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%% section 9 AN ALLMULTIPLICITY RESULT [slacpub7191009 in slacpub7191009: slacpub71910010]
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\section{\usemenu{slacpub7191::context::slacpub7191009}{AN ALLMULTIPLICITY RESULT}}\label{section::slacpub7191009}
We had previously obtained\cite{12}, with Dave Dunbar,
an explicit formula for the oneloop amplitude,
in the $N=4$ supersymmetric gauge theory, with an
{\it arbitrary\/} number of external gluons with the helicity configuration
of the ParkeTaylor treelevel amplitudes~\cite{13,14},
namely all but two gluons
carrying the same helicity. (Let us take the majority to have positive
helicity, the two oppositehelicity gluons to have negative helicity;
exchanging these two amounts to a complex conjugation.)
Here, we will outline the derivation of this result
using
the unitaritybased method.
Such results demonstrate the power of the
method, because they would require the computation of an infinite
number of Feynman diagrams in the traditional approach.
This amplitude is uniquely determined by its cuts, and can be expressed
entirely in terms of scalar box integrals. There are two kinds of cuts
to consider, one where both oppositehelicity gluons are one the same
side of the cut, the other where one such gluon appears on each side of
the cut. The former cut receives contributions only from gluons crossing
the cut, the latter from fermions and scalars as well; but it turns out
that after summing over the $N=4$ supermultiplet (using supersymmetry
Ward identities~\cite{15}), the structure of the
second type of cut is similar to that of the first. It thus suffices
to consider the first kind.
Let us suppose that the oppositehelicity gluons are on the left side
of the cut (drawn vertically through the diagram). Then all the external
legs on the righthand side of the cut have positive helicity; as a
result, the tree amplitude on the righthand side will vanish unless
both legs crossing the cut on the right have negative helicity. Since
we adopt the convention that all momenta are directed inwards into
an amplitude, the cutcrossing legs flip sign as we cross the cut to
the left. We must flip their helicities as well, which tells us that
all legs of the tree amplitude to the left of the cut have
positive helicity, except the
two external legs carrying negative helicity.
Thus both tree amplitudes on either side of the cut have the helicity
configuration of the ParkeTaylor amplitudes~\cite{13,14},
so that simple all$n$
formul\ae\ are available for them. Most of the factors in these
amplitudes are independent of the momenta crossing the cut, and so
can be pulled out in front of the cut integral~(\docLink{slacpub7191007.tcx}[cutEqn]{4}). Indeed,
most of these factors reassemble into the tree amplitude itself, so
that (up to a phase) we obtain
\begin{eqnarray}
&& \hskip 15mm
A_n^\tree\;\int d^D{\rm LIPS}\;\nonumber\\
&&\hskip 10mm
\times{\spa{(m_11)}.{m_1}\spa{\ell_1}.{\ell_2}^2 \spa{m_2}.{(m_2+1)}
\over \spa{(m_11)}.{\ell_1}\spa{\ell_1}.{m_1}\spa{m_2}.{\ell_2}
\spa{\ell_2}.{(m_2+1)}}
\end{eqnarray}
where $(m_11,m_1)$ and $(m_2,m_2+1)$ are the pairs of legs adjacent to
the cut.
The four factors in the denominator give rise to four propagators, which
along with the two propagators crossing the cut means that for an
arbitrary number of external gluons, we need consider only a hexagon
integral. Indeed, it turns out to be a rather special hexagon integral;
rearranging the spinor products in its numerator, it decomposes into
a sum of one and twomass scalar boxes. From this decomposition and
the uniqueness theorem, we then obtain the complete oneloop
$N=4$ supersymmetric amplitude.
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