HPS and beam polarization

Michel GuidalIPN Orsay

HPS collaboration meeting17/06/2014

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)ÞThrough interference with the trident amplitude,

s ++ s -

s +- s -

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 = g or A’ (only timelike)

from T. Beranek and M. Vanderhaeghen arXiv:1303.2540 [hep-ph](Relation to M. Guidal and M. Vanderhaeghen (Double DVCS) Phys.Rev.Lett. 90 (2003) 012001)

(with antisymmetrization)

Kinematics

At fixed beam energy, there are 8 independent variables:

Ee’, qe’ , fe’ , qA’ , fA’ , Me+e-, qcm , qcm

fA’

fe’

Theoretical beam spin asymmetries

qcm=0 deg, fcm=0 degqcm=20 deg, fcm=70 degqcm=70 deg, fcm=170 degqcm=170 deg, fcm=250 deg

NOANTISYMMETRIZATION

Ebeam=2.2 GeV, qe=0.5 deg, Ee’=1 GeV, qA’=2 deg, Me+e-=50 MeV MA’=50 MeV, e=10-2 (a’/a=10-4)

fA’ (deg.)

qcm=0 deg, fcm=0 degqcm=20 deg, fcm=70 degqcm=70 deg, fcm=170 degqcm=170 deg, fcm=250 deg

WITHANTISYMMETRIZATION

Ebeam=2.2, GeV qe=0.5 deg, Ee’=1 GeV, qA’=2 deg, Me+e-=50 MeV MA’=50 MeV, e=10-2 (a’/a=10-4)

Theoretical beam spin asymmetries

fA’ (deg.)

What happens to the BSA when one integratesover the 8 kinematic variables ?

Complicated numerical problem:8 variables, structures and peaks,…

Monte-Carlo integration, ~stable results with 1011 events (around mA’ ; less if only background),24 hours with use of IN2P3 grid

qv (« vertical » angle)

fh (« horizontal » angle)

y

x

qe’ , fe’ , Me+e- , Ee+ , qv (e+), fh (e+), qv (e-), fh (e-),

In the following, we will use the 8 independent variables:

-.015 (rad)

-.05 (rad)

-.06 (rad)

.015 (rad)

.06 (rad)

.05 (rad)

MA’=50 MeV, e=10-2 G ~ 10 eV(a’/a=10-4)

Integrating over:

qe’ [0,p] (rad)fe’ [0,2p] (rad)Me+e- 1 keV around MA’

Ee+ [300,2000] (MeV)qv (e-) [-.06,-.015]+ [.015,.06] (rad)fh (e-) [-.05,.05] (rad)qv (e+) [.015,.06] (rad)fh (e+) [-.05,.05] (rad)

qv (« vertical » angle)

fh (« horizontal » angle)

y

x

qv (e+) [.015,.06]

A «bit more» (~3/1000) e+ on the left side than on the right side

Count rates/Statistics

• N(A’)~4.1011 (arb. units proportional to cross section)

• N(backgr)~1.5 1011 (arb. units proportional to cross section)

• DN(A’)~ 8.108 BSA ~ 8.108/4.1011 ~ 2/1000

If one takes the fh (e+) bin where BSA peaks (i.e. fh (e+) ~2 deg.):

backgr

A’

Count rates/Statistics

• N(A’)~4.1011 (arb. units proportional to cross section)

• N(backgr)~1.5 1011 (arb. units proportional to cross section)

• DN(A’)~ 8.108 BSA ~ 8.108/4.1011 ~ 2/1000

This all over a DMe+e=1 keV bin !

If one takes the fh (e+) bin where BSA peaks (i.e. fh (e+) ~2 deg.):

Count rates/Statistics

• N(A’)~4.1011 (arb. units proportional to cross section)

• N(backgr)~1.5 1011 (arb. units proportional to cross section)

• DN(A’)~ 8.108 BSA ~ 8.108/4.1011 ~ 2/1000

This all over a DMe+e=1 keV bin !

If one takes the fh (e+) bin where BSA peaks (i.e. fh (e+) ~2 deg.):

If one scales over a DMe+e=1 MeV bin: • N(A’)~4.1011 (arb. units proportional to cross section)

• N(backgr)~1.5 1011 X 103 (arb. units proportional to cross section)• DN(A’)~ 8.108 BSA ~ 8.108/1.5 1014 ~ 5/106

Count rates/Statistics

• N(A’)~4.1011 (arb. units proportional to cross section)

• N(backgr)~1.5 1011 (arb. units proportional to cross section)

• DN(A’)~ 8.108 BSA ~ 8.108/4.1011 ~ 2/1000

This all over a DMe+e=1 keV bin !

If one takes the fh (e+) bin where BSA peaks (i.e. fh (e+) ~2 deg.):

If one scales over a DMe+e=1 MeV bin: • N(A’)~4.1011 (arb. units proportional to cross section)

• N(backgr)~1.5 1011 X 103 (arb. units proportional to cross section)• DN(A’)~ 8.108 BSA ~ 8.108/1.5 1014 ~ 5/106

And we expect 107 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 andall HPS acceptance, identify a particular cornerof the phase space where BSA is of the orderof the percent and, if doing so, one goes from107 to 104 events… (one can always dream )

Stay tuned !