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\begin{document}

\begin{flushright}
GEF-Th-3/2002\\
\end{flushright}

\bigskip\bigskip
\begin{center}
{\large \bf The Transversity Function\\
and Double Spin Azimuthal Asymmetry
\\
in Semi-Inclusive Pion Leptoproduction}
\end{center}
\vspace{12pt}

\begin{center}
 {\bf Elvio Di Salvo\\}

 {Dipartimento di Fisica and I.N.F.N. - Sez. Genova, Via Dodecaneso, 33 \\-
 16146 Genova, Italy\\}
\end{center}

\vspace{10pt}
\begin{center} {\large \bf Abstract}

We show that the transverse momentum dependent transversity function is 
proportional to the longitudinal polarization of a quark in a transversely 
polarized proton. This result suggests an alternative, convenient method for 
determining transversity, without knowing unusual fragmentation functions. The 
method consists of measuring the double spin azimuthal asymmetry in 
semi-inclusive pion leptoproduction by a transversely polarized proton target. 
The asymmetry, which is twist 3, is 
estimated to be more than $10\%$ under the most favourable conditions. The 
experiment we suggest is feasible at facilities like DESY and CERN. 

\end{center}


\vspace{10pt}

\centerline{PACS numbers: 13.85.Qk, 13.88.+e}

\newpage
\section{Introduction}
$~~~~$ For several years high energy spin physicists have been concentrating 
their efforts on determining the quark transversity distribution[1-5], which 
appears a particularly difficult task[6-10]. Indeed, different observables have 
been singled out, which are sensitive to this quantity; among them the 
Drell-Yan double spin asymmetry\cite{ba} and the interference fragmentation 
functions\cite{ja1}.    

For the moment the most promising experiments in this sense are those realized 
or planned by SMC\cite{smc} and HERMES\cite{her,he1} collaborations. These
experiments are based on the Collins effect\cite{coll} and consist of measuring 
the azimuthal single spin asymmetry\cite{ja01,mt,ko,tmp,km1,ef} in 
semi-inclusive pion electroproduction by a 
longitudinally\cite{her} or transversely\cite{he1} polarized proton target. Up 
to now such experiments have provided a rough evaluation\cite{ef}
of the transversity function, $h_1$. The 
single spin asymmetry is sensitive to the product $h_1(x) c(z)$, where 
$c$ is the azimuthal asymmetry fragmentation function of a transversely 
polarized quark into a pion\cite{ja01,ef}, and, as usual, $x$ and $z$ are the 
longitudinal fractional momenta, respectively, of the active quark and of the 
pion with respect to the fragmenting quark. As claimed by Jaffe\cite{ja01}, 
this may become the "classic" way of determining the proton transversity 
distribution functions, provided $c(z)$ is known to some precision and is not 
too small. But at present we know very little about this function\cite{ef}. 
Analogous considerations could be done about the method suggested by Jaffe and 
Ji\cite{jj2} (JJ). They propose to measure the double spin asymmetry 
in a semi-inclusive deep inelastic scattering (SIDIS) experiment of the type 
\begin{equation}
\vec{\ell} p^{\uparrow} \to \ell' \pi X,
\label{r1}
\end{equation}
where $\vec{\ell}$ is a longitudinally polarized charged lepton and 
$p^{\uparrow}$ a transversely polarized proton target. The asymmetry is defined 
as
\begin{equation}
A(|{\bf k}|; Q, \nu; \Pi_{\parallel}) = \frac{d\sigma_{\uparrow \rightarrow} - 
d\sigma_{\uparrow \leftarrow}}{d\sigma_{\uparrow \rightarrow} + 
d\sigma_{\downarrow \leftarrow}}.
\label{as1}
\end{equation}
Here, as usual, $\nu$ is the lepton energy transfer and $Q^2$ = $-q^2$, $q$ 
being the four-momentum transfer. Furthermore ${\bf k}$ is the momentum of the 
initial lepton and $\Pi_{\parallel}$ the component of the final pion momentum 
along the momentum transfer. Lastly
$d\sigma_{\uparrow\rightarrow}$ and $d\sigma_{\uparrow\leftarrow}$ are the
polarized differential cross sections for reaction (\ref{r1}), integrated over 
the transverse momentum of the final pion with respect to the momentum 
transfer; arrows indicate the proton and lepton polarization. Asymmetry 
(\ref{as1}) includes the product $h_1(x){\hat e}(z)$\cite{jj2}, where 
${\hat e}(z)$ is the twist-3 fragmentation function of the pion. The 
extraction of $h_1$ depends again critically on an unknown function.

If the cross section is not integrated over the transverse momentum of the 
final pion, reaction (\ref{r1})\cite{km,ko} and the analogous one with a 
longitudinally polarized target\cite{km,ko,dmo} exhibit an azimuthal asymmetry. 
This is particularly suitable for determining the transverse momentum dependent 
(t.m.d.) distribution functions\cite{mt,ko,km,tm,bdr}, defined as "new" by 
Kotzinian and Mulders\cite{km} (KM): see also Mulders and Tangerman\cite{mt} 
(MT). In particular, the double spin azimuthal asymmetry arising from a 
tranversely polarized target has been found\cite{km} to be sensitive to the 
"new" function $g_{1T}$, proportional to the longitudinal quark polarization 
in a transversely polarized proton. 

The aim of this paper is to re-examine such azimuthal double spin asymmetries.
We derive the differential cross section for reaction (\ref{r1}), starting
from the definition of t.m.d. transversity function as given by Jaffe and 
Ji\cite{jj} (JJ1), {\it i. e.}, 
\begin{equation}
\delta q_{\perp}(x, {\bf p}_{\perp}) = \sum_{T=\pm1/2} 2T 
q_T(x,{\bf p}_{\perp}), \label{ttr}
\end{equation}
where $q_T(x,{\bf p}_{\perp})$ is the probability density to find, in a 
transversely polarized proton, a quark whose spin is parallel ($T$ = 1/2) or 
opposite ($T$ = -1/2) to the proton spin. This amounts to taking, instead of 
the usual helicity representation, a canonical one, such that the quantization 
axis is along the proton transverse polarization. The asymmetry we calculate 
turns out to coincide with the one by KM, provided we identify $g_{1T}$ with 
$\delta q_{\perp}$. We shall prove this identity for massless quarks, which 
amounts to saying that, owing to transverse momentum, a quark in a transversely 
polarized proton has a longitudinal polarization, related to transversity. 
Therefore $\delta q_{\perp}$ - denoted as $h_{1T}$ by MT - plays a major role 
in the azimuthal asymmetry of reaction 
(\ref{r1}). Moreover, as we shall see, this distribution is somewhat relevant 
also in the case of a longitudinally polarized target. All this suggests an 
alternative, convenient method for determining the 
transversity. Indeed, as a consequence of our result, $\delta q_{\perp}$ 
contributes not only to the to the chiral-odd component of the t.m.d.
correlation matrix\cite{mt,tm,bmt}, but also to its chiral-even part. Therefore 
this distribution function may be coupled - and this is the case of the 
azimuthal double spin asymmetry - to a chiral-even fragmentation function, 
which is generally easier to determine than a chiral-odd one. 
This result is not completely surprising, since we have shown in a previous 
paper\cite{dis} that the inclusive muon pair production from singly polarized 
proton-hadron collisions, at a fixed transverse momentum of the pair with 
respect to the initial beams, causes a muon polarization sensitive to $\delta 
q_{\perp}$. But it is well-known\cite{coll} that SIDIS is kinematically 
isomorphic to Drell-Yan. Therefore we expect an analogous effect in reaction 
(\ref{r1}), provided we fix the pion direction. 

In this paper we calculate the improved parton model contribution and evaluate 
the $Q^{-1}$ power corrections, but we do not take into account the QCD 
evolution effects. Sect. 2 is 
dedicated to the derivation of the formulae for the cross section and for the 
asymmetry we are interested in. In sect. 3 we make some remarks, concerning 
power corrections and comparison with other authors; in particular, we 
illustrate some consequences of the identity $g_{1T}$ = $\delta q_{\perp}$ in 
the chiral limit, which we prove in the appendix. In sect. 4 we 
outline some methods for inferring the transversity from the azimuthal 
asymmetry. Lastly sect. 5 is devoted to numerical estimates and to a conclusion.

