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\begin{document}

\title{Zero-norm States and Reduction of Stringy Scattering Amplitudes}
\author{Jen-Chi Lee}
\email{jcclee@cc.nctu.edu.tw}
\affiliation{Department of Electrophysics, National Chiao-Tung University, Hsinchu,
Taiwan, R.O.C. \ \ }
\date{\today }

\begin{abstract}
We give a systematic method to generate two types of zero-norm states in the
old covariant first quantized (OCFQ) spectrum of open bosonic string.
Zero-norm states up to the fourth massive level and general formulas of some
zero-norm tensor states at an arbitrary mass level are calculated. On-shell
Ward identities generated by zero-norm states can then be used to express
the stringy scattering amplitudes of the degenerate lower spin propagating
states in terms of those of higher spin propagating states at each fixed
mass level. This decoupling phenomenon is, in contrast to Gross's
high-energy symmetries, valid to \emph{all} energy. As examples, we
explicitly demonstrate this stringy phenomenon up to the fourth massive
level (spin-five), which justifies the calculation of two other previous
approaches based on the massive worldsheet sigma-model and Witten's string
field theory (WSFT).
\end{abstract}

\maketitle

\section{Introduction}

The theory of string, as a consistent quantum theory, has no free parameter
and an infinite number of states. It is thus conceivable that there exists
huge hidden symmetry group which is responsible for the ultraviolet
finiteness of the theory. In fact, it was conjectured by Gross\cite{1} more
than a decade ago that an infinite broken gauge symmetry get restored at
energy much higher than the Planck energy. Moreover, he showed that, for the
closed string, there existed an infinite number of linear relations between
the scattering amplitudes of different string states that were valid order
by order and were of the \emph{identical form} in string perturbation theory
as $\alpha ^{\prime }$\ goes to infinity. As a result, the scattering
amplitudes of all string states can be expressed in terms of, say, the
dilaton amplitudes. A similar result was presented in Ref \cite{2} for the
open string case. Soon later it was discovered that\cite{3} the equations of
motion for massive background fields of the degenerate positive-norm
propagating states can be expressed in terms of those of higher spin states
at each fixed mass level. This decoupling phenomenon was argued to be arisen
from the existence of two types of zero-norm states with the same Young
representations as those of the degenerate positive-norm states in the OCFQ
spectrum. This was demonstrated by using massive worldsheet sigma-model
approach in the lowest order weak field approximation but valid to all
orders in $\alpha ^{\prime }$\ , and thus was, in contrast to Gross's
result, valid to \emph{all} energy. To compare with the usual sigma-model
loop approximation, this result was argued to be a sigma-model n+1 loop
result for the n-th massive level (spin-n+1)\cite{3,4,5}. This calculation
applied to both open and closed string cases. In a recent paper\cite{6}, the
same decoupling phenomenon was demonstrated by using WSFT for the open
string case up to the spin-five level. It was shown that the background
fields of these degenerate positive-norm states can be gauged to the higher
rank fields at the same mass level.

In this paper, we will derive this interesting stringy decoupling phenomenon
from a third and more direct method, namely, the S-matrix approach. The key
was to explicitly calculate both types of zero-norm states\cite{7} in the
OCFQ spectrum. An infinite number of linear relations between string
scattering amplitudes of different string states with the same momenta at
each fixed mass level can then be written down\cite{8}. These relations, or
on-shell Ward identities are, as in Gross's case, valid order by order and
are of the identical form in string perturbation theory since zero-norm
states should be decoupled from the string amplitudes at each order of
string perturbation theory. They can be used to express the scattering
amplitudes of the degenerate lower spin propagating states in terms of those
of higher spin propagating states at each fixed mass level, thus reduce the
number of independent amplitudes. These Ward identities and thus the
resulting decoupling phenomenon are, in contrast to Gross's high-energy
symmetries, valid to \emph{all} energy except only for each \textit{fixed}
mass level. Presumably, the mass gap approaches zero if we take the
high-energy limit of these Ward identities, and the number of independent
amplitude reduces to one as in Gross's result. Since these relations between
the string amplitudes are valid order by order and are of the identical
form, it should be good enough to show only the string-tree level
amplitudes. The high-energy limit of string G-loop amplitudes was calculated
in Ref \cite{1,2} by the saddle-point method.

The powerfulness of zero-norm states and their direct relation to spacetime $%
w_{\infty }$ symmetry and Ward identities\cite{9} of toy 2d string model had
been stressed in Ref\cite{10}. A general formula of zero-norm states at an
arbitrary mass level with Polyakov's momentum was given in terms of Schur
Polynomials. These zero-norm states were shown to carry the charges of $%
w_{\infty }$ symmetry, which was used to determine the tachyon scattering
amplitudes \textit{without} any integration. On the other hand, the explicit
form of some massive string-tree level amplitudes of 26d string had been
calculated by the present author\cite{8}. Some low-lying zero-norm states
were also calculated and their corresponding massive Ward identities were
used to reduce the number of independent string amplitudes. However, a
general method to generate zero-norm states, similar to the one used in Ref%
\cite{10} for toy 2d string, is still lacking mostly due to the high
dimensionality of spacetime D = 26. Thus the reduction of the amplitudes can
only be shown at the third massive level (spin-four). Also, the subtlety of
the scalar state at this level remained. With the help of a simplified
method to construct stringy \emph{positive}-norm vertex operators\cite{11},
we invent a systematic method in this paper to construct stringy \emph{zero}%
-norm states. As examples, we calculate all relevant zero-norm states up to
the spin-five level. General\texttt{\ }formulas of some zero-norm tensor
states at an arbitrary mass level will also be given. We then use these
calculated zero-norm states and their corresponding Ward identities to
explicitly show the reduction of stringy scattering amplitudes of degenerate
positive-norm propagating states up to the spin-five level. This justifies
two previous independent calculations based on the massive worldsheet
sigma-model approach \cite{3} and WSFT approach\cite{6}.

\section{Calculations of zero-norm states}

The vertex operator of a physical state of open bosonic string

\begin{equation}
\left| \Psi \right\rangle =\sum C_{\mu _{1}...\mu _{m}}\alpha _{-n_{1}}^{\mu
_{1}}...\alpha _{-n_{m}}^{\mu _{m}}\left| 0;k\right\rangle ,[\alpha
_{m}^{\mu },\alpha _{n}^{\nu }]=m\eta ^{\mu \nu }\delta _{m+n}
\end{equation}%
is given by\cite{12}

\begin{equation}
\Psi (z)=\sum C_{\mu _{1}...\mu _{m}}N_{m}:\prod (\partial
_{z}^{n_{j}}x^{\mu _{j}})e^{ik\cdot X(z)}:
\end{equation}%
, where $N_{m}=i^{M}\prod \{(n_{j}-1)!\}^{-1}$. In the OCFQ spectrum,
physical states in eq. (1) are subject to the following Virasoro conditions

\begin{equation}
(L_{0}-1)\left| \Psi \right\rangle =0,L_{1}\left| \Psi \right\rangle
=L_{2}\left| \Psi \right\rangle =0  \tag{3a,b}
\end{equation}
, where

\begin{equation}
L_{m}=\frac{1}{2}\sum_{-\infty }^{\infty }:\alpha _{m-n}\cdot \alpha _{n}: 
\tag{4}
\end{equation}%
and $\alpha _{0}\equiv k$. The solutions of eqs. (3a,b) include
positive-norm propagating states and two types of zero-norm states which
were neglected in the most literature. They are\cite{13}

