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\begin{document}
\title{EXPLICIT PROOF THAT ELECTROPRODUCTION OF TRANSVERSELY POLARIZED
MESONS
VANISHES IN PERTURBATIVE QCD}
\author{Pervez Hoodbhoy\thanks{%
Permanent Address: Department of Physics, Quaid-e-Azam University,
Islamabad
45320, Pakistan}}
\address{Department of Physics \\
University of Maryland \\
College Park, Maryland 20742. }
\date{UMD PP\#02-007 ~~~ August 2001}
\maketitle



\begin{abstract}
By means of an explicit one-loop calculation, it is shown that the leading
twist contribution to the exclusive electroproduction of transversely
polarized vector mesons from the nucleon vanishes. This confirms the
all-orders proof by Collins and Diehl.
\end{abstract}

\pacs{xxxxxx}

\tighten

\vspace{0.2in}

\narrowtext

Electroproduction of vector mesons from a nucleon by a highly virtual
longitudinally-polarized photon \cite{ryskin,diehl} is a process that is
computable in perturbative QCD and, apparently, capable of accessing
off-forward parton distributions\cite{ji,rad}. Among others, Collins,
Frankfurt and Strikman \cite{collins} showed that the reaction amplitude
for the diffractive meson electroproduction can be factorized into a form
involving convolutions of the off-forward parton distributions and meson
wave functions with hard scattering coefficients. Of particular interest
is the electroproduction of transversely polarized vector mesons because
this may provide a handle to access various twist-2 chiral-odd off-forward
parton distribution functions. (We refer to Refs. \cite{hoodji,Diehl2} for
a categorization of twist-two off-diagonal distribution functions.)
Experimentally, chiral-odd distributions are notoriously difficult to
measure because they can make non-vanishing contributions only if matched
with some other chiral-odd quantities. As noted in Ref. \cite{collins},
the leading twist wave function of transversely polarized vector mesons is
chirally odd. Thus there arises the possibility of accessing chiral-odd
off-forward parton distributions by studying the production of
transversely polarized vector mesons.

%%%%%%%%%%%%%%%%%%% Figure 1 %%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[tbh]
\begin{center}
\label{FIG.1}
\epsfig{figure=FIG1.eps,height=5.0cm}
\end{center}
\caption{Factorized vector meson electroproduction amplitude.}
\end{figure}

Unfortunately, it turns out that amplitudes associated with matching
chiral-odd off-forward parton distributions with chiral-odd vector meson
wave functions vanish \cite{man} at leading order in strong coupling
constant, $\alpha _{s}$. More interestingly, Diehl, Gousset and Pire \cite
{diehl} presented a proof based on chiral invariance that these hard
coefficients vanish to {\it all orders} in perturbation theory. This proof
is a direct consequence of the identity, 
\begin{equation}
\gamma ^{\mu }\sigma ^{\rho \lambda }\gamma _{\mu }=0.  \label{sig1}
\end{equation}
Here the $\sigma ^{\rho \lambda }$-matrix comes either from the density
matrix associated with the off-forward quark helicity-flip distribution in
the vector meson (see Fig.2a) or from that associated with the chiral-odd
light-cone wave function of the nucleon (see Fig.2b), while the two
$\gamma $%
-matrices sandwiching $\sigma ^{\rho \lambda }$ correspond to the hard
gluon
scattering exchange. In fact, attaching any number of gluon lines to the
quark lines in the basic diagram leaves the conclusion unchanged: the hard
coefficients vanish to {\it all orders} in perturbation theory.


\begin{figure}[h]
\begin{center}
\label{FIG.2}
\epsfig{figure=FIG2.eps,height=5.0cm}
\end{center}
\caption{Tree-level hard partonic scattering diagrams.}
\end{figure}

