two photon exchange and transverse spin asymmetries in the...

Two photon exchange and transverse spinasymmetries in the A4 experiment.

David Balaguer Rios

PAVI09Bar Harbor

A4 CollaborationInstitut für Kern Physik, Mainz

June 25, 2009

David Balaguer Rios, June 25, 2009 1/39

Introduction

Experimental set up

Data analysis

Results H2 data

Results D2 data

Summary and outlook

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Outline

Introduction

David Balaguer Rios, June 25, 2009 3/39

Two photon exchange

◮ R = GpE/Gp

M discrepancy

◮ 2γ exchange amplitude A2γ

◮ Target and beam normal SSA sensitive to Im(A2γ).

◮ Dispersion relations between imaginary and real part.

◮ PVA experiments set up: transverse beam spin asymmetry A⊥

David Balaguer Rios, June 25, 2009 4/39

Transverse spin asymmetry

Am⊥ =

σ↑ − σ↓

σ↑ + σ↓= A⊥(θ)~Pe ·

~S = A⊥ cos φ

~k

~k′

θe

φ-π2

~S⊥ =~k×~k′

|~k×~k′|

~P

φ

David Balaguer Rios, June 25, 2009 5/39

Transverse spin asymmetry

Am⊥ =

σ↑ − σ↓

σ↑ + σ↓= A⊥(θ)~Pe ·

~S = A⊥ cos φ

~k

~k′

θe

φ-π2

~S⊥ =~k×~k′

|~k×~k′|

~P

φ

David Balaguer Rios, June 25, 2009 6/39

Outline

Experimental set up

David Balaguer Rios, June 25, 2009 7/39

MAMI floor plan

David Balaguer Rios, June 25, 2009 8/39

Polarized electron beam source

David Balaguer Rios, June 25, 2009 9/39

Calorimeter, target and luminosity monitors

◮ Detector covers 2πφ.

◮ Counts singleevents andmeasures energy.

◮ Target of liquid hy-drogen/deuterium.

David Balaguer Rios, June 25, 2009 10/39

David Balaguer Rios, June 25, 2009 11/39

David Balaguer Rios, June 25, 2009 11/39

David Balaguer Rios, June 25, 2009 11/39

David Balaguer Rios, June 25, 2009 11/39

Plastic scintillators

◮ Plastic scintillators detect charged particles. Neutral particlesnot detected.

◮ 72 plastic scintillators: two rings of 36 withoverlap.

◮ One scintillator covers two frames: 14 modules.

David Balaguer Rios, June 25, 2009 12/39

Medusa

David Balaguer Rios, June 25, 2009 13/39

Measurements performed at backward angles

Target Energy (MeV) angle (deg) Q2 (GeV/c)2

H2 315.1 MeV 140 − 150 0.23D2 315.1 MeV 140 − 150 0.23

David Balaguer Rios, June 25, 2009 14/39

Outline

Data analysis

David Balaguer Rios, June 25, 2009 15/39

Energy spectra for H2 target

◮ Counts and energy: energy spectrum histograms every 5 min.

◮ Two histograms for each polarization state.

◮ Scintillators: a histogram for charged particles and one forneutral paticles.

David Balaguer Rios, June 25, 2009 16/39

Energy spectra for H2 target

◮ Counts and energy: energy spectrum histograms every 5 min.

◮ Two histograms for each polarization state.

◮ Scintillators: a histogram for charged particles and one forneutral paticles.

David Balaguer Rios, June 25, 2009 16/39

Energy spectra for H2 target

◮ Counts and energy: energy spectrum histograms every 5 min.

◮ Two histograms for each polarization state.

◮ Scintillators: a histogram for charged particles and one forneutral paticles.

David Balaguer Rios, June 25, 2009 16/39

Energy spectra for H2 target

◮ Separation of the elastic peak in the charged particles spectrum

◮ Still neutral background from γ → e−e+

David Balaguer Rios, June 25, 2009 17/39

Energy spectra for H2 target

◮ Separation of the elastic peak in the charged particles spectrum

◮ Still neutral background from γ → e−e+

David Balaguer Rios, June 25, 2009 17/39

Understanding the energy spectrum

◮ Monte Carlo Geant4 simulation: e− processes and γs

◮ Background from Al walls: measurement with empty target

◮ Agreement above 125 MeV

David Balaguer Rios, June 25, 2009 18/39

Background obtained from neutral spectrum

◮ γ background asymmetry?