\section{Cross section and azimuthal asymmetry}
\subsection{Cross section}

We calculate  the SIDIS differential cross section in the framework of a 
QCD-improved parton model\cite{si}. In the laboratory frame the differential 
cross section reads, in one-photon exchange approximation,
\begin{equation}
d\sigma = {1 \over {4 |{\bf k}| M}} {{e^4} \over {Q^4}} L_{\mu\nu} H^{\mu\nu} 
\ d\Gamma,
\label{dsg}
\end{equation}
where $M$ is the proton rest mass and $L_{\mu\nu}$ ($H_{\mu\nu}$) the leptonic 
(hadronic) tensor. $d\Gamma$, the phase space element, reads
\begin{equation}
d\Gamma = \frac{1}{(2\pi)^6} d^4 k ~ \delta(k^2) ~ \theta(k_0) ~~ 
d^4 P ~ \delta(P^{2}-m_{\pi}^2) ~ \theta(P_0). 
\label{mps}
\end{equation}
Here $k$ and $P$ are, respectively, the four-momenta of the initial lepton and 
of the pion, whose rest mass is $m_{\pi}$. 
The leptonic tensor is, in the massless approximation, 
\begin{equation}
L_{\mu\nu} = \frac{1}{4} Tr[\rlap/k(1+\lambda_{\ell}\gamma_5)
\gamma_{\mu}\rlap/k'\gamma_{\nu}],
\label{lept}
\end{equation}
$\lambda_{\ell}$ being the helicity of the initial lepton and $k'$ 
= $k-q$ the four-momentum of the final lepton. Trace calculation yields
\begin{equation}
L_{\mu\nu} = k_{\mu} k'_{\nu} + k'_{\mu} k_{\nu} - g_{\mu\nu} k \cdot k' + 
i \lambda_{\ell}\varepsilon_{\alpha\mu\beta\nu} k^{\alpha} k^{'\beta}.
\label{lept1}
\end{equation}
As regards the hadronic tensor, 
the generalized factorization theorem\cite{qi1,tm,bo} in the covariant 
formalism\cite{land} yields, at zero order in the QCD coupling constant, 
\begin{eqnarray}
H_{\mu\nu} &=& \frac{1}{3} \sum_{f=1}^6 e_f^2 \int d\Gamma_q \varphi^{f} (p';P) 
h^f_{\mu\nu} (p, p'; S), \label{hadt} 
\\
h^f_{\mu\nu} &=& \sum_Lq^f_L(p) Tr (\rho^L \gamma_{\mu}\rho' \gamma_{\nu}).
\label{form0} 
\end{eqnarray}
Here the factor 1/3 comes from color averaging in the elementary scattering 
process and $f$ runs over the three light flavors ($u,d,s$) and antiflavors 
($\bar{u},\bar{d},\bar{s}$),  $e_1$ = $-e_4$ = 2/3, $e_2$ = $e_3$ = $-e_5$ = 
$-e_6$ = -1/3. $p$ and $p'$ are respectively the four-momenta of 
the active parton before and after being struck by the virtual photon. $S$ is 
the Pauli-Lubanski (PL) four-vector of the proton. $q^f_L$ is the probability 
density function of finding a quark 
(or an antiquark) in a pure spin state, whose third component along the proton
polarization is $L$. Analogously $\varphi^f$ is the fragmentation function of a 
quark of four-momentum $p'$ into a pion of four-momentum $P$. Moreover
\begin{equation}
d\Gamma_q = \frac{1}{(2\pi)^2} d^4 p ~ \delta(p^2) \theta(p_{0}) 
d^4 p' ~ \delta(p^{'2}) \theta(p'_{0}) ~~ \delta^4(p'-p-q),
\label{qps}
\end{equation}
the active parton being taken on shell and massless. Lastly the $\rho$'s are 
the spin density matrices of the initial and final active parton, 
{\it i. e.}\cite{dis},
\begin{equation}
\rho^L = {1 \over 2} \rlap/p [1 + 2 L\gamma_5 (\lambda + 
\rlap/\eta)] \ ~~~~~~ {\mathrm and} ~~~~~~~ \
\rho' = {1 \over 2} \rlap/p'. \label{dens}
\end{equation}
Here $2L\eta$ is the transverse PL four-vector of the active parton, while 
$\lambda$ is the longitudinal component of the quark spin vector. Formulae 
(\ref{dens}) are consistent with the Politzer theorem\cite{po} in the parton 
model approximation. These imply, together with eq. (\ref{form0}), 
that $\eta$ does not contribute to $h^f_{\mu\nu}$.
For later convenience we re-write this last tensor as 
\begin{equation}
h^f_{\mu\nu} = \frac{1}{4}\left[q^f(p) s_{\mu\nu}+\lambda\delta q^f(p) 
a_{\mu\nu}\right]. \label{form1}
\end{equation}
Here
\begin{equation}
s_{\mu\nu} = Tr(\rlap/p \gamma_{\mu}\rlap/p'\gamma_{\nu}), \ ~~~~~ \ ~~~~~~ \ 
a_{\mu\nu} = Tr(\gamma_5\rlap/p \gamma_{\mu}\rlap/p'\gamma_{\nu}).
\label{form13}
\end{equation}
Moreover $q^f(p)$ = $\sum_{L=\pm1/2}q^f_L(p)$ is the unpolarized quark 
distribution function and $\delta q^f(p)$ = $\sum_{L=\pm1/2}2Lq^f_L(p)$.
$\lambda$ is a Lorentz scalar, such that $|\lambda|$ $\leq$ 1. 
If we neglect the parton transverse momentum,
the only way of constructing such a quantity with the available vectors is
\begin{equation}
\lambda = \lambda_{\parallel}  = {\bf S} \cdot \frac{\bf q}{|{\bf q}|} = 
\frac{-S \cdot q}{\sqrt{\nu^2+Q^2}}. \label{elic}
\end{equation}
Here we have exploited the fact that $\nu$ is a Lorentz scalar and that in the 
laboratory frame $q$ $\equiv$ $(\nu,{\bf q})$ and $S$ $\equiv$ $(0,{\bf S})$,
where ${\bf S}$ is the proton spin vector, ${\bf S}^2$ = 1.
$\lambda_{\parallel}$ can be viewed as the helicity of the proton in a 
frame moving along ${\bf q}$. Now, in order to take into account the
transverse momentum, we have to adopt a frame where the proton momentum is 
large in comparison to $M$\cite{ale}. But, in this more refined approximation, 
we are still faced with the problem of defining $\lambda$ in a Lorentz 
invariant way. As we are going to show, the only way to do this is to consider 
the Breit frame, that is, where the virtual photon has four-momentum $q$ = 
$(0, {\bf q}_B)$, with $|{\bf q}_B|$ = $Q$. In this frame the proton momentum 
is $-\frac{1}{2x} {\bf q}_B$, therefore the active parton carries a momentum 
${\bf p}_B = -\frac{1}{2} {\bf q}_B + {\bf p}_{\perp}$, where, as usual, $x = 
Q^2/2M\nu$ is the longitudinal fractional momentum and ${\bf p}_{\perp}$ the 
transverse momentum with respect to ${\bf q}_B$. 
We decompose the proton spin vector ${\bf S}$ into a longitudinal and a 
transverse component, {\it i. e.},
\begin{equation}
{\bf S} = \lambda_{\parallel}\frac{\bf q}{|{\bf q}|} + {\bf S}_{\perp}, \ 
~~~~~~ \ ~~~~~ \ {\bf S}_{\perp} \cdot {\bf q} = 0. \label{rel111}
\end{equation}
But the average helicity of the quark is independent of the quantization axis, 
therefore the decomposition (\ref{rel111}) implies, together with eq. 
(\ref{elic}), 
\begin{equation}
\lambda\delta q^f(x, {\bf p}_{\perp}^2) =
\lambda_{\parallel}\delta q_{\parallel}^f(x, {\bf p}_{\perp}^2)+
\lambda_{\perp}\delta q_{\perp}^f(x, {\bf p}_{\perp}).\label{aver}
\end{equation}
Here 
\begin{equation}
\lambda_{\perp} = \frac{{\bf S}_{\perp}\cdot {\bf p}_B}{|{\bf p}_B|} \simeq
\frac{2{\bf S}\cdot {\bf p}_{\perp}}{Q} \label{helic33}
\end{equation}
and $\delta q_{\parallel}^f(x, {\bf p}_{\perp}^2)$ (denoted as $g_{1L}(x, {\bf 
p}_{\perp}^2)$ by MT) is the t.m.d. helicity distribution function. 
Now we carry on the integration (\ref{hadt}) over the time and 
longitudinal components of $p$, taking the $z$-axis opposite to ${\bf q}$. We 
get, in the light cone formalism,
\begin{equation}
H_{\mu\nu} = \frac{1}{4\pi^2 Q^2} \sum_{f=1}^6 e_f^2 \int d^2 p_{\perp} 
\varphi^{f} (z, {\bf P}^{2}_{\perp}) h^f_{\mu\nu} (x, {\bf p}_{\perp}; 
{\bf S}), 
\label{hadt2}
\end{equation}
where the tensor $h^f_{\mu\nu}$ (see eq. (\ref{form1})) reads\cite{jj}, after 
insertion of eq. (\ref{aver}),
\begin{equation}
h^f_{\mu\nu} = \frac{1}{4}\left\{q^f(x, {\bf p}_{\perp}^2) s_{\mu\nu}+
\left[\lambda_{\parallel}\delta q_{\parallel}^f(x, {\bf p}_{\perp}^2)+
\lambda_{\perp}\delta q_{\perp}^f(x, {\bf p}_{\perp})\right] a_{\mu\nu}\right\}.
\label{elemt}
\end{equation}
Here $z$ = $(P_{\parallel}+P_0)/(2|{\bf p}'|)$ is 
the longitudinal fractional momentum of the pion resulting from fragmentation 
of the struck parton, whose momentum is ${\bf p}'$. We have defined $P_0$ = 
$\sqrt{m_{\pi}^2+{\bf P}^2}$, $P_{\parallel}$ = ${\bf P}\cdot{\bf p}'/|{\bf 
p}'|$ and ${\bf P}_{\perp}$ = ${\bf P}-P_{\parallel}{\bf p}'/|{\bf p}'|$, 
${\bf P}$ being the pion momentum in the laboratory frame. 
Denoting by ${\bf \Pi}_{\perp}$ the transverse 
momentum of the pion with respect to the photon momentum, we get
\begin{equation}
{\bf P}_{\perp} = {\bf \Pi}_{\perp}-z{\bf p}_{\perp}.
\label{rel002}
\end{equation}
Therefore, if we keep ${\bf \Pi}_{\perp}$ fixed, ${\bf P}_{\perp}$ depends on 
${\bf p}_{\perp}$. Since we want to pick up a pion resulting from fragmentation 
of the active quark, we pick up events such that $|{\bf \Pi}_{\perp}|$ $<<$ 
$|{\bf P}|$. 

We notice that, although we have chosen a 
particular frame - coincident with the one adopted by Feynman\cite{fey} -, the 
tensor (\ref{elemt}) is covariant. Indeed, the spatial direction of the virtual 
photon could also be defined covariantly by means of the four-momenta of the 
photon and of the proton\cite{km1}. 

\subsection{Azimuthal asymmetry}
In order to calculate the asymmetry $A(|{\bf k}|; Q, \nu; {\bf P})$ - defined 
analogously to (\ref{as1}), but keeping the pion momentum ${\bf P}$ fixed - we 
have to substitute the leptonic tensor (\ref{lept1}) and the hadronic tensor 
(\ref{hadt2}) into the cross section (\ref{dsg}), taking into account relations 
(\ref{elemt}) and (\ref{helic33}). The result is 
\begin{equation}
A(|{\bf k}|; Q,\nu; {\bf P}) = {\cal F} ~~ 
\frac{\sum_{f=1}^6 e_f^2\delta Q^f}
{\sum_{f=1}^6 e_f^2 Q^f}, 
~~~~ \ ~~~~
{\cal F} = \frac{k_+k'_- - k_-k'_+}{k_+k'_- + k_-k'_+}. \label{assidis} 
\end{equation} 
${\cal F}$ is the depolarization of the virtual photon with respect to 
the parent lepton\cite{km}. Moreover we have introduced the quantities
\begin{eqnarray}
Q^f &=& Q^f(x,z,{\bf \Pi}_{\perp}^2) = \int d^2 p_{\perp} q^f(x, 
{\bf p}_{\perp}^2) \varphi^f (z, {\bf P}^{2}_{\perp}), \label{qf}
\\
\delta Q^f &=& \delta Q_{\parallel}^f(x,z,{\bf \Pi}_{\perp}^2) +  
{\bf \Pi}_{\perp}\cdot {\bf S} 
\delta Q_{\perp}^f(x,z,{\bf \Pi}_{\perp}), \label{dqf}
\end{eqnarray}
\begin{eqnarray}
\delta Q_{\parallel}^f(x,z,{\bf \Pi}_{\perp}^2) &=& \lambda_{\parallel}
\int d^2 p_{\perp} \delta q^f_{\parallel} (x, {\bf p}_{\perp}^2) \varphi^f (z, 
{\bf P}^{2}_{\perp}), \label{dqf0}
\\
\delta Q_{\perp}^f(x,z,{\bf \Pi}_{\perp}) {{\bf \Pi}_{\perp}\cdot {\bf S}} &=& 
\int d^2 p_{\perp} \lambda_{\perp}
\delta q^f_{\perp} (x, {\bf p}_{\perp}) \varphi^f (z, {\bf P}^{2}_{\perp}).
\label{dqf1}
\end{eqnarray}
The pseudoscalar character of $\delta Q^f$ (eq. (\ref{dqf})) follows from
assuming a massless lepton: indeed, the expression we have deduced for the 
asymmetry (see the first eq. (\ref{assidis})) holds in any frame where the 
lepton mass is negligible and the lepton helicity has the same value as in the 
laboratory frame. Below we shall show that $\delta Q^f$ is twist 3. 
From formula (\ref{dqf}) we deduce that, in order to maximize the 
contribution of $\delta q^f_{\perp}$ to our asymmetry, one has 
to take the vector ${\bf \Pi}_{\perp}$ parallel to or opposite
to ${\bf S}$, that is, to select pions whose momenta lie in the (${\bf q}$, 
${\bf S}$) plane. Furthermore the second term of eq. (\ref{dqf}) is especially 
sensitive to $\delta q^f_{\perp} (x, {\bf p}_{\perp})$ if
${\bf q}\cdot{\bf S}$ = 0. In this situation the first term of eq. 
(\ref{dqf}) - and more generally the JJ asymmetry - vanishes. 
Therefore events such that the lepton scattering plane is perpendicular to the 
proton polarization are particularly relevant to our aims.