\bigskip

There may, in spite of the above, exist a way out of this conclusion. It
was
observed by Hoodbhoy and Lu \cite{Lu} that because chiral invariance is
anomalously broken in QCD, a non-zero hard coefficient may nevertheless
exist. In d dimensions, the identity in Eq.(\ref{sig1}) is replaced by,

\[
\gamma ^{\mu }\sigma ^{\rho \lambda }\gamma _{\mu }=(d-4)\sigma ^{\rho
\lambda }. 
\]
\label{sig2}
Thus, the Dirac trace for the quark loop will be of order $\varepsilon $,
where $d=(4+2\varepsilon ).$ Beyond the leading order, $1/\varepsilon $
divergences from poles in the loop integrals occur that cancel the factor
of 
$\varepsilon $ in the numerator. Thus some non-vanishing terms may still
survive, and this seemed to indicate that non-zero hard-scattering
coefficients exist. However, it was pointed out by Collins and Diehl \cite
{ColDiehl} that Hoodbhoy and Lu \cite{Lu} had neglected to subtract the
one-loop evolution of the nucleon and meson, which is necessary to
correctly
derive the hard coefficients of scattering. In this Brief Report, the
above
omission will be rectified and it will be seen that indeed an exact
cancellation occurs between the overall amplitude and the parts coming
from
hadronic evolution.

\bigskip

Consider the process shown in Fig.1, 
\[
\gamma ^{\ast }(q,e_{L})+N(P,S)\rightarrow V(K,e_{T})+N(P^{\prime
},S^{\prime }), 
\]
where the first and second symbols in the parentheses stands for the
particle momentum and spin vector, respectively. As usual, we define the
average momentum and momentum difference for the initial and final-state
nucleons: 
\begin{eqnarray}
&&\bar{P}=\frac{1}{2}(P^{\prime }+P)\ , \\
&&\Delta =P^{\prime }-P\ .
\end{eqnarray}
It is most convenient to work in the frame in which $\bar{P}$ and $q$ are
collinear with each other and put them in the third direction. In such
light-cone dominated scattering processes, it is usual to introduce two
conjugate light-like vectors $p^{\mu }$ and $n^{\mu }$ in the third
direction with $p^{2}=n^{2}=0$ and $p\cdot n=1$. Correspondingly, using
Ji's
parameterization \cite{ji}, the relevant momenta can be written as
follows: 
\begin{eqnarray}
&&q^{\mu }=-2\xi p^{\mu }+\nu n^{\mu }\ , \\
&&\bar{P}^{\mu }=p^{\mu }+\frac{\bar{M}^{2}}{2}n^{\mu }\ , \\
&&\Delta ^{\mu }=-2\xi (p^{\mu }-\frac{\bar{M}^{2}}{2}n^{\mu })+\Delta
_{\perp }^{\mu }\ ,
\end{eqnarray}
with $\nu =Q^{2}/(4\xi )$, $Q^{2}=-q^{2}$ and $\bar{M}^{2}=M^{2}-\Delta
^{2}/4$. With this choice of coordinates, the longitudinal polarization
vector of the virtual photon reads, 
\begin{equation}
e_{L}^{\mu }=\frac{1}{Q}\left( 2\xi p^{\mu }+\nu n^{\nu }\right) \ .
\end{equation}
At lowest twist we can safely approximate the particle momenta as follows: 
\begin{eqnarray}
&&P^{\mu }=(1+\xi )p^{\mu }+\cdots \ , \\
&&P^{\prime \mu }=(1-\xi )p^{\mu }+\cdots \ , \\
&&K^{\mu }=\nu n^{\mu }+\cdots \ .
\end{eqnarray}

