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\begin{document}
\title{Massive Gauge Field Theory Without Higgs Mechanism\\
IV. Illustration of Unitarsity}
\author{Jun-Chen Su}
\address{Center for Theorectical Physics, Department of Physics,\\
Jilin University, Changchun 130023,\\
People's Republic of China}
\maketitle

\begin{abstract}
To illustrate the unitarity of the massive gauge field theory described in
the foregoing papers, we calculate the imaginary part of two-gauge boson and
fermion-antifermion scattering amplitudes in the fourth order perturbative
approximation. Through these calculations, we find that for a given process,
if all the diagrams are taken into account, the contributions arising from
the unphysical intermediate states to the amplitudes are eventually
cancelled out. Thus, the unitarity of the S-matrix is perfectly ensured.

PACS:11.15-q,12.38.t
\end{abstract}

\setcounter{section}{1}

\section*{1.Introduction}

In the preceding paper (which will be referred to as paper III), it was
proved that the S-matrix given by our theory is independent of the gauge
parameter. The gauge-independence of S-matrix is, usually, not considered to
be a sufficient condition of the unitarity of a theory. Therefore, to
demonstrate the unitarity, it must be checked that for a given process,
whether the contributions arising from unphysical intermediate states to the
S-matrix element are cancelled in a perturbative calculation . Historically,
as mentioned in paper I, several attempts$^{[1]-[7]}$ of establishing the
massive gauge theory without Higgs bosons were eventually negated. The
reason for this partly is due to that the theories were criticized to suffer
from the difficulty of unitarity$^{[8]-[15]}$. Whether our theory is unitary
in perturbative calculations? That just is the question we want to answer in
this paper.

To display the unitarity, we will compute the imaginary parts of two-gauge
boson and fermion-antifermion scattering amplitudes in the perturbative
approximation of order $g^4$. The imaginary part of an amplitude can be
evaluated by the following formula$^{[16]}$%
\begin{equation}
2ImT_{ab}={\sum_c}T_{ac}T_{bc}^{*}
\end{equation}
which was derived from the unitarity condition of S-matrix: $S$ $%
S^{+}=S^{+}S=1$ and the definition: $S=1+iT$. we would like to emphasize
that the above formula holds provided that the intermediate states \{c\}
form a complete set. This means that when we use this formula to evaluate
the imaginary part of an amplitude, we have to work in Feynman gauge. In
this gauge, the gauge boson propagator and the ghost particle one are given
in the form 
\begin{equation}
iD_{\mu \nu }^{ab}(k)=\frac{-i\delta ^{ab}g_{\mu \nu }}{k^2-M^2+i\varepsilon 
}
\end{equation}
\begin{equation}
i\Delta ^{ab}(k)=\frac{-i\delta ^{ab}}{k^2-M^2+i\varepsilon }
\end{equation}
(see Eqs.(4.17) and (4.18) of paper I). where M denotes the gauge boson mass.

In Eq.(1.2), the unit tensor $g_{\mu \nu }$ represents the completeness of
the gauge boson intermediate states of polarization. 
\begin{equation}
g_{\mu \nu }={\sum_{\lambda =0}^3}e_\mu ^\lambda (k)e_{\lambda \nu
}(k)=P_{\mu \nu }(k)+Q_{\mu \nu }(k)
\end{equation}
where $P_{\mu \nu }(k)$ and $Q_{\mu \nu }(k)$ are the transverse and
longitudinal projectors, respectively. On the mass-shell, they are expressed
as 
\begin{equation}
P_{\mu \nu }(k)=g_{\mu \nu }-k_\mu k_\nu /M^2,Q_{\mu \nu }(k)=k_\mu k_\nu
/M^2
\end{equation}
It is noted here that in some previous works,$^{[13][14]}$ the Landau gauge
propagators were chosen at beginning to examine the unitarity through
calculation of the imaginary part of transition amplitudes. This procedure,
we think, is not reasonable and can not give a correct result in any case .
This is because that in the Landau gauge, the gauge boson propagator, only
includes the transverse projector $P_{\mu \nu }(k)$ which does not represent
a complete set of the intermediate polarized states as seen from Eq.(1.4).
Usually, the RHS of Eq.(1.) is calculated by using the Landau-Cutkosky(L-C)
rule$^{[16],[17]}$. By this rule, the intermediate propagators should be
replaced by their imaginary parts 
\begin{equation}
Im(k_i^2-M^2+i\varepsilon )^{-1}=-\pi \delta (k_i^2-M^2)\theta (k_0)
\end{equation}

Utilizing the L-C rule to calculate the imaginary parts of two-boson and
fermion-antifermion scattering amplitudes, we find, the unitarity of our
theory is no problems. A key point to achieve this conclusion is how to deal
with the loop diagram given by the gauge boson four-line vertex which was
considered to give no contribution to the S-matrix element in the previous
investigations$^{[2][11-14]}$. This diagram can be viewed as a limit of the
loop diagram formed by the gauge boson three-line vertices when one internal
line in the latter loop is shrunk into a point. In this way, we are able to
isolate from the former loop the term contributed from the unphysical
intermediate states which just guarantees the cancellation of the unphysical
amplitudes.

The rest of this paper is arranged as follows. In Sect.2, we sketch the
unitarity of the S-matrix elements of order $g^2$. In Sect.3, we describe
the calculations of the imaginary part of the two-gauge boson scattering
amplitude in the perturbative approximation of order $g^4$ and show how the
unitarity is ensured. In Sect.4, the same thing will be done for the
fermion-antifermion scattering. The last section serves to make comments and
discussions. In Appendix, we will discuss the sign of imaginary parts of the
loop diagrams by a rigorous calculation.

\setcounter{section}{2}

\section*{2. Unitarity of The Tree Diagrams of Order $g^2$}

\setcounter{equation}{0} For tree diagrams of order $g^2$, the unitarity of
their transition amplitudes is directly assured by the on-mass shell
condition. To illustrate this point, we discuss the fermion-antifermion and
two-gauge boson scattering taking place in the S-channel as shown in
Figs.(1) and (2)

For the fermion-antifermion scattering, the S-matrix element may be written
as 
\begin{equation}
T_{fi}=ig^2j^\mu (p_1,p_2)D_{\mu \nu }(k)j^\nu (p_1^{\prime },p_2^{\prime })
\end{equation}
where 
\begin{equation}
j^\mu (p_1,p_2)=\overline{\upsilon }(p_2)\frac{\lambda ^a}2\gamma ^\mu u(p_1)
\end{equation}
and 
\begin{equation}
D_{\mu \nu }(k)=\frac{g_{\mu \nu }-k_\mu k_\nu /k^2}{k^2-M^2+i\varepsilon }+%
\frac{\alpha k_\mu k_\nu /k^2}{k^2-\alpha M^2+i\varepsilon }
\end{equation}
Noticing $k=p_1+p_2=p_1^{\prime }+p_2^{\prime }$ and employing Dirac
equation, it is easy to see 
\begin{equation}
k_\mu j^\mu (p_1,p_2)=0
\end{equation}
Therefore, the longitudinal term $k_\mu k_\nu /k^2$ in the propagator does
not contribute to the S-matrix in the approximation of order $g^2$. In other
words, the unphysical pole $k^2=0$ does not appear in the scattering
amplitude.

For the process depicted in Fig.(2), the transition amplitude is 
\begin{equation}
\begin{array}{c}
T_{fi}=ig^2f^{abc}f^{a^{\prime }b^{\prime }c}e^\mu (k_1)e^\nu (k_2)e^{\mu
^{\prime }}(k_1^{\prime })^{*}e^{\nu ^{\prime }}(k_2^{\prime })^{*} \\ 
\times \Gamma _{\mu \nu \lambda }(k_1,k_2,q)D^{\lambda \lambda ^{\prime
}}(q)\Gamma _{\mu ^{\prime }\nu ^{\prime }\lambda ^{\prime }}(k_1^{\prime
},k_2^{\prime },q)
\end{array}
\end{equation}
where 
\begin{equation}
\Gamma _{\mu \nu \lambda }(k_1,k_2,q)=g_{\mu \nu }(k_1-k_2)_\lambda +g_{\nu
\lambda }(k_2+q)_\mu -g_{\lambda \mu }(k_1+q)_\nu
\end{equation}
and $e^\mu (k)$ stands for the gauge boson wave function satisfying 
\begin{equation}
k_\mu e^\mu (k)=0
\end{equation}
The transversality of the polarized states and the relation $%
q=k_1+k_2=k_1^{\prime }+k_2^{\prime }$ directly lead to 
\begin{equation}
e^\mu (k_1)e^\nu (k_2)\Gamma _{\mu \nu \lambda }(k_1,k_2,q)q^\lambda =0
\end{equation}
This equality, analogous to Eq.(2.4), guarantees the removal of the
unphysical pole from the S-matrix element written in Eq.(2.5).

Similarly, for the t-channel and u-channel diagrams, it is easy to verify
that the equalities in Eqs.(2.4) and (2.8) hold as well. These equalities
ensure the S-matrix elements for these diagrams and other processes such as
a fermion and an antifermion annihilate into two bosons to be also unitary.

The fact that the term $k_\mu k_\nu /k^2$ in the propagator gives no
contribution to the S-matrix elements means that the S-matrix is
gauge-independent at tree level. Although the Feynman gauge is only needed
to be considered in calculation of the S-matrix elements, the fact mentioned
above allows us to write the intermediate states as transverse ones. When we
evaluate the imaginary part of the transition amplitudes by the L-C rule,
these intermediate states will be put on the mass shell.

\setcounter{section}{3}

\section*{3.Unitarity of Two Boson Scattering Amplitude of Order $g^4$}

\setcounter{equation}{0}

In the preceding section. it was shown that in the lowest approximation of
perturbation, the unitarity is no problem. How is it for higher order
perturbative approximations? To answer this question, in this section, we
investigate the unitarity of the two-boson scattering amplitude given in the
order of $g^4$. For this purpose, we only need to consider the diagrams
shown in Figs.(3) and (4) and evaluate imaginary parts of the amplitudes of
these diagrams.

Fig.(3) contains eleven diagrams of gauge boson intermediate states. Except
for the last diagram shown in Fig.(1k), the other diagrams all have
two-gauge boson intermediate states. If the unitarity condition is
satisfied, in the amplitudes given by these diagrams, the unphysical parts
arising from the longitudinally polarized intermediate states should be
cancelled by the amplitudes of the five ghost diagrams depicted in Fig(4).
To demonstrate this point, in the following, we separately calculate the
imaginary parts of the amplitudes of all the diagrams in Figs.(3) and (4).