Since the products $k_+ k'_-$ and $k_- k'_+$ 
are invariant under boosts along the $z$-axis, we calculate them in the 
laboratory frame, where 
\begin{equation}
k_{\pm} = \frac{|{\bf k}|}{\sqrt{2}}(1\pm cos\beta), ~~~~~~ \ ~~~~~ \ 
k'_{\pm} = \frac{|{\bf k'}|}{\sqrt{2}}[1\pm cos(\theta+\beta)].
\label{pmc}
\end{equation} 
Here ${\bf k}' = {\bf k}-{\bf q}$ is the final lepton momentum. Moreover
$\beta$ and $\theta$ are, respectively, the angle between ${\bf k}$ and {\bf q}
and between ${\bf k}$ and ${\bf k}'$:  
\begin{equation}
|{\bf q}| cos\beta = |{\bf k}| - |{\bf k}'| cos\theta. \label{angg}
\end{equation} 
Now we consider the scaling limit, {\it i. e.},
$Q^2 \to \infty$, $\nu \to \infty$, $Q^2/2M\nu \to x$.
Since $Q^2 \simeq 2 |{\bf k}| |{\bf k}'| (1-cos\theta)$,
$\theta$ tends to zero in that limit, as well as $\beta$: 
\begin{equation}
\theta \simeq \frac{M}{Q} \frac{y}{x(1-y)^{1/2}}, ~~~~ \ ~~~~
\beta \simeq \theta\frac{1-y}{y}, ~~~~ \ ~~~~ y = \frac{\nu}{|{\bf k}|}. 
\label{rell2}
\end{equation} 
Then the second eq. (\ref{assidis}) and eq. (\ref{elic}) yield, respectively,
\begin{equation}
{\cal F} = \frac{y(2-y)}{1+(1-y)^2}, ~~~~ \ ~~~~ \lambda_{\parallel} = 
\frac{1-y}{y}sin\theta cos\phi, \label{kfac}
\end{equation} 
where $\phi$ is the azimuthal angle between the (${\bf k},{\bf k}'$) plane and 
the (${\bf k},{\bf S}$) plane. Therefore $\delta Q_{\parallel}^f$, eq. 
(\ref{dqf0}), is 
twist 3, as follows from the second eq. (\ref{kfac}) and from the first eq. 
(\ref{rell2}). But also the second term of eq. (\ref{dqf}) is twist 3, as 
is immediate to check. Therefore our asymmetry is twist 3. 

\section{Remarks}
At this point some remarks are in order.