\begin{equation}
\text{Type I}:L_{-1}\left| x\right\rangle ,\text{ where }L_{1}\left|
x\right\rangle =L_{2}\left| x\right\rangle =0,\text{ }L_{0}\left|
x\right\rangle =0;  \tag{5}
\end{equation}

\begin{equation}
\text{Type II}:(L_{-2}+\frac{3}{2}L_{-1}^{2})\left\vert \widetilde{x}%
\right\rangle ,\text{ where }L_{1}\left\vert \widetilde{x}\right\rangle
=L_{2}\left\vert \widetilde{x}\right\rangle =0,\text{ }(L_{0}+1)\left\vert 
\widetilde{x}\right\rangle =0.  \tag{6}
\end{equation}%
While type I states have zero-norm at any spacetime dimension, type II
states have zero-norm \textit{only} at D=26. The existence of type II
zero-norm states signals the importance of zero-norm states in the structure
of the theory of string. It is straightforward to solve positive-norm state
solutions of eq. (3a, b) for some low-lying states, but soon becomes
practically unmanageable. The authors of Ref\cite{11} gave a simple
prescription to solve the positive-norm state solutions of eq. (3a, b). The
strategy is to apply the Virasoro conditions only to purely transverse
states, so that the zero-norm states will be got rid of at the very
beginning. This prescription simplified a lot of computation although some
complexities remained for low spin states at higher levels. Our aim here, on
the contrary, is to generate zero-norm states in eqs. (5) and (6), so that
all physical state solutions of eq. (3) will be completed. Let's first
assume we are given all positive-norm state solutions of some low-lying
levels. It is interesting to see the similarity between eqs. (3a, b) and
eqs. (5) and (6) for $\left\vert x\right\rangle $ and $\left\vert \widetilde{%
x}\right\rangle $. The only difference is the \textquotedblright mass
shift\textquotedblright\ of $L_{0}$ equations. As is well-known, the $L_{1%
\text{ }}$and $L_{2}$ equations give the transverse and traceless conditions
on the spin polarization. It turns out that, in many cases, the $L_{1\text{ }%
}$and $L_{2}$ equations will not refer to the $L_{0}$ equation or
on-mass-shell condition. In these cases, a positive-norm state solution for $%
\left\vert \Psi \right\rangle $ at mass level $n$ will give a zero-norm
state solution $L_{-1}\left\vert x\right\rangle $ at mass level $n+1$ simply
by taking $\left\vert x\right\rangle =\left\vert \Psi \right\rangle $ and
shifting $k^{2}$ by one unit. Similarly, one can easily get a type II
zero-norm state $(L_{-2}+\frac{3}{2}L_{-1}^{2})\left\vert \widetilde{x}%
\right\rangle $ at mass level $n+2$ simply by taking $\left\vert \widetilde{x%
}\right\rangle =\left\vert \Psi \right\rangle $ and shifting $k^{2}$ by two
units. For those cases where $L_{1\text{ }}$and $L_{2}$ equations do refer
to $L_{0}$ equation, our prescription needs to be modified. We will give
some examples to illustrate this method. Note that once we generate a
zero-norm state, it soon becomes \ a candidate of physical state $\left\vert
\Psi \right\rangle $ to generate two new zero-norm states at even higher
levels.

1. The \ first zero-norm state begin at $k^{2}=0$. This state is suggested
from the positive-norm tachyon state $\left\vert 0,k\right\rangle $ with $%
k^{2}$ $=2.$ Taking $\left\vert x\right\rangle =\left\vert 0,k\right\rangle $
and shifting $k^{2}$ by one unit to $k^{2}=0$, we get a type I zero-norm
state.

\begin{equation}
L_{-1\text{ }}\left\vert x\right\rangle =k\cdot \alpha _{-1}\left\vert
0,k\right\rangle ;\left\vert x\right\rangle =\left\vert 0,k\right\rangle
,-k^{2}=M^{2}=0.  \tag{7}
\end{equation}

2. At the first massive level $k^{2}=-2,$ tachyon suggests a type II
zero-norm state

\begin{equation}
(L_{-2}+\frac{3}{2}L_{-1}^{2})\left| \widetilde{x}\right\rangle =[\frac{1}{2}%
\alpha _{-1}\cdot \alpha _{-1}+\frac{5}{2}k\cdot \alpha _{-2}+\frac{3}{2}%
(k\cdot \alpha _{-1})^{2}]\left| 0,k\right\rangle ;\left| \widetilde{x}%
\right\rangle =\left| 0,k\right\rangle ,-k^{2}=2.  \tag{8}
\end{equation}

Positive-norm massless vector state suggests a type I zero-norm state

\begin{equation}
L_{-1}\left\vert x\right\rangle =[\theta \cdot \alpha _{-2}+(k\cdot \alpha
_{-1})(\theta \cdot \alpha _{-1})]\left\vert 0,k\right\rangle ;\left\vert
x\right\rangle =\theta \cdot \alpha _{-1}\left\vert 0,k\right\rangle
,-k^{2}=2,\theta \cdot k=0.  \tag{9}
\end{equation}%
However, massless singlet zero-norm state (7) does not give a type I
zero-norm state at the first massive level $k^{2}=-2$ since $L_{1\text{ }}$
equation on state (7) refers to $L_{0}$ equation, $k^{2}=0.$ This means that 
$L_{1}$ will not annihilate state (7) if one shifts the mass to $k^{2}=-2.$

3. At the second massive level $k^{2}=-4,$ positive-norm massless vector
state suggests a type II zero-norm state

\begin{eqnarray}
(L_{-2}+\frac{3}{2}L_{-1}^{2})\left\vert \widetilde{x}\right\rangle &=&\{%
\frac{5}{2}\theta \cdot \alpha _{-3}+\frac{1}{2}(\alpha _{-1}\cdot \alpha
_{-1})(\theta \cdot \alpha _{-1})+\frac{5}{2}(k\cdot \alpha _{-2})(\theta
\cdot \alpha _{-1})  \notag \\
&&+\frac{3}{2}(k\cdot \alpha _{-1})^{2}(\theta \cdot \alpha _{-1})+3(k\cdot
\alpha _{-1})(\theta \cdot \alpha _{-2})\}\left\vert 0,k\right\rangle ; 
\notag \\
\left\vert \widetilde{x}\right\rangle &=&\theta \cdot \alpha _{-1}\left\vert
0,k\right\rangle ,-k^{2}=4,k\cdot \theta =0.  \TCItag{10}
\end{eqnarray}%
However, massless singlet zero-norm state (7) does not give a type II
zero-norm state at mass level $k^{2}=-4$ for the same reason stated after eq
(9). Positive-norm spin-two state at $k^{2}=-2$ suggests a type I zero-norm
state

\begin{eqnarray}
L_{-1}\left\vert x\right\rangle &=&[2\theta _{\mu \nu }\alpha _{-1}^{\mu
}\alpha _{-2}^{\nu }+k_{\lambda }\theta _{\mu \nu }\alpha _{-1}^{\lambda \mu
\nu }]\left\vert 0,k\right\rangle ;\left\vert x\right\rangle =\theta _{\mu
\nu }\alpha _{-1}^{\mu \nu }\left\vert 0,k\right\rangle ,-k^{2}=4,  \notag \\
k\cdot \theta &=&\eta ^{\mu \nu }\theta _{\mu \nu }=0,\theta _{\mu \nu
}=\theta _{\nu \mu }  \TCItag{11}
\end{eqnarray}%
, where $\alpha _{-1}^{\lambda \mu \nu }\equiv \alpha _{-1}^{\lambda }\alpha
_{-1}^{\mu }\alpha _{-1}^{\nu }.$ Similar notations will be used in the rest
of this paper. Vector zero-norm state with $k^{2}=-2$ in eq (9) does not
give a type I zero-norm state for the same reason stated after eq (9). In
this case, however, one can modify $\left\vert x\right\rangle $ to be