By the L-C rule, the diagrams in Figs.(1a-1j) can be given by folding the
tree diagrams shown in Fig.(5) with their conjugates. Through the folding,
we have two times of Figs.(1a-1f) and one time of Figs.(1g)-(1j). Noticing
that each of the diagrams in Figs.(1g)-(1j) has a symmetry factor $\frac 12$%
, we can write the imaginary part of Figs.(1a)-(1j) as follows 
\begin{equation}
\begin{array}{c}
2ImT_1^{ab,a^{\prime }b^{\prime }}(p_1p_2;p_1^{\prime }p_2^{\prime })=\frac 1%
2\int d\tau T_{\mu \nu }^{abcd}(p_1,p_2,k_1,k_2) \\ 
\times T_{\mu ^{\prime }\nu ^{\prime }}^{a^{\prime }b^{\prime
}cd*}(p_1^{\prime },p_2^{\prime };k_1,k_2)g^{\mu \mu ^{\prime }}g^{\nu \nu
^{\prime }}
\end{array}
\end{equation}
where $d\tau $ designates the volume element of two-particle phase space 
\begin{equation}
\begin{array}{c}
d\tau =\frac{d^4k_1}{(2\pi )^4}\frac{d^4k_2}{(2\pi )^4}(2\pi )^4\delta
^4(p_1+p_2-k_1-k_2)\pi \delta (k_1^2-M^2)\theta (k_1^0) \\ 
\times \pi \delta (k_2^2-M^2)\theta (k_2^0)
\end{array}
\end{equation}
and 
\begin{equation}
\begin{array}{c}
T_{\mu \nu }^{abcd}(p_1,p_2;k_1,k_2)={\sum_{i=1}^4}T_{\mu \nu
}^{(i)abcd}(p_1,p_2;k_1,k_2) \\ 
=-ig^2e^\rho (p_1)e^\sigma (p_{2)}{\sum_{i=1}^4}T_{\rho \sigma \mu \nu
}^{(i)abcd}(p_1,p_2;k_1,k_2)
\end{array}
\end{equation}
$T_{\mu \nu }^{(i)abcd}(p_1,p_2;k_1,k_2)(i=1,2,3,4)$ stand for the matrix
elements of Figs.(5a)-(5d) respectively.

In light of Feynman rules and setting 
\begin{equation}
C_1=f^{ace}f^{bde},C_2=f^{ade}f^{bce},C_3=f^{abe}f^{cde}
\end{equation}
and 
\begin{equation}
q_1=p_1-k_1=k_2-p_2,q_2=p_1-k_2=k_1-p_2,q_3=p_1+p_2=k_1+k_2
\end{equation}
the functions $T_{\rho \sigma \mu \nu }^{(i)abcd}(p_1,p_2;k_1,k_2)$ may be
expressed as follows.

For Fig.(5a), 
\begin{equation}
\begin{array}{c}
T_{\rho \sigma \mu \nu }^{(1)abcd}(p_1,p_2;k_1,k_2)=\frac{C_1}{%
q_1^2-M^2+i\varepsilon }\Gamma _{\rho \mu \lambda }^{(1)}(p_1,k_1,q_1) \\ 
\times \Gamma _{\sigma \nu }^{(1)\lambda }(p_2,k_2,q_1)
\end{array}
\end{equation}
where

\begin{equation}
\Gamma _{\rho \mu \lambda }^{(1)}(p_1,k_1,q_1)=g_{\rho \mu
}(k_1+p_1)_\lambda +g_{\mu \lambda }(q_1-k_1)_\rho -g_{\lambda \rho
}(q_1+p_1)_\mu
\end{equation}
\begin{equation}
\Gamma _{\sigma \nu \lambda }^{(1)}(p_2,k_2,q_1)=g_{\sigma \nu
}(p_2+k_2)_\lambda -g_{\nu \lambda }(q_1+k_2)_\sigma +g_{\lambda \sigma
}(q_1-p_2)_\nu
\end{equation}
For Fig.(5b), 
\begin{equation}
\begin{array}{c}
T_{\rho \sigma \mu \nu }^{(2)abcd}(p_1,p_2;k_1,k_2)=\frac{C_2}{%
q_2^2-M^2+i\varepsilon }\Gamma _{\rho \nu \lambda }^{(2)}(p_1,k_2,q_2)\times
\\ 
\Gamma _{\sigma \mu }^{(2)\lambda }(p_2,k_1,q_1)
\end{array}
\end{equation}
where 
\begin{equation}
\begin{array}{c}
\Gamma _{\rho \nu \lambda }^{\left( 2\right) }(p_{1,}k_2,q_2)=g_{\rho \nu
}(p_1+k_2)_\lambda +g_{\nu \lambda }(q_2-k_2)_\rho -g_{\lambda \rho
}(q_2+p_1)_\nu \\ 
\\ 
\Gamma _{\sigma \mu \lambda }^{(2)}(p_2,k_1,q_2)=g_{\mu \sigma
}(p_2+k_1)_\lambda -g_{\mu \lambda }(k_1+q_2)_\sigma +g_{\lambda \sigma
}(q_2-p_2)_\mu
\end{array}
\end{equation}
For Fig.(5c), 
\begin{equation}
\begin{array}{c}
T_{\rho \sigma \mu \nu }^{(3)abcd}(p_1,p_2;k_1,k_2)=\frac{C_3}{%
q_3^2-M^2+i\varepsilon }\Gamma _{\rho \sigma \lambda }^{(3)}(p_1,p_2,q_3) \\ 
\times \Gamma _{\mu \nu }^{(3)\lambda }(k_1,k_2,q_3)
\end{array}
\end{equation}
where 
\begin{equation}
\Gamma _{\rho \sigma \lambda }^{(3)}(p_1,p_2,q_3)=g_{\rho \sigma
}(p_1-p_2)_\lambda +g_{\sigma \lambda }(p_2+q_3)_\rho -g_{\lambda \rho
}(q_3+p_1)_\sigma
\end{equation}
\begin{equation}
\Gamma _{\mu \nu \lambda }^{(3)}(k_1,k_2,q_3)=g_{\mu \nu }(k_2-k_1)_\lambda
-g_{\nu \lambda }(k_2+q_3)_\mu +g_{\lambda \mu }(q_3+k_1)_\nu
\end{equation}
For Fig.(5d) 
\begin{equation}
T_{\rho \sigma \mu \nu }^{(4)abcd}(p_1,p_2;k_1,k_2)=C_1\gamma _{\rho \sigma
\mu \nu }^{(1)}+C_2\gamma _{\rho \sigma \mu \nu }^{(2)}+C_3\gamma _{\rho
\sigma \mu \nu }^{(3)}
\end{equation}
where 
\begin{equation}
\gamma _{\rho \sigma \mu \nu }^{(1)}=g_{\rho \sigma }g_{\mu \nu }-g_{\sigma
\mu }g_{\rho \nu }
\end{equation}
\begin{equation}
\gamma _{\rho \sigma \mu \nu }^{(2)}=g_{\rho \sigma }g_{\mu \nu }-g_{\rho
\mu }g_{\sigma \nu }
\end{equation}
\begin{equation}
\gamma _{\rho \sigma \mu \nu }^{(3)}=g_{\rho \mu }g_{\sigma \nu }-g_{\rho
\nu }g_{\sigma \mu }
\end{equation}
The expressions in Eqs.(3.1),(3.2),(3.6),(3.9) and (3.12), as indicated in
the Introduction, are all given in the Feynman gauge. When the intermediate
states $g^{\mu \mu ^{\prime }}$ and $g^{\nu \nu ^{\prime }}$ are decomposed
into physical and unphysical parts in accordance with Eqs.(1.4) and (1.5),
Eq.(3.1) will be represented as 
\begin{equation}
\begin{array}{c}
2ImT_1=\frac 12\int d\tau T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu ^{\prime
}}^{a^{\prime }b^{\prime }cd*}[P^{\mu \mu ^{\prime }}(k_1)P^{\nu \nu
^{\prime }}(k_2)+Q^{\mu \mu ^{\prime }}(k_1)g^{\nu \nu ^{\prime }} \\ 
+g^{\mu \mu ^{\prime }}Q^{\nu \nu ^{\prime }}(k_2)-Q^{\mu \mu ^{\prime
}}(k_1)Q^{\nu \nu ^{\prime }}(k_2)]
\end{array}
\end{equation}
We see, except for the first term, the other terms are all related to the
unphysical intermediate states. These terms should be cancelled out in the
total amplitude. In the following, we calculate these terms separately. In
the calculations, we note, the transversality of the polarization vectors
(see Eq.(2.7)), the relations written in Eq.(3.5) and the on shell property
of the momenta $p_1,p_2,k_1$ and $k_2$ will be often used.

\section*{ A.Calculation of $T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu
^{\prime }}^{a^{\prime }b^{\prime }cd*}Q^{\mu \mu ^{\prime }}(k_1)g^{\nu \nu
^{\prime }}$}

According to the definition in Eq.(3.3). we need to calculate the
contractions $k_1^\mu T_{\rho \sigma \mu \nu
}^{(i)abcd}(p_1,p_2;k_1,k_2)(i=1,2,3,4)$. From Eqs.(3.7) and (3.8). we find 
\begin{equation}
k_1^\mu \Gamma _{\rho \mu \lambda }^{(1)}(p_1,k_1,q_1)=-q_{1\rho
}q_{1\lambda }+g_{\lambda \rho }(q_1^2-M^2)
\end{equation}
and 
\begin{equation}
q_1^\lambda \Gamma _{\sigma \nu \lambda }^{(1)}(p_2,k_2,q_1)=-k_{2\nu
}q_{1\sigma }
\end{equation}
Using these equalities, from Eq.(3.6), we obtain 
\begin{equation}
k_1^\mu T_{\rho \sigma \mu \nu }^{(1)abcd}(p_1,p_2;k_1,k_2)=C_1k_{2\nu
}S_{\rho \sigma }^{(1)}(q_1)+C_1\Gamma _{\sigma \nu \rho }^{(1)}(p_2,k_2,q_1)
\end{equation}
where 
\begin{equation}
S_{\rho \sigma }^{(1)}(q_1)=\frac{q_{1\rho }q_{1\sigma }}{%
q_1^2-M^2+i\varepsilon }
\end{equation}

Similarly, from Eqs.(3.10) and (3.11), one can get 
\begin{equation}
k_1^\mu \Gamma _{\sigma \mu \lambda }^{(2)}(p_2,k_1,q_2)=-q_{2\sigma
}q_{2\lambda }+g_{\lambda \sigma }(q_2^2-M^2)
\end{equation}
and 
\begin{equation}
q_2^\lambda \Gamma _{\rho \nu \lambda }^{(2)}(p_1,k_2,q_2)=-q_{2\rho
}k_{2\nu }
\end{equation}
Based on these equalities, it is found form Eq.(3.9) 
\begin{equation}
k_1^\mu T_{\rho \sigma \mu \nu }^{(2)abcd}(p_1,p_2;k_1,k_2)=C_2k_{2\nu
}S_{\rho \sigma }^{(2)}(q_2)+C_2\Gamma _{\rho \nu \sigma }^{(2)}(p_1,k_2,q_2)
\end{equation}
where 
\begin{equation}
S_{\rho \sigma }^{(2)}(q_2)=\frac{q_{2\rho }q_{2\sigma }}{%
q_2^2-M^2+i\varepsilon }
\end{equation}