\bigskip

The basic idea of factorization for vector meson electroproduction is
illustrated in Fig.1. According to the factorization theorem, the dominant
mechanism is a single quark scattering process. The reaction amplitude is
approximated as a product of three components: the hard partonic
scattering,
the non-perturbative matrix associated with the nucleon, and the matrix
associated with the vector meson production. The active quark has to come
back into the nucleon blob after experiencing the hard scattering. On the
nucleon side, the initial and final quarks carry momenta $(x+\xi )p$ and
$%
(x-\xi )p$, respectively. By decomposing the nucleon matrix one has, 
\begin{eqnarray}
&&\int \frac{d\lambda }{2\pi }e^{i\lambda x}\langle P^{\prime }S^{\prime
}|%
\bar{\psi}_{\alpha }(-\frac{1}{2}\lambda n)\psi _{\beta
}(\frac{1}{2}\lambda
n)|PS\rangle  \nonumber \\
&=&\frac{\sigma _{\beta \alpha }^{\rho \lambda }}{8}[\bar{U}(P^{\prime
}S^{\prime })H_{T}(x,\xi )\sigma _{\rho \lambda }U(PS)+\cdots ]  \nonumber
\\
&\equiv &\frac{i}{4}(\rlap/p\rlap/e_{T})_{\beta \alpha }F_{N}(x,\xi
,Q^{2}).
\end{eqnarray}
To save space, only one of the four twist-2 chiral-odd off-forward parton
distributions have been displayed\cite{Diehl2} in the second equation
above%
\footnote{%
Recently Diehl \cite{Diehl2} noticed that the quark and gluon
helicity-flip
distributions first identified in ref.\cite{hoodji} needed to be
augmented,
doubling the number of such distributions. For our purposes here, this
difference is immaterial.}; there is no loss of generality since the
remaining tensor structures have identical transformation properties. In
the
above, $\alpha $ and $\beta $ are the quark spinor indices. Flavor and
color
indices have been suppressed. Also suppressed is the gauge link operator
in
the definition of the matrix elements. For convenience, the calculations
were performed in the Feynman gauge. On the side of the vector meson
production, the collinear momenta that the quark and antiquark carry can
be
parameterized as $(\frac{1}{2}+z)\nu n$ and $(\frac{1}{2}-z)\nu n$
respectively. Similarly, one can write down the following decomposition
for
the non-perturbative matrix associated with the vector meson production, 
\begin{equation}
\int {\frac{d\lambda }{2\pi }}e^{-i\lambda z}\langle 0|\bar{\psi}_{\beta
}(-%
\frac{1}{2}\lambda \bar{n})\psi _{\alpha }(\frac{1}{2}\lambda \bar{n}%
)|K,e_{T}\rangle =\frac{\sigma _{\alpha \beta }^{\rho \lambda }}{2}%
F_{V}(z,Q^{2})e_{T\lambda }^{\ast }K_{\rho }+\cdots \ ,
\end{equation}
where $F_{V}$ is a twist-two chiral-odd vector meson wave function.

\bigskip

It is convenient to write the scattering amplitude as a perturbation series in 
$\alpha _{s}$ in the following form,\bigskip

\begin{equation}
{\cal A}=\sum_{n=0}\alpha _{s}^{n}{\cal A}^{(n)}=\left( \frac{e}{Q}\right)
\int_{-1}^{+1}dx\int_{-\frac{1}{2}}^{+\frac{1}{2}}dz\frac{F_{N}(x,\xi
,Q^{2})F_{V}(z,Q^{2})}{(x-\xi +i\epsilon )(\frac{1}{2}-z)}\sum_{n=0}\alpha
_{s}^{n}S^{(n)}
\end{equation}
\bigskip There is no diagram at zeroth order, so $S^{(0)}=0.$

\bigskip

At the tree level there are two Feynman diagrams for the hard partonic
scattering as shown in Fig.2a-2b. This corresponds to the fact that either
before, or after, it is struck by the virtual photon, the active quark
must
undergo a hard scattering to adjust its momentum so as to form the
final-state vector meson. (Remember that in our chosen frame, both
initial-state and final-state nucleons move in the third plus direction,
while the vector meson goes in the opposite direction.) The sum of the two
diagrams in Fig.2 gives the $O(\alpha _{s})$ term, 
\begin{equation}
S^{(1)}=2C_{F}\pi (4-d)=2C_{F}\pi (-2\varepsilon ).
\end{equation}
Calculation of the one-loop term, of $O(\alpha _{s}^{2})$ is more
complicated. We work with renormalized perturbation theory, so all the
self-energy and vertex corrections are understood to be accompanied by the
corresponding ultraviolet counter-terms. It is preferable to group the
diagrams by their colour structure, and the task is to calculate 
\begin{equation}
S^{(2)}=\sum_{i}C_{i}f_{i}(x,\xi ,z)
\end{equation}
where $C_{i}$ is the color factor and $i$ runs over all distinct one-loop
Feynman diagrams, illustrated in Figs.3-5. Results for each of the
different
sets of diagrams is discussed next.

