hep-ph0207089.tex
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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 favorable 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}
$~~~~$The transversity function, usually denoted as $h_1(x)$, is one 
of the three leading twist distributions of the quark inside the 
nucleon, together with the unpolarized distribution, $q(x)$, and with 
the longitudinally polarized one, $\Delta q(x)$. In an infinite momentum 
frame where the nucleon spin is perpendicular to its momentum, one has
\begin{equation}
h_1(x) = q_{\uparrow}(x) - q_{\downarrow}(x), \label{p1}
\end{equation} 
where $q_{\uparrow(\downarrow)}(x)$ is the probability density to find a quark 
whose spin is parallel (opposite) to the 
nucleon spin. Important information on quark dynamics inside the 
nucleon can be extracted from transversity. For example, the difference 
$h_1(x)-\Delta q(x)$ is quite sensitive to the orbital angular momentum of 
the quarks. This is why high energy spin physicists have been concentrating 
their efforts for several years in determining transversity[1-5].
However, this appears a particularly difficult task[6-10], owing to 
the chiral odd character of $h_1$, unlike $q$ and $\Delta q$. Different
observables sensitive to transversity have been singled out.
Among them we recall the Drell-Yan (DY) double spin asymmetry 
\cite{rs,ba} and the single \cite{ja1,her,he1,smc} and 
double \cite{sof1,ans1} spin asymmetry in semi-inclusive deep inelastic 
scattering (SIDIS). 

For the moment SIDIS reactions -  realized or planned by 
HERMES\cite{her,he1}, SMC\cite{smc} and COMPASS collaborations - 
appear most promising
for determining $h_1$. These reactions are of the type
\begin{equation}
\ell \vec{p} \to \ell' h X,
\label{r1}
\end{equation}
where $\ell(\ell')$ is a charged lepton, $\vec{p}$ a polarized proton 
target and $h$ a hadron. If $h$ is a spinning particle whose 
polarization can be determined (e. g., a $\Lambda$), reaction 
(\ref{r1}) is kinematically isomorphic to the double spin DY process. 
However, as we shall discuss in the following, it is more 
convenient to detect a pion in the final state. In this case we are 
faced with a single spin asymmetry, therefore we have to exploit
a possible azimuthal asymmetry\cite{ja01,mt,ko,tmp,km1,ef} of reaction 
(\ref{r1}). Indeed, taking a frame whose origin coincides with the 
location of the proton target and whose $z$-axis is along the momentum 
transfer of the lepton, the angular distribution of the pion may present,
in principle, an azimuthal asymmetry with respect to the plane passing 
through the proton spin and through the $z$-axis. Incidentally, such kind 
of asymmetry may be sensitive to various transverse momentum dependent 
(t.m.d.) distributions\cite{mt,ko,km,dmo,tm,bdr}, defined as "new" by 
Kotzinian and Mulders\cite{km} (KM); see also Mulders and Tangerman\cite{mt} 
(MT). In order to extract distributions from azimuthal
asymmetries, it is convenient to define weighted asymmetries\cite{km1,km}, 
that is, ratios of the type
\begin{equation}
\langle A_w\rangle = \frac{\int w d\sigma}{\int d\sigma}. \label{w1}
\end{equation}
Here $w$ is a given function of the azimuthal angle $\phi$ of the 
final hadron $h$, 
over which we perform the integrations indicated in formula (\ref{w1}). 
In the reaction considered the azimuthal asymmetry is caused by the Collins
effect\cite{col}, consisting of an interference
term\cite{ter}. This effect is described by a T-odd (and chiral odd) t.m.d. 
fragmentation function, say $\kappa$. The corresponding weighted asymmetry
results to be sensitive to the product $h_1(x) c(z)$, where $c$ = 
$\int w \kappa d\phi$ is the so-called Collins fragmentation 
function\cite{col}. As claimed by Jaffe\cite{ja01}, 
this may become the "classic" way of determining the proton transversity 
distribution, provided $c(z)$ is known to some precision and is not 
too small. But at present we know very little about this 
function, moreover the methods suggested for inferring it from data are 
complicated and require an adequate statistics\cite{ef,ja01}. 