Along the same line, we can derive from Eqs.(3.13) and (3.14) that 
\begin{equation}
k_1^\mu \Gamma _{\mu \nu \lambda }^{(3)}(k_1,k_2,q_3)=k_{1\nu }q_{3\lambda
}+k_{1\lambda }k_{2\nu }-g_{\nu \lambda }(q_3^2-M^2)
\end{equation}
and 
\begin{equation}
q_3^\lambda \Gamma _{\rho \sigma \lambda
}^{(3)}(p_1,p_2,q_3)=(k_1+k_2)^\lambda \Gamma _{\rho \sigma \lambda
}^{(3)}(p_1,p_2,q_3)=0
\end{equation}
thereby, we get from Eq.(3.12) 
\begin{equation}
k_1^\mu T_{\rho \sigma \mu \nu }^{(3)abcd}(p_1,p_2;k_1,k_2)=C_3k_{2\nu
}S_{\rho \sigma }^{(3)}(p_1,p_2,k_1)-C_3\Gamma _{\sigma \nu \rho
}^{(3)}(p_1,p_2,q_3)
\end{equation}
where 
\begin{equation}
S_{\rho \sigma }^{(3)}(p_1,p_2,k_1)=\frac 1{q_2^2-M^2+i\varepsilon }%
k_1^\lambda \Gamma _{\rho \sigma \lambda }^{(3)}(p_1,p_2,q_3)
\end{equation}

In addition, from Eq.(3.15), we may write 
\begin{equation}
\begin{array}{c}
k_1^\mu T_{\rho \sigma \mu \nu }^{(4)abcd}(p_1,p_2;k_1,k_2)=C_1\gamma _{\rho
\sigma \nu }^{(1)}(k_1)+C_2\gamma _{\rho \sigma \nu }^{(2)}(k_1) \\ 
+C_3\gamma _{\rho \sigma \nu }^{(3)}(k_1)
\end{array}
\end{equation}
where 
\begin{equation}
\gamma _{\rho \sigma \nu }^{(1)}(k_1)=k_1^\mu \gamma _{\rho \sigma \mu \nu
}^{(1)}=g_{\rho \sigma }k_{1\nu }-g_{\rho \nu }k_{1\sigma }
\end{equation}
\begin{equation}
\gamma _{\rho \sigma \nu }^{(2)}(k_1)=k_1^\mu \gamma _{\rho \sigma \mu \nu
}^{(2)}=g_{\rho \sigma }k_{1\nu }-g_{\sigma \nu }k_{1\rho }
\end{equation}
\begin{equation}
\gamma _{\rho \sigma \nu }^{(3)}(k_1)=k_1^\mu \gamma _{\rho \sigma \mu \nu
}^{(3)}=g_{\sigma \nu }k_{1\rho }-g_{\rho \nu }k_{1\sigma }
\end{equation}

Summing up the results denoted in Eqs.(3.22),(3.26),(3.30) and (3.32) and
noticing Eq.(3.3), we have 
\begin{equation}
k_1^\mu T_{\rho \sigma \mu \nu }^{abcd}(p_1,p_2;k_1,k_2)=-ig^2e^\rho
(p_1)e^\sigma (p_2)[k_{2\nu }S_{\rho \sigma }^{abcd}+G_{\rho \sigma \nu
}^{abcd}]
\end{equation}
where 
\begin{equation}
S_{\rho \sigma }^{abcd}={\sum_{i=1}^{3} }C_iS_{\rho \sigma }^{(i)}
\end{equation}
and 
\begin{equation}
G_{\rho \sigma \nu }^{abcd}={\sum_{i=1}^{3}}C_iG_{\rho \sigma \nu }^{(i)}
\end{equation}
in which 
\begin{equation}
G_{\rho \sigma \nu }^{(1)}=\Gamma _{\sigma \nu \rho
}^{(1)}(p_2,k_2,q_1)+\gamma _{\rho \sigma \nu }^{(1)}(k_1)
\end{equation}
\begin{equation}
G_{\rho \sigma \nu }^{(2)}=\Gamma _{\rho \nu \sigma
}^{(2)}(p_1,k_2,q_2)+\gamma _{\rho \sigma \nu }^{(2)}(k_2)
\end{equation}
\begin{equation}
G_{\rho \sigma \nu }^{(3)}=-\Gamma _{\rho \sigma \nu
}^{(3)}(p_1,p_2,q_3)+\gamma _{\rho \sigma \nu }^{(3)}(p_1,p_2,k_1)
\end{equation}
Employing the expressions given in Eqs.(3.8),(3.10),(3.13),(3.29) and
(3.33)-(3.35), it is not difficult to find the following relation 
\begin{equation}
G_{\rho \sigma \nu }^{(1)}=-G_{\rho \sigma \nu }^{(2)}=-G_{\rho \sigma \nu
}^{(3)}
\end{equation}

Now, let us look at the color factors. According to the expression 
\begin{equation}
f^{abc}f^{cde}=\frac 2N(\delta _{ac}\delta _{bd}-\delta _{ad}\delta
_{bc})+(d_{ace}d_{bde}-d_{bce}d_{ade})
\end{equation}
and defining 
\begin{equation}
\begin{array}{c}
\beta _1=\frac 2N\delta _{ab}\delta _{cd}+d_{abe}d_{cde} \\ 
\beta _2=\frac 2N\delta _{ad}\delta _{bc}+d_{ade}d_{bce} \\ 
\beta _3=\frac 2N\delta _{ac}\delta _{bd}+d_{ace}d_{bde}
\end{array}
\end{equation}
we may write 
\begin{equation}
C_1=\beta _1-\beta _2,C_2=\beta _1-\beta _3,C_3=\beta _3-\beta _2
\end{equation}

Substitution of Eq.(3.45) into Eq.(3.38) and use of Eq.(3.42) directly lead
to 
\begin{equation}
\begin{array}{c}
G_{\rho \sigma \nu }^{abcd}=\beta _1(G_{\rho \sigma \nu }^{(1)}+G_{\rho
\sigma \nu }^{(2)})-\beta _2(G_{\rho \sigma \nu }^{(1)}+G_{\rho \sigma \nu
}^{(3)}) \\ 
+\beta _3(G_{\rho \sigma \nu }^{(3)}-G_{\rho \sigma \nu }^{(2)})=0
\end{array}
\end{equation}
This result makes Eq.(3.36) reduce to 
\begin{equation}
k_1^\mu T_{\mu \nu }^{abcd}=k_{2\nu }S^{abcd}
\end{equation}
where

\begin{equation}
S^{abcd}=-ig^2e^\rho (p_1)e^\sigma (p_2)S_{\rho \sigma }^{abcd}
\end{equation}
By using Eq.(3.47) and noticing $k_2^2=M^2,$ we finally obtain 
\begin{equation}
T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu ^{\prime }}^{a^{\prime }b^{\prime
}cd^{*}}Q^{\mu \mu ^{\prime }}(k_1)g^{\nu \nu ^{\prime
}}=S^{abcd}S^{a^{\prime }b^{\prime }cd*}
\end{equation}

It is emphasized that from the above derivation, we see, the four-line
vertex diagram in Fig.(5d) plays an essential role to give the relation in
Eq.(3.42) and hence to guarantee the cancellation of the second terms in
Eq.(3.22),(3.26) and (3.30) which are free from the poles at $q_i^2=M^2$, as
shown in Eq.(3.46).

\section*{ B.Calculation of $T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu
^{\prime }}^{a^{\prime }b^{\prime }cd^{*}}g^{\mu \mu ^{\prime }}Q^{\nu \nu
^{\prime }}(k_2)$}

The procedure of calculating $T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu
^{\prime }}^{a^{\prime }b^{\prime }cd^{*}}g^{\mu \mu ^{\prime }}Q^{\nu \nu
^{\prime }}(k_2)$ completely parallels to that described in the former
subsection.

From Eqs.(3.7) and (3.8), it follows that 
\begin{equation}
k_2^\nu \Gamma _{\sigma \nu \lambda }^{(1)}(p_2,k_2,q_1)=-q_{1\sigma
}q_{1\lambda }+g_{\lambda \sigma }(q_1^2-M^2)
\end{equation}
and 
\begin{equation}
q_1^\lambda \Gamma _{\rho \mu \lambda }^{(2)}(p_1,k_1,q_{1)}=-k_{1\mu
}q_{1\rho }
\end{equation}
These equalities allow us to get from Eq.(3.6) that 
\begin{equation}
k_2^\nu T_{\rho \sigma \mu \nu }^{(1)abcd}(p_1,p_2;k_1,k_2)=C_1k_{1\mu
}S_{\rho \sigma }^{(1)}+C_1\Gamma _{\rho \mu \sigma }^{(1)}(p_1,k_1,q_1)
\end{equation}
Based on the equalities 
\begin{equation}
k_2^\nu \Gamma _{\rho \nu \lambda }^{(2)}(p_1,k_2,q_2)=-q_{2\rho
}q_{2\lambda }+g_{\rho \lambda }(q_2^2-M^2)
\end{equation}
and 
\begin{equation}
q_2^\lambda \Gamma _{\sigma \mu \lambda }^{(2)}(p_2,k_1,q_2)=-q_{2\sigma
}k_{1\mu }
\end{equation}
which are derived Eqs.(3.10) and (3.11), it is found 
\begin{equation}
k_2^\nu T_{\rho \sigma \mu \nu }^{(2)abcd}(p_1,p_2;k_1,k_2)=C_2k_{1\mu
}S_{\rho \sigma }^{(2)}+C_2\Gamma _{\sigma \mu \rho }^{(2)}(p_2,k_2,q_2)
\end{equation}
By making use of the equality 
\begin{equation}
k_2^\nu \Gamma _{\mu \nu \lambda }^{(3)}(k_1,k_2,q_3)=-k_{2\mu }q_{3\lambda
}-k_{1\mu }k_{2\lambda }+g_{\mu \lambda }(q_3^2-M^2)
\end{equation}
which is derived from Eq.(3.14) and Eq.(3.29), we have 
\begin{equation}
k_2^\nu T_{\rho \sigma \mu \nu }^{(3)abcd}(p_1,p_2;k_1,k_2)=C_3k_{1\mu
}S_{\rho \sigma }^{(3)}+C_3\Gamma _{\rho \sigma \mu }^{(3)}(p_1,p_2,q_3)
\end{equation}
From Eq.(3.15), it is clear 
\begin{equation}
\begin{array}{c}
k_2^\nu T_{\rho \sigma \mu \nu }^{(4)abcd}(p_1,p_2;k_1,k_2)=C_1\gamma _{\rho
\sigma \mu }^{(1)}(k_2) \\ 
+C_2\gamma _{\rho \sigma \mu }^{(2)}(k_2)+C_3\gamma _{\rho \sigma \mu
}^{(3)}(k_2)
\end{array}
\end{equation}
where 
\begin{equation}
\gamma _{\rho \sigma \mu }^{(1)}(k_2)=k_2^\nu \gamma _{\rho \sigma \mu \nu
}^{(1)}=g_{\rho \sigma }k_{2\mu }-g_{\sigma \mu }k_{2\rho }
\end{equation}
\begin{equation}
\gamma _{\rho \sigma \mu }^{(2)}{}(k_2)=k_2^\nu \gamma _{\rho \sigma \mu \nu
}^{(2)}=g_{\rho \sigma }k_{2\mu }-g_{\rho \mu }k_{2\sigma }
\end{equation}
\begin{equation}
\gamma _{\rho \sigma \mu }^{(3)}{}(k_2)=k_2^\nu \gamma _{\rho \sigma \mu \nu
}^{(3)}=g_{\rho \mu }k_{2\sigma }-g_{\sigma \mu }k_{2\rho }
\end{equation}
Combining Eqs.(3.52),(3.55),(3.57) and (3.58), we obtain 
\begin{equation}
k_2^\nu T_{\rho \sigma \mu \nu }^{abcd}(p_1,p_2;k_1,k_2)=-ig^2e^\rho
(p_1)e^\sigma (p_2)[k_{1\mu }S_{\rho \sigma }^{abcd}+\widetilde{G}_{\rho
\sigma \mu }^{abcd}]
\end{equation}
where $S_{\rho \sigma }^{abcd}$ was defined in Eq.(3.37) and 
\begin{equation}
\widetilde{G}_{\rho \sigma \mu }^{abcd}(p_1,p_2;k_1,k_2)={\sum_{i=1}^3}C_i%
\widetilde{G}_{\rho \sigma \mu }^{(i)}
\end{equation}
in which 
\begin{equation}
\widetilde{G}_{\rho \sigma \mu }^{(1)}=\Gamma _{\rho \mu \sigma
}^{(1)}(p_1,k_1,q_1)+\gamma _{\rho \sigma \mu }^{(1)}(k_2)
\end{equation}
\begin{equation}
\widetilde{G}_{\rho \sigma \mu }^{(2)}=\Gamma _{\sigma \mu \rho
}^{(2)}(p_2,k_2,q_2)+\gamma _{\rho \sigma \mu }^{(3)}(k_2)
\end{equation}
\begin{equation}
\widetilde{G}_{\rho \sigma \mu }^{(3)}=\Gamma _{\rho \sigma \mu
}^{(3)}(p_1,p_2,q_3)+\gamma _{\rho \sigma \mu }^{(3)}(k_2)
\end{equation}
Similar to Eq.(3.42), one may find 
\begin{equation}
\widetilde{G}_{\rho \sigma \mu }^{(1)}=-\widetilde{G}_{\rho \sigma \mu
}^{(2)}=-\widetilde{G}_{\rho \sigma \mu }^{(3)}
\end{equation}
These relations and those given in Eq.(3.45) also lead Eq.(3.63) to vanish 
\begin{equation}
G_{\rho \sigma \mu }^{abcd}(p_1,p_2;k_1,k_2)=0
\end{equation}
Thus, Eq.(3.62) becomes 
\begin{equation}
k_2^\nu T_{\mu \nu }^{abcd}(p_1,p_2;k_1,k_2)=k_{1\mu }S^{abcd}
\end{equation}
Thereby, we have 
\begin{equation}
T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu ^{\prime }}^{a^{\prime }b^{\prime
}cd*}g^{\mu \mu ^{\prime }}Q^{\nu \nu ^{\prime }}(k_2)=S^{abcd}S^{a^{\prime
}b^{\prime }cd*}
\end{equation}

