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HPS and beam polarization Michel Guidal IPN Orsay HPS collaboration meeting 17/06/2014
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Any perspective for HPS ? The QED (trident) background amplitude is purely real => No sensitivity to beam spin The JLab beam comes polarized to a high degree for (almost) free The A’ amplitude has an imaginary part (width) Through interference with the trident amplitude, there should be a beam spin asymmetry (BSA) Spin observables are well-known to be sensitive to small effects An asymmetry (ratio) doesn’t need the perfect knowledge of the normalization and shape of the QED background In principle, any non-zero asymmetry would sign an A’ Motivation
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Tree-level diagrams for ep->epe + e - V = or A’ (only timelike) from T. Beranek and M. Vanderhaeghen arXiv:1303.2540 [hep-ph]arXiv:1303.2540 (Relation to M. Guidal and M. Vanderhaeghen (Double DVCS) Phys.Rev.Lett. 90 (2003) 012001) (with antisymmetrization)
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Kinematics At fixed beam energy, there are 8 independent variables: E e’ e’, e’, A’, A’, M e+e-, cm, cm A’ e’
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Theoretical beam spin asymmetries cm =0 deg, cm =0 deg cm =20 deg, cm =70 deg cm =70 deg, cm =170 deg cm =170 deg, cm =250 deg NO ANTISYMMETRIZATION E beam =2.2 GeV, e =0.5 deg, E e’ =1 GeV, A’ =2 deg, M e+e- =50 MeV M A’ =50 MeV, =10 -2 ( ’/ =10 -4 ) A’ (deg.)
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cm =0 deg, cm =0 deg cm =20 deg, cm =70 deg cm =70 deg, cm =170 deg cm =170 deg, cm =250 deg WITH ANTISYMMETRIZATION E beam =2.2, GeV e =0.5 deg, E e’ =1 GeV, A’ =2 deg, M e+e- =50 MeV M A’ =50 MeV, =10 -2 ( ’/ =10 -4 ) Theoretical beam spin asymmetries A’ (deg.)
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What happens to the BSA when one integrates over the 8 kinematic variables ? Complicated numerical problem: 8 variables, structures and peaks,… Monte-Carlo integration, ~stable results with 10 11 events (around m A’ ; less if only background), 24 hours with use of IN2P3 grid
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v (« vertical » angle) h (« horizontal » angle) y x e’, e’, M e+e-, E e+, v (e+), h (e+), v (e-), h (e-), In the following, we will use the 8 independent variables: -.015 (rad) -.05 (rad) -.06 (rad).015 (rad).06 (rad).05 (rad)
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M A’ =50 MeV, =10 -2 ~ 10 eV ( ’/ =10 -4 )
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Integrating over: e’ [0, ](rad) e’ [0,2 ](rad) M e+e- 1 keV around M A’ E e+ [300,2000] (MeV) v (e-) [-.06,-.015]+ [.015,.06] (rad) h (e-) [-.05,.05] (rad) v (e+) [.015,.06] (rad) h (e+) [-.05,.05] (rad)
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v (« vertical » angle) h (« horizontal » angle) y x v (e+) [.015,.06] A «bit more» (~3/1000) e+ on the left side than on the right side
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Count rates/Statistics N(A’)~4.10 11 (arb. units proportional to cross section) N(backgr)~1.5 10 11 (arb. units proportional to cross section) N(A’)~ 8.10 8 BSA ~ 8.10 8 /4.10 11 ~ 2/1000 If one takes the h (e+) bin where BSA peaks (i.e. h (e+) ~2 deg.):
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backgr A’
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Count rates/Statistics N(A’)~4.10 11 (arb. units proportional to cross section) N(backgr)~1.5 10 11 (arb. units proportional to cross section) N(A’)~ 8.10 8 BSA ~ 8.10 8 /4.10 11 ~ 2/1000 This all over a M e+e =1 keV bin ! If one takes the h (e+) bin where BSA peaks (i.e. h (e+) ~2 deg.):
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Count rates/Statistics N(A’)~4.10 11 (arb. units proportional to cross section) N(backgr)~1.5 10 11 (arb. units proportional to cross section) N(A’)~ 8.10 8 BSA ~ 8.10 8 /4.10 11 ~ 2/1000 This all over a M e+e =1 keV bin ! If one takes the h (e+) bin where BSA peaks (i.e. h (e+) ~2 deg.): If one scales over a M e+e =1 MeV bin: N(A’)~4.10 11 (arb. units proportional to cross section) N(backgr)~1.5 10 11 X 10 3 (arb. units proportional to cross section) N(A’)~ 8.10 8 BSA ~ 8.10 8 /1.5 10 14 ~ 5/10 6
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Count rates/Statistics N(A’)~4.10 11 (arb. units proportional to cross section) N(backgr)~1.5 10 11 (arb. units proportional to cross section) N(A’)~ 8.10 8 BSA ~ 8.10 8 /4.10 11 ~ 2/1000 This all over a M e+e =1 keV bin ! If one takes the h (e+) bin where BSA peaks (i.e. h (e+) ~2 deg.): If one scales over a M e+e =1 MeV bin: N(A’)~4.10 11 (arb. units proportional to cross section) N(backgr)~1.5 10 11 X 10 3 (arb. units proportional to cross section) N(A’)~ 8.10 8 BSA ~ 8.10 8 /1.5 10 14 ~ 5/10 6 And we expect 10 7 events in a 1 MeV bin for HPS…
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Doesn’t look too good… But maybe still some hope: Instead of integrating over all 8 variables and all HPS acceptance, identify a particular corner of the phase space where BSA is of the order of the percent and, if doing so, one goes from 10 7 to 10 4 events… (one can always dream ) Stay tuned !
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