Analogous considerations could be done about the method proposed by Jaffe and 
Ji\cite{jj2} (JJ). This consists of measuring the double spin asymmetry 
in a SIDIS reaction of the type (\ref{r1}), where the initial 
lepton is longitudinally polarized, the proton target is transversely 
polarized and the final hadron is a pion, {\it i. e.}, 
\begin{equation}
\vec{\ell} p^{\uparrow} \to \ell' \pi X.  \label{r2}
\end{equation}
The corresponding 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 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{r2}) integrated over 
the azimuthal angle of the pion, arrows indicating the proton and lepton 
polarization. Asymmetry 
(\ref{as1}) is sensitive to 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{r2})\cite{km,ko} exhibits an azimuthal asymmetry. 
This is sensitive\cite{km} 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 an azimuthal double spin 
asymmetry.
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}
Here, analogously to formula (\ref{p1}), $q_T(x,{\bf p}_{\perp})$  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. That is, instead of the usual helicity 
representation, we consider a canonical one, such that the quantization 
axis is taken along the proton polarization. The asymmetry we calculate 
turns out to coincide with the one by KM, provided we identify \footnote{up to
a normalization constant, see appendix and subsect. 2.4} $g_{1T}$ 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{r2}), where the target is 
transversely polarized. 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 chiral-odd component of the t.m.d.
correlation matrix\cite{mt,tm,bmt}, but also to its chiral-even part. Therefore 
this distribution may be coupled 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 DY from singly polarized 
proton-hadron collisions - which is kinematically isomorphic to 
reaction (\ref{r2}) - produces a muon polarization sensitive to $\delta 
q_{\perp}$.

Drawbacks, common to any SIDIS reaction, have been pointed out 
recently\cite{bhs,ter,ko2,gr}. In fact, the asymmetries measured in these
reactions are likely to 
derive non-negligible contributions from spurious effects, caused by 
interactions\cite{bhs,ter} or correlations\cite{gr} between the active quark and 
the spectator partons, or by fragmentation from the target remnant diquark,
or by light cluster decays\cite{ko2}. Such effects have to be taken into
account by means of suitable fracture functions\cite{ter}.

Sect. 2 is dedicated to the derivation of the formulae for 
the cross section and for the asymmetry we are interested in. Moreover we 
illustrate some consequences of the identity $g_{1T}$ = $\delta q_{\perp}$ in 
the chiral limit, which we prove in the appendix. In sect. 3 we 
discuss how to infer the transversity from data of azimuthal 
double spin asymmetry, taking into account the spurious effects. In
sect. 4 we compare our method with others. Lastly 
sect. 5 is devoted to a short conclusion.

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

We calculate the differential cross section for reaction (\ref{r2}) in the 
framework of a QCD-improved parton model\cite{si}. For the moment we neglect 
the contribution of the fracture function, which will be discussed in 
sect. 3. 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 final 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}
$\lambda_{\ell}$ and $k$ = $k'+q$ ($k^{2}$ = 0) being respectively the helicity
 and the four-momentum of the initial 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. $q^f_L(p)$  is the probability
density of finding a quark (or an antiquark)
with four-momentum $p$, whose spin is parallel 
($L= 1/2$) or antiparallel ($L= -1/2$) to the proton spin. Analogously
$\varphi^f(p')$ 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 and $\delta q^f(p)$ = $\sum_{L=\pm1/2}2Lq^f_L(p)$.
$\lambda$ is a Lorentz invariant, pseudoscalar quantity, 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 where the proton is moving along ${\bf q}$. 
In this connection we observe that the spatial direction of 
the virtual photon can also be defined covariantly by means of the four-momenta
of the photon and of the proton\cite{km1}. 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}. To this end we consider 
the Breit frame - coincident with the one adopted by Feynman\cite{fey} -
where the virtual photon has a 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 spin of the quark is independent of the quantization axis, 
therefore the decomposition (\ref{rel111}) implies
\begin{equation}
\lambda\delta q^f(x, {\bf p}_{\perp}) =
\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}
moreover $\delta q_{\parallel}^f(x, {\bf p}_{\perp}^2)$ (denoted as 
$g_{1L}(x, {\bf p}_{\perp}^2)$ by MT) is the t.m.d. helicity distribution. It 
is interesting to write the relationships of $\delta q_{\parallel}^f$ 
and $\delta q_{\perp}^f$ to the usual distributions:  
\begin{equation}
\Delta q^f(x) = \int d^2 p_{\perp} \delta q_{\parallel}^f(x, {\bf 
p}_{\perp}^2); ~~~~~ \ ~~~ 
h_1^f(x) = \int d^2p_{\perp}  \delta q_{\perp}^f(x, {\bf 
p}_{\perp}). \label{uds}
\end{equation}
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 $h^f_{\mu\nu}$ reads, after 
insertion of eq. (\ref{aver}) into eq. (\ref{form1}),
\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}
$q^f$ and $\varphi^f$ are related, respectively, to the usual 
unpolarized distribution and to the pion fragmentation 
function, {\it i. e.}, 
\begin{equation}
q^f(x) = \int d^2p_{\perp} q^f(x, {\bf p}_{\perp}^2), ~~~~~ \ ~~~
D^f(z) = \int d^2P_{\perp}  \varphi^f(z, {\bf P}_{\perp}^2). 
\end{equation}
Moreover $z$ = $(P_{\parallel}+P_0)/(2|{\bf p}'|)$ 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 
${\bf p}_{\perp}$.