\section*{ Calculation of $T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu ^{\prime
}}^{a^{\prime }b^{\prime }cd*}Q^{\mu \mu ^{\prime }}(k_1)Q^{\nu \nu ^{\prime
}}(k_2)$}

To calculate $T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu ^{\prime }}^{a^{\prime
}b^{\prime }cd*}Q^{\mu \mu ^{\prime }}(k_1)Q^{\nu \nu ^{\prime }}(k_2)$, it
is necessary to calculate $k_1^\mu k_2^\nu T_{\mu \nu }^{abcd}$. This may be
done in several ways. For example, we may simply contract Eq.(3.67) with
vector $k_1^\mu $ to give 
\begin{equation}
k_1^\mu k_2^\nu T_{\mu \nu }^{abcd}(p_1,p_2;k_1,k_2)=M^2S^{abcd}
\end{equation}
Certainly, paralleling to the procedure shown in the foregoing subsections,
we may firstly compute $k_1^\mu k_2^\nu T_{\rho \sigma \mu \nu }^{(i)abcd}$.
For instance, contracting Eq.(3.56) with $k_1^\mu $, we derive 
\begin{equation}
k_1^\mu k_2^\nu \Gamma _{\mu \nu \lambda }^{(3)}(k_1,k_2,q_3)=-k_1\cdot
k_2q_{3\lambda }-M^2k_{2\lambda }+k_{1\lambda }(q_3^2-M^2)
\end{equation}
From the above equality, noticing the identity in Eq.(3.29), it follows 
\begin{equation}
k_1^\mu k_2^\nu T_{\rho \sigma \mu \nu }^{(3)abcd}=C_3M^2S_{\rho \sigma
}^{(3)}+C_3k_1^\lambda \Gamma _{\rho \sigma \lambda }^{(3)}(p_1,p_2,q_3)
\end{equation}
The other terms can be given by contracting Eqs.(3.52), (3.55) and (3.58)
with $k_1^\mu $. Summing all these terms, one can exactly obtain the result
as written in Eq.(3.69). Employing Eq.(3.69), we get 
\begin{equation}
T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu ^{\prime }}^{a^{\prime }b^{\prime
}cd*}Q^{\mu \mu ^{\prime }}(k_1)Q^{\nu \nu ^{\prime
}}(k_2)=S^{abcd}S^{a^{\prime }b^{\prime }cd*}
\end{equation}
Up to the present, the last three terms in Eq.(3.19) have been calculated.
Inserting Eqs.(3.49),(3.68) and (3.72) into Eq.(3.19), we arrive at 
\begin{equation}
\begin{array}{c}
2ImT_1=\frac 12\int d\tau T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu ^{\prime
}}^{a^{\prime }b^{\prime }cd*}P^{\mu \mu ^{\prime }}(k_1)P^{\nu \nu ^{\prime
}}(k_2) \\ 
+\frac 12\int d\tau S^{abcd}S^{a^{\prime }b^{\prime }cd*}
\end{array}
\end{equation}
The second term in the above needs to be cancelled by the ghost diagrams.

\section*{ D. The Imaginary part of the Ghost Diagrams}

The ghost diagrams in Fig.(4) can be given by folding the three tree
diagrams plotted in Fig.(6) with their conjugates. The folding gives two
times of Figs.(4a)-(4d) and one time of Fig.(4e). Considering that the
symmetry factor of Fig.(4e) is $1$ other than $\frac 12$, the imaginary part
of the transition amplitude of Fig.(4) may be represented as 
\begin{equation}
\begin{array}{c}
2ImT_2^{aba^{\prime }b^{\prime }}=-\frac 12\int d\tau
T^{abcd}(p_1,p_2;k_1,k_2)T^{a^{\prime }b^{\prime }cd*}(p_1^{\prime
},p_2^{\prime };k_1,k_2) \\ 
-\frac 12\int d\tau T^{(3)abcd}(p_1,p_2;k_1,k_2)T^{(3)a^{\prime }b^{\prime
}cd*}(p_1^{\prime },p_2^{\prime };k_1,k_2)
\end{array}
\end{equation}
where 
\begin{equation}
T^{abcd}(p_1,p_2;k_1,k_2)={\sum_{i=1}^{3} }T^{(i)abcd}(p_1,p_2;k_1,k_2)
\end{equation}
$T^{(i)abcd}(p_1,p_2;k_1,k_2)$ represent the matrix elements of
Figs.(6a)-(6c), and the minus sign is inherent for the ghost loops.

According to the Feynman rules and considering the transversality of the
polarization states, it is clear that 
\begin{equation}
T^{(i)abcd}(p_1,p_2;k_1,k_2)=S^{(i)abcd}
\end{equation}
where 
\begin{equation}
S^{(i)abcd}=-ig^2e^\rho (p_1)e^\sigma (p_2)S_{\rho \sigma }^{(i)}
\end{equation}
the $S_{\rho \sigma }^{(i)}(i=1,2,3)$ were defined in Eqs.(3.23),(3.27) and
(3.31) respectively. In accordance with Eq.(3.76), Eq.(3.74) can be
expressed as 
\begin{equation}
2ImT_2=-\frac 12\int d\tau S^{abcd}S^{a^{\prime }b^{\prime }cd*}-\frac 12%
\int d\tau S^{(3)abcd}S^{(3)a^{\prime }b^{\prime }cd*}
\end{equation}
When adding Eq.(3.78) to Eq.(3.73), we see, the second term in Eq.(3.73) is
just cancelled by the first term in Eq.(3.78). However, still remains the
second term in Eq.(3.78) which represents half of the contribution of the
loop diagram in Fig.(4e) to the imaginary part of the amplitude, We are
particularly interested in the fact that the first term in Eq.(3.78)
contains the entire contributions from the ghost diagrams in Figs.(4a)-(4d)
and half of the contribution of the diagram in Fig.(4e). They completely
eliminate the unphysical part of the amplitudes given by Figs.(3a)-(3j),
needless to introduce any extra scalar particle for this elimination. How to
understand the remaining term in Eq.(3.78)? The occurrence of this term in
the sum of the amplitudes given in Eqs.(3.73) and (3.78) is due to that the
loop diagrams in Figs.(3g) and (4e) have different symmetry factors.
Therefore, only half of Fig.(4e) is needed to cancel the unphysical part of
Fig.(3g). It is reminded that until now, the loop diagram in Fig.(3k) has
not been considered. This diagram, as Fig.(3g), has also a symmetry factor $%
\frac 12$ and, as indicated soon later, gives a nonvanishing contribution to
the scattering amplitude and its imaginary part in the case of massive gauge
theory. This contribution, of course, includes a part arising from the
unphysical intermediate states which needs to be cancelled by the
corresponding ghost diagram as well. From the theoretical logic, it is
conceivable that the second term in Eq.(3.78) just serves to cancel the
unphysical part of the diagram in Fig.(3k).

\section*{ E.The Imaginary Part of The Diagram in Fig.(3k)}

How to evaluate the imaginary part of the amplitude of Fig.(3k) by the L-C
rule? This seems to be a difficult problem because we are not able to divide
the diagram into two parts by cutting the internal boson line of the closed
loop in Fig.(3k). However, we observe that when letting one boson line of
the closed loop in Fig.(3g) shrink into a point, Fig.(3g) will convert to
Fig.(3k). This graphically intuitive observation suggests that the amplitude
given by Fig.(3k) can be treated as a limit of the amplitude of Fig.(3g)
when setting the momentum of one propagator in the loop shown in Fig.(3g)
tend to infinity. In this way, we can isolate from Fig.(3k) the unphysical
contribution which looks like to be given by two- particle intermediate
states and hence is able to compare with the second term in Eq.(3.78). It is
obvious that the difference between the both diagrams in Figs.(3k) and (3g)
only lies in their loops, one of which is formed by the four-line vertex
(See Fig.(7a) and another by the three-line vertex (see Fig.(7b)).
Therefore, it is only necessary to compare expressions of the two loops and
establish a connection between them.