(i) We write, in quite a general way,
\begin{equation}
\delta q^f_{\perp}  = \delta q^f_{e\perp} + \delta q^f_{o\perp},  
\label{symde}
\end{equation}
where the indices ~ {\it e} ~ and ~ {\it o} ~ mean, respectively, ~~ {\it even} 
~~ and ~ {\it odd} ~ terms ~~
under time ~~ reversal\cite{anm}. The T-odd term  $\delta q^f_{o\perp}$
corresponds to the so-called Sivers effect\cite{si,anm}.
Invariance under parity, together with a 
rotation by $\pi$ around the proton momentum, implies
\begin{equation}
\delta q^f_{e\perp} (x, {\bf p}_{\perp}) = \delta q^f_{e\perp} 
(x, -{\bf p}_{\perp}), ~~~~~ \ ~~~~~ \ ~~~~~~ \delta q^f_{o\perp} (x, 
{\bf p}_{\perp}) = -\delta q^f_{o\perp} (x, -{\bf p}_{\perp}).
\label{symde1}
\end{equation}
From these relations it follows that, if we integrate the asymmetry 
(\ref{assidis}) over ${\bf \Pi}_{\perp}$, the second term of eq. (\ref{dqf}) 
derives its contribution solely from the T-odd term.
On the other hand, upon integration, the first term of eq. (\ref{dqf}) 
goes over into $\lambda_{\parallel}\Delta q^f(x)D^f(z)$,
corresponding to the "kinematic" twist-3 term of the numerator in the JJ 
asymmetry\cite{jj2}. Here $\Delta q^f(x)$ is the helicity distribution function 
and $D^f(z)$ the usual fragmentation function of the pion. The above mentioned
numerator includes also a "dynamic" twist-3 term of the type 
$\lambda_{\parallel} [\frac{1}{x}h_1^f(x)\frac{1}{z}{\hat e}^f(z)+
g^f_T(x)D^f(z)]
$\cite{jj2}, where $g^f_T(x)$ is the transverse spin distribution function.

(ii) The asymmetry derives contributions also from the one-gluon 
exchange\cite{qi1} terms, demanded by gauge invariance. More precisely, a gluon 
emitted by spectator partons may interact with the active quark {\it a)} before 
or {\it b)} after being struck by the photon. These amplitudes interfere with 
those in which the active parton interacts with the sole photon, giving rise to 
"dynamic" twist-3 contributions\cite{bmt,dis} (see also refs. \cite{br,co2}). 
In our approach these contributions are conveniently 
calculated in the light cone gauge. The terms of type {\it a)}, which may be 
also deduced from the equations of motion\cite{bmt}, result in the distribution 
$\delta g^f_T(x,{\bf p}_{\perp})$\cite{bmt,mt}, whose integral over the 
transverse momentum is $g^f_{T}(x)$. Calculations\cite{dis} within the model 
proposed by Qiu and Sterman\cite{qi1} assure that such contributions are about 
$10\%$ of the parton model term.

(iii) Formulae (\ref{assidis}) to (\ref{dqf1}) and the first eq. (\ref{kfac}) 
hold true 
independent of the orientation of the proton spin. However, if ${\bf S}$ is not 
oriented perpendicularly to the lepton beam, $\lambda_{\parallel}$ does not 
decrease with $Q$. Therefore if, {\it e. g.}, the proton is polarized 
longitudinally, the asymmetry has still a contribution sensitive to the t.m.d. 
tranversity function, which, however, risks to be masked by the JJ 
term\cite{dmo}. We shall see in 
the next section a method for extracting $\delta q^f_{\perp}$ under such 
unfavourable conditions, as, {\it e. g.}, in the HERMES experiment\cite{her}.

(iv) The hadronic tensor (\ref{hadt2}), which we 
have derived starting from the definition of transversity by JJ1, turns out to 
coincide with the tensor found by KM, provided we assume $g^f_{1T}$ $\propto$ 
$\delta q^f_{\perp}$, with an undetermined proportionality constant. In the
appendix we prove this relation, showing that the constant may be chosen so as 
to identify $g^f_{1T}$ with the t.m.d. transversity function for massless 
quarks. This result may be read as follows. Owing to the intrinsic transverse 
momentum, a quark in a transversely polarized proton has a nonvanishing 
longitudinal polarization, related to $\delta q^f_{\perp}$ according to the 
definition by JJ1, but also described by $g^f_{1T}$ according to the 
parametrizations by MT and by Ralston and Soper\cite{rs}. As a consequence of 
this result, all KM's considerations and deductions on $g^f_{1T}$ 
- like, {\it e. g.}, its relation with $g_2^f$ - can be applied to $\delta 
q^f_{\perp}$. The above identification implies that, although the transversity 
is a typical chiral-odd distribution, the t.m.d. transversity function has a 
chiral-even component, coincident with $g^f_{1T}$, owing to the 
non-collinearity of the quark with respect to the proton momentum. It is just 
this chiral-even component that appears in formula (\ref{dqf1}).
Therefore, according to chirality conservation, our asymmetry formula 
(\ref{assidis}) - unlike those previously considered 
for determining the transversity function\cite{ja01,ja1,ef,jj2} - 
does not contain any chiral-odd (and therefore unusual) distribution or 
fragmentation functions. Indeed, the t.m.d. functions $q^f$ and $\varphi^f$, 
involved in the asymmetry, can be parametrized in a well defined way. 
Incidentally, in the appendix we establish other useful connections among the 
transverse momentum dependent distribution functions defined by MT (see also
ref.\cite{dis2}).

\section{Extracting transversity from data}
In this section we discuss how to extract $\delta q^f_{\perp}$ from data. To 
this end we may recur either to a best fit with a suitable parametrization or 
to the method of the weighted asymmetries\cite{km,km1}. Here we examine both 
options. 

\subsection{Gaussian parametrization}
A frequently used parametrization of the t.m.d. unpolarized distributions and 
fragmentation functions consists of\cite{mt,tm,bo} 
\begin{eqnarray}
q^f(x,{\bf p}_{\perp}^2) &=& (a/\pi)q^f(x) exp(-a {\bf p}_{\perp}^2), \label{qu}
\\
\varphi^f(z,{\bf P}_{\perp}^2) &=& (a_{\pi}/\pi)D^f(z) exp(-a_{\pi} \label{fi}
{\bf P}_{\perp}^2).
\end{eqnarray}
Here $q^f(x)$ is the usual unpolarized distribution function and $a$ $\sim$ 
$0.53$ $(GeV/c)^{-2}$, as results from Drell-Yan\cite{cm}. $a_{\pi}$ may be 
determined from observation of two-jet events in $e^+e^- \to \pi X$. As regards 
the t.m.d. transversity, the transverse spin induces an anisotropy around the 
direction of the proton momentum, so that it looks appropriate to set
\begin{equation}
\delta q^f_{\perp}(x,{\bf p}_{\perp}) = (\sqrt{ab}/\pi)h_1^f(x) exp(-a 
p_1^2-b p_2^2). \label{gaus}
\end{equation}
Here 
\begin{equation}
p_1 = {\bf p}_{\perp}\cdot {\bf s}\times {\bf t}, \ ~~~~~~ \ ~~~~~~ \ 
p_2 = {\bf p}_{\perp}\cdot {\bf s}, \label{pii}
\end{equation}
where ${\bf s}$ = ${\bf S}_{\perp}/|{\bf S}_{\perp}|$ and 
${\bf t}$ = ${\bf q}/|{\bf q}|$. $h^f_1$ may be parametrized according to the
suggestion of ref.\cite{bl}, {\it i. e.},
\begin{equation}
h^f_1(x) = \frac{1}{2} 
N\frac{x^{\alpha}(1-x)^{\beta}}{\alpha^{\alpha}\beta^{\beta}}
(\alpha+\beta)^{\alpha+\beta}\left[q(x)+\Delta q(x)\right],
\end{equation}
where $N$, $\alpha$, $\beta$ and $b$ are free parameters, with $|N|\leq 1$. 