We notice that, although we have chosen a particular frame, the 
result (\ref{hadt2}) is apparently covariant. 

\subsection{Azimuthal asymmetry}
In order to calculate the double spin azimuthal asymmetry $A(|{\bf k}|; Q,
\nu; {\bf P})$ for reaction (\ref{r2}) - 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 eq. (\ref{dqf}) implies that $\delta Q^f$
is especially sensitive to $\delta q^f_{\perp} (x, {\bf p}_{\perp})$ if
${\bf q}\cdot{\bf S}$ = 0. Indeed, 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. 

\subsection{Numerical estimates}
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 
${\bf \Pi}_{\perp}$ is parallel (or antiparallel) to ${\bf S}$. Moreover, eqs. 
(\ref{dqf}) to (\ref{dqf1}) and eqs. (\ref{kfac}) suggest that $y$  
should be chosen as close as possible to 1. Assuming the transverse momentum 
distributions involved in asymmetry (\ref{assidis})
to be of the Gaussian type\footnote{see eqs. (\ref{qu}) to 
(\ref{gaus}) below}, with equal widths of $\sim$ 0.85 $GeV/c$, and 
setting $|{\bf \Pi}_{\perp}|\simeq 1$ $GeV$ and $Q$ = 2.5 $GeV$, 
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}, $|R| = (50\pm 30)\%$. 

\subsection{Remarks}

At this point some remarks are in order.

(i) Parity invariance and a rotation of the reference frame by $\pi$ around 
the proton spin imply
\begin{equation}
\delta q^f_{\perp} (x, {\bf p}_{\perp}) = \delta q^f_{\perp} 
(x, -{\bf p}_{\perp}), \label{symde}
\end{equation}
the typical property of a T-even function\cite{anm}. This, in turn,
implies that, if we integrate the SIDIS differential cross section - 
and therefore the hadronic tensor (\ref{hadt2}) - over ${\bf \Pi}_{\perp}$, 
the second term of eq. (\ref{dqf}) vanishes. 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}. 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} (see introduction), where $g^f_T(x)$ is 
the transverse spin distribution. This term does not appear 
in asymmetry (\ref{assidis}), since we limit ourselves to the 
contributions of the QCD parton model.

(ii) Eqs. (\ref{assidis}) to (\ref{dqf1}) and the first eq. (\ref{kfac}) 
hold true for any
orientation of the proton spin. However, if ${\bf S}$ is not 
perpendicular to the lepton beam, $\lambda_{\parallel}$ does not 
decrease with $Q$. Therefore if, {\it e. g.}, the proton is polarized 
longitudinally, the asymmetry derives still a contribution from the t.m.d. 
tranversity function, which, however, risks to be taken over 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}.

(iii) 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\footnote{according to the normalization constant chosen by KM for 
$g^f_{1T}$}
\begin{equation}
g^f_{1T} = \frac{2M}{Q} \delta q^f_{\perp}. \label{relkm}
\end{equation}
This relationship is proven in appendix, together with other useful 
connections
among the t.m.d. distributions defined by MT (see also ref.\cite{dis2}).
Therefore KM's considerations and deductions about $g^f_{1T}$ 
- like, {\it e. g.}, its relation with $g_2^f$ - can be applied to $\delta 
q^f_{\perp}$.