The expression of the loop in Fig.(7a) is 
\begin{equation}
\Pi _{\lambda \lambda ^{\prime }}^{(1)ab}(q)=-g^2f^{acd}f^{bcd}[g_{\lambda
\lambda ^{\prime }}g_{\mu \nu }-g_{\lambda \mu }g_{\lambda ^{\prime }\nu
}]\int \frac{d^4k}{(2\pi )^4}\frac{g^{\mu \nu }}{(k^2-M^2+i\varepsilon )}
\end{equation}
The imaginary part of the above function will be exactly calculated in
Appendix (see Eqs.(A.2) and (A.16)). The result is 
\begin{equation}
Im\Pi _{\lambda \lambda ^{\prime }}^{(1)ab}(q)=-\frac{3g^2}{(4\pi )^2}%
f^{acd}f^{bcd}g_{\lambda \lambda ^{\prime }}\int_0^\infty \frac{dx}{x^2}%
sin(xM^2)
\end{equation}
Clearly, it does not vanish when the mass M is not equal to zero. The
expression of Fig.(7b) will be written in the form 
\begin{equation}
\Pi _{\lambda \lambda ^{\prime }}^{(2)ab}(q)=\int \frac{d^4k_1}{(2\pi )^4}%
\frac{d^4k_2}{(2\pi )^4}(2\pi )^4\delta ^4(k_1+k_2-q)\widetilde{\Pi }%
_{\lambda \lambda ^{\prime }}^{(2)ab}(k_1,k_2,q)
\end{equation}
where 
\begin{equation}
\begin{array}{c}
\widetilde{\Pi }_{\lambda \lambda ^{\prime }}^{(2)ab}(k_1,k_2,q)=\frac 12%
g^2f^{acd}f^{bcd}\Gamma _{\mu \nu \lambda }(k_1,k_2,q) \\ 
\times \Gamma _{\mu ^{\prime }\nu ^{\prime }\lambda ^{\prime
}}(k_2,k_2,q)D^{\mu \mu ^{\prime }}(k_1)D^{\nu \nu ^{\prime }}(k_2)
\end{array}
\end{equation}
in which the propagator $D^{\mu \nu }(k)$ was given in Eq.(2.3) with the
gauge parameter $\alpha =1$ and the vertex $\Gamma _{\mu \nu \lambda
}(k_1,k_2,q)$ was defined in Eq.(2.6). Let us take the limit :$\left|
k_{2\mu }\right| \rightarrow \infty $. In this limit, the product of the
propagator $D^{\nu \nu ^{\prime }}(k_2)$ and the vertices will approach to 
\begin{equation}
\begin{array}{c}
D^{\nu \nu ^{\prime }}(k_2)\Gamma _{\mu \nu \lambda }(k_1,k_2,q)\Gamma _{\mu
^{\prime }\nu ^{\prime }\lambda ^{\prime }}(k_1,k_2,q) \\ 
\rightarrow -\frac 1{k_2^2}g^{\nu \nu ^{\prime }}[g_{\mu \nu }k_{2\lambda
}-g_{\lambda \nu }k_{2\mu }][g_{\mu ^{\prime }\nu ^{\prime }}k_{2\lambda
^{\prime }}-g_{\lambda ^{\prime }\nu ^{\prime }}k_{2\mu ^{\prime }}] \\ 
=-\frac 1{k_2^2}[g_{\mu \mu ^{\prime }}k_{2\lambda }k_{2\lambda ^{\prime
}}+g_{\lambda \lambda ^{\prime }}k_{2\mu }k_{2\mu ^{\prime }}-g_{\mu \lambda
^{\prime }}k_{2\lambda }k_{2\mu ^{\prime }}-g_{\lambda \mu ^{\prime
}}k_{2\mu }k_{2\lambda ^{\prime }}]
\end{array}
\end{equation}
If the tensor $k_{2\mu }k_{2\nu }/k_2^2$ behaves in such a way in the limit 
\begin{equation}
k_{2\mu }k_{2\nu }/k_2^2\rightarrow g_{\mu \nu }
\end{equation}
(this limit will be justified in the Appendix), then we find 
\begin{equation}
\widetilde{\Pi }_{\lambda \lambda ^{\prime }}^{(2)ab}(k_1,k_2,q)\text{ }_{%
\overrightarrow{\left| k_{2\mu }\right| \rightarrow \infty }\text{ }%
}g^2f^{acd}f^{bcd}(g_{\lambda \lambda ^{\prime }}g_{\mu \nu }-g_{\lambda \mu
}g_{\lambda ^{\prime }\nu })\frac{g^{\mu \nu }}{k_1^2-M^2+i\varepsilon }
\end{equation}
and hence 
\begin{equation}
\left| \Pi _{\lambda \lambda _{}^{\prime }}^{(2)ab}(q)\right| \text{ }_{%
\overrightarrow{\left| k_{2\mu }\right| \rightarrow \infty }\text{ }}\left|
\Pi _{\lambda \lambda ^{\prime }}^{(1)ab}(q)\right|
\end{equation}
Particularly, in the physical region, the sign of the imaginary part of the
amplitude $\Pi _{\lambda \lambda ^{\prime }}^{(2)ab}(q)$ is the same as the
corresponding part for the amplitude $\Pi _{\lambda \lambda ^{\prime
}}^{(1)ab}(q)$, as will be demonstrated in the Appendix. In view of these,
the imaginary part of Fig.(3k) may equivalently be replaced by the imaginary
part of Fig.(3g) in the limit $\left| k_{2\mu }\right| \rightarrow \infty $. 
\begin{equation}
\begin{array}{c}
2ImT_3^{aba^{\prime }b^{\prime }}=\frac 12\int d\tau {\lim_{\left| k_{2\mu
}\right| \rightarrow \infty }}T_{\mu \nu }^{(3)abcd}T_{\mu ^{\prime }\nu
^{\prime }}^{(3)a^{\prime }b^{\prime }cd*}[P^{\mu \mu ^{\prime }}(k_1) \\ 
\times P^{\nu \nu ^{\prime }}(k_2)+Q^{\mu \mu ^{\prime }}(k_1)g^{\nu \nu
^{\prime }}+g^{\mu \mu ^{\prime }}Q^{\nu \nu ^{\prime }}(k_2)-Q^{\mu \mu
^{\prime }}(k_1)Q^{\nu \nu ^{\prime }}(k_2)]
\end{array}
\end{equation}
The first term in the above only concerns the physical intermediate states.
we do not pursue here what the limit looks like because it is of no
importance at present. we are interested in examining the other three terms.
Look at the expression given in Eq.(3.28). The first term in it can be
ignored due to the equality in Eq.(3.29). The last term can also be
neglected comparing to the second term in the limit $\left| k_{2\mu }\right|
\rightarrow \infty $. Thus, Eq.(3.30) will be reduced to 
\begin{equation}
k_1^\mu T_{\rho \sigma \mu \nu }^{(3)abcd}\rightarrow C_3k_{2\nu }S_{\rho
\sigma }^{(3)}
\end{equation}
where $S_{\rho \sigma }^{(3)}$ was defined in Eq.(3.23) and is irrelevant to 
$k_2$. Eq.(3.88) immediately gives rise to 
\begin{equation}
{\lim_{\left| k_{2\mu }\right| \rightarrow \infty }}T_{\mu \nu
}^{(3)abcd}T_{\mu ^{\prime }\nu ^{\prime }}^{(3)a^{\prime }b^{\prime
}cd*}Q^{\mu \mu ^{\prime }}(k_1)g^{\nu \nu ^{\prime
}}=S^{(3)abcd}S^{(3)a^{\prime }b^{\prime }cd*}
\end{equation}
where $S^{(3)abcd}$ was defined in Eq.(3.77) and the compatibility of the
on- shell condition $k_2^2=M^2$ with the limit $\left| k_{2\mu }\right|
\rightarrow \infty $ has been noticed .

By the same reason as stated above, only the second term in Eq.(3.56) should
be considered in the limit. Therefore, Eq.(3.57) is approximated to 
\begin{equation}
k_2^\nu T_{\rho \sigma \mu \nu }^{\left( 3\right) abcd}\rightarrow
-C_3k_{1\mu }\widetilde{S}_{\rho \sigma }^{(3)}
\end{equation}
where 
\begin{equation}
\widetilde{S}_{\rho \sigma }^{(3)}={\lim_{\left| k_{2\mu }\right|
\rightarrow \infty }}\frac{k_2^\lambda \Gamma _{\rho \sigma \lambda
}^{(3)}(p_1,p_2;q_3)}{q_3^2-M^2+i\varepsilon }
\end{equation}
With the result in Eq.(3.90). the third term in Eq.(3.87) becomes 
\begin{equation}
{\lim_{\left| k_{2\mu }\right| \rightarrow \infty }}T_{\mu \nu
}^{(3)abcd}T_{\mu ^{\prime }\nu ^{\prime }}^{(3)a^{\prime }b^{\prime
}cd*}g^{\mu \mu ^{\prime }}Q^{\nu \nu ^{\prime }}(k_2)=\widetilde{S}%
^{(3)abcd}\widetilde{S}^{(3)a^{\prime }b^{\prime }cd^{*}}
\end{equation}
where 
\begin{equation}
\widetilde{S}^{(3)abcd}=-ig^2e^\rho (p_1)e^\sigma (p_2)C_3\widetilde{S}%
_{\rho \sigma }^{(3)}
\end{equation}
Similarly, in the limit$\left| k_{2\mu }\right| \rightarrow \infty $, we can
neglect the first term (due to Eq.(3.29)) and the last terms in Eq.(3.70).
The second term in Eq.(3.70) permits us to rewrite Eq.(3.71) in the form 
\begin{equation}
k_1^\mu k_2^\nu T_{\rho \sigma \mu \nu }^{(3)abcd}\rightarrow -C_3M_2%
\widetilde{S}_{\rho \sigma }^{(3)}
\end{equation}
From this result, it is clear to see 
\begin{equation}
{\lim_{\left| k_{2\mu }\right| \rightarrow \infty }}T_{\mu \nu
}^{(3)abcd}T_{\mu ^{\prime }\nu ^{\prime }}^{(3)a^{\prime }b^{\prime
}cd*}Q^{\mu \mu ^{\prime }}(k_1)Q^{\nu \nu ^{\prime }}(k_2)=\widetilde{S}%
^{(3)abcd}\widetilde{S}^{(3)a^{\prime }b^{\prime }cd*}
\end{equation}

On inserting Eqs.(3.89),(3.92) and (3.95) into Eq.(3.87), we see, the last
two terms in Eq.(3.87) cancel with each other. As a result, we have 
\begin{equation}
\begin{array}{c}
2ImT_3^{ab}=\frac 12\int d\tau {\lim_{\left| k_{2\mu }\right| \rightarrow
\infty } }T_{\mu \nu }^{(3)abcd}T_{\mu ^{\prime }\nu ^{\prime
}}^{(3)a^{\prime }b^{\prime }cd*}P^{\mu \mu ^{\prime }}(k_1) \\ 
\times P^{\nu \nu ^{\prime }}(k_2)+\frac 12\int d\tau
S^{(3)abcd}S^{(3)a^{\prime }b^{\prime }cd*}
\end{array}
\end{equation}
The second term above is just cancelled by the second term in Eq.(3.78).

Combining the results given in Eqs.(3.73),(3.78) and (3.96), we obtain the
total amplitude as follows 
\begin{equation}
\begin{array}{c}
2ImT_{}^{aba^{\prime }b^{\prime }}=\sum_{i=1}^32ImT_{\left( i\right)
}^{abcd}=\frac 12\int d\tau T_{\mu \nu }^{abcd}T_{\mu ^{\prime }\nu ^{\prime
}}^{a^{\prime }b^{\prime }cd*}P^{\mu \mu ^{\prime }}(k_1)P^{\nu \nu ^{\prime
}}(k_2) \\ 
+\frac 12\int d\tau {\lim_{\left| k_{2\mu }\right| \rightarrow \infty }}
T_{\mu \nu }^{(3)abcd}T_{\mu ^{\prime }\nu ^{\prime }}^{(3)a^{\prime
}b^{\prime }cd*}P^{\mu \mu ^{\prime }}(k_1)P^{\nu \nu ^{\prime }}(k_2)_{}
\end{array}
\end{equation}
which is only related to the physical intermediate states. Thus, the proof
of the unitarity is accomplished.