\begin{equation}
\text{Ansatz: }\left\vert x\right\rangle =[\theta \cdot \alpha
_{-2}+a(k\cdot \alpha _{-1})(\theta \cdot \alpha _{-1})]\left\vert
0,k\right\rangle ;-k^{2}=4,\theta \cdot k=0  \tag{12}
\end{equation}%
, where $a$ is an undetermined constant. $L_{0}$ equation is then trivially
satisfied and $L_{1},$ $L_{2}$ equations give $a=\frac{1}{2}$. This gives a
type I zero-norm state

\begin{eqnarray}
L_{-1}\left| x\right\rangle  &=&[\frac{1}{2}(k\cdot \alpha _{-1})^{2}(\theta
\cdot \alpha _{-1})+2\theta \cdot \alpha _{-3}+\frac{3}{2}(k\cdot \alpha
_{-1})(\theta \cdot \alpha _{-2})  \notag \\
&&+\frac{1}{2}(k\cdot \alpha _{-2})(\theta \cdot \alpha _{-1})]\left|
0,k\right\rangle ;-k^{2}=4,\theta \cdot k=0.  \TCItag{13}
\end{eqnarray}%
Similarly, we modify the singlet zero-norm state with $k^{2}=-2$ in eq (8)
to be

\begin{equation}
\text{Ansatz: }\left| x\right\rangle =[\frac{5}{2}ak\cdot \alpha _{-2}+\frac{%
1}{2}b\alpha _{-1}\cdot \alpha _{-1}+\frac{3}{2}c(k\cdot \alpha
_{-1})^{2}]\left| 0,k\right\rangle ;-k^{2}=4  \tag{14}
\end{equation}%
, where $a$, $b$ and $c$ are undetermined constants. $L_{1}$ and $L_{2}$
equations give

\begin{equation}
5a+b+3k^{2}c=0,5k^{2}a+13b+\frac{3}{2}k^{2}c=0.  \tag{15}
\end{equation}%
For $k^{2}=-4$, we \ have $a:b:c=5:9:\frac{17}{6}$. This gives a type I
zero-norm state

\begin{eqnarray}
L_{-1}\left\vert x\right\rangle &=&[\frac{17}{4}(k\cdot \alpha _{-1})^{3}+%
\frac{9}{2}(k\cdot \alpha _{-1})(\alpha _{-1}\cdot \alpha _{-1})+9(\alpha
_{-1}\cdot \alpha _{-2})  \notag \\
&&+21(k\cdot \alpha _{-1})(k\cdot \alpha _{-2})+25(k\cdot \alpha
_{-3})]\left\vert 0,k\right\rangle ;  \TCItag{16} \\
-k^{2} &=&4.  \notag
\end{eqnarray}%
This completes the four zero-norm states at the second massive level. Note
that state (16) was calculated in Ref\cite{7} without modification. The
coefficients there thus need to be corrected although the main results
remain valid. It is important to note that the Young tableau of zero-norm
states at level $M^{2}=4$ are the sum of those of all physical states at two
lower levels, $M^{2}=2$ and $M^{2}=0$, \textit{except }the singlet zero-norm
state. \textit{This rule turns out to be true for any higher massive level,
and gives us a very simple way to tabulate Young diagrams of all zero-norm
states at any mass level \ }$n$\textit{\ given those of positive-norm states
at lower levels constructed by the simplified method in Ref\cite{11}.} This
is the main theme of the method of our construction here. Furthermore,
according to the decoupling conjecture\cite{3}, to pick up those decoupled
positive-norm states at level $n$, it is good enough to know the Young
tableau of all zero-norm states at level $n$ rather than calculating their
detailed forms. Our construction of zero-norm states thus offers a very
economical way to determine the decoupled states.

4. Similar method can be used to calculate zero-norm states at level $%
M^{2}=6.$ We will just list those which are relevant for the discussion in
section III. They are (from now on, unless otherwise stated, each spin
polarization in is assumed to be transverse, traceless and is symmetric with
respect to each group of indices as in Ref \cite{11})

\begin{equation}
L_{-1}\left\vert x\right\rangle =\theta _{\mu \nu \lambda }(k_{\beta }\alpha
_{-1}^{\mu \nu \lambda \beta }+3\alpha _{-1}^{\mu \nu }\alpha _{-2}^{\lambda
})\left\vert 0,k\right\rangle ;\left\vert x\right\rangle =\theta _{\mu \nu
\lambda }\alpha _{-1}^{\mu \nu \lambda }\left\vert 0,k\right\rangle , 
\tag{17}
\end{equation}

\begin{equation}
L_{-1}\left\vert x\right\rangle =[k_{\lambda }\theta _{\mu \nu }\alpha
_{-1}^{\mu _{\lambda }}\alpha _{-2}^{\nu }+2\theta _{\mu \nu }\alpha
_{-1}^{\mu }\alpha _{-3}^{\nu }\left\vert 0,k\right\rangle ;\left\vert
x\right\rangle =\theta _{\mu \nu }\alpha _{-1}^{\mu }\alpha _{-2}^{\nu
}\left\vert 0,k\right\rangle ,\text{ where }\theta _{\mu \nu }=-\theta _{\nu
\mu },  \tag{18}
\end{equation}

\begin{eqnarray}
L_{-1}\left\vert x\right\rangle &=&[2\theta _{\mu \nu }\alpha _{-2}^{\mu \nu
}+4\theta _{\mu \nu }\alpha _{-1}^{\mu }\alpha _{-3}^{\nu }+2(k_{\lambda
}\theta _{\mu \nu }+k_{(\lambda }\theta _{\mu \nu )})\alpha _{-1}^{\lambda
\mu }\alpha _{-2}^{\nu }+\frac{2}{3}k_{\lambda }k_{\beta }\theta _{\mu \nu
}\alpha _{-1}^{\mu \nu \lambda \beta }]\left\vert 0,k\right\rangle ;  \notag
\\
\left\vert x\right\rangle &=&[2\theta _{\mu \nu }\alpha _{-1}^{\mu }\alpha
_{-2}^{\nu }+\frac{2}{3}k_{\lambda }\theta _{\mu \nu }\alpha _{-1}^{\mu \nu
\lambda }]\left\vert 0,k\right\rangle  \TCItag{19}
\end{eqnarray}%
and

\begin{eqnarray}
(L_{-2}+\frac{3}{2}L_{-1}^{2})\left\vert \widetilde{x}\right\rangle
&=&[3\theta _{\mu \nu }\alpha _{-2}^{\mu \nu }+8\theta _{\mu \nu }\alpha
_{-1}^{\mu }\alpha _{-3}^{\nu }+(k_{\lambda }\theta _{\mu \nu }+\frac{15}{2}%
k_{(\lambda }\theta _{\mu \nu )})\alpha _{-1}^{\lambda \mu }\alpha
_{-2}^{\nu }  \notag \\
&&+(\frac{1}{2}\eta _{\lambda \beta }\theta _{\mu \nu }+\frac{3}{2}%
k_{\lambda }k_{\beta }\theta _{\mu \nu })\alpha _{-1}^{\mu \nu \lambda \beta
}]\left\vert 0,k\right\rangle ;  \notag \\
\left\vert \widetilde{x}\right\rangle &=&\theta _{\mu \nu }\alpha _{-1}^{\mu
\nu }\left\vert 0,k\right\rangle .  \TCItag{20}
\end{eqnarray}

