HPS and beam polarization Michel Guidal IPN Orsay HPS collaboration meeting 17/06/2014

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

Tree-level diagrams for ep->epe + e - V =  or A’ (only timelike) from T. Beranek and M. Vanderhaeghen arXiv: [hep-ph]arXiv: (Relation to M. Guidal and M. Vanderhaeghen (Double DVCS) Phys.Rev.Lett. 90 (2003) ) (with antisymmetrization)

Kinematics At fixed beam energy, there are 8 independent variables: E e’  e’,  e’,  A’,  A’, M e+e-,  cm,  cm  A’  e’

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.)

 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.)

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 events (around m A’ ; less if only background), 24 hours with use of IN2P3 grid

 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: (rad) -.05 (rad) -.06 (rad).015 (rad).06 (rad).05 (rad)

M A’ =50 MeV,  =10 -2  ~ 10 eV (  ’/  =10 -4 )

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)

 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

Count rates/Statistics N(A’)~ (arb. units proportional to cross section) N(backgr)~ (arb. units proportional to cross section)  N(A’)~ BSA ~ / ~ 2/1000 If one takes the  h (e+) bin where BSA peaks (i.e.  h (e+) ~2 deg.):

backgr A’

Count rates/Statistics N(A’)~ (arb. units proportional to cross section) N(backgr)~ (arb. units proportional to cross section)  N(A’)~ BSA ~ / ~ 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.):

Count rates/Statistics N(A’)~ (arb. units proportional to cross section) N(backgr)~ (arb. units proportional to cross section)  N(A’)~ BSA ~ / ~ 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’)~ (arb. units proportional to cross section) N(backgr)~ X 10 3 (arb. units proportional to cross section)  N(A’)~ BSA ~ / ~ 5/10 6

Count rates/Statistics N(A’)~ (arb. units proportional to cross section) N(backgr)~ (arb. units proportional to cross section)  N(A’)~ BSA ~ / ~ 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’)~ (arb. units proportional to cross section) N(backgr)~ X 10 3 (arb. units proportional to cross section)  N(A’)~ BSA ~ / ~ 5/10 6 And we expect 10 7 events in a 1 MeV bin for HPS…

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 !