We note here that the results given in this subsection rely on how to
correctly treat the limit procedure. As will be shown in the Appendix, the
limit given in Eq.(3.84) is the only choice when the relation in Eq.(3.29)
is considered. Similarly, to obtain the desirable limiting results presented
in Eqs.(3.89), (3.92) and (3.95) , the reasonable expressions in Eqs.(3.30),
(3.57) and (3.71) are necessary to be used. In addition, we mention that for
proving the unitarity, the diagrams involving fermion intermediate states
were not considered because the fermion intermediate state is already
physical.

\setcounter{section}{4}

\section*{4.Unitarity of Fermion-Antifermion Scattering Amplitude of Order $%
g^4$}

\setcounter{equation}{0}

In this section, to illustrate the unitarity of the theory further. we plan
to evaluate the imaginary part of the fermion-antifermion scattering
amplitude in the perturbative approximation of order $g^4$. For this
purpose, it is only necessary to consider the diagrams shown in Fig.(8).

The diagrams in Figs.(8a)-(8e) can be reconstructed by folding the tree
diagrams in Figs.(9a)-(9c) with their conjugates. Since the folding gives
two times of Figs.(8a)-(8d) and one time of Fig.(8e) which possesses a
symmetry factor $\frac 12$, the imaginary parts of the amplitudes given by
Figs.(8a)-(8e) may be represented as 
\begin{equation}
\begin{array}{c}
2ImT_1=\frac 12\int d\tau T_{\mu \nu }^{ab}T_{\mu ^{\prime }\nu ^{\prime
}}^{ab*}g^{\mu \mu ^{\prime }}g^{\nu \nu ^{\prime }} \\ 
=\frac 12\int d\tau T_{\mu \nu }^{ab}T_{\mu ^{\prime }\nu ^{\prime
}}^{ab*}[P^{\mu \mu ^{\prime }}(k_1)P^{\nu \nu ^{\prime }}(k_2)+Q^{\mu \mu
^{\prime }}(k_1)g^{\nu \nu ^{\prime }} \\ 
+g^{\mu \mu ^{\prime }}Q^{\nu \nu ^{\prime }}(k_2)-Q^{\mu \mu ^{\prime
}}(k_1)Q^{\nu \nu ^{\prime }}(k_2)]
\end{array}
\end{equation}
where  
\begin{equation}
T_{\mu \nu }^{ab}(p_1,p_2;k_1,k_2)={\sum_{i=1}^3}T_{\mu \nu
}^{(i)ab}(p_1,p_2;k_1,k_2)
\end{equation}
$T_{\mu \nu }^{(i)ab}(p_1,p_2;k_1,k_2)$ denote the matrix elements of
Figs.(9a)-(9c) respectively. According to the Feynman rules. they can be
written as 
\begin{equation}
T_{\mu \nu }^{(1)ab}(p_1,p_2;k_1,k_2)=-ig^2\overline{v}(p_2)\frac{\lambda ^b}%
2\gamma _\nu \frac{\not p_1-\not k_1+m}{(p_1-k_1)^2-m^2+i\varepsilon }\frac{%
\lambda ^a}2\gamma _\mu u(p_1)
\end{equation}
\begin{equation}
T_{\mu \nu }^{(2)ab}(p_1,p_2;k_1,k_2)=-ig^2\overline{v}(p_2)\frac{\lambda ^a}%
2\gamma _\mu \frac{\not k_1-\not p_2+m}{(k_1-p_2)^2-m^2+i\varepsilon }\frac{%
\lambda ^b}2\gamma _\nu u(p_1)
\end{equation}
\begin{equation}
T_{\mu \nu }^{(3)ab}(p_1,p_2;k_1,k_2)=-\frac{g^2f^{abc}}{q^2-M^2+i%
\varepsilon }\Gamma _{\mu \nu \lambda }(k_1,k_2,q)\overline{v}(p_2)\frac{%
\lambda ^c}2\gamma ^\lambda u(p_1)
\end{equation}
where $\Gamma _{\mu \nu \lambda }(k_1,k_2,q)$ was defined in Eq.(2.6).

For evaluating the second term in Eq.(4.1), we have to compute the
contraction of $T_{\mu \nu }^{(i)ab}$ with $k_1^\mu $. By applying Dirac
equation, the on-mass shell condition and the relation $q=k_1+k_2=p_1+p_2$,
one may get 
\begin{equation}
\begin{array}{c}
k_1^\mu [T_{\mu \nu }^{(1)ab}+T_{\mu \nu }^{(2)ab}]=-ig^2\overline{v}(p_2)[%
\frac{\lambda ^a}2,\frac{\lambda ^b}2]\gamma _\nu u(p_1) \\ 
=g^2f^{abc}\overline{v}(p_2)\frac{\lambda ^c}2\gamma _\nu u(p_1)
\end{array}
\end{equation}
\begin{equation}
k_1^\mu T_{\mu \nu }^{(3)ab}=-g^2f^{abc}\overline{v}(p_2)\frac{\lambda ^c}2%
\gamma _\nu u(p_1)+k_{2\nu }S^{ab}
\end{equation}
where 
\begin{equation}
S^{ab}=\frac{g^2f^{abc}}{2p_1\cdot p_2+M^2}\overline{v}(p_2)\frac{\lambda ^c}%
2\not k_1u(p_1)
\end{equation}
Adding Eq.(4.7) to Eq.(4.6), we find 
\begin{equation}
k_1^\mu T_{\mu \nu }^{ab}=k_{2\nu }S^{ab}
\end{equation}
As is seen, there is a cancellation among the diagrams in Figs.(9a)-(9c).
From Eq.(4.9), one may derive 
\begin{equation}
T_{\mu \nu }^{ab}T_{\mu ^{\prime }\nu ^{\prime }}^{ab*}Q^{\mu \mu ^{\prime
}}(k_1)g^{\nu \nu ^{\prime }}=S^{ab}S^{ab*}
\end{equation}

Let us turn to calculate the third term in Eq.(4.1). Along the same line
stated above, one may get 
\begin{equation}
k_2^\mu [T_{\mu \nu }^{(1)ab}+T_{\mu \nu }^{(2)ab}]=-g^2f^{abc}\overline{v}%
(p_2)\frac{\lambda ^c}2\gamma _\mu u(p_1)
\end{equation}
and 
\begin{equation}
k_2^\nu T_{\mu \nu }^{(3)ab}=g^2f^{abc}\overline{v}(p_2)\frac{\lambda ^c}2%
\gamma _\mu u(p_1)+k_{1\mu }\widetilde{S}^{ab}
\end{equation}
where 
\begin{equation}
\widetilde{S}^{ab}=\frac{-g^2f^{abc}}{q^2-M^2+i\varepsilon }\overline{v}(p_2)%
\frac{\lambda ^c}2\not k_2u(p_1)
\end{equation}
From the equality 
\begin{equation}
\overline{v}(p_2)\frac{\lambda ^c}2(\not k_1+\not k_2)u(p_1)=\overline{v}%
(p_2)\frac{\lambda ^c}2(\not p_1+\not p_2)u(p_1)=0
\end{equation}
it follows that 
\begin{equation}
\widetilde{S}^{ab}=S^{ab}
\end{equation}
Adding Eq.(4.11) to Eq.(4.9) and noticing Eq.(4.14), we have 
\begin{equation}
k_2^\nu T_{\mu \nu }^{ab}=k_{1\mu }S^{ab}
\end{equation}
This result gives rise to 
\begin{equation}
T_{\mu \nu }^{ab}T_{\mu ^{\prime }\nu ^{\prime }}^{ab*}g^{\mu \mu ^{\prime
}}Q^{\nu \nu ^{\prime }}(k_2)=S^{ab}S^{ab*}
\end{equation}

For evaluating the last term in Eq.(4.1). we may use the following
equalities which are obtained by contracting Eqs.(4.10) and (4.11) with $%
k_1^\mu $%
\begin{equation}
k_1^\mu k_2^\nu [T_{\mu \nu }^{(1)ab}+T_{\mu \nu }^{(2)ab}]=-g^2f^{abc}%
\overline{v}(p_2)\frac{\lambda ^c}2\not k_1u(p_1)
\end{equation}
and 
\begin{equation}
k_1^\mu k_2^\nu T_{\mu \nu }^{(3)ab}=g^2f^{abc}\overline{v}(p_2)\frac{%
\lambda ^c}2\not k_1u(p_1)+M^2\widetilde{S}^{ab}
\end{equation}

These equalities and Eq.(4.15) lead to 
\begin{equation}
k_1^\mu k_2^\nu T_{\mu \nu }^{ab}=M^2S^{ab}
\end{equation}
This result allows us to give the last term in Eq.(4.1) in the form 
\begin{equation}
T_{\mu \nu }^{ab}T_{\mu ^{\prime }\nu ^{\prime }}^{ab*}Q^{\mu \mu ^{\prime
}}(k_1)Q^{\nu \nu ^{\prime }}(k_2)=S^{ab}S^{ab*}
\end{equation}

Substituting Eqs.(4.9), (4.17) and (4.21) in Eq.(4.1), we arrive at 
\begin{equation}
2ImT_1=\frac 12\int d\tau T_{\mu \nu }^{ab}T_{\mu ^{\prime }\nu ^{\prime
}}^{ab*}P^{\mu \mu ^{\prime }}(k_1)P^{\nu \nu ^{\prime }}(k_2)+\frac 12\int
d\tau S^{ab}S^{ab*}
\end{equation}

The ghost diagram in Fig.(8f) can be given by folding the tree diagram in
Fig.(9d) with its conjugate. Therefore, the imaginary part of Fig.(8f) can
be written as 
\begin{equation}
2ImT_2=-\int d\tau S^{ab}S^{ab*}
\end{equation}
In complete analogy with the two-boson scattering discussed in the preceding
section, the second term in Eq.(4.22) can only cancel half of the above
amplitude. The reason for this still is due to the difference between the
symmetry factors of Figs.(8e) and (8f). To achieve a complete cancellation,
it is necessary to consider the contribution of the diagram in Fig.(8g).
This diagram can also be treated as a limit of the diagram in Fig.(8e) when
the momentum of one internal line in the loop tends to infinity, 
\begin{equation}
\begin{array}{c}
2ImT_3=\frac 12\int d\tau {\lim_{\left| k_{2\mu }\right| \rightarrow \infty
} }T_{\mu \nu }^{(3)ab}T_{\mu ^{\prime }\nu ^{\prime }}^{(3)ab*}[P^{\mu \mu
^{\prime }}(k_1)P^{\nu \nu ^{\prime }}(k_2) \\ 
+Q^{\mu \mu ^{\prime }}(k_1)g^{\nu \nu ^{\prime }}+g^{\mu \mu ^{\prime
}}Q^{\nu \nu ^{\prime }}(k_2)-Q^{\mu \mu ^{\prime }}(k_1)Q^{\nu \nu ^{\prime
}}(k_2)]
\end{array}
\end{equation}
In the limit:$\left| k_{2\mu }\right| \rightarrow \infty $, comparing to the
second terms in Eqs.(4.7), (4.12) and (4.17), the first terms in these
equations can be ignored. Thus, Eqs.(4.7), (4.12) and (4.17) respectively
reduce to 
\begin{equation}
k_1^\mu T_{\mu \nu }^{(3)ab}\approx k_{2\nu }S^{ab}
\end{equation}
\begin{equation}
k_2^\mu T_{\mu \nu }^{(3)ab}\approx k_{1\mu }\widetilde{S}^{ab}
\end{equation}
\begin{equation}
k_1^\mu k_2^\nu T_{\mu \nu }^{(3)ab}\approx M^2\widetilde{S}^{ab}
\end{equation}
On inserting these expressions into Eq.(4.24), we have 
\begin{equation}
\begin{array}{c}
2ImT_3=\frac 12\int d\tau {\lim_{\left| k_{2\mu }\right| \rightarrow \infty
} }T_{\mu \nu }^{(3)ab}T_{\mu ^{\prime }\nu ^{\prime }}^{(3)ab*}P^{\mu \mu
^{\prime }}(k_1)P^{\nu \nu ^{\prime }}(k_2) \\ 
+\frac 12\int d\tau S^{ab}S^{ab*}
\end{array}
\end{equation}
Thus, as shown before, it is indeed possible to find a way which allows us
to isolate from Fig.(8g) the unphysical part of the amplitude like the
second term in Eq.(4.28).