◮ From experimental spectrumof neutral particles

◮ Model to obtain γ background

◮ Parameters

◮ shift δ◮ scaling factor ǫ

David Balaguer Rios, June 25, 2009 19/39

Background obtained from neutral spectrum

◮ Scaling shifting modelagrees with simulatedbackground:

◮ above 125 MeV

◮ energy range of ourinterest

David Balaguer Rios, June 25, 2009 20/39

Extraction of the asymmetry from the spectra

◮ f =Nback

e

Nbacke + Nel

e

◮ Cuts applied f = 5, 9, 16%

◮ Ae =N+

e − N−

e

N+e + N−

e

◮ Aγ =N+

γ− N−

γ

N+γ

+ N−

γ

◮ Arawphys =

Ae − f Aγ

1 − f

David Balaguer Rios, June 25, 2009 21/39

Extraction of the asymmetry from the spectra

◮ f =Nback

e

Nbacke + Nel

e

◮ Cuts applied f = 5, 9, 16%

◮ Ae =N+

e − N−

e

N+e + N−

e

◮ Aγ =N+

γ− N−

γ

N+γ

+ N−

γ

◮ Arawphys =

Ae − f Aγ

1 − f

David Balaguer Rios, June 25, 2009 21/39

Extraction of the asymmetry from the spectra

◮ f =Nback

e

Nbacke + Nel

e

◮ Cuts applied f = 5, 9, 16%

◮ Ae =N+

e − N−

e

N+e + N−

e

◮ Aγ =N+

γ− N−

γ

N+γ

+ N−

γ

◮ Arawphys =

Ae − f Aγ

1 − f

David Balaguer Rios, June 25, 2009 21/39

Extraction of the asymmetry from the spectra

◮ f =Nback

e

Nbacke + Nel

e

◮ Cuts applied f = 5, 9, 16%

◮ Ae =N+

e − N−

e

N+e + N−

e

◮ Aγ =N+

γ− N−

γ

N+γ

+ N−

γ

◮ Arawphys =

Ae − f Aγ

1 − f

David Balaguer Rios, June 25, 2009 21/39

Extraction of the asymmetry from the spectra

◮ f =Nback

e

Nbacke + Nel

e

◮ Cuts applied f = 5, 9, 16%

◮ Ae =N+

e − N−

e

N+e + N−

e

◮ Aγ =N+

γ− N−

γ

N+γ

+ N−

γ

◮ Arawphys =

Ae − f Aγ

1 − f

David Balaguer Rios, June 25, 2009 21/39

Extraction of the asymmetry from the spectra

◮ f =Nback

e

Nbacke + Nel

e

◮ Cuts applied f = 5, 9, 16%

◮ Ae =N+

e − N−

e

N+e + N−

e

◮ Aγ =N+

γ− N−

γ

N+γ

+ N−

γ

◮ Arawphys =

Ae − f Aγ

1 − f

David Balaguer Rios, June 25, 2009 21/39

Combination of asymmetries

◮ Detector divided in sectors: azimuthal modulation of A⊥

◮ A⊥ averaged over the scattering angle θ

David Balaguer Rios, June 25, 2009 22/39

Outline

Results H2 data

David Balaguer Rios, June 25, 2009 23/39

Raw asymmetry dependence on φ / H2

Q2 = 0.23 (GeV/c)2

A4 backwards kinematics

f = 9%Aexp = A⊥ cos(φ + δ) + p

(deg)φ

0 50 100 150 200 250 300 350

pp

m

-60

-40

-20

0

20

40

60

80Chi2/ndf = 3.03/5

Prob = 0.70

A = (-66.41 +- 4.92) ppm

d = (5.09 +- 4.24) deg

p = (5.15 +- 3.39) ppm

David Balaguer Rios, June 25, 2009 24/39

GVZ test of systematics

GVZ = OUT

(deg)φ

0 50 100 150 200 250 300 350

pp

m

-60

-40

-20

0

20

40

60

80

100

Chi2/ndf = 1.07/5

Prob = 0.96

A = (-71. 58 +/- 6.90) ppm

d = (6.93 +/- 5.52) deg

p = (11.74 +/- 4.76) ppm

GVZ = IN

(deg)φ

0 50 100 150 200 250 300 350

pp

m

-60

-40

-20

0

20

40

60

80 Chi2/ndf = 2.69/5

Prob = 0.75

A = (61.17 +/- 7.00) ppm

d = (2.90 +/- 6.56) deg

p = (1.63 +/- 4.83) ppm

David Balaguer Rios, June 25, 2009 25/39

Preeliminary results for H2 data

◮ 5 inner rings: 730 crystals.

◮ 50 h of data taking.

◮ Altogether 1.8 · 1011 elastic events.

◮ Effective Pe = 70.0%, error ∆(Pe) = 4%

At A4 Backward kinematics and Q2 = 0.23 (GeV/c)2

A⊥ = (−86.65 ± 3.35 ± 3.46) ppm

David Balaguer Rios, June 25, 2009 26/39

Outline

Results D2 data

David Balaguer Rios, June 25, 2009 27/39

Neutral and charged particles spectra for D2

David Balaguer Rios, June 25, 2009 28/39

Neutral and charged particles spectra for D2

David Balaguer Rios, June 25, 2009 28/39

Raw asymmetry dependence on φ / D2

Q2 = 0.23 (GeV/c)2

A4 backwards kinematics

f = 16%Aexp = A⊥ cos(φ + δ) + p

(deg)φ

0 50 100 150 200 250 300 350

pp

m

-40

-20

0

20

40

Chi2/ndf = 4.23/5

Prob = 0.52

A = (-41.02 +/- 3.71) ppm

d = (-5.25 +/- 5.15) deg

p = (3.01 +/- 2.55) ppm

David Balaguer Rios, June 25, 2009 29/39

GVZ test of systematics

GVZ = OUT

(deg)φ

0 50 100 150 200 250 300 350

pp

m

-40

-20

0

20

40

60

Chi2/ndf = 7.19/5

Prob = 0.21

A = (-39.79 +/- 5.37) ppm

d = (3.21 +/- 7.71) deg

p = (0.97 +/- 3.70) ppm

GVZ = IN

(deg)φ

0 50 100 150 200 250 300 350

pp

m

-60

-40

-20

0

20

40

60 Chi2/ndf = 0.9/5

Prob = 0.97

A = ( 44.96 +/- 8.95)