\subsection{Weighted asymmetries}
A weighted asymmetry is defined as\cite{km,km1}
\begin{equation}
\langle A_w\rangle = \frac{\int W({\bf \Pi}_{\perp})d\sigma}{\int d\sigma}.
\label{wei}
\end{equation}
Here $W$ is a given weight function of the transverse momentum 
${\bf \Pi}_{\perp}$, over which we make the integrations indicated in eq. 
(\ref{wei}). We present two kinds of 
weight functions, suitable for extracting the t.m.d. transversity function from 
the asymmetry we have proposed.
\subsubsection{The KM weight function}
For the reaction we are studying, KM have proposed $W({\bf \Pi}_{\perp})$ = $2 
{\bf \Pi}_{\perp}\cdot{\bf S}/M$, where $M$ is the proton rest mass. 
Then, according to the formulas of sect. 2, we get
\begin{equation}
\langle A_w\rangle = {\cal F}
\frac{\sum_{f=1}^6 e^2_f h^{f(1)}_1(x) D^f(z)}{\sum_{f=1}^6
e^2_f q^f(x) D^f(z)}, \label{kma}
\end{equation}
where $h^{f(1)}_1(x)$ = $2/QM$ $\int d^2p_{\perp}({\bf p}_{\perp}\cdot {\bf 
S})^2\delta q^f_{\perp}$. This quantity is quite analogous to the one defined 
by KM, {\it i. e.}, 
\begin{equation}
g^{f(1)}_{1T}(x) = \frac{1}{2M^2} \int d^2p_{\perp} {\bf p}_{\perp}^2 g^f_{1T}.
\label{kmg}
\end{equation}
KM determine $\sum_{f=1}^6 
e_f^2g^{f(1)}_{1T}(x)$ (see fig. 2 of that paper). But in appendix we show 
that, according to the normalization adopted by KM, $g^f_{1T}$ = $2M/Q ~ \delta 
q^f_{\perp}$. We substitute this equality into eq. (\ref{kmg}) and assume 
the parametrization (\ref{gaus}) for $\delta q^f_{\perp}$, with $b$ = $a$. 
Taking into account the KM deduction, we get an approximate evaluation of 
$h_1$ for a single quark. The behavior is similar to the bag model prediction 
(see, {\it e. g.}, JJ1), especially at small $x$, and the accord between the
two calculations can be made quantitative for $Q$ of order 12 to 15 $GeV$.