Note that $\left| x\right\rangle $ in eq (19) has been modified as we did
for eq (12). To further illustrate our method, we calculate the type I
singlet zero-norm state from eq (16) as following

\begin{eqnarray}
\text{Ansatz} &:&\left| x\right\rangle =[a(k\cdot \alpha _{-1})^{3}+b(k\cdot
\alpha _{-1})(\alpha _{-1}\cdot \alpha _{-1})+c(k\cdot \alpha _{-1})(k\cdot
\alpha _{-2})  \notag \\
&&+d(\alpha _{-1}\cdot \alpha _{-2})+f(k\cdot \alpha _{-3})\left|
0,k\right\rangle ;  \notag \\
-k^{2} &=&6.  \TCItag{21}
\end{eqnarray}

The $L_{1}$ and $L_{2}$ equations can be easily used to determine $%
a:b:c:d:f=4:7:29:21:51.$ This gives the type I singlet zero-norm state

\begin{eqnarray}
L_{-1}\left\vert x\right\rangle &=&[a(k\cdot \alpha _{-1})^{4}+b(k\cdot
\alpha _{-1})^{2}(\alpha _{-1}\cdot \alpha _{-1})+(2b+d)(k\cdot \alpha
_{-1})(\alpha _{-1}\cdot \alpha _{-2})  \notag \\
&&+(c+3a)(k\cdot \alpha _{-1})^{2}(k\cdot \alpha _{-2})+c(k\cdot \alpha
_{-2})^{2}+d(\alpha _{-2}\cdot \alpha _{-2})+b(k\cdot \alpha _{-2})(\alpha
_{-1}\cdot \alpha _{-1})  \notag \\
&&+(2c+f)(k\cdot \alpha _{-3})(k\cdot \alpha _{-1})+2d(\alpha _{-1}\cdot
\alpha _{-3})+3f(k\cdot \alpha _{-4})]\left\vert 0,k\right\rangle ,  \notag
\\
-k^{2} &=&6.  \TCItag{22}
\end{eqnarray}

5. We list relevant zero-norm states at level $M^{2}=8$ from the known
positive-norm states and zero-norm states at level $M^{2}=4,6.$ They are

\begin{equation}
L_{-1}\left\vert x\right\rangle =(k_{\beta }\theta _{\mu \nu \lambda \gamma
}\alpha _{-1}^{\mu \nu \lambda \gamma \beta }+4\theta _{\mu \nu \lambda
\gamma }\alpha _{-1}^{\mu \nu \lambda }\alpha _{-2}^{\gamma }\left\vert
0,k\right\rangle ;\left\vert x\right\rangle =\theta _{\mu \nu \lambda \gamma
}\alpha _{-1}^{\mu \nu \lambda \gamma }\left\vert 0,k\right\rangle , 
\tag{23}
\end{equation}

\begin{eqnarray}
L_{-1}\left\vert x\right\rangle &=&\theta _{\mu \nu \lambda }[\frac{3}{4}%
k_{\beta }k_{\gamma }\alpha _{-1}^{\mu \nu \lambda \gamma \beta }+3k_{\beta
}\alpha _{-1}^{\mu \nu \beta }\alpha _{-2}^{\lambda }+3k_{\beta }\alpha
_{-1}^{(\mu \nu \lambda }\alpha _{-2}^{\beta )}+6\alpha _{-1}^{(\mu }\alpha
_{-2}^{\nu \lambda )}+6\alpha _{-1}^{(\mu \nu }\alpha _{-3}^{\lambda )}] 
\notag \\
\left\vert 0,k\right\rangle ;\left\vert x\right\rangle &=&\theta _{\mu \nu
\lambda }(\frac{3}{4}k_{\beta }\alpha _{-1}^{\mu \nu \lambda \beta }+3\alpha
_{-1}^{\mu \nu }\alpha _{-2}^{\lambda })\left\vert 0,k\right\rangle , 
\TCItag{24}
\end{eqnarray}

\begin{eqnarray}
(L_{-2}+\frac{3}{2}L_{-1}^{2})\left| \widetilde{x}\right\rangle  &=&\theta
_{\mu \nu \lambda }[(\frac{3}{2}k_{\beta }k_{\gamma }+\frac{1}{2}\eta
_{\gamma \beta })\alpha _{-1}^{\mu \nu \lambda \beta \gamma }+k_{\gamma }(%
\frac{1}{2}\alpha _{-1}^{\mu \nu \lambda }\alpha _{-2}^{\gamma }+8\alpha
_{-1}^{(\mu \nu \lambda }\alpha _{-2}^{\gamma )})  \notag \\
&&+3\alpha _{-1}^{(\mu }\alpha _{-2}^{\nu \lambda )}+6\alpha _{-1}^{(\mu \nu
}\alpha _{-3}^{\lambda )}]\left| 0,k\right\rangle ;  \notag \\
\left| \widetilde{x}\right\rangle  &=&\theta _{\mu \nu \lambda }\alpha
_{-1}^{\mu \nu \lambda }\left| 0,k\right\rangle ,  \TCItag{25}
\end{eqnarray}

\begin{eqnarray}
L_{-1}\left| x\right\rangle &=&\theta _{\mu \nu ,\lambda }(k_{\gamma }\alpha
_{-1}^{\gamma \mu \nu }\alpha _{-2}^{\lambda }+2\alpha _{-1}^{\mu }\alpha
_{-2}^{\nu \lambda }+2\alpha _{-1}^{\mu \nu }\alpha _{-3}^{\lambda })\left|
0,k\right\rangle ;  \notag \\
\left| x\right\rangle &=&\theta _{\mu \nu ,\lambda }\alpha _{-1}^{\mu \nu
}\alpha _{-2}^{\lambda }\left| 0,k\right\rangle ,\text{ where }\theta _{\mu
\nu ,\lambda }\text{ is mixed symmetric,}  \TCItag{26}
\end{eqnarray}

\begin{eqnarray}
L_{-1}\left| x\right\rangle &=&\theta _{\mu \nu }(\frac{3}{4}k_{\beta
}k_{\lambda }\alpha _{-1}^{\beta \lambda \mu }\alpha _{-2}^{\nu
}+4k_{\lambda }\alpha _{-1}^{\lambda \mu }\alpha _{-3}^{\nu }+\frac{3}{4}%
k_{\lambda }\alpha _{-1}^{\mu }\alpha _{-2}^{\nu \lambda }+2\alpha
_{-2}^{\mu }\alpha _{-3}^{\nu }+6\alpha _{-1}^{\mu }\alpha _{-4}^{\nu
})\left| 0,k\right\rangle ;  \notag \\
\left| x\right\rangle &=&(\frac{3}{4}k_{\lambda }\alpha _{-1}^{\lambda \mu
}\alpha _{-2}^{\nu }+2\alpha _{-1}^{\mu }\alpha _{-3}^{\nu })\left|
0,k\right\rangle ,\text{ where }\theta _{\mu \nu }=-\theta _{\nu \mu }, 
\TCItag{27}
\end{eqnarray}%
and

\begin{eqnarray}
(L_{-2}+\frac{3}{2}L_{-1}^{2})\left\vert \widetilde{x}\right\rangle
&=&\theta _{\mu \nu }[(\frac{3}{2}k_{\gamma }k_{\lambda }+\frac{1}{2}\eta
_{\gamma \lambda })\alpha _{-1}^{\gamma \lambda \mu }\alpha _{-2}^{\nu
}+6k_{\lambda }\alpha _{-1}^{\lambda \mu }\alpha _{-3}^{\nu }+\frac{5}{2}%
k_{\lambda }\alpha _{-1}^{\mu }\alpha _{-2}^{\nu \lambda }  \notag \\
+2\alpha _{-2}^{\mu }\alpha _{-3}^{\nu }+\alpha _{-1}^{\mu }\alpha
_{-4}^{\nu }]\left\vert 0,k\right\rangle ,\left\vert \widetilde{x}%
\right\rangle &=&\theta _{\mu \nu }\alpha _{-1}^{\mu }\alpha _{-2}^{\nu
}\left\vert 0,k\right\rangle ,\text{ where }\theta _{\mu \nu }=-\theta _{\nu
\mu }.  \TCItag{28}
\end{eqnarray}%
Note that the modified method was used in eqs (24) and (27).

