Beam Single Spin Asymmetries in Electron-Proton Scattering · p 2 p 1 p 2+q 1 q 1 p 4 q 2 k p 3...

Beam Single Spin Asymmetries in Electron-ProtonScattering

Pratik SachdevaWashington University in St. Louis

Mentors:Wally Melnitchouk

Jefferson Laboratory Theory Center

Peter BlundenUniversity of Manitoba

July 23, 2014

Outline

1 IntroductionBeam Single Spin Asymmetries (BSSA)Transverse and Normal PolarizationsMotivation: Q-weak Collaboration

2 Elastic ScatteringBeam Normal SSABeam Transverse Single Spin AsymmetryCombination of Asymmetries

3 Inelastic Hadronic Intermediate StateM ∗

γMγγ InterferenceLeptonic and Hadronic TensorsNormal BSSA

4 Conclusion

Pratik Sachdeva Beam Single Spin Asymmetries in Electron-Proton Scattering 2 / 26

Introduction Beam Single Spin Asymmetries (BSSA)

IntroductionBeam Single Spin Asymmetries (BSSA)

Polarized electron beam (longitudinal, transverse, normal), butunpolarized target.

In general the beam asymmetry, B, is

B = σ↑ − σ↓σ↑ + σ↓

BSSA disappear for one-photon exchange.

B = σ↑ − σ↓σ↑ + σ↓

What exchanges can cause BSSA?

Consider the scattering amplitude:

|M |2 = |Mγ + MZ + Mγγ + · · · |2

= |Mγ |2 + M ∗γMZ + M ∗

γMγγ + · · ·

At the Born level, the spin orientation has no effect on |Mγ |2, so theasymmetry vanishes.Higher order effects result in asymmetries dependent on whether thespin is transversely, normally, or longitudinally polarized.

|M |2 = |Mγ + MZ + Mγγ + · · · |2

= |Mγ |2 + M ∗γMZ + M ∗

γMγγ + · · ·

|M |2 = |Mγ + MZ + Mγγ + · · · |2

= |Mγ |2 + M ∗γMZ + M ∗

γMγγ + · · ·

At the Born level, the spin orientation has no effect on |Mγ |2, so theasymmetry vanishes.

Higher order effects result in asymmetries dependent on whether thespin is transversely, normally, or longitudinally polarized.

|M |2 = |Mγ + MZ + Mγγ + · · · |2

= |Mγ |2 + M ∗γMZ + M ∗

γMγγ + · · ·

Introduction Transverse and Normal Polarizations

Transverse and Normal Polarization

Transverse asymmetry Bt behaves like cosφs and normal asymmetryBn behaves like sinφs .

Figure 2. Electron-Proton Scattering

Introduction Motivation: Q-weak Collaboration

Motivation: Q-weak Collaboration

Precise measurement of the weak charge of the proton, QPW , using

asymmetry produced by longitudinally polarized electrons.

Transversely polarized electrons: BSSA has azimuthal (φ) dependence(Buddhini Waidyawansa, CIPANP 2012 Conference).

Concerned about possible phase offsets due to unconsidered BSSAeffects.

asymmetry produced by longitudinally polarized electrons.Transversely polarized electrons: BSSA has azimuthal (φ) dependence(Buddhini Waidyawansa, CIPANP 2012 Conference).

Approach

Elastic Scattering Contribution

Interference of one- and two-photon exchange.Interference of one-photon and Z exchange.

Hadronic Inelastic Intermediate