Summing the results denoted in Eqs.(4.22), (4.23) and (4.28). we obtain the
imaginary part of the total amplitude such that 
\begin{equation}
\begin{array}{c}
2ImT=\frac 12\int d\tau T_{\mu \nu }^{ab}T_{\mu ^{\prime }\nu ^{\prime
}}^{ab*}P^{\mu \mu ^{\prime }}(k_1)P^{\nu \nu ^{\prime }}(k_2) \\ 
+\frac 12\int d\tau {\lim_{\left| k_{2\mu }\right| \rightarrow \infty }}%
T_{\mu \nu }^{\left( 3\right) ab}T_{\mu ^{\prime }\nu ^{\prime }}^{\left(
3\right) ab*}P^{\mu \mu ^{\prime }}(k_1)P^{\nu \nu ^{\prime }}(k_2)
\end{array}
\end{equation}
in which the unphysical contributions are all cancelled. Thus, the unitarity
is ensured.

\setcounter{section}{5}

\section*{5.Comments and Discussions}

\setcounter{equation}{0}

In the previous sections, the unitarity of our theory has been illustrated
by evaluating the imaginary parts of two-boson and fermion-antifermion
scattering amplitudes up to the fourth order perturbation. The imaginary
parts of the amplitudes were calculated by means of the L-C rule. In this
kind of calculation, as emphasized in the Introduction, we have to work in
the Feynman gauge because the formula used requires the intermediate states
must being complete. Although the unitarity of the S-matrix elements is
proved in the Feynman gauge, it would be true for other gauges since it was
exactly proved in paper III that the S-matrix is gauge-independent.

As mentioned in the Introduction, In the previous works of examining the
unitarity of some kinds of massive gauge theories$^{[13][14]}$, the boson
propagator in the Landau gauge 
\begin{equation}
D_{\mu \nu }(k)=\frac{g_{\mu \nu }-k_\mu k_\nu /k^2}{k^2-M^2+i\varepsilon }
\end{equation}
was chosen to calculate the imaginary part of scattering amplitudes by the
L-C rule. For such a calculation, we note, the Landau gauge is , actually,
not suitable. Since the intermediate states characterized by the transverse
projector appearing in the numerator of the above propagator does not form a
complete set. The unsuitability of the procedure may be seen from the
massless gauge theory. The unitarity of the theory was exampled by computing
the imaginary part of the fermion-antifermion scattering amplitude of order $%
g^4$ in the Feynman gauge$^{[18]}$. However, if one tries to perform the
proof in the Landau gauge, he could not get a reasonable result. The
unsuitability may also be seen from the fact that the longitudinal projector 
$k_\mu k_\nu /k^2$ in Eq.(5.1) could not be given an unambiguous definition
on the mass shell since the momentum k on the mass shell becomes an
isotropic vector. For the massive gauge theory, the propagator in Eq.(5.1)
can be divided into two parts 
\begin{equation}
D_{\mu \nu }(k)=\widetilde{D}_{\mu \nu }(k)+\frac{k_\mu k_\nu }{M^2k^2}
\end{equation}
where 
\begin{equation}
\widetilde{D}_{\mu \nu }(k)=\frac{g_{\mu \nu }-k_\mu k_{\nu /M^2}}{%
k^2-M^2+i\varepsilon }
\end{equation}
In the literature$^{[2][14]}$, the first term in Eq.(5.2) was viewed as the
physical part of the propagator in Eq.(5.1), i.e. the propagator which is
given in the unitary gauge and represents a spin-one particle, while, the
second term in Eq.(5.2) was thought of the unphysical part representing a
spin-zero particle. The latter term must be eliminated by the ghost particle
and some others in the S-matrix element. Otherwise, the theory was viewed as
nonunitary. The above points of view are questionable. Firstly, we note that
the Landau gauge propagator shown in Eq.(5.1) precisely represents the
off-shell intermediate state of the spin-one particle. It could not be able
to simultaneously represent two particles of different spins. This point may
also be seen from the imaginary part of the propagator. According to the L-C
rule, the imaginary parts of the both propagators shown in Eqs.(5.1) and
(5.3) are equal to each other. There is no place for the second term of
Eq.(5.2) in the imaginary part. Therefore. this term appears to be useless
in the calculation of the imaginary part of S-matrix elements, contradicting
the original anticipation that it should take part in the cancellation with
the ghost terms. Next, suppose the second term in Eq.(5.2) could be
eliminated in the full S-matrix element, one can not avoid the problem of
unrenormalizability caused by the first term in Eq.(5.2) owing to the bad
ultraviolet behavior of the term $k^\mu k^\nu /m^2$ $^{[15]}$. The question
still concerns the understanding of the propagator $\widetilde{D}_{\mu \nu
}(k)$. As one knows. this propagator is a direct result of the massive
Yang-Mills Lagrangian without constraints which was often called the
Lagrangian given in the unitary gauge$^{[8][18]}$. This Lagrangian, as
indicated in our preceding papers, does not give a complete description of
the massive gauge field dynamics. Therefore, the propagator derived from it
can not be considered to be physical. In view of this , we can say , the
conventional viewpoint that the so-called unitary gauge is a physical gauge,
actually, is an ill-concept. Nevertheless, that propagator was often quoted
as a starting point in the previous literature of discussing the unitarity
problem$^{[8]-[10]}$. In Refs.(9) and (10). the authors investigated the
tree-unitarity conditions of the theories involving fermions, gauge bosons
and scalar particles and concluded that the unique unitary theory is of
Higgs type, i.e. the spontaneously broken gauge theory. it is pointed out
that this conclusion is only drawn from the Lagrangian without imposing
necessary constraints on it. From this Lagrangian, one can only derive the
propagator as written in Eq.(5.3). Just such a propagator was used in their
derivation of the transition amplitude between the longitudinally polarized
gauge boson states. In the amplitude, there appear a series of terms which
violate the unitarity of the S-matrix at high energy regime. These terms
need to be cancelled by introducing a certain scalar particles. As we have
argued before, the Lagrangian without constraints is not complete, the
propagator in Eq.(5.3) is wrong and the longitudinal polarization physically
is absent for the massive gauge bosons. Therefore, the above conclusion is
questionable, at least, not universal. Certainly, it can not exclude the
possibility of setting up an unitary gauge theory without involving the
Higgs boson in it.

In general, for examining the unitarity of a massive gauge theory, it is
only necessary to evaluate the S-matrix element between the physical
transversely polarized states. In this way, it was shown in Sections (2) and
(3) that the unitarity is well satisfied. Particularly, the calculation in
Sect.3 indicates that except for the diagrams involving the closed loops,
there is a natural cancellation among the contributions coming from the
unphysical gauge boson and ghost particle intermediate states for all the
other diagrams, without the help of any scalar particle. This result and the
theoretical logic strongly suggest that the same cancellation is bound to
happen for the loop diagrams. To achieve this cancellation, the loop diagram
formed by the gauge boson four-line vertex is necessary to be considered and
recast in the form as if it is given by the two-particle intermediate states
so as to be able to compare with the contributions given by the other loop
diagrams. For this purpose, we proposed in Sect.3 a reasonable limit
procedure which allows us to reach the cancellation mentioned above. The
results we obtained are undoubtedly correct , though, some profound problems
concerning the limit procedure still need to pursue further. It should be
noted that in all the previous investigations$^{[11][13][14]}$ on the
unitarity problem, the diagram involving the loop given by the boson
four-line vertex such as Figs.(3k),(7c) and (8g) was never taken into
account in the cancellation of the unphysical amplitudes. Moreover, in the
work by Mohapatra et al $^{[11]}$, the authors only considered the tree
diagrams without concerning loop diagrams. From the calculations described
in Sections(3) and (4), it is clearly seen that the loop diagrams play an
essential role to guarantee the cancellation of the unphysical part of the
amplitudes and hence the unitarity of the S-matrix element. That is why we
said in the Introduction that the previous proofs are not complete and the
conclusion is not faithful. Particularly, as was indicated in paper I, the
theories presented in Refs.(2) and (6) which were pointed out to be
non-unitary by Mohapatra et al and others$^{[3,11.14]}$, are not correct
because the Feynman rule concerning the closed ghost loop has an extra
factor $\frac 12$ other than 1 as given in our theory. In this paper, the
unitarity has been proved in the perturbation approximation up to the order
of $g^4$. For higher order approximations, we believe that for a given
process, if all the diagrams are taken into account and treated
appropriately, the unitarity would be proved to be no problem.

\begin{center}
{\bf ACKNOWLEDGMENTS }
\end{center}

The author wishes to think professor Shi-Shu Wu and professor Bing-Lin Young
( Iowa State University) for helpful discussions. This project was supported
in part by National Natural Science Foundation of China.

