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Hadron Beams

Consider first a proton (or, with trivial modifications, any other baryon or antibaryon).
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If the initiator parton is a $\u $ or $\d $ quark, it is assumed to be a valence quark, and therefore leaves behind a diquark beam remnant, i.e. either a $\u\d $ or a $\u\u $ diquark, in a colour antitriplet state. Relative probabilities for different diquark spins are derived within the context of the non-relativistic SU(6) model, i.e. flavour SU(3) times spin SU(2). Thus a $\u\d $ is $3/4$ $\u\d _0$ and $1/4$ $\u\d _1$, while a $\u\u $ is always $\u\u _1$.
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An initiator gluon leaves behind a colour octet $\u\u\d $ state, which is subdivided into a colour triplet quark and a colour antitriplet diquark. SU(6) gives the appropriate subdivision, $1/2$ of the time into $\u + \u\d _0$, $1/6$ into $\u + \u\d _1$ and $1/3$ into $\d + \u\u _1$.
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A sea quark initiator, such as an $\mathrm{s}$, leaves behind a $\u\u\d\overline{\mathrm{s}}$ four-quark state. The PDG flavour coding scheme and the fragmentation routines do not foresee such a state, so therefore it is subdivided into a meson plus a diquark, i.e. $1/2$ into $\u\overline{\mathrm{s}}+ \u\d _0$, $1/6$ into $\u\overline{\mathrm{s}}+ \u\d _1$ and $1/3$ into $\d\overline{\mathrm{s}}+ \u\u _1$. Once the flavours of the meson are determined, the choice of meson multiplet is performed as in the standard fragmentation description.
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Finally, an antiquark initiator, such as an $\overline{\mathrm{s}}$, leaves behind a $\u\u\d\mathrm{s}$ four-quark state, which is subdivided into a baryon plus a quark. Since, to first approximation, the $\mathrm{s}\overline{\mathrm{s}}$ pair comes from the branching $\mathrm{g}\to \mathrm{s}\overline{\mathrm{s}}$ of a colour octet gluon, the subdivision $\u\u\d + \mathrm{s}$ is not allowed, since it would correspond to a colour-singlet $\mathrm{s}\overline{\mathrm{s}}$. Therefore the subdivision is $1/2$ into $\u\d _0\mathrm{s}+ \u $, $1/6$ into $\u\d _1\mathrm{s}+ \u $ and $1/3$ into $\u\u _1\mathrm{s}+ \d $. A baryon is formed among the ones possible for the given flavour content and diquark spin, according to the relative probabilities used in the fragmentation. One could argue for an additional weighting to count the number of baryon states available for a given diquark plus quark combination, but this has not been included.

One may note that any $\u $ or $\d $ quark taken out of the proton is automatically assumed to be a valence quark. Clearly this is unrealistic, but not quite as bad as it might seem. In particular, one should remember that the beam-remnant scenario is applied to the initial-state shower initiators at a scale of $Q_0 \approx 1$ GeV and at an $x$ value usually much larger than the $x$ at the hard scattering. The sea quark contribution therefore normally is small.

For a meson beam remnant, the rules are in the same spirit, but somewhat easier, since no diquark or baryons need be taken into account. Thus a valence quark (antiquark) initiator leaves behind a valence antiquark (quark), a gluon initiator leaves behind a valence quark plus a valence antiquark, and a sea quark (antiquark) leaves behind a meson (which contains the partner to the sea parton) plus a valence antiquark (quark).


next up previous contents
Next: Photon Beams Up: Beam Remnants Previous: Beam Remnants   Contents
Stephen_Mrenna 2012-10-24