\section{Extracting transversity from data}
In this section we discuss how to extract $\delta q^f_{\perp}$ from data. To 
this end we illustrate two different methods: a best fit to the asymmetry, with 
a suitable parametrization for $\delta q^f_{\perp}$, and the use of weighted 
asymmetries\cite{km,km1}. Furthermore we suggest how to take into account the 
spurious effects, that is, contributions to asymmetry (\ref{assidis}) not 
related to $\delta q^f_{\perp}$.

\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  $a$ $\sim$ 
$1.38$ $(GeV/c)^{-2}$, as results from DY\cite{cm}, and $a_{\pi}$ may be 
determined from two-jet events in $e^+e^- \to \pi X$. As regards 
the t.m.d. transversity, it looks appropriate to set\cite{ss}
\begin{equation}
\delta q^f_{\perp}(x,{\bf p}_{\perp}) = 
(a/\pi)h_1^f(x) exp(-a {\bf p}_{\perp}^2). \label{gaus}
\end{equation}
Here $h^f_1$ 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^f(x)+\Delta q^f(x)\right],
\end{equation}
where $N$, $\alpha$, $\beta$ and $b$ are free parameters, with $|N|\leq 1$. 

\subsection{Weighted asymmetries}

\subsubsection{The KM weight function}
For the reaction we are studying, KM have defined a weighted asymmetry 
of the type (\ref{w1}), suggesting the weight function $w({\bf \Pi}_{\perp})$ 
= $2{\bf \Pi}_{\perp}\cdot{\bf S}/M$. Inserting this function and 
eq. (\ref{dsg}) into eq. (\ref{w1}), and taking into account eqs. 
(\ref{lept1}) and (\ref{hadt2}), we get
\begin{equation}
\langle A_w\rangle = z{\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 
\begin{equation}
h^{f(1)}_1(x) = \frac{2}{QM} \int d^2p_{\perp}({\bf p}_{\perp}\cdot {\bf 
S})^2\delta q^f_{\perp}. \label{was}
\end{equation}
The expression KM obtain for the weighted asymmetry differs from 
eq. (\ref{kma}) by the substitution $h^{f(1)}_1(x)$ by 
$g^{f(1)}_{1T}(x)$, where 
\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}
But eq. (\ref{relkm}) implies $g^{f(1)}_{1T}(x)$ = $h^{f(1)}_1(x)$,
so that the KM asymmetry turns out to coincide with eq. (\ref{kma}). 
Moreover the numerical evaluation of $g_{1T}$ that KM infer from E143 
data\cite{sla} allows, by means of eqs. (\ref{was}) and (\ref{gaus}), 
to evaluate $h_1$. This behaves quite similarly to the bag model
prediction (see, {\it e. g.}, JJ1), especially at small $x$. 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 assume for $\varphi^f$ 
and $\delta q^f_{\perp}$ expansions of the type
\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(p). \label{exp}
\end{equation}
Here $P$ = $({\bf P}_{\perp}^2)^{1/2}$, $p$ = $({\bf p}_{\perp}^2)^{1/2}$
and 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 $\xi = id /dP$. 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 3.1. $c^f_K(z)$ 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 (|{\bf \Pi}_{\perp}|).
\label{wei2}
\end{equation}
We denote by $\langle A_w^K \rangle$ the weighted asymmetries obtained by 
substituting the functions (\ref{wei2}) into eq. (\ref{w1}). Taking into account 
eqs. (\ref{exp}) and
once more eqs. (\ref{dsg}), (\ref{lept1}) and (\ref{hadt2}), we get
\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}(z){\tilde c}^f_L(x), \label{ls}
\end{equation}
where
\begin{equation}
M^f_{KL}(z) = \frac{4}{QM}\int d^2P_{\perp}{\bf P}_{\perp} \cdot {\bf S} 
\Phi_K(P) \int d^2p_{\perp}{\bf p}_{\perp} \cdot {\bf S} {\tilde \Phi}_L(p) 
\varphi^f \left[z,({\bf P}_{\perp}-z {\bf p}_{\perp})^2\right]. 
\label{matr}
\end{equation}
The linear system (\ref{ls}) can be solved with respect to ${\tilde c}^f_L(x)$,
provided we assume the 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}(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 $x$-interval 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 $f$ run over three flavors.
Once the coefficients ${\tilde c}^f_L(x)$ have been determined for 
any $f$ and for a sufficiently large number of $L$, we may determine 
$h_1^f$ thanks to the second eq. (\ref{exp}) and to the second eq. (\ref{uds}). 
The method we have just suggested is somewhat similar to the purity 
method used by HERMES in splitting the single flavor contribution\cite{rk} 
in longitudinally polarized distributions (see also 
refs.\cite{ko2,gr,fs,cl,df,ni}).