6. Finally, we calculate general formulas of some zero-norm tensor states at
an arbitrary mass level by making use of general formulas of some
positive-norm states listed in Ref\cite{11}.

a. 
\begin{equation}
L_{-1}\theta _{\mu _{1}...\mu _{m}}\alpha _{-1}^{\mu _{1}...\mu
_{m}}\left\vert 0,k\right\rangle =\theta _{\mu _{1}...\mu _{m}}(k_{\lambda
}\alpha _{-1}^{\lambda \mu _{1}...\mu _{m}}+m\alpha _{-2}^{\mu _{1}}\alpha
_{-1}^{\mu _{2}...\mu _{m}})\left\vert 0,k\right\rangle  \tag{29}
\end{equation}%
, where $-k^{2}=M^{2}=2m,m=0,1,2,3...$. For example, $m=0,1$ give eqs (7)
and (9).

b. 
\begin{eqnarray}
&&(L_{-2}+\frac{3}{2}L_{-1}^{2})\theta _{\mu _{1}...\mu _{m}}\alpha
_{-1}^{\mu _{1}...\mu _{m}}\left\vert 0,k\right\rangle  \notag \\
&=&\{\theta _{\mu _{1}...\mu _{m}}[(\frac{3}{2}k_{\nu }k_{\lambda }+\frac{1}{%
2}\eta _{\nu \lambda })\alpha _{-1}^{\nu \lambda \mu _{1}...\mu _{m}}+\frac{3%
}{2}m(m-1)\alpha _{-2}^{\mu _{1}\mu _{2}}\alpha _{-1}^{\mu _{3}...\mu _{m}} 
\notag \\
&&+(1+3m)\alpha _{-1}^{\mu _{1}...\mu _{m-1}}\alpha _{-3}^{\mu _{m}}]+[\frac{%
3}{2}(m+1)k_{(\lambda }\theta _{\mu _{1}...\mu _{m})}+\frac{3}{2}mk_{\mu
_{m}}\theta _{\mu _{1}...\mu _{m-1\lambda })}]  \notag \\
&&\alpha _{-1}^{\mu _{1}...\mu _{m}}\alpha _{-2}^{\lambda }\}\left\vert
0,k\right\rangle  \TCItag{30}
\end{eqnarray}%
, where $-k^{2}=M^{2}=2m+2,m=0,1,2...$. For example, $m=0,1$give eqs (8) and
(10).

c. 
\begin{eqnarray}
&&L_{-1}\theta _{\mu _{1}...\mu _{m-2},\mu _{m-1}}\alpha _{-1}^{\mu
_{1}...\mu _{m-2}}\alpha _{-2}^{\mu _{m-1}}\left\vert 0,k\right\rangle 
\notag \\
&=&\theta _{\mu _{1}...\mu _{m-2},\mu _{m-1}}[k_{\lambda }\alpha
_{-1}^{\lambda \mu _{1}...\mu _{m-2}}\alpha _{-2}^{\mu _{m-1}}+(m-2)\alpha
_{-1}^{\mu _{1}...\mu _{m-3}}\alpha _{-2}^{\mu _{m-2}\mu _{m}}  \notag \\
&&+2\alpha _{-1}^{\mu _{1}...\mu _{m-2}}\alpha _{-2}^{\mu _{m-1}}]\left\vert
0,k\right\rangle ,\hspace{0.1cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{......%
\rule[0.01cm]{0.18cm}{0cm}}}\hspace{-0.025cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}  \TCItag{31}
\end{eqnarray}%
where $-k^{2}=M^{2}=2m,m=3,4,5...$. For example, $m=3,4$give eqs (18) and
(26).

d. 
\begin{eqnarray}
&&(L_{-2}+\frac{3}{2}L_{-1}^{2})\theta _{\mu _{1}...\mu _{m-2},\mu
_{m-1}}\alpha _{-1}^{\mu _{1}...\mu _{m-2}}\alpha _{-2}^{\mu
_{m-1}}\left\vert 0,k\right\rangle  \notag \\
&=&\theta _{\mu _{1}...\mu _{m-2},\mu _{m-1}}[(\frac{3}{2}k_{\lambda }k_{\nu
}+\frac{1}{2}\eta _{\lambda \nu })\alpha _{-1}^{\mu _{1}...\mu _{m-2}\lambda
\nu }\alpha _{-2}^{\mu _{m-1}}+6k_{\lambda }\alpha _{-1}^{\mu _{1}...\mu
_{m-2}\lambda }\alpha _{-3}^{\mu _{m-1}}  \notag \\
&&+(\frac{3}{2}m-2)k_{\lambda }\alpha _{-1}^{\mu _{1}...\mu _{m-2}}\alpha
_{-2}^{\mu _{m-1\lambda }}+2(m-2)\alpha _{-1}^{\mu _{1}...\mu _{m-3}}\alpha
_{-2}^{\mu _{m-2}}\alpha _{-3}^{\mu _{m-1}}+11\alpha _{-1}^{\mu _{1}...\mu
_{m-2}}\alpha _{-4}^{\mu _{m-1}}  \notag \\
&&+k_{\lambda }\alpha _{-1}^{\mu _{1}...\mu _{m-3}\lambda }\alpha _{-2}^{\mu
_{m-2}\mu _{m-1}}+(m-3)\alpha _{-1}^{\mu _{1}...\mu _{m-4}}\alpha _{-2}^{\mu
_{m-3}\mu _{m-2}\mu _{m-1}}]\left\vert 0,k\right\rangle ,\hspace{0.1cm}%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{%
-0.04cm}\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in%
}\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{......%
\rule[0.01cm]{0.18cm}{0cm}}}\hspace{-0.025cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}  \TCItag{32}
\end{eqnarray}%
, where $-k^{2}=M^{2}=2m+2,m=3,4,5...$. For example, $m=3$ gives eq (28).