\subsubsection{Harmonic oscillator weight functions}
 More refined information on the t.m.d. transversity function could be 
extracted from data by using a different set of weight functions, 
inspired to the Gaussian parametrization. We expand $\varphi^f$ 
and $\delta q^f_{\perp}$ as a series of eigenfunctions of the harmonic 
oscillator, {\it i. e.},
\begin{equation}
\varphi^f(z,{\bf P}_{\perp}^2) = \sum_K c^f_K(z) \Phi_K(P), 
~~~~~~ \ ~~~~~~~ \ \delta q^f_{\perp}(x,{\bf p}_{\perp}) = \sum_K {\tilde 
c}^f_K(x) {\tilde \Phi}_K({\tilde p}).
\end{equation}
Here $P$ = ${(\bf P}_{\perp}^2)^{1/2}$ and ${\tilde p}$ = $\sqrt{b/a 
p_1^2+p_2^2}$, where $p_1$ and $p_2$ are given by eqs. (\ref{pii}). The 
$\Phi_K(P)$ are the eigenfunctions of the equation 
\begin{equation}
\left[a_{\pi}^2P^2+\frac{1}{4}\xi^2-a_{\pi}(K+\frac{1}{2})\right]\Phi_K(P) = 0, 
\end{equation}
where $\xi$ is 
the variable conjugate to $P$. The ${\tilde \Phi}_K$ fulfil an analogous 
equation, with $a$ instead of $a_{\pi}$. These functions have been chosen in 
such a way that the terms with $K$ = 0 reproduce the parametrizations of 
subsection 4.1. $c^f_K(z)$ and ${\tilde c}^f_K(x)$ are real coefficients. 
The set of weight functions we propose is
\begin{equation}
W_K = 2\frac{{\bf \Pi}_{\perp}\cdot {\bf S}}{M} \Phi_K (P).
\label{wei2}
\end{equation}
Denoting by $\langle A_w^K \rangle$ the quantity obtained by substituting eq. 
(\ref{wei2}) into (\ref{wei}), eq. (\ref{kma}) implies
\begin{equation}
\langle A_w^K\rangle \sum_{f=1}^6e_f^2q^f(x)D^f(z) = {\cal F} 
\sum_{f=1}^6\sum_L e_f^2M^f_{KL}(x,z){\tilde c}^f_L(x), \label{ls}
\end{equation}
where
\begin{equation}
M^f_{KL}(x,z) = \frac{2}{M}\int d^2P_{\perp} \Phi_K(P)\varphi (z, P^2) 
\int d^2p_{\perp} p_1^2 {\tilde \Phi}_L({\tilde p}).  
\label{matr}
\end{equation}
The linear system (\ref{ls}) can be solved with respect to ${\tilde c}^f_L(x)$,
provided we assume dominance of some fragmentation mechanism\cite{jj2}, which
reduces the sum over $f$ to a single term. Alternatively, if data 
relative to asymmetry for $\pi^{\pm}$ and to $K$-mesons are simultaneously 
available, the system (\ref{ls}) can be written as
\begin{equation}
\langle A_w^{KF}\rangle \sum_{f=1}^6e_f^2q^f(x)D_F^f(z) = {\cal F} 
\sum_{f=1}^6\sum_L e_f^2M^{fF}_{KL}(x,z){\tilde c}^f_L(x), \label{ls1}
\end{equation}
where $F$ runs over the final hadrons $\pi^+$, $\pi^-$ and $K$. This new system 
can be solved if we make some assumptions, so as to reduce the sum over $f$ to 
three or less terms. For example, we may consider separately small and large 
$x$. In the latter region we may neglect sea contribution and the system is 
overdetermined, since $f$ runs over two flavors. For small $x$, where the sea 
prevails, we may solve the system by assuming a relation between quark and 
antiquark distributions,
in such a way that $f$ runs over three flavors. The parameter $b$ may be 
determined so as to minimize, {\it e. g.}, $\sum_{L>0} |{\tilde c}^f_L(x)|$.

The method of weighted cross sections washes out the unwanted JJ contribution, 
therefore it is particularly suitable in an experiment, like HERMES\cite{her},
where the target is longitudinally polarized.

\section{Results and conclusion}
Now we calculate the order of magnitude of the asymmetry (\ref{assidis}) 
under optimal conditions. To this end, first of all, according to the 
considerations of subsection 2.2, we take into account events such that the 
azimuthal angle $\phi$ (see the second eq. (\ref{kfac})) is about $\pi/2$, and 
such that 
${\bf \Pi}_{\perp}$ is parallel (or antiparallel) to ${\bf S}$. Moreover, eqs. 
(\ref{dqf}) to (\ref{dqf1}) and the first eq. (\ref{kfac}) suggest that $y$ and 
$z$ should be chosen as close as possible to 1. As regards the functions 
involved in our asymmetry, we assume the parametrizations (\ref{qu}) to 
(\ref{gaus}), with $a_{\pi}$ =$b$ = $a$ for the sake of simplicity. Lastly we 
set $|{\bf \Pi}_{\perp}|\simeq 1$ $GeV$ and $Q$ = 2.5 $GeV$. With such inputs, 
the asymmetry (\ref{assidis}) results in $A \sim 0.4 R$, where $R = 
h_1^f(x)/q^f(x)$ has been determined by HERMES\cite{her}, $|R| = (50\pm 30)\%$. 

To conclude, we have shown (see appendix) that the t.m.d. distribution 
function $g_{1T}$ by RS and MT turns out to coincide with the t.m.d. 
transversity function, $\delta q_{\perp}$, in the chiral limit. Therefore 
$\delta q_{\perp}$ may be coupled to a chiral-even 
fragmentation function, in particular to the twist-two, unpolarized t.m.d. 
fragmentation function. This result suggests an alternative, convenient method 
for extracting $h_1$, circumventing the usual drawbacks that plague 
determination of transversity. Specifically, we propose to measure the 
the double spin azimuthal asymmetry in pion semi-inclusive leptoproduction.
For reasonable values of $Q^2$ (4 to 10 $GeV^2$), and under the most favourable 
kinematic conditions, the order of magnitude of 
the asymmetry is estimated to be at least $\sim 10\%$. The suggested experiment 
could be performed at facilities like CERN (COMPASS coll.) and DESY (HERMES 
coll.), where similar asymmetry measurements are being realized or planned. 
As a last comment, our results confirm the crucial role, on the one hand, of 
the intrinsic transverse momentum\cite{mt,ko,ale,anm} and, on the other hand,
of polarized SIDIS 
experiments\cite{fs,cl}, in extracting quark distribution functions.
\vskip 0.25in
\centerline{\bf Acknowledgments}
The author is thankful to his friends M. Anselmino, P. Mulders, J. Soffer and 
O. Teryaev for useful and stimulating discussions.

\vskip 0.40in
\setcounter{equation}{0}
\renewcommand\theequation{A.\arabic{equation}}

\appendix{\large \bf Appendix}

Here we establish some relationships among transverse momentum dependent 
(t.m.d.) distribution functions, for which we adopt the notations by Mulders 
and Tangerman\cite{mt} (MT) (see also Ralston and Soper\cite{rs} (RS) and 
other authors\cite{ko,km}). In particular we show that, in the chiral limit, 
$g_{1T}$ equals $h_{1T}$ ($\delta q_{\perp}$ in the present paper). First of 
all, we write the quark correlation matrix in 
the QCD parton model. Secondly we compare it with the general expression of the 
correlation matrix. As a result we find some relationships among t.m.d. 
distribution 
functions, which turn out to hold true even taking into account quark-gluon 
interactions and renormalization effects. Lastly, we give an alternative 
proof of the relationship between $g_{1T}$ and $h_{1T}$ in the chiral limit.

\vskip 0.30in
{\bf A.1 - Correlation matrix in QCD parton model}
\vskip 0.20in
{\it A.1.1 - The correlation matrix}
\vskip 0.15in
We define the correlation matrix as 
\begin{equation}
\Phi(x, p_{\perp}; P,S) = \int  dp^-\int\frac{d^4y}{(2\pi)^4} e^{ipy} 
\langle P,S| \psi(y) \bar{\psi}(0)|P,S\rangle, \label{corr}
\end{equation}
assuming the light cone gauge $A^+$ = 0. 
Here $P$ and $S$ are, respectively, the four-momentum and the Pauli-Lubanski 
(PL) four-vector of the proton, with $S^2$ = $-1$; $\psi$ is the quark field. 
$p$ is the quark four-momentum, whose transverse component with respect to the 
proton momentum is $p_{\perp}$, {\it i. e.},
\begin{eqnarray}
p &=& p^+n_++p_{\perp} +p^-n_-, \ ~~~~ \ ~~~~~ \ ~~~~
P = P^+n_+ +P^-n_-,\label{imp}
\\
n_+^2 &=& n_-^2 = 0, \ ~~~~ \ ~~~~~ \ ~~~~ \ ~~~~ \ ~~~~~ \ ~~~~
n_+\cdot n_- = 1.
\end{eqnarray}
Moreover $x$ = $p^+/P^+$ is the light cone fraction of the quark momentum.

We take a frame such that the proton has a large momentum, ${\cal P}$, much 
greater than its rest mass $M$, and it is transversely polarized. Moreover we 
choose the $z$-axis along the proton momentum and 
the $y$-axis along the proton polarization, so that
\begin{eqnarray}
P &\equiv& (\sqrt{M^2+{\cal P}^2}, 0, 0, {\cal P}), \ ~~~~ \ ~~~~~ \ ~~~~ \
~~ S\equiv (0,0,1,0),
\\
n_{\pm} &\equiv& \frac{1}{\sqrt{2}}(1, 0, 0, \pm1), \ ~~~~ \ ~~~~~ \ ~~~~ \ 
~~~~ \ ~~~ p_{\perp} \equiv (0, {\bf p}_{\perp}),
\\
{\bf p}_{\perp} &\equiv& (p_1, p_2, 0), \ ~~~~ \ ~~~~~ \ ~~~~ \ ~~~~ \ ~~~~ \ 
~~~~ \ |{\bf p}_{\perp}| = O(M). 
\end{eqnarray}
From now on this frame will be called ${\cal P}$-frame. For the sake of 
simplicity, we exclude T-odd terms in the correlation matrix (\ref{corr}).
\vskip 0.20in
{\it A.1.2 - QCD-improved parton model}
\vskip 0.15in
The correlation matrix for free, on-shell\cite{tm} quarks in a transversely 
polarized proton reads

\begin{equation} 
\Phi^{free}_{\perp} = \sum_{T=\pm 1/2} q_T (x, {\bf p}_{\perp}) 
\frac{1}{2}(\rlap/p+m)(1+2T\gamma_5\rlap/S_q). \label{d2}
\end{equation} 
Here $m$ and $p$ are, respectively, the rest mass and the four-momentum of the
quark, such that $p^2$ = $m^2$. $2T S_q$ is the {\it quark} PL vector, with 
$S_q^2$ = $-1$. $q_T (x, {\bf p}_{\perp})$ is the probability density of 
finding a quark with its spin aligned with ($T$ = 1/2) or opposite to 
($T$ = -1/2) the proton spin. 

In the {\it quark} rest frame - named $q$-frame from now on - we have $S_q$ = 
$S^{(0)}_q$ = $S$ $\equiv$ $(0, 0, 1, 0)$\cite{ale}, taking the axes of the 
$q$-frame parallel to those of the ${\cal P}$-frame. But the quark has a 
nonzero momentum with respect to the proton: choosing a frame at rest with 
respect to the proton, whose axes are parallel to those of the 
${\cal P}$-frame, we get
\begin{equation} 
{\bf p}_r \equiv (p_1, p_2, p_3),  ~~~~~ \ ~~~~~ \ ~~~~ p_3 = O(M).
\label{p-rel} 
\end{equation} 
In the $q$-frame, we decompose $S_q$ = $S^{(0)}_q$ = $S$ into a transverse and 
a longitudinal component with respect to the quark momentum, {\it i. e.},
\begin{equation} 
S^{(0)}_q = S = \Sigma_{\perp} cos \theta' + \nu sin \theta'.
\end{equation} 
Here 
\begin{equation} 
sin\theta' = sin\theta sin\phi, ~~~ \ ~~~ 
sin\theta = \frac{|{\bf p}_{\perp}|}{|{\bf p}|}, 
~~~ \ ~~~ sin\phi = \frac{-p_{\perp}\cdot S}{|{\bf p}_{\perp}|}, \label{tetp}
\end{equation} 
\begin{eqnarray} 
{\bf p} &\equiv& ({\bf p}_{\perp}, x{\cal P}), \ ~~~~ \ ~~~~~ \ 
 \nu \equiv (0, {\bf t}), ~~~~~ \ ~~~~~ \
\Sigma_{\perp} \equiv (0, {\bf n}), \label{deff}
\\
{\bf t} &\equiv& (sin \theta cos \phi, sin \theta sin \phi, cos \theta),
\\
{\bf n} &\equiv& (cos \theta cos \phi, cos \theta sin \phi, -sin \theta).
\end{eqnarray} 
In order to calculate $S_q$ in the ${\cal P}$-frame, we perform a boost along 
the quark momentum. This boost leaves $\Sigma_{\perp}$ invariant and transforms 
$\nu$ into ${\tilde p}/m$, where 
\begin{equation} 
{\tilde p} \equiv \left(|{\bf p}|, E_q {\bf t}\right), \ ~~~~ \ ~~~~~~ \ ~~~~ 
E_q = \sqrt{m^2+{\bf p}^2}. 
\end{equation} 
As a result we get
\begin{equation} 
S_q = S + \left[\frac{p}{m} - (\delta+\nu) \right] sin\theta'. \label{boost} 
\end{equation} 
Here we have defined
\begin{equation} 
\delta = \frac{m}{\sqrt{2}x{\cal P}} n'_- \left[1 + O({\cal P}^{-2})\right], 
~~ \ ~~~~~~ \
~~~ n'_{\pm} \equiv \frac{1}{\sqrt{2}}(1,\pm {\bf t}). \label{vers}
\end{equation} 
We stress that it is essential that the relative momentum of the quark with 
respect to the proton - see eq. (\ref{p-rel}) - is nonzero; otherwise $x$ would 
be fixed and equal to $m/M$\cite{ale}. 
Substituting eq. (\ref{boost}) into eq. (\ref{d2}), and taking into account 
the definitions (\ref{deff}) and (\ref{vers}) of $\nu$ and $\delta$, we get
\begin{eqnarray} 
\Phi^{free}_{\perp} &=& \frac{1}{2}q(x, {\bf p}_{\perp}^2)(\rlap/p+m) \nonumber 
\\
&+& \frac{1}{2} \delta q_{\perp}(x, 
{\bf p}_{\perp}) \gamma_5 \left\{ \frac{1}{2} [\rlap/S,\rlap/p] + 
\rlap/p sin\theta' - A + mB\right\} + O({\cal P}^{-1}). \label{for2}
\end{eqnarray}
Here we have set
\begin{equation} 
q(x, {\bf p}_{\perp}) = \sum_{T=\pm 1/2} q_T (x, {\bf p}^2_{\perp}) ~~~~~~ \ 
~~~~~~~ \ \delta q_{\perp}(x, {\bf p}_{\perp}) = \sum_{T=\pm 1/2} 
2T q_T (x, {\bf p}_{\perp}),
\end{equation} 
\begin{equation} 
A = E_q\frac{1}{2} [\rlap/n'_+,\rlap/n'_-]sin\theta',
\end{equation} 
\begin{equation} 
B = \rlap/S + \frac{1}{\sqrt{2}}\left\{\rlap/n'_-\left(1 - \frac{m}{|{\bf 
p}|}\right)-\rlap/n'_+ + \frac{1}{\sqrt{2}}[\rlap/n'_+,\rlap/n'_-]\right\}
sin\theta'.
\end{equation} 
Moreover we have exploited the relations $-\rlap/p \rlap/S$ =
$1/2 [\rlap/S,\rlap/p] - p\cdot S$, $p\cdot S$ = $p_{\perp}\cdot S$ and 
$\nu$ = $\frac{1}{\sqrt{2}}(n'_+-n'_-)$. 
\vskip 0.30in
{\bf A.2 - Parametrization of the correlation matrix}
\vskip 0.20in
{\it A.2.1 - The general formula}
\vskip 0.15in
Now we parametrize the correlation matrix of a transversely polarized proton in 
the ${\cal P}$-frame. We normalize this matrix in such a way that, 
at the leading twist, in the case 
of an unpolarized proton, integrating over the quark tranverse momentum, it
coincides with the usual quark density matrix times the 
unpolarized quark density. Taking into account the operators of the Dirac 
algebra\cite{tm,mt,bdr}, we get
\begin{equation}
\Phi_{\perp} = \Phi_{0a}+\Phi_{0b}+\Phi_1+\Phi_2. \label{param}
\end{equation}
Here
\begin{eqnarray}
\Phi_{0a} &=& \frac{1}{\sqrt{2}}x{\cal P}\left( f_1\rlap/n_+ + 
\lambda_{\perp}g_{1T}\gamma_5\rlap/n_+ + \frac{1}{2}h_{1T}\gamma_5
[\rlap/S,\rlap/n_+]\right) \nonumber
\\ 
&+& \frac{1}{4\sqrt{2}}\lambda_{\perp} h^{\perp}_{1T} \gamma_5 
[\rlap/p_{\perp},\rlap/n_+],\label{par0} 
\\
\Phi_{0b} &=& \frac{1}{2}\left(f_1^{\perp}+\lambda_{\perp}g^{\perp}_T\gamma_5
\right)\rlap/p_{\perp} \nonumber
\\
&+& \frac{1}{4}\lambda_{\perp} \left(h_T^{\perp} \gamma_5 [\rlap/S,
\rlap/p_{\perp}] + h_T \mu \gamma_5 [\rlap/n_-,\rlap/n_+]\right), \label{par0b} 
\\
\Phi_1 &=& \frac{1}{2}M \left(e + g_T\gamma_5 \rlap/S\right), \label{par1}
\end{eqnarray}
while $\Phi_2$ contains terms of $O({\cal P}^{-1})$.
We have set 
\begin{equation}
\lambda_{\perp} = -S\cdot p_{\perp}/\mu
\end{equation}
and $\mu$ is an arbitrary constant, which we shall 
fix below. Moreover the distributions involved, for 
which we have used the notations of MT, are functions of the Bjorken variable 
$x$ and of the intrinsic transverse momentum ${\bf p}_{\perp}$. 
The term 
\begin{equation}
\Phi_0 = \Phi_{0a}+\Phi_{0b} \label{fi0}
\end{equation}
is interaction independent.
This is evident for $\Phi_{0a}$, which consists of twist-2 operators. 
We shall show that also $\Phi_{0b}$ shares this feature, 
although the operators involved are classified as "twist-3".
\vskip 0.20in
{\it A.2.2 - Comparison with the QCD parton model}
\vskip 0.15in
We equate the coefficients of the independent Dirac operators in eqs. 
(\ref{for2}) and (\ref{param}), taking into
account the first eq. (\ref{imp}), which reads, in the ${\cal P}$-frame,
\begin{equation} 
p = \sqrt{2}x{\cal P} n_+ +p_{\perp} +O\left({\cal P}^{-1}\right). \label{rel02}
\end{equation}
We get
\begin{eqnarray} 
f_1 &=& f_1^{\perp} = q, \label{rel2}
\\
\lambda_{\perp} h_{T}^{\perp} &=& sin\theta' \delta q_{\perp},
\\
\lambda_{\perp} h_{1T}^{\perp} &=& (1-\epsilon_1) sin\theta' \delta q_{\perp}, 
\\ 
\mu\lambda_{\perp} h_T &=& (1-\epsilon_1) sin\theta' E_q \delta q_{\perp}.
\\
\lambda_{\perp} g_{1T} &=& (1-\epsilon_2) sin\theta'\delta q_{\perp}, 
\label{rel23} 
\\
\lambda_{\perp} g_T^{\perp} &=& (1-\epsilon_3) sin\theta'\delta q_{\perp}.
 \label{rel24}
\end{eqnarray} 
Here $\epsilon_1 = m/E_q$, $\epsilon_2 = m/2x{\cal P}$ and $\epsilon_3 = 
m/2|{\bf p}|$ are the correction terms to the chiral limit, which are
generally small for light quarks. The terms of order $O\left[(m^2+{\bf 
p}_{\perp}^2)/{\cal P}^2\right]$ have been neglected. As regards $\mu$, RS and 
MT have set it equal to $M$. We require the various 
functions to be normalized, in the chiral limit, as $\delta q_{\perp}$, which 
is a difference of two probability densities. Therefore we assume 
\begin{equation}
\lambda_{\perp} = sin\theta', \label{lambd}
\end{equation}
which implies $\mu = |{\bf p}|$. In particular, this choice leads to the 
relationship
\begin{equation} 
g_{1T} = (1-\epsilon_2) h_{1T}.\label{hel22} 
\end{equation} 
In the chiral limit, $sin\theta'h_{1T}$ = $sin\theta'g_{1T}$ is the average 
helicity of a quark in a transversely polarized proton. 

