$\displaystyle\rightarrow$
$\displaystyle\left(b_{\frac{-1}{2}}^{P}\alpha_{-1}^{L}+b_{\frac{-1}{2}}^{L}\alpha_{-1}^{P}-Mb_{\frac{-3}{2}}^{L}\right)\left|0;k\right\rangle$
$\displaystyle=$
$\displaystyle\left(2b_{\frac{-1}{2}}^{L}\alpha_{-1}^{L}-Mb_{\frac{-3}{2}}^{L}\right)\left|0;k\right\rangle\rightarrow\frac{2}{M^{2}}+1.$
(3.82)

So the ratios of the hard SSA in the NS sector at this mass level are

$\displaystyle\left\langle
b_{\frac{-1}{2}}^{T},\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T}\right\rangle:\left\langle
b_{\frac{-1}{2}}^{T},\left(2b_{\frac{-1}{2}}^{L}\alpha_{-1}^{L}-Mb_{\frac{-3}{2}}^{L}\right)\right\rangle:\left\langle
b_{\frac{-1}{2}}^{T_{i}},\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T_{j}}\right\rangle:\left\langle
b_{\frac{-1}{2}}^{T_{k}},b_{\frac{-1}{2}}^{L}b_{\frac{-1}{2}}^{T}b_{\frac{-1}{2}}^{T_{l}}\right\rangle$
$\displaystyle=-2k_{3}^{T}E^{2}B:-2(\frac{2}{M^{2}}+1)k_{3}^{T}E^{2}B:\delta_{ij}2k_{3}^{T}E^{2}B:\delta_{lk}\frac{-2k_{3}^{T}E^{2}}{M}B=1:2:-\delta_{ij}:\frac{\delta_{lk}}{\sqrt{2}}.$
(3.83)

In sum, in the NS sector one gets

$1+1+7+7=16$ (3.84)

HSSA in Eq.(3.83). This result agrees with those of hard massive PFSSA in the
R sector calculated recently PFSSA .

##  IV Discussion

One main result of this work is the derivation of the superstring version of
Eq.(2.24), namely, the ratios in Eq.(3.78). The important motivation to do
this derivation is of course due to the recent calculation of the hard PFSSA
in PFSSA .

Both the effects of ”worldsheet fermion exchange” susy  and $\mathit{no}$
”inter-particle gauge transformation” of the NS sector shown in this paper are
related to the existence of R sector for the case of superstring. This means
that it is important to include R sector PFSSA  when considering high energy
symmetry of superstring theory. Unfortunately, the construction of the general
massive fermion string vertex operators is difficult at this stage. However,
the results of this paper seem to indicate that spacetime SUSY of the spectrum
may help to provide useful information.

###### Acknowledgements.

This work is supported in part by the Ministry of Science and Technology
(MoST) and S.T. Yau center of National Chiao Tung University (NCTU), Taiwan.

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