The weighted cross sections $\langle A_w^K\rangle$ wash out the unwanted JJ 
contribution, therefore they are particularly suitable in an experiment, like 
HERMES\cite{her}, where the target is longitudinally polarized.

\subsection{Spurious effects}
Asymmetry (\ref{assidis}), as well as any asymmetry connected to a SIDIS 
experiment, is sensitive to processes which have nothing to do with the 
distribution we want to determine. Such processes, whose contributions may 
be perhaps reduced, but not 
eliminated by means of kinematic cuts\cite{ko2}, are essentially of two kinds: 

~(i) interactions\cite{bhs} and/or correlations\cite{gr} between the 
active quark and the spectator partons;

(ii) fragmentation of the target remnant diquark or decay of a light 
cluster\cite{ko2}.

Gluon exchange between the active quark and the spectator partons 
is demanded by gauge invariance\cite{qi1}. A gluon 
emitted by spectator partons may interact with the active quark {\it a)} before 
or {\it b)} after being struck by the photon. The amplitude for such a
process interferes with the amplitude for the active parton to interact
with the sole photon, giving rise to a 
"dynamic"twist-3 contribution\cite{bmt,dis} (see also refs. 
\cite{bhs,co2}). In our case this term must be T-even, since 
T-odd functions\cite{bhs,co2} are involved only in single spin 
asymmetries. 
In particular, the type-{\it a)} contribution corresponds to 
$g^f_T(x,{\bf p}_{\perp})$, 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.
 
The spurious fragmentation effects (ii)\cite{ko2} and quark-spectator 
correlations\cite{gr} are nonperturbative and can be taken into account 
according to a strategy similar to the one followed by Kotzinian\cite{ko2}. The 
method consists of a simulation program based on the Lund fragmentation model. 
The simulation\cite{rk2,rk3} reproduces the data of unpolarized 
SIDIS, but exhibits deviations from the purity assumption\cite{rk,fs,cl,df,ni} 
that the
final hadron observed come exclusively from the active quark fragmentation.  

Such deviations are suitably described by fracture functions\cite{ter}. In 
our case the asymmetry (\ref{assidis}) has to be modified into
\begin{equation}
A'(|{\bf k}|; Q,\nu; {\bf P}) = {\cal F} ~~ 
\frac{\sum_{f=1}^6\left[e_f^2\delta Q^f +\delta {\cal M}^f \right]}
{\sum_{f=1}^6 \left[e_f^2 Q^f + {\cal M}^f\right]}. \label {assmod} 
\end{equation}
Here ${\cal M}^f$ and $\delta {\cal M}^f$ are respectively the unpolarized and 
polarized fracture functions, which depend on $x$, $z$ and ${\bf \Pi}_{\perp}$. 
As in the case of longitudinal polarization\cite{ko2}, the cross section for 
reaction (\ref{r2}) can be calculated by means of the simulation program, 
assuming parametrizations (\ref{qu}) to (\ref{gaus}) for the nonperturbative 
functions involved, and using some model calculation for $h_1^f$\cite{jj,ef,bl}. 
Since the program is endowed with a pointer, which shows the origin of the final 
hadrons\cite{ko2}, one can determine the fracture functions by selecting
the spurious events that result from the simulation, and fitting them
by means of a suitable parametrization, {\it e. g.}, $\delta {\cal M}^f$ = $C 
\delta Q^f$, $C$ being a constant.

Incidentally, in the case of longitudinal polarization, these spurious
effects might be responsible for the discrepancy between the recent HERMES
results\cite{rk} and expectations from the so-called "spin 
crisis"\cite{ko2}.   