\vskip 0.20in
{\it A.2.3 - Equations of motion}
\vskip 0.15in

Owing to the Politzer theorem\cite{po}, the Fourier transform of $\Phi_{\perp}$ 
must fulfil the equation of motion (e.o.m.) for a Dirac particle interacting 
with the gluon field. We set $\Phi_{\perp}$ = $\Phi^{free}_{\perp}+
\Phi^{int}_{\perp}$. Since the term $\Phi^{free}_{\perp}$ (eq. (\ref{d2})) 
fulfils the Dirac equation for a plane wave, the e.o.m. implies that 
$\Phi^{int}_{\perp}$ depends on the quark-gluon interaction and is of 
order $g{\cal P}^{-1}$\cite{bmt,bq}, $g$ being the strong interaction coupling 
constant. But $\Phi^{free}_{\perp}$ includes the term $\Phi_0$ (eq. 
(\ref{fi0})), with the constraints (\ref{rel2})-(\ref{rel24}). On the other 
hand, $\Phi_{\perp}-\Phi_0$ includes the interaction dependent term 
$\Phi^{int}_{\perp}$ and is orthogonal to $\Phi_0$: 
$Tr\left[\Phi_0(\Phi_{\perp}-\Phi_0)\right]$ = 0. Therefore $\Phi_0$ 
is interaction independent and the relationships (\ref{rel2}) to (\ref{rel24}), 
although deduced from the naive parton model\cite{tm}, hold true even after 
inserting interactions. In particular the term $\Phi_{0b}$, although made up 
with twist-3 operators, is interaction independent and should be classified as 
a "kinematic" higher twist term.

The Politzer theorem survives renormalization and off-shell effects\cite{po}. 
Therefore relationships (\ref{rel2}) to (\ref{rel24}) have a quite general 
validity; in particular they hold also when QCD evolution is taken into account.
In the next subsection we shall give an alternative 
proof of this fact as regards relationship (\ref{hel22}) in the chiral limit. 
 