e. 
\begin{eqnarray}
&&L_{-1}\theta _{\mu _{1}...\mu _{m-4},\mu _{m-3}\mu _{m-2}}(\alpha
_{-1}^{\mu _{1}...\mu _{m-4}}\alpha _{-2}^{\mu _{m-3}\mu _{m-2}}-\frac{4}{3}%
\alpha _{-1}^{\mu _{1}...\mu _{m-3}}\alpha _{-3}^{\mu _{m-2}})  \notag \\
&=&\theta _{\mu _{1}...\mu _{m-4},\mu _{m-3}\mu _{m-2}}[k_{\lambda }\alpha
_{-1}^{\lambda \mu _{1}...\mu _{m-4}}\alpha _{-2}^{\mu _{m-3}\mu
_{m-2}}+(m-4)\alpha _{-1}^{\mu _{1}...\mu _{m-3}}\alpha _{-2}^{\mu _{m-4}\mu
_{m-3}\mu _{m-2}}  \notag \\
&&+\frac{16}{3}\alpha _{-1}^{\mu _{1}...\mu _{m-4}}\alpha _{-3}^{\mu
_{m-3}}\alpha _{-2}^{\mu _{m-2}}+\frac{4}{3}k_{\lambda }\alpha
_{-1}^{\lambda \mu _{1}...\mu _{m-3}}\alpha _{-3}^{\mu _{m-2}}+4\alpha
_{-1}^{\mu _{1}...\mu _{m-3}}\alpha _{-4}^{\mu _{m-4}}],\hspace{0.1cm}%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{%
-0.04cm}\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in%
}\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{......%
\rule[0.01cm]{0.18cm}{0cm}}}\hspace{-0.025cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}  \TCItag{33}
\end{eqnarray}%
where $-k^{2}=M^{2}=2m,m=5,6...$.

f. The zero-norm states of eq (29) can be used to generate new type I
zero-norm states by the modified method as following\bigskip 
\begin{eqnarray}
&&L_{-1}\theta _{\mu _{1}...\mu _{m}}(\frac{m}{m+1}k_{\lambda }\alpha
_{-1}^{\lambda \mu _{1}...\mu _{m}}+\alpha _{-2}^{\mu _{1}}\alpha _{-1}^{\mu
_{2}...\mu _{m}})\left\vert 0,k\right\rangle  \notag \\
&=&[\frac{m}{m+1}k_{\nu }k_{\lambda }\theta _{\mu _{1}...\mu _{m}}\alpha
_{-1}^{\nu \lambda \mu _{1}...\mu _{m}}+m(k_{(\lambda }\theta _{\mu
_{1}...\mu _{m)}}+k_{\lambda }\theta _{\mu _{1}...\mu _{m}})\alpha
_{-2}^{\mu _{1}}\alpha _{-1}^{\lambda \mu _{2}...\mu _{m}}  \notag \\
&&+m(m-1)\theta _{\mu _{1}...\mu _{m}}\alpha _{-2}^{\mu _{1}\mu _{2}}\alpha
_{-1}^{\mu _{3}...\mu _{m}}+2m\theta _{\mu _{1}...\mu _{m}}\alpha _{-3}^{\mu
_{1}}\alpha _{-1}^{\mu _{2}...\mu _{m}}]\left\vert 0,k\right\rangle , 
\TCItag{34}
\end{eqnarray}%
where $-k^{2}=M^{2}=2m+2,m=1,2,3...$. For example, $m=1,2$ and 3 give eqs
(13), (19) and (24). Note that the coefficient of the first term in eq(34)
has been modified to $\frac{m}{m+1}.$Similarly, new type II zero-norm states
can also be constructed.

These are examples of some higher spin zero-norm states at an arbitrary mass
level. As in the case of positive-norm states, the complexity of the
calculation increases when calculating low spin zero-norm states for higher
levels. Fortunately, for our purpose in this paper, it is usually good
enough to calculate higher spin zero-norm states as it will become clear in
the next section. For those formulas with transverse trace\cite{11}

\begin{equation}
\eta _{\mu \nu }^{T}=\eta _{\mu \nu }-k_{\mu }k_{\nu }/k^{2},  \tag{35}
\end{equation}%
the modified method should be used, and we have no general formulas for them.

Each zero-norm state calculated in this section corresponds to an on-shell
Ward identity, which can be easily written down. As an interesting example%
\cite{8}, we write down the inter-particle Ward identity for two states at
the second massive level

\begin{equation}
(\frac{1}{2}k_{\mu }k_{\nu }\theta _{\lambda }+2\eta _{\mu \nu }\theta
_{\lambda })T^{(\mu \nu \lambda )}+9k_{\mu }\theta _{\nu }T^{[\mu \nu
]}+12\theta _{\mu }T^{\mu }=0.  \tag{36}
\end{equation}%
The zero-norm state responsible for this Ward identity is obtained by
antisymmetrizing those terms which involve $\alpha _{-1}^{\mu }\alpha
_{-2}^{\nu }$ in the original type I and type II vector zero-norm states in
eqs (13) and (10). In the above equation $k_{\mu }$ is the momentum of the
zero-norm state and $T^{\prime }s$ are defined to be n-point functions
containing one-point to be $\alpha _{-1}^{\mu \nu \lambda },$ $\alpha
_{-1}^{[\mu }\alpha _{-2}^{\nu ]}$ and $\alpha _{-3}^{\mu }$ respectively.
Note that this Ward identity is valid to \emph{all} energy. It is valid
order by order and is automatically of the identical form in string
perturbation theory. The high-energy limit of eq (36) was recently discussed
in Ref[14].

\section{Reduction of degenerate state's amplitude}

The decoupling of degenerate positive-norm states was first discovered in Ref%
\cite{3} by using generalized sigma-model approach. This stringy phenomenon
begins to show up at spin-four level of open bosonic string. The explicit
form of four positive-norm states at level four can be found in Ref\cite{12}%
. According to the decoupling conjecture, the spin-two and the scalar
positive-norm states should be decoupled. That is, their amplitudes are
determined from those of two other higher spin states. Let's begin the
discussion by first making an important observation. The most general form
of a physical state at this level can be written as

\begin{equation}
(A_{\mu \nu \lambda \gamma }\alpha _{-1}^{\mu \nu \lambda \gamma }+B_{\mu
\nu \lambda }\alpha _{-1}^{\mu \nu }\alpha _{-2}^{\lambda }+C_{\mu \nu
}\alpha _{-2}^{\mu \nu }+D_{\mu \nu }\alpha _{-1}^{\mu }\alpha _{-3}^{\nu
}+E_{\mu }\alpha _{-4}^{\mu })\left\vert 0,k\right\rangle .  \tag{37}
\end{equation}%
Decompose A, B,... as the tensor products of the subspace spanned by $k$ and 
$E^{\perp },$ the subspace orthogonal to $k$. For example, A can be
decomposed as