\renewcommand{\thesection}{\Alph{section}} \setcounter{section}{1} %
\setcounter{section}{1} \setcounter{equation}{0}

\begin{center}
{\bf APPENDIX :Examination of Signs of The Imaginary Parts of The Loop
Diagrams }
\end{center}

It was mentioned in Sect.(3) that the imaginary part of the matrix element
of the loop in Fig.(7a) has the same sign as that given by the loop in
Fig.(7b) at least for the large momentum $k_2$. To convince ourself of this
point. we investigate the imaginary part of the loops in a parametric
representation. In this representation, the propagator will be expressed in
the form $^{[16]}$%
\begin{equation}
\frac 1{k_i^2-M^2+i\varepsilon }=-i{\int _0 ^\infty }d\alpha _ie^{i\alpha
_i(k_i^2-M^2+i\varepsilon )}
\end{equation}
With this representation, the matrix element shown in Eq.(3.79) for Fig.(7a)
may be rewritten as 
\begin{equation}
\Pi _{\lambda \lambda ^{\prime }}^{(1)ab}(q)=-3g^2f^{abc}f^{bcd}g_{\mu \nu
}J^{(1)}
\end{equation}
where 
\begin{equation}
\begin{array}{c}
J^{(1)}=\int \frac{d^4k}{(2\pi )^4}\frac 1{k^2-M^2+i\varepsilon } \\ 
=-i {\int _0 ^\infty }d\alpha \int \frac{d^4k}{(2\pi )^4}e^{i\alpha
(k^2-M^2+i\varepsilon )} \\ 
=-\frac 1{(4\pi )^2}{\int _0 ^\infty }\frac{d\alpha }{\alpha ^2}e^{-i\alpha
(M^2-i\varepsilon )}
\end{array}
\end{equation}
In the above, we have used the representation in Eq.(A.1) and the formula of
Fresnel integral$^{[16]}$%
\begin{equation}
\int \frac{d^4k}{(2\pi )^4}e^{i\alpha k^2}=\frac{-i}{(4\pi \alpha )^2}
\end{equation}

For the loop in Fig.(7b), we confine ourselves to investigate its expression
in the limit of large momentum $k_2$ for the purpose of comparing to the one
given in Eqs.(A.2) and (A.3). As stated in Eqs.(3.83)-(3.86), in order to
convert the matrix element of Fig.(7b) to the one for Fig.(7a), it is
necessary to take the approximate expression denoted in Eq.(3.83) and the
limit assumed in Eq.(3.84) which is applied to Eq.(3.83). To achieve such a
conversion, as easily verified, instead of Eq.(3.83), we may simply take an
equivalent approximate expression such that 
\begin{equation}
g^{\mu \mu ^{\prime }}g^{\nu \nu ^{\prime }}\Gamma _{\mu \nu \lambda
}(k_1,k_2,q)\Gamma _{\mu ^{\prime }\nu ^{\prime }\lambda ^{\prime
}}(k_1,k_2,q)\rightarrow 6k_{2\lambda }k_{2\lambda ^{\prime }}
\end{equation}
With this expression, Eqs.(3.81) and (3.82) may be rewritten as 
\begin{equation}
\Pi _{\lambda \lambda ^{\prime }}^{(2)ab}(q)\approx
3g^2f^{acd}f^{bcd}J_{\lambda \lambda ^{\prime }}^{(2)}(q)
\end{equation}
where 
\begin{equation}
\begin{array}{c}
J_{\lambda \lambda ^{\prime }}^{(2)}(q)=\int \frac{d^4k_1}{(2\pi )^4}\frac{%
d^4k_2}{(2\pi )^4}(2\pi )^4\delta ^4(q-k_1-k_2) \\ 
\times \frac{k_{2\lambda }k_{2\lambda ^{\prime }}}{(k_1^2-M^2+i\varepsilon
)(k_2^2-M^2+i\varepsilon )}
\end{array}
\end{equation}
Employing the parametrization given in Eq.(A.1) and the Fourier
representation of the $\delta $-function, Eq.(A.7) reads 
\begin{equation}
\begin{array}{c}
J_{\lambda \lambda ^{\prime }}^{(2)}(q)= {\int _0 ^\infty }d\alpha _1d\alpha
_2\int d^4z\int \frac{d^4k_1}{(2\pi )^4}e^{i\alpha _1k_1^2-\frac i4\frac{z^2%
}{\alpha _1}-iqz} \\ 
\times \partial _\lambda ^z\partial _{\lambda ^{\prime }}^z\int \frac{d^4k_2%
}{(2\pi )^4}e^{i\alpha _2k_2^2-\frac i4\frac{z^2}{\alpha _2}-i(\alpha
_1+\alpha _2)(M^2-i\varepsilon )}
\end{array}
\end{equation}
Upon completing the integrations over $k_1,k_2$ and z by using the formula
of Fresnel integral, we have 
\begin{equation}
\begin{array}{c}
J_{\lambda \lambda ^{\prime }}^{(2)}(q)=\frac i{(4\pi )^2} {\int _0 ^\infty }%
\frac{d\alpha _1d\alpha _2}{(\alpha _1+\alpha _2)^2}[g_{\lambda \lambda
^{\prime }}\frac i{2\alpha _2}-\frac 1{4\alpha _2^2}\partial _\lambda
^q\partial _{\lambda ^{\prime }}^q] \\ 
\times e^{iQ(q,\alpha )}
\end{array}
\end{equation}
where 
\begin{equation}
Q(q,\alpha )=\frac{\alpha _1\alpha _2}{\alpha _1+\alpha _2}q^2-(\alpha
_1+\alpha _2)(M^2-i\varepsilon )
\end{equation}
Inserting the identity 
\begin{equation}
{\int _0 ^\infty }dx\delta (x-\alpha _1-\alpha _2)=1
\end{equation}
into Eq.(A.9) and then making the transformation $\alpha _i\rightarrow
x\alpha _i$, one can write 
\begin{equation}
\begin{array}{c}
J_{\lambda \lambda ^{\prime }}^{(2)}(q)=\frac i{(4\pi )^2} {\int _0 ^1}%
d\alpha _1d\alpha _2\delta (1-\alpha _1-\alpha _2) \\ 
{\int _0 ^\infty }\frac{dx}x(\frac i{2x}g_{\lambda \lambda ^{\prime
}}+\alpha _1^2q_\lambda q_{\lambda ^{\prime }})e^{ixQ(q,\alpha )}
\end{array}
\end{equation}
Noticing the equalities shown in Eqs.(3.29) and (4.14), the second term in
the parenthesis, actually, can be ignored. Thus, the function $J_{\lambda
\lambda ^{\prime }}^{(2)}(q)$ is only proportional to the unit tensor $%
g_{\lambda \lambda ^{\prime }}$. This result precisely justifies the limit
taken in Eq.(3.84). On substituting Eq.(A.12) in Eq.(A.6), and performing
the integration over $\alpha _2$, one gets 
\begin{equation}
\Pi _{\lambda \lambda ^{\prime }}^{(2)ab}(q)=-3g^2f^{acd}f^{bcd}g_{\lambda
\lambda ^{\prime }}J^{(2)}(q)
\end{equation}
where 
\begin{equation}
J^{(2)}(q)=\frac 1{(4\pi )^2}{\int _0 ^1 }d\alpha {\int _0 ^\infty }\frac{dx%
}{2x^2}e^{ixQ(q,\alpha )}
\end{equation}
in which 
\begin{equation}
Q(q,\alpha )=\alpha (1-\alpha )q^2-M^2
\end{equation}

Now, we are in position to examine the imaginary parts of the functions $\Pi
_{\lambda \lambda ^{\prime }}^{(1)ab}$ and $\Pi _{\lambda \lambda ^{\prime
}}^{(2)ab}$. Firstly, we write down the imaginary parts of the functions $%
J^{(1)}$ and $J^{(2)}$, 
\begin{equation}
ImJ^{(1)}=\frac 1{(4\pi )^2}{\int_0^\infty }\frac{dx}{x^2}\sin (xM^2)
\end{equation}
and 
\begin{equation}
ImJ^{(2)}=\frac 1{(4\pi )^2}{\int_0^1}d\alpha {\int_0^\infty }\frac{dx}{2x^2}%
\sin \{x[\alpha (1-\alpha )q^2-M^2]\}
\end{equation}
For the integral over x, obviously, the major contribution arises from the
integrand at the neighborhood of the origin. Therefore 
\begin{equation}
ImJ^{(1)}\geq 0
\end{equation}
As for the $ImJ^{(2)}$, the integral over $\alpha $ may be estimated by
taking the mean value $\frac 12$ of the variable $\alpha $ in the integrand,
Thus 
\begin{equation}
ImJ^{(2)}\approx \frac 1{(4\pi )^2}{\int_0^\infty }\frac{dx}{2x^2}\sin [x(%
\frac 14q^2-M^2)]
\end{equation}
It is well known that in the physical region, 
\begin{equation}
q^2\geq 4M^2
\end{equation}
where $q^2=4M^2$ is the starting point of a cut which is the solution of the
following Landau equations$^{[16]}$%
\begin{equation}
\begin{array}{c}
\lambda _1(k^2-M^2)=0 \\ 
\lambda _2[(q-k)^2-M^2]=0 \\ 
\lambda _1k_\mu -\lambda _2(q-k)_\mu =0
\end{array}
\end{equation}
In view of Eq.(A.20), we may conclude 
\begin{equation}
ImJ^{(2)}\geq 0
\end{equation}
The results in Eqs.(A.18) and (A.22) straightforwardly lead to that the
imaginary parts of the $\Pi _{\lambda \lambda ^{\prime }}^{(1)ab}(q)$ and $%
\Pi _{\lambda \lambda ^{\prime }}^{(2)ab}(q)$ have the same sign as we see
from Eq.(A.2) and (A.13).

At last , we would like to mention the imaginary part of the loop in
Fig.(7c). The expression of the loop is 
\begin{equation}
\Pi _{\lambda \lambda ^{\prime }}^{(3)ab}(q)=-g^2f^{acd}f^{bcd}J_{\lambda
\lambda ^{\prime }}^{(3)}(q)
\end{equation}
where 
\begin{equation}
\begin{array}{c}
J_{\lambda \lambda ^{\prime }}^{(3)}(q)=-\int \frac{d^4k_1}{(2\pi )^4}\frac{%
d^4k_2}{(2\pi )^4}(2\pi )^4\delta ^4(q-k_1-k_2) \\ 
\frac{k_{1\lambda }k_{2\lambda ^{\prime }}}{(k_1^2-M^2+i\varepsilon
)(k_2^2-M^2+i\varepsilon )}
\end{array}
\end{equation}
Completely following the procedure formulated in Eqs.(A.8)-(A.12), one may
derive 
\begin{equation}
\begin{array}{c}
J_{\lambda \lambda ^{\prime }}^{(3)}(q)=\frac i{(4\pi )^2} {\int _0 ^1}%
d\alpha {\int _0 ^\infty }\frac{dx}x[\frac i{2x}g_{\lambda \lambda ^{\prime
}}-\alpha (1-\alpha )q_\lambda q_{\lambda ^{\prime }}] \\ 
\times e^{ixQ(q,\alpha )}
\end{array}
\end{equation}
Neglecting the second term containing $q_\lambda q_{\lambda ^{\prime }}$ and
then substituting the above equation into Eq.(A.23), we can write 
\begin{equation}
\Pi _{\lambda \lambda ^{\prime }}^{(3)ab}(q)=-g^2f^{acd}f^{bcd}g_{\lambda
\lambda ^{\prime }}J^{(3)}(q)
\end{equation}
where 
\begin{equation}
J^{(3)}(q)=-\frac 1{(4\pi )^2}{\int _0 ^1}d\alpha {\int _0 ^\infty }\frac{dx%
}{2x^2}e^{ix[\alpha (1-\alpha )q^2-M^2]}
\end{equation}
Clearly, the sign of the imaginary part $ImJ^{(3)}(q)$ is opposite to the $%
ImJ^{(2)}(q)$. Therefore, the imaginary part of the $\Pi _{\lambda \lambda
^{\prime }}^{(3)ab}(q)$ and $\Pi _{\lambda \lambda ^{\prime }}^{(2)ab}(q)$
have opposite signs.

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\end{document}