\vskip 0.30in
{\bf A.3 - Relationship between $h_{1T}$ and $g_{1T}$}
\vskip 0.20in
{\it A.3.1 - Proof}
\vskip 0.15in

We consider the projection
\begin{equation} 
x{\cal P} \lambda_{\perp} g_{1T}  = -\frac{1}{2}Tr\left[\Phi\gamma_5
\rlap/n_-\right] = -\frac{1}{2}{\cal H} \left[\langle P,S|{\bar 
\psi}(0)\gamma_5\rlap/n_-\psi(y)|P,S\rangle\right]. \label{g1tdef}
\end{equation} 
Here we have defined the functional
\begin{equation} 
{\cal H}\left[\rho(y)\right] = \int  dp^-\int\frac{d^4y}{(2\pi)^4} e^{ipy} 
\rho(y).
\end{equation} 
On the other hand,
\begin{equation} 
x{\cal P} h_{1T} = \frac{1}{4} Tr \left\{\Phi_0\gamma_5[\rlap/S,
\rlap/n_-] \right\} = \frac{1}{4} 
{\cal H} \left[\langle P,S|{\bar \psi}(0)\gamma_5[\rlap/S,\rlap/n_-]
\psi(y)|P,S\rangle\right]. \label{h1def}
\end{equation} 
Now we decompose the quark field into the eigenstates of the canonical 
representation where the quantization axis is taken along the proton 
polarization:
\begin{equation} 
\psi(y) = \psi'_{\uparrow} (y) + \psi'_{\downarrow}(y) + \psi''(y).
\end{equation} 
Here $\psi'$ and $\psi''$ denote respectively the "good" and "bad" component 
of the quark field, {\it i. e.},
\begin{equation} 
\psi'_{\uparrow(\downarrow)} (y) = \frac{1}{4} 
(1\pm\gamma_5\rlap/S) \rlap/n_+\rlap/n_-\psi(y), \ ~~~~~~ \ ~~~~~~
\psi''(y) = \frac{1}{2} \rlap/n_-\rlap/n_+\psi(y).
\end{equation} 
The field may also be decomposed into chirality eigenstates, {\it i. e.},
\begin{equation} 
\psi(y) = \psi'_R (y) + \psi'_L(y) + \psi''(y),
\ ~~~~~~ \ ~~~~~~
\psi'_{R(L)} (y) = \frac{1}{4} (1\pm\gamma_5)\rlap/n_+\rlap/n_- \psi(y).
\end{equation} 
Then
\begin{eqnarray} 
{\bar \psi}(0)\gamma_5\rlap/n_-\psi(y) &=& -\frac{1}{\sqrt{2}}
\left[\psi^{'\dagger}_R(0) \psi'_R (y) - 
\psi^{'\dagger}_L (0) \psi'_L (y) \right],\label{long}
\\
\frac{1}{2}{\bar \psi}(0)\gamma_5[\rlap/S,\rlap/n_-]\psi(y) &=& 
+\frac{1}{\sqrt{2}}\left[\psi^{'\dagger}_{\uparrow} (0) \psi'_{\uparrow} (y) - 
\psi^{'\dagger}_{\downarrow} (0) \psi'_{\downarrow} (y)\right].\label{tras}
\end{eqnarray} 
Substituting eqs. (\ref{long}) and (\ref{tras}) respectively into 
(\ref{g1tdef}) and (\ref{h1def}), we get 
\begin{eqnarray} 
\lambda_{\perp} g_{1T} (x,p_{\perp}) &=& q^{\Uparrow}_R (x,p_{\perp}) -
q^{\Uparrow}_L (x,p_{\perp}), \label{g1tp}  
\\
h_{1T} (x,p_{\perp}) &=& q^{\Uparrow}_{\uparrow} (x,p_{\perp}) -
q^{\Uparrow}_{\downarrow} (x,p_{\perp}). \label{h1} 
\end{eqnarray} 
Here we have set
\begin{eqnarray} 
q^{\Uparrow}_R &=& \frac{1}{2\sqrt{2}x{\cal P}} {\cal H} 
\left[\langle P,S|\psi^{'\dagger}_R (0) \psi'_R (y)|P,S\rangle\right], 
\label{dde}
\\
q^{\Uparrow}_{\uparrow} &=& \frac{1}{2\sqrt{2}x{\cal P}} {\cal H} 
\left[\langle P,S|\psi^{'\dagger}_{\uparrow} (0) \psi'_{\uparrow}
(y)|P,S\rangle\right] \label{ddf}
\end{eqnarray} 
and we have defined analogously $q^{\Uparrow}_L$ and 
$q^{\Uparrow}_{\downarrow}$. 
From eqs. (\ref{dde}) and (\ref{ddf}) it results that $q^{\Uparrow}_{R(L)}$ is 
the probability of finding the quark in a chirality state, whereas 
$q^{\Uparrow}_{\uparrow(\downarrow)}$ is the probability for a quark to be in a 
state of the canonical represention defined above. In this representation the 
spin density matrix of a quark in a transversely polarized proton reads 
\begin{equation}
\rho = q^{\Uparrow}_{\uparrow} |\uparrow\rangle\langle\uparrow| +
q^{\Uparrow}_{\downarrow} |\downarrow\rangle\langle\downarrow| \label{dm}.
\end{equation} 
On the other hand, the helicity operator can be written as
\begin{equation}
\Lambda = |+\rangle\langle +| - |-\rangle\langle -|, 
\end{equation} 
where $|\pm\rangle$ denote the helicity states.
Therefore the 
average helicity of a quark in a transversely polarized proton results in 
\begin{eqnarray} 
\langle\lambda_{\Uparrow}\rangle &=& Tr(\rho\Lambda)
= q^{\Uparrow}_{\uparrow} \left[|\langle +|\uparrow\rangle|^2 - 
|\langle -|\uparrow\rangle|^2\right] + q^{\Uparrow}_{\downarrow} 
\left[|\langle +|\downarrow\rangle|^2 - |\langle -|\downarrow\rangle|^2\right] 
\nonumber
\\
&=& \left(q^{\Uparrow}_{\uparrow} - q^{\Uparrow}_{\downarrow} \right) 
sin\theta' = h_{1T} sin\theta'.\label{fond}
\end{eqnarray} 
But $\langle\lambda_{\Uparrow}\rangle$ equals $q^{\Uparrow}_R  - 
q^{\Uparrow}_L$ in the chiral limit. Therefore eqs. (\ref{fond}),
(\ref{dde}) and (\ref{ddf}) imply
\begin{equation} 
{\cal H}\left[ \langle P,S|{\bar\psi}(0)O\psi(y)|P,S\rangle\right] = 0
\ ~~~~~ \ {\mathrm for} ~~~ \ ~~~~~ \ m = 0, \label{chir10}
\end{equation}
having set
\begin{equation} 
O = \gamma_5\left\{\rlap/n_- - \frac{1}{2} 
sin\theta'[\rlap/S,\rlap/n_-]\right\}.
\end{equation}

Therefore, in the chiral limit, 
relationship (\ref{hel22}) holds true independent of renormalization and 
off-shell effects. 
\vskip 0.20in
{\it A.3.2 - Discussion}
\vskip 0.15in
The result we have just found appears in contrast with the behavior of the 
ordinary distribution functions, that is, those integrated over the transverse 
momentum. Indeed, $g_1(x)$, a chiral even function, is independent of $h_1(x)$, 
which is chiral odd. Therefore
a measure for determining $g_1(x)$ automatically excludes the possibility of 
inferring $h_1(x)$ and {\it vice-versa}; moreover $g_1(x)$ is coupled to a 
polarized gluon distribution, whereas $h_1(x)$ is not.

But transverse momentum attenuates these differences. A quark polarized 
perpendicularly to the proton momentum has a nonzero helicity if its transverse 
momentum is different from zero. And also the converse is true: a 
longitudinally polarized quark contributes to the proton transversity if it has 
a nonvanishing ${\bf p}_{\perp}$. Quantitatively, in a transversely polarized 
proton, a massless quark, whose average spin component along the proton spin is 
$h_{1T}(x,{\bf p}_{\perp})$, has an average helicity $sin\theta' h_{1T}(x,{\bf 
p}_{\perp})$, which implies $h_{1T}(x,{\bf p}_{\perp})$ = $g_{1T}(x,{\bf 
p}_{\perp})$ in the chiral limit. The fact that $h_{1T}$ is associated to 
a chiral odd operator and $g_{1T}$ to a chiral even one simply means that the 
same function can be deduced from different types of experiment: either from 
single polarization, exploiting the Collins effect\cite{coll}, or from double 
polarization, as suggested in the present paper and by Kotzinian and 
Mulders\cite{km}. The effects 
of transverse momentum imply also that, in the QCD evolution equations, 
$h_{1T}$ is coupled to the t.m.d. gluon transverse polarization, $\delta 
g_{\perp}$. In fact, in a transversely polarized proton, gluons with nonzero 
transverse momentum have a nonvanishing transverse and longitudinal 
polarization, as well as quarks; therefore 
they give rise to longitudinally polarized quark-antiquark pairs, which, in 
turn, owing to transverse momentum, contribute to the t.m.d. tranversity 
function $h_{1T}$. The coupling between $\delta g_{\perp}$ and $h_{1T}$ is 
washed out by integration over transverse momentum.

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\end{document}