\bigskip

\begin{figure}
\label{FIG.3}
\epsfig{figure=FIG3.eps,height=8.0cm}
\caption{ One-loop corrections to the hard partonic scattering with a
three-gluon vertex.}
\end{figure}


The diagrams shown in Fig.3 are characteristic of the three-gluon vertex
and
possess a common color factor of $C_{{\rm fig.3}}=(N_{c}^{2}-1)/4$. After
cancelling ultraviolet divergences through counter-terms, only the
infrared
divergences make contributions. The individual diagrams (plus their
respective counterterms) yield,

\begin{eqnarray}
&&f_{{\rm 3a}}=-2-\frac{1}{(\frac{1}{2}+z)}\log (\frac{1}{2}-z)-\frac{2\xi
}{%
\xi +x}\log \frac{\xi -x}{2\xi }\ , \\
&&f_{{\rm 3b}}=-1-\frac{1}{(\frac{1}{2}+z)}\log (\frac{1}{2}-z), \\
&&f_{{\rm 3c}}=-1-\frac{2\xi }{\xi +x}\log \frac{\xi -x}{2\xi }\ , \\
&&f_{{\rm 3d}}=-2, \\
&&f_{{\rm 3e}}=-2\ .
\end{eqnarray}
\bigskip Summing over all the five diagrams in Fig.3, one has 
\begin{equation}
\sum\limits_{{\rm fig.3}}C_{i}f_{i}(\xi ,x,z)=-\frac{N_{c}^{2}-1}{4}\left[
8+%
\frac{2}{(\frac{1}{2}+z)}\log (\frac{1}{2}-z)+\frac{4\xi }{\xi +x}\log
\frac{%
\xi -x}{2\xi }\right] \ .  \label{three}
\end{equation}

\bigskip

Fig.4 contains a group of diagrams that have a common color factor but
vanish. The first three drop out simply because their Dirac traces vanish
even in the $(4+2\varepsilon )$ space. The last two do not contribute
because their vertex corrections contain no infrared divergences, while
the
ultraviolet divergences are canceled by the counter-terms.

\begin{figure}
\label{FIG.4}
\epsfig{figure=FIG4.eps,height=8.0cm}
\caption{A group of one-loop diagrams that vanish individually.}
\end{figure}


\bigskip

Shown in Fig.5 are another group of diagrams that share the common color
factor, $C_{{\rm fig.5}}=-(N_{c}^{2}-1)/4N_{c}^{2}$. Some diagrams in
this group require lengthy calculation. After considerable algebra, the
contributions from individual diagrams were evaluated to be,

\begin{figure}
\label{FIG.5}
\epsfig{figure=FIG5.eps,height=10.0cm}
\caption{A group of one-loop diagrams that share a common color factor.}
\end{figure}