\begin{eqnarray}
A_{\mu \nu \lambda \gamma } &=&\epsilon _{\mu \nu \lambda \gamma
}+k_{(\gamma }\epsilon _{\mu \nu \lambda )}+,k_{(\mu }k_{\nu }\overline{%
\epsilon }_{\lambda \gamma )}+\eta _{(\mu \nu }\epsilon _{\lambda \gamma
)}+k_{(\mu }\eta _{\nu \lambda }\overline{\epsilon }_{\gamma )}  \notag \\
&&+k_{(\mu }k_{\nu }k_{\lambda }\epsilon _{\gamma )}+k_{(\mu }k_{\nu
}k_{\lambda }k_{\gamma )}\widehat{\epsilon }+k_{(\mu }k_{\nu }\eta _{\lambda
\gamma )}\overline{\epsilon }+\eta _{(\mu \nu }\eta _{\lambda \gamma
)}\epsilon  \TCItag{38}
\end{eqnarray}%
, where all $\epsilon _{\mu \nu ...}$ are traceless symmetric tensors over $%
E^{\perp }$. To get rid of zero-norm states which do not contribute to the
amplitude, the prescription given by Ref \cite{11} was to suppress all terms
having $k^{\prime }s$ and \textit{simultaneously} replace $\eta _{\mu \nu }$
by the \textit{transverse trace} $\eta _{\mu \nu }^{T}$ defined in eq (35).
So, in addition to $\epsilon _{\mu \nu \lambda \gamma },$ we end up \ with
only two \textquotedblright trace terms\textquotedblright\ in eq (38). Now
according to eq (2), the vertex operator corresponding to $\alpha _{-1}^{\mu
\nu \lambda \gamma }$ is :$\partial x^{\mu }\partial x^{\nu }\partial
x^{\lambda }\partial x^{\gamma }e^{ik\cdot x}:$. Due to the factorization
structure of this tensor vertex, the amplitude corresponding to two trace
terms, $\eta _{(\mu \nu }^{T}\epsilon _{\lambda \gamma )}$ and $\eta _{(\mu
\nu }^{T}\eta _{\lambda \gamma )}^{T}\epsilon ,$ are fixed by that of $%
\epsilon _{\mu \nu \lambda \gamma }$. This means that given the on-shell
amplitude of $\epsilon _{\mu \nu \lambda \gamma },$ the on-shell amplitude
of $A_{\mu \nu \lambda \gamma },$ $T_{\mu \nu \lambda \gamma },$ is fixed. $%
T^{\mu \nu \lambda \gamma }$ is defined to be the four-point function
containing the \textit{rank}-four tensor : $\partial x^{\mu }\partial x^{\nu
}\partial x^{\lambda }\partial x^{\gamma }e^{ik\cdot x}:$and three tachyons.
In fact, $T_{\mu \nu \lambda \gamma }$ can be explicitly calculated to be

\begin{eqnarray}
T^{\mu \nu \lambda \gamma } &=&\frac{\Gamma (-\frac{s}{2}-1)\Gamma (-\frac{t%
}{2}-1)}{\Gamma (\frac{u}{2}+2)}[(\frac{s^{2}}{4}-s)(\frac{s^{2}}{4}%
-1)k_{3}^{\mu }k_{3}^{\nu }k_{3}^{\lambda }k_{3}^{\gamma }-t(\frac{t^{2}}{4}%
-1)(s+2)k_{1}^{(\mu }k_{1}^{\nu }k_{1}^{\lambda }k_{3}^{\gamma )}  \notag \\
&&+\frac{3st}{2}(\frac{s}{2}+1)(\frac{t}{2}+1)k_{1}^{(\mu }k_{1}^{\nu
}k_{3}^{\lambda }k_{3}^{\gamma )}-s(\frac{s^{2}}{4}-1)(t+2)k_{1}^{(\mu
}k_{3}^{\nu }k_{3}^{\lambda }k_{3}^{\gamma )}  \notag \\
&&+(\frac{t^{2}}{4}-t)(\frac{t^{2}}{4}-1)k_{1}^{\mu }k_{1}^{\nu
}k_{1}^{\lambda }k_{1}^{\gamma }],  \TCItag{39}
\end{eqnarray}
where $s=-(k_{1}+k_{2})^{2},t=-(k_{2}+k_{3})^{2},$ and $u$ =$%
-(k_{1}+k_{3})^{2}$are the Mandelstam variables. We have chosen the second
state to be the tensor and have done the SL(2,R) gauge fixing by setting $%
x_{1}=0,0\leq x_{2}\leq 1,x_{3}=1,x_{4}=\infty .$ One easily see from eq(39)
that the trace parts of $T^{\mu \nu \lambda \gamma }$ contain no new
amplitude other than those of the spin part $\epsilon _{\mu \nu \lambda
\gamma }.$ This result can be easily generalized to amplitudes containing
more than one tensor. This is because there is no contribution of terms
resulting from contraction within the vertex: $\partial x^{\mu }\partial
x^{\nu }\partial x^{\lambda }\partial x^{\gamma }e^{ik\cdot x}:$ when doing
the amplitude calculation due to the normal ordering of the tensor vertex
operator. We conclude that knowing the amplitude of $\epsilon _{\mu \nu
\lambda \gamma }$ is equivalent to knowing $T^{\mu \nu \lambda \gamma }.$
Similar results apply to tensors B, C, ... in eq(37).

With the observation discussed above in mind, we now discuss the decoupling
phenomenon at level four. It was pointed out\cite{3} that the positive-norm
spin-two state can be gauged to a gauge which contains only $\alpha
_{-1}^{\mu \nu \lambda \gamma }$ and $\alpha _{-1}^{\mu \nu }\alpha
_{-2}^{\lambda }$ terms by making use of the gauge transformations induced
by the type I and the type II spin-two zero-norm states, eqs (19) and (20).
Also,the amplitude corresponds to $\alpha _{-1}^{(\mu \nu }\alpha
_{-2}^{\lambda )}$ is fixed by that of $\alpha _{-1}^{\mu \nu \lambda \gamma
}$due to the existence of the totally symmetric spin-three zero-norm state.
Our observation \ stated above then implies that the amplitude of
positive-norm spin-two state is fixed by those of positive-norm spin-four
and mixed-symmetric spin-three states. Take a representative of the
positive-norm scalar state to be\cite{12}

\begin{eqnarray}
&&[-(\eta _{\mu \nu }+\frac{13}{3}k_{\mu }k_{\nu })\alpha _{-2}^{\mu \nu
}+-i(\frac{20}{9}k_{\mu }k_{\nu }k_{\rho }+\frac{2}{3}k_{\mu }\eta _{\nu
\rho }+\frac{13}{3}k_{\rho }\eta _{\mu \nu })\alpha _{-1}^{\mu \nu }\alpha
_{-2}^{\rho }  \notag \\
&&+(\frac{23}{81}k_{\mu }k_{\nu }k_{\rho }k_{\sigma }+\frac{32}{27}k_{\mu
}k_{\nu }\eta _{\rho \sigma }+\frac{19}{18}\eta _{\mu \nu }\eta _{\rho
\sigma })\alpha _{-1}^{\mu \nu \rho \sigma }]\left\vert 0,k\right\rangle . 
\TCItag{40}
\end{eqnarray}%
It turns out that one can't gauge away the first term in eq (40) by using
the gauge transformations induced by the two singlet zero-norm states as in
the case of positive-norm spin-two state. However, since the amplitude
corresponding to $\alpha _{-2}^{\mu \nu }$ has been fixed by those of higher
spin states, we conclude that the positive-norm scalar state amplitude is
again fixed by those of higher spin states. This concludes the justification
of decoupling conjecture for spin-four level.