d = (-13.3 +/- 11.4) deg

p = (-10.2 +/- 6.2) ppm

David Balaguer Rios, June 25, 2009 30/39

Preeliminary results for D2 data

◮ 5 inner rings: 730 crystals.

◮ 60 h of data taking.

◮ Altogether 2.5 · 1011 elastic events

◮ Effective Pe = 78.4%, error ∆(Pe) = 4%

At A4 Backward kinematics and Q2 = 0.23 (GeV/c)2

A⊥ = (−57.60 ± 2.54 ± 3.07) ppm

David Balaguer Rios, June 25, 2009 31/39

Systematic errors

Systematic error contributionPolarization 2.30 ppm

False asymmetries 0.64 ppmModel parameter δ 1.3 ppmModel parameter ǫ 0.25 ppm

Target density 0.03 ppmSpin deviation 1.4 ppm

Aluminium —Pile-up —

Nonlinearity LuMo —Total 3.07 ppm

David Balaguer Rios, June 25, 2009 32/39

Transverse spin asymmetry in neutralbackground

f = 12%

(deg)φ

0 50 100 150 200 250 300 350

pp

m

-40

-30

-20

-10

0

10

20

30

40

50Chi2/ndf = 4.2/5

Prob = 0.53

A = ( -36.6 +/- 3.0) ppm

d = (-0.36 +/- 4.68) deg

p = ( 4.22 +/- 2.06) ppm

f = 16%

(deg)φ

0 50 100 150 200 250 300 350

pp

m

-40

-20

0

20

40

60

Chi2/ndf = 1.45/5

Prob = 0.92

A = ( -48.65 +/- 2.65) ppm

d = (-1.68 +/- 2.87) deg

p = ( 2.01 +/- 1.68) ppm

f = 20%

(deg)φ

0 50 100 150 200 250 300 350

pp

m

-40

-20

0

20

40

60Chi2/ndf = 3.08/5

Prob = 0.69

A = ( -53.75 +/- 2.08) ppm

d = (-3.33 +/- 2.21) deg

p = ( -0.25 +/- 1.43) ppm

◮ Aback⊥ is large and depends on

the cut:

◮ A⊥ sensitive to backgroundsubtraction

David Balaguer Rios, June 25, 2009 33/39

Summary of asymmetries

Target elastic backgroundH2 −86.65 ± 3.35 ppm −100.60 ± 2.78 ppmD2 −57.60 ± 2.54 ppm −62.37 ± 2.28 ppm

David Balaguer Rios, June 25, 2009 34/39

Comparision of measured and calculated A⊥

Calculation of A⊥ for proton 1

-140

-120

-100

-80

-60

-40

-20

0

0 20 40 60 80 100 120 140 160 180

An

[ppm

]

θlab[deg]

1Resonance estimates for single spin asymmetries in elasticelectron-nucleon scattering. B. Pasquini et al. physical review c 70. 045206(2004)

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Outline

Summary and outlook

David Balaguer Rios, June 25, 2009 36/39

Summary

◮ Transverse spin asymmetry with H2 and D2 at Q2 = 0.23(GeV/c)2 and backward angles.

◮ Detector of plastic scintillators to separate neutral background

◮ Understanding the energy spectrum and mixed γ background

◮ Modulation of the transverse spin asymmetry on cos φ

◮ Combination of whole data. Preliminary results

◮ Comparision of the measured transverse spin asymmetries withthe calculations.

David Balaguer Rios, June 25, 2009 37/39

Outlook

◮ Analysis in progress: sensitivity of A⊥ to background subtraction.

◮ Investigation the transverse spin asymmetry in the neutralbackground.

◮ Analysis of data at backward angles with H2 and D2 andQ2 = 0.35 (GeV/c)2

David Balaguer Rios, June 25, 2009 38/39

Previous A⊥ at A4 forward kinematics

570 MeV

-70

-60

-50

-40

-30

-20

-10

0

0 20 40 60 80 100 120 140 160 180

An

[ppm

]

θlab[deg]

855 MeV

-25

-20

-15

-10

-5

0

5

0 20 40 60 80 100 120 140 160 180

An

[ppm

]

θlab[deg]

Target l-H2

A⊥(Q2 = 0.11(GeV/c)2) = (−8.59 ± 0.89 ± 0.75) ppm

A⊥(Q2 = 0.23(GeV/c)2) = (−8.52 ± 2.31 ± 0.87) ppm

David Balaguer Rios, June 25, 2009 39/39