\begin{eqnarray}
f_{{\rm 5a}} &=&-1-\frac{\frac{1}{2}-z}{\frac{1}{2}+z}\log \left[
\frac{1}{2}%
-z\right] \ , \\
f_{{\rm 5b}} &=&-1-\frac{\xi -x}{\xi +x}\log \frac{\xi -x}{2\xi }\ , \\
f_{{\rm 5c}} &=&\frac{1}{\varepsilon _{I}}-2+\log \left[
\frac{(\frac{1}{2}%
-z)(\xi -x)}{2\xi }\frac{Q^{2}e^{\gamma }}{4\pi \mu ^{2}}\right] \ , 
\nonumber \\
f_{{\rm 5d}} &=&\frac{1}{\varepsilon _{I}}-2+\log \left[
\frac{(\frac{1}{2}%
-z)(\xi -x)}{2\xi }\frac{Q^{2}e^{\gamma }}{4\pi \mu ^{2}}\right] \ , \\
f_{{\rm 5e}} &=&-\frac{1}{\varepsilon _{I}}-\frac{1}{\frac{1}{2}+z}\log (%
\frac{1}{2}-z)-\log \left[ \frac{(\frac{1}{2}+z)^{2}(\xi -x)}{2\xi
}\frac{%
Q^{2}e^{\gamma }}{4\pi \mu ^{2}}\right] \ , \\
f_{{\rm 5f}} &=&-\frac{1}{\varepsilon _{I}}-\frac{2\xi }{\xi +x}\log
\frac{%
\xi -x}{2\xi }-\log \left[ \frac{(\frac{1}{2}-z)(\xi +x)^{2}}{(2\xi
)^{2}}%
\frac{Q^{2}e^{\gamma }}{4\pi \mu ^{2}}\right] , \\
f_{{\rm 5g}} &=&+\frac{1}{\varepsilon
_{I}}-1+\frac{\frac{1}{2}-z}{\frac{1}{2%
}+z}\log (\frac{1}{2}-z)+\log \left[ \frac{(z+\frac{1}{2})(\xi +x)}{2\xi
}%
\frac{Q^{2}e^{\gamma }}{4\pi \mu ^{2}}\right] , \\
f_{{\rm 5h}} &=&+\frac{1}{\varepsilon _{I}}-1+\frac{\xi -x}{\xi +x}\log 
\frac{\xi -x}{2\xi }+\log \left[ \frac{(z+\frac{1}{2})(\xi +x)}{2\xi
}\frac{%
Q^{2}e^{\gamma }}{4\pi \mu ^{2}}\right] , \\
f_{{\rm 5i}} &=&-\frac{2}{\varepsilon
_{I}}-\frac{\frac{1}{2}-z}{\frac{1}{2}%
+z}\log (\frac{1}{2}-z)-\frac{\xi -x}{\xi +x}\log \frac{\xi -x}{2\xi
}-2\log %
\left[ \frac{(1-z)(\xi -x)}{2\xi }\frac{Q^{2}e^{\gamma }}{4\pi \mu ^{2}}%
\right] 
\end{eqnarray}
\bigskip 
\noindent
where $1/\varepsilon _{I}$ is the infrared pole, $\mu ^{2}$ is the scale
parameter in the dimensional regularization scheme, and $\gamma$ is the
Euler constant. Summing over all the diagrams in Fig.5 we have,
\begin{equation}
\sum\limits_{{\rm fig.5}}C_{i}f_{i}(\xi,x,z)=\frac{N_{c}^{2}-1}{4N_{c}^{2}}%
\left[ 8+\frac{2}{\frac{1}{2}+z}\log (\frac{1}{2}-z)+\frac{4\xi }{\xi +x}%
\log \frac{\xi -x}{2\xi }\right].
\label{five}
\end{equation}

\bigskip

At this stage, we comment on the one-loop self-energy corrections for the
hard scattering partonic processes. In renormalized perturbation theory,
one
need not consider the self-energy insertions either on the incoming or
outgoing quark lines. Instead, we need to take the diagrams in Fig.2, but
in 
$(4+2\varepsilon )$ dimensions, and include a factor of $Z_{F}^{-1/2}=1-%
\frac{\alpha _{s}}{2\pi }C_{F}\frac{1}{d-4}$ for each external quark line
of
the hard scattering part. The ultraviolet pole in $Z_{F}$ can be
compensated
by the $\varepsilon $ factor from the tree-level trace. This is exactly
where the ultraviolet divergences make their contribution. Consequently, 
\begin{equation}
\sum_{{\rm tree}}C_{i}f_{i}(\xi ,x,z)=2C_{F}^{2}.  \label{self}
\end{equation}
where $C_{F}=(N_{c}^{2}-1)/(2N_{c})$. On the other hand, diagrams with a
self-energy insertion onto an internal line do not contribute because they
have no infrared divergences.