It is worth noticing that the weighted asymmetries wash out in part such effects,
for example the light cluster decays, which are isotropic. 

\section{Comparison with other methods}

Extraction of $h_{1}$ from SIDIS 
single spin asymmetry is affected by drawbacks similar to those 
described in subsect. 3.3. In particular this asymmetry derives a 
contribution from the Sivers effect\cite{si}, which could 
be even greater than the term we are interested in. In fact the Sivers 
function is chiral even and therefore is coupled to the usual t.m.d. 
fragmentation function; this is considerably larger than $\kappa$, the
chiral odd and  T-odd fragmentation function coupled to $h_{1T}$ in the 
single spin asymmetry (see sect. 1).
Moreover there are suggestions that this single spin asymmetry, usually 
classified as twist 2\cite{col}, could be instead twist 3\cite{dis3,ter}, 
as well as asymmetry (\ref{assidis}). Indeed, this latter is twist 3 
(see subsect. 2.2), while it was 
considered twist 2 in previous papers\cite{mt,tm,km}. The reason for 
 this discrepancy is explained in appendix. There we discuss a factor
 $\mu^{-1}$, introduced for dimensional reasons, where $\mu$ is an 
 undetermined energy scale.
 This is usually assumed equal to $M$\cite{rs,mt},
whereas a correct normalization of the t.m.d. distributions as
 probability densities demands $\mu = Q/2$.
Lastly, in our method, the complication of inferring $h_{1}$ from a 
number of weighted asymmetries, as described in subsect. 3.2, is largely
compensated by the fact that $g_{1T}$ is coupled to the ordinary fragmentation
function of the pion, whose 
parameters are well determined, contrary to the Collins function.

Among the other methods proposed for extracting transversity, we 
recall proton-proton collisions, with muon pairs or 
hadrons (pions or lambdas) in the final state\cite{sof2}, and SIDIS with a 
lambda in the final state\cite{sof1,ans1}.
But asymmetries in proton-proton collisions
 have been shown to be rather insensitive to $h_{1}$\cite{sof2}. 
 Concerning SIDIS, string fragmentation into a lambda is much
 more difficult than pion fragmentation; this not only makes statistics 
 poorer, but also increases the drawbacks described in subsect. 
 3.3\cite{ter}.  

\section{Conclusion}

To conclude, $\delta q_{\perp}$ may be coupled to a chiral-even 
fragmentation function, in particular to the twist-2, unpolarized t.m.d. 
fragmentation function of the final hadron observed. This suggests an 
alternative, convenient method 
for extracting $h_1$, circumventing some of the usual drawbacks that plague 
determination of transversity. Specifically, we propose to measure 
the double spin azimuthal asymmetry in semi-inclusive pion leptoproduction.
For reasonable values of $Q^2$ (4 to 10 $GeV^2$), and under the most favourable 
kinematic conditions,  
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 of the intrinsic 
transverse momentum\cite{mt,ko,ale,anm} and of polarized SIDIS 
experiments\cite{fs,cl,df,ni} in extracting quark distributions.
\vskip 0.25in
\centerline{\bf Acknowledgments}
The author is thankful to his friends M. Anselmino, A. Di Giacomo, E. 
Leader, P. Mulders 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.) distributions, 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 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. We give also an alternative 
proof of the relationship between $g_{1T}$ and $h_{1T}$ in the chiral limit.
Lastly, we shortly discuss our result.

