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%% subsection 4.1 Polarized production [slac-pub-7129-0-0-4-1 in slac-pub-7129-0-0-4: ^slac-pub-7129-0-0-4 >slac-pub-7129-0-0-4-2]
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\sectionLink{slac-pub-7129-0-0-4}{slac-pub-7129-0-0-4}{Above: 4. $\psi'$ and $J/\psi$ Polarization}%
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\subsection{\usemenu{slac-pub-7129::context::slac-pub-7129-0-0-4-1}{Polarized production}}\label{subsection::slac-pub-7129-0-0-4-1}
For arguments sake, let us consider the production of a $\psi'$ in
a polarization state $\lambda$. This state can be reached through
quark-antiquark pairs in various spin and orbital angular momentum states,
and we are led to consider the intermediate quark-antiquark pair
as a coherent superposition of these states. Because of parity
and charge conjugation symmetry, intermediate states with
different spin $S$ and angular momentum $L$ can not
interfere\footnote{Technically, this means that NRQCD matrix elements
with an odd number of derivatives or spin matrices vanish if the
quarkonium is a $C$ or $P$ eigenstate.},
so that the only non-trivial situation occurs for ${}^3 P_J$-states,
i.e. $S=1$, $L=1$.
In \cite{9} it is assumed that intermediate states with
different $J J_z$, where $J$ is total angular momentum do not
interfere, so that the production cross section can be expressed
as the sum over $J J_z$ of the amplitude squared for production
of a color octet quark-antiquark pair in a ${}^3 P_{J J_z}$ state
times the amplitude squared for its transition into the $\psi'$.
The second factor can be inferred from spin symmetry to be
a simple Clebsch-Gordon coefficient so that
\begin{equation}
\label{cho}
\sigma^{(\lambda)}_{\psi'} \sim \sum_{J J_z}
\sigma(\bar{c} c[{}^3 P^8_{J J_z}])\,|\langle J J_z|1
(J_z-\lambda);1\lambda\rangle|^2\,.
\end{equation}
\noindent We will show that this equation is incompatible with spin
symmetry which requires interference of intermediate
states with different $J$.
A simple check can be obtained by applying (\docLink{slac-pub-7129-0-0-4.tcx}[cho]{21}) to
the calculation of the gluon fragmentation function into
longitudinally polarized $\psi'$. Since the fragmentation functions
into quark-antiquark pairs in a ${}^3 P^8_{J J_z}$ state
follow from \cite{28} by a change of color factor, the
sum in (\docLink{slac-pub-7129-0-0-4.tcx}[cho]{21}) can be computed. The result not only
differs from the fragmentation function obtained in \cite{16}
but contains an infrared divergence which
can not be absorbed into another NRQCD matrix element.
To see the failure of (\docLink{slac-pub-7129-0-0-4.tcx}[cho]{21}) more clearly we return to the
NRQCD factorization formalism. After Fierz rearrangement of color
and spin indices as explained in \cite{5}, the cross
section can be written as
\begin{equation}
\label{factor}
\sigma^{(\lambda)} \sim H_{ai;bj}\cdot S_{ai;bj}^{(\lambda)}\,.
\end{equation}
\noindent In this equation $H_{ai;bj}$ is the hard scattering cross
section, and $S_{ai;bj}$ is the soft (non-per\-tur\-ba\-tive) part that
describes the `hadronization' of the color octet quark pair into
a $\psi'$ plus light hadrons. Note that the
statement of factorization entailed in this equation occurs only on
the cross section and not on the amplitude level. The indices $ij$ and
$ab$ refer to spin and angular momentum in a Cartesian basis $L_a S_i$
($a,i=1,2,3=x,y,z$). Since spin-orbit coupling is
suppressed by $v^2$ in the NRQCD Lagrangian, $L_z$ and $S_z$ are good
quantum numbers. In the specific situation we are considering,
the soft part is simply given by (the notation follows
\cite{5,16})
\begin{equation}
S_{ai;bj}^{(\lambda)}=
\langle 0|\chi^\dagger\sigma_i T^A\left(-\frac{i}{2}
\stackrel{\leftrightarrow}{D}_a\right)
\psi\,{a_{\psi'}^{(\lambda)}}^\dagger
a_{\psi'}^{(\lambda)}\,\psi^\dagger\sigma_j T^A\left(-\frac{i}{2}
\stackrel{\leftrightarrow}{D}_b\right)\chi|0\rangle\,,
\end{equation}
\noindent where $a_{\psi'}^{(\lambda)}$ destroys a $\psi'$
in an out-state with
polarization $\lambda$. To evaluate this matrix element at leading order
in $v^2$, we may use spin symmetry. Spin symmetry tells us that
the spin of the $\psi'$ is aligned with the spin of the $\bar{c} c$ pair,
so $S_{ai;bj}^{(\lambda)}\propto
{\epsilon^i}^*(\lambda)\epsilon^j(\lambda)$. Now all
vectors $S^{(\lambda)}_{ai;bj}$
can depend on have been utilized, and thus by rotational invariance, only the Kronecker
symbol is left to tie up $a$ and $b$. The overall normalization is
determined by taking appropriate contractions, and we obtain
\begin{equation}
\label{decomp}
S_{ai;bj}^{(\lambda)}=
\langle {\cal O}^{\psi^\prime}_8(^3\!P_0)\rangle\,
\delta_{ab}\,{\epsilon^i}^*(\lambda)\epsilon^j(\lambda)\,.
\end{equation}
\noindent This decomposition tells us that to calculate
the polarized production
rate we should project the hard scattering amplitude onto states
with definite $S_z=\lambda$ and $L_z$, square the
amplitude, and then sum over
$L_z$ ($\sum_{L_z}\epsilon_a(L_z)
\epsilon_b(L_z)=\delta_{ab}$ in the rest frame). In other words, the
soft part is diagonal in the $L_z S_z$ basis.
It is straightforward to transform to the $J J_z$ basis. Since
$J_z=L_z+S_z$, there is no interference between intermediate states
with different $J_z$. To see this we write, in obvious
notation,
\begin{equation}
\label{factornew}
\sigma^{(\lambda)} \sim \sum_{J J_z;J' J_z^\prime}
H_{J J_z;J' J_z^\prime}\cdot S_{J J_z;J' J_z^\prime}^{(\lambda)}\,,
\end{equation}
\noindent and using (\docLink{slac-pub-7129-0-0-4.tcx}[decomp]{24}) obtain,
\begin{equation}
S_{J J_z;J' J_z^\prime}^{(\lambda)} =
\langle {\cal O}_8^{\psi'} ({}^3 P_0)\rangle
\sum_M \langle 1M;1\lambda|J J_z\rangle\langle J' J_z^\prime|1 M;1 \lambda
\rangle\,,
\end{equation}
\noindent which is diagonal in $(J J_z)(J' J_z^\prime)$ only after
summation over $\lambda$ (unpolarized production). In general, the
off-diagonal matrix elements cause interference of the following
$J J_z$ states: $00$ with $20$, $11$ with $21$ and $1(-1)$ with $2 (-1)$.
While the diagonal elements agree with (\docLink{slac-pub-7129-0-0-4.tcx}[cho]{21}), the off-diagonal
ones are missed in (\docLink{slac-pub-7129-0-0-4.tcx}[cho]{21}).
To assess the degree of transverse $\psi'$ (direct $J/\psi$)
polarization at moderate $p_t$,
the calculation of \cite{9} should be redone with the correct
angular momentum projections.