\bigskip

We have by now exhausted all the one-loop diagrams for the hard scattering
process. Combining Eqs. (\ref{three}), (\ref{five}) and (\ref{self}), the
following compact expression for the scattering amplitude upto $O(\alpha
_{s}^{2})$ emerges: 
\begin{equation}
S^{(0)}+S^{(1)}+S^{(2)}=-4\alpha _{s}C_{F}\pi \varepsilon -2\alpha
_{s}^{2}C_{F}^{2}\left[ 3+\frac{\log
(\frac{1}{2}-z)}{(\frac{1}{2}+z)}+\frac{%
2\xi }{\xi +x}\log \frac{\xi -x}{2\xi }\right] \ .
\end{equation}

\bigskip

Let us finally turn to the issue of factorization and extraction of the
hard scattering coefficient. Schematically, the result of the above
calculation for the amplitude can be written as,

\begin{equation}
{\cal A=}H\ast F_{N}\ast F_{V}.  \label{adef}
\end{equation}
where $H$ is the hard-scattering function and $\ast $ denotes convolution.
We have already perturbatively evaluated ${\cal A}$ up to $O(\alpha
_{s}^{2}) $. Each of the 3 quantities on the rhs of the above equation can
be expanded as well: 
\begin{eqnarray}
H &=&H^{(0)}+\alpha _{s}H^{(1)}+\alpha _{s}^{2}H^{(2)}+\cdots \\
F_{N} &=&F_{N}^{(0)}+\alpha _{s}F_{N}^{(1)}+\cdots \\
F_{V} &=&F_{V}^{(0)}+\alpha _{s}F_{V}^{(1)}+\cdots
\end{eqnarray}
The hard coefficients $H^{(n)}$ are to be regarded as unknowns, to be
determined by substituting the power series for ${\cal A,}~H,~F_{N},\ $and
$%
F_{V}$ into Eq. (\ref{adef}).\ 

\bigskip

From Ref.\cite{hoodji} (see also \cite{belitsky}), the first-order
modification of the nucleon distribution is\footnote{%
There is a typographical error in Eq.9 of Ref.\cite{hoodji}. The correct
expression involves $\int dy$ everywhere and not $\int \frac{dy}{y}$ .}, 
\begin{eqnarray}
F_{N}^{(1)}{(x,\xi ,Q^{2})} &=&{\frac{1}{2\pi \varepsilon }}C_{F}\left[ {%
\frac{3}{2}}+\int_{\xi }^{x}{\frac{dy}{y-x-i\epsilon }}+\int_{-\xi }^{x}{%
\frac{dy}{y-x-i\epsilon }}\right] F_{N}^{(0)}(x,\xi ,Q^{2})+  \nonumber \\
&&{\frac{1}{2\pi \varepsilon }}C_{F}\left[ \theta (x-\xi
)\int_{x}^{1}{dy}{%
\frac{x-\xi }{y-\xi }}+\theta (x+\xi )\int_{x}^{1}{dy\frac{x+\xi }{y+\xi
}}%
~~\right.  \nonumber \\
&&\left. -\text{ }\theta (\xi -x)\int_{-1}^{x}{dy\frac{x-\xi }{y-\xi }}%
-\theta (-\xi -x)\int_{-1}^{x}{dy\frac{x+\xi }{y+\xi }}\right] {\frac{%
F_{N}^{(0)}(y,\xi ,Q^{2})}{y-x+i\epsilon }},
\end{eqnarray}
A similar calculation gives the first-order modification to the meson
distribution,

\begin{eqnarray}
F_{V}^{(1)}{(z,Q^{2})} &=&{\frac{1}{2\pi \varepsilon }}C_{F}\left[
{\frac{3}{%
2}}+\int_{-\frac{1}{2}}^{z}{\frac{dy}{y-z+i\epsilon
}}+\int_{\frac{1}{2}}^{z}%
{\frac{dy}{y-z+i\epsilon }}\right] F_{V}^{(0)}(z,Q^{2})~~+  \nonumber \\
&&{\frac{1}{2\pi \varepsilon }}C_{F}\left[
\int_{-\frac{1}{2}}^{z}dy{\frac{%
\frac{1}{2}-z}{(z-y-i\epsilon
)(\frac{1}{2}-y)}}+\int_{z}^{\frac{1}{2}}dy{%
\frac{\frac{1}{2}+z}{(z-y+i\epsilon )(\frac{1}{2}+y)}}\right]
F_{V}^{(0)}(z,Q^{2})
\end{eqnarray}