The positive-norm states at level five were calculated in Ref\cite{11} to be

\begin{equation}
\epsilon _{\mu \nu \lambda \beta \gamma }\alpha _{-1}^{\mu \nu \lambda \beta
\gamma }\left\vert 0,k\right\rangle ,\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},  \tag{41}
\end{equation}

\begin{equation}
\epsilon _{\mu \nu \lambda ,\beta }\alpha _{-1}^{\mu \nu \lambda }\alpha
_{-2}^{\beta }\left\vert 0,k\right\rangle ,\hspace{0.1cm}\raisebox{0.06in}{%
\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.025cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}  \tag{42}
\end{equation}

\begin{equation}
\epsilon _{\mu ,\nu \lambda }(\alpha _{-1}^{\mu }\alpha _{-2}^{\nu \lambda }-%
\frac{4}{3}\alpha _{-1}^{\mu \nu }\alpha _{-3}^{\lambda })\left\vert
0,k\right\rangle ,\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}%
\hspace{-0.094in}\hspace{-0.04cm}\raisebox{-.047in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}\hspace{-0.006in}\hspace{0.02cm}%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}  \tag{43}
\end{equation}

\begin{equation}
\lbrack \frac{4}{5!(D+5)}\epsilon _{\mu \nu \lambda }\eta _{\beta \gamma
}^{T}\alpha _{-1}^{\mu \nu \lambda \beta \gamma }+\epsilon _{\mu \nu \lambda
}(\alpha _{-1}^{\mu }\alpha _{-2}^{\nu \lambda }-\frac{4}{3}\alpha
_{-1}^{\mu \nu }\alpha _{-3}^{\lambda })]\left\vert 0,k\right\rangle ,%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},  \tag{44}
\end{equation}

\begin{equation}
\lbrack \frac{5}{6(D+1)}\eta _{(\mu \nu }^{T}\epsilon _{\lambda )\beta
}\alpha _{-1}^{\mu \nu \lambda }\alpha _{-2}^{\beta }+\epsilon _{\mu \nu
}(\alpha _{-2}^{\mu }\alpha _{-3}^{\nu }-\frac{1}{2}\alpha _{-1}^{\mu
}\alpha _{-4}^{\nu })]\left\vert 0,k\right\rangle ,\text{ }%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{%
-0.04cm}\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\text{\ and } 
\tag{45}
\end{equation}

\begin{eqnarray}
&&[\frac{D-2}{80(D+3)}\eta _{(\mu \nu }^{T}\eta _{\lambda \beta
}^{T}\epsilon _{\gamma )}\alpha _{-1}^{\mu \nu \lambda \beta \gamma }+(\eta
_{\mu \nu }^{T}\epsilon _{\lambda }-\frac{1}{2}(D-1)\epsilon _{(\mu }\eta
_{\nu )\lambda }^{T})\alpha _{-1}^{\mu \nu }\alpha _{-3}^{\lambda } 
\TCItag{46} \\
&&+\frac{3}{4}(D\epsilon _{\mu }\eta _{\nu \lambda }^{T}-\eta _{\mu (\nu
}^{T}\epsilon _{\lambda )})\alpha _{-1}^{\mu }\alpha _{-2}^{\nu \lambda
})]\left| 0,k\right\rangle ,\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}.  \notag
\end{eqnarray}%
According to our decoupling conjecture, states (44), (45) and (46) should be
decoupled. Note that states (26) and (43) are different in the $\alpha
_{i}^{\prime }s$ operator content although they share the same Young
diagram. One corresponds to $\alpha _{-1}^{\mu }\alpha _{-2}^{\nu \lambda }$
and the other $\alpha _{-1}^{\mu \nu }\alpha _{-3}^{\lambda }$ or vice
versa. With the explicit form of zero-norm states calculated in section III,
we can now justify the decoupling conjecture at level five. The terms $%
\alpha _{-1}^{(\mu }\alpha _{-2}^{\nu \lambda )}$ and $\alpha _{-1}^{(\mu
\nu }\alpha _{-3}^{\lambda )}$ in eq (44) can be gauged away by zero-norm
states in eqs (24) and (25), and the amplitude corresponding to $\alpha
_{-1}^{(\mu \nu \lambda }\alpha _{-2}^{\beta )}$ is fixed by that of $\alpha
_{-1}^{\mu \nu \lambda \beta \gamma }$ through zero-norm state in eq (23)
and our observation discussed after eq (38). Thus the amplitude of state
(44) is fixed by those of states (41) and (42). Now turn to state (45). The
terms $\alpha _{-2}^{[\mu }\alpha _{-3}^{\nu ]}$ and $\alpha _{-1}^{[\mu
}\alpha _{-4}^{\nu ]}$ can be gauged away by zero-norm states in eqs (27)
and (28), the amplitudes corresponding to $\alpha _{-1}^{(\mu }\alpha
_{-2}^{\nu \lambda )}$ and $\alpha _{-1}^{(\mu \nu }\alpha _{-3}^{\lambda )}$
are fixed by those of states in eqs (41) and (42) through zero-norm states
in eqs (24) and (25). Finally the amplitude of mixed-symmetric $\alpha
_{-1}^{\mu }\alpha _{-2}^{\nu \lambda }$ (or $\alpha _{-1}^{\mu \nu }\alpha
_{-3}^{\lambda }$) is fixed by those of states (41), (42) and (43). Thus the
amplitude of state (45) is fixed by those of states (41), (42) and (43).
Similar analysis shows that the amplitude of state (46) is again fixed by
those of states (41), (42) and (43). This completes the justification of our
decoupling conjecture at level five.

The decoupling calculation presented in this paper by the S-matrix approach
can be easily generalized to the closed string theory by making use of the
simple relation between closed and open string amplitudes in Ref \cite{15}.
A similar generalization to the closed string theory can also be done for
the massive worldsheet sigma-model approach. Our calculation in this section
justifies two previous independent calculations based on the massive
worldsheet sigma-model approach \cite{3} and WSFT approach\cite{6}.

\section{Acknowledgments}

I would like to thank Physics Departments of National Taiwan University and
Simon-Fraser University, where part of this work was completed during my
sabbatical visits. This work is supported in part by a grant of National
Science Council and a travelling fund of government of Taiwan, R. O. C.

\appendix

\section{ \ \ \ \ \ \ }

The Young tabulations of all physical states solutions of eq(3) up to level
six, including two types of zero-norm state solutions of eqs (5) and (6),
are listed in the following table

\begin{tabular}{|l|l|l|}
\hline
massive level & positive-norm states & zero-norm states \\ \hline
$M^{2}=-2$ & $\bullet $ &  \\ \hline
$M^{2}=0$ & $\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}$ & $%
\bullet $ (singlet) \\ \hline
$M^{2}=2$ & $\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}$ & $\raisebox{0.06in}{%
\fbox{\rule[0.04cm]{0.04cm}{0cm}}},$ $\bullet $ \\ \hline
$M^{2}=4$ & $\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}$ & $%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},2\times \raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\bullet $ \\ \hline
$M^{2}=6$ & $\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\bullet $ & $\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}},2\times %
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},3\times \raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},$ $2\times \bullet $ \\ \hline
$M^{2}=8$ & $\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{%
\fbox{\rule[0.04cm]{0.04cm}{0cm}}}$ & $\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},2\times \raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},2\times \raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}},4\times %
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},5\times \raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},3\times \bullet $ \\ \hline
$M^{2}=10$ & 
\begin{tabular}{l}
$\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{%
\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},$ \\ 
$\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}%
\hspace{-0.04cm}\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{%
-0.04cm}\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}},%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.28cm}%
\raisebox{-.150in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}},2\times %
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\bullet $%
\end{tabular}
& 
\begin{tabular}{l}
$\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{%
\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},2\times \raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},3\times \raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.006in}%
\hspace{-0.006in}\hspace{0.02cm}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},$ \\ 
$4\times \raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}%
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},7\times \raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}},4\times \raisebox{0.06in}{\fbox{%
\rule[0.04cm]{0.04cm}{0cm}}}\hspace{-0.094in}\hspace{-0.04cm}%
\raisebox{-.047in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}},9\times %
\raisebox{0.06in}{\fbox{\rule[0.04cm]{0.04cm}{0cm}}},$ 5$\times \bullet $%
\end{tabular}
\\ \hline
\end{tabular}

Note that the Young tabulations of zero-norm states at level n are the sum
of all physical states at levels n-1 and n-2 except the singlet zero-norm
states.

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\end{document}