\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,
\\
n_+\cdot p_{\perp}  &=& n_-\cdot p_{\perp} = 0.
\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, say ${\cal P}$, 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. Now we define a {\it quark} rest 
frame\cite{ale} - to be named $q$-frame from now on - whose axes are parallel to 
those of the ${\cal P}$-frame. In this frame we have $S_{q}$ = $S^{(0)}_q$ = $S$
$\equiv$ $(0,0,1,0)$. Decomposing $S^{(0)}_q$ = $S$ into a transverse and 
a longitudinal component with respect to the quark momentum, we get
\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} &\simeq& ({\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 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}^2) = \sum_{T=\pm 1/2} q_T (x, {\bf p}_{\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
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 
the leading twist contribution of $\Phi$ for 
an unpolarized proton, when integrated over ${\bf p}_{\perp}$, 
coincide with the product of the usual density matrix times the 
distribution of an 
unpolarized quark. 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 \label{ms}
\end{equation}
and $\mu$ is an undetermined energy scale, 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 0.30in
{\bf A.3 - Relationship between $h_{1T}$ and $g_{1T}$}
\vskip 0.20in
{\it A.3.1 - 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 in the ${\cal P}$-frame reads
\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 between two probability densities. Therefore we assume 
\begin{equation}
\lambda_{\perp} = sin\theta', \label{lambd}
\end{equation}
which, according to eqs. (\ref{tetp}) and (\ref{ms}), 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.3.2 - 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 
$\tilde{\Phi}_0$, corresponding to eq. 
(\ref{fi0}) with the constraints (\ref{rel2})-(\ref{rel24}). On the other 
hand, $\Phi_{\perp}-\tilde{\Phi}_0$ includes the interaction dependent term 
$\Phi^{int}_{\perp}$ and is orthogonal to $\tilde{\Phi}_0$: 
$Tr\left[\tilde{\Phi}_0(\Phi_{\perp}-\tilde{\Phi}_0)\right]$ = 0. 
Therefore $\tilde{\Phi}_0$ 
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}, 
which, on the other hand, do not cause any mixing among the mutually orthogonal
Dirac operators into which we have decomposed the correlation matrix.  
Therefore the relationships (\ref{rel2}) to (\ref{rel24}) have a quite general 
validity; in particular they hold also when QCD evolution is taken into account.
 
\vskip 0.20in
{\it A.3.3 - An alternative proof of relationship (\ref{hel22})}
\vskip 0.15in

Eqs. (\ref{param}) and (\ref{corr}) imply 
\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} 
having 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} 
Similarly,
\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 a canonical 
representation, in which 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 
$q^{\Uparrow}_{\downarrow}$. 
Eqs. (\ref{dde}) and (\ref{ddf}) imply 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} 
$|\uparrow(\downarrow)>$ denoting the two states of the canonical 
representation.
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 two 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. Then 
\begin{equation}
q^{\Uparrow}_R  - q^{\Uparrow}_L = (q^{\Uparrow}_{\uparrow} -
q^{\Uparrow}_{\downarrow}) sin \theta'.
\end{equation} 
The definitions (\ref{dde}) and (\ref{ddf})
and equations (\ref{long}) and (\ref{tras}) 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 0.30in
{\bf A.4 - Discussion}
\vskip 0.20in
The result we have just found appears in contrast with the behavior of the 
ordinary distributions. Indeed, $g_1(x)$, a chiral even function, is independent
of $h_1(x)$, which is chiral odd. Therefore an experiment for determining 
$g_1(x)$ automatically excludes the possibility of 
inferring $h_1(x)$ and {\it vice-versa}.
But transverse momentum attenuates this difference. A quark polarized 
perpendicularly to the proton momentum has a nonzero helicity if its transverse 
momentum is different from zero.
%Incidentally, 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{col}, or from double 
polarization, as suggested in the present paper and by Kotzinian and 
Mulders\cite{km}.

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Phys. B {\bf 375} (1992) 527

\bibitem{am} X. Artru and M. Mekhfi: Z. Phys. C - Particles and Fields {\bf 45} 
(1990) 669

\bibitem{ja01} R.L. Jaffe: Phil. Trans. Roy. Soc. Lond. A {\bf 359} (2001) 391

\bibitem{cpr} J. Cortes, B. Pire and J. Ralston: Z. Phys. C - Particles and 
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\bibitem{ji} X. Ji: Phys Lett. B {\bf 284} (1992) 137

\bibitem{ja1} R.L. Jaffe, X. Jin and J. Tang: Phys. Rev. Lett. {\bf 80} (1998) 
1166; Phys. Rev. D {\bf 57} (1998) 5920

\bibitem{ja2} R.L Jaffe: "2nd Topical Workshop on Deep Inelastic Scattering off
polarized targets: Theory meets Experiment (Spin 97)", Proceedings eds. J. 
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\bibitem{tm} R.D. Tangerman and P.J. Mulders: Phys. Rev. D {\bf 51} (1995) 3357

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\end{thebibliography}