\bigskip

With all ingredients now present, we can complete the calculation. Since
$%
{\cal A}^{(0)}$ vanishes, $H^{(0)}$ is also zero. $H^{(1)}$ is
proportional
to $\varepsilon $, but this is cancelled by the $\varepsilon ^{-1}$ in $%
F_{N}^{(1)}$ and $F_{V}^{(1)}$ and we find that,

\begin{eqnarray}
\alpha _{s}^{2}H^{(1)}\ast F_{N}^{(1)}\ast F_{V}^{(0)} &=&-2\alpha
_{s}^{2}C_{F}^{2}\left(
\frac{e}{Q}\right) \int_{-1}^{+1}dx\int_{-\frac{1}{2}%
}^{+\frac{1}{2}}dz\frac{F_{N}^{(0)}(x,\xi ,Q^{2})F_{V}^{(0)}(z,Q^{2})}{%
(x-\xi +i\epsilon )(\frac{1}{2}-z)}\left[ \frac{3}{2}+\frac{2\xi }{\xi
+x}%
\log \frac{\xi -x}{2\xi }\right]  \\
\alpha _{s}^{2}H^{(1)}\ast F_{N}^{(0)}\ast F_{V}^{(1)} &=&-2\alpha
_{s}^{2}C_{F}^{2}\left(
\frac{e}{Q}\right) \int_{-1}^{+1}dx\int_{-\frac{1}{2}%
}^{+\frac{1}{2}}dz\frac{F_{N}^{(0)}(x,\xi ,Q^{2})F_{V}^{(0)}(z,Q^{2})}{%
(x-\xi +i\epsilon )(\frac{1}{2}-z)}\left[ \frac{3}{2}+\frac{\log
(\frac{1}{2}%
-z)}{(\frac{1}{2}+z)}\right] .
\end{eqnarray}
The sum of the above two terms precisely equals ${\cal A}^{(2)}.$ Since, 
\begin{equation}
{\cal A}^{(2)}{\cal =}H^{(2)}\ast F_{N}^{(0)}\ast F_{V}^{(0)}+H^{(1)}\ast
F_{N}^{(1)}\ast F_{V}^{(0)}+H^{(1)}\ast F_{N}^{(0)}\ast F_{V}^{(1)},
\end{equation}
it follows that the one-loop hard scattering coefficient vanishes, ${\cal
H}%
^{(2)}=0$. 

In conclusion, the general proof by Collins and Diehl \cite{ColDiehl} appears to be
correct and certainly holds at one-loop. However, it is also fairly complicated and the
explicit calculation presented here shows exactly how it works out at leading order.
For all this, it is still puzzling why the proof should work to all orders, given that
no fundamental symmetry of QCD is being violated by the process under consideration.
(Chiral symmetry is not fundamental!). No other process in QCD seems to share this
property. The good news is that there will not be any leading twist chiral-odd
contaminations in the measurement of chiral-even distributions. The bad news is that,
at leading twist, it is impossible to access the chiral-odd parton distributions by
means of vector meson electroproduction.

\bigskip 

\acknowledgments

I would like to thank Wei Lu for his collaboration in the initial stages 
of the work, Xiangdong Ji for valuable suggestions and encouragement, and 
Andrei Belitsky for a discussion. This work is supported in part by funds 
provided by the U.S. National Science Foundation under grant no. .

\begin{description}
\item  

\item  
\end{description}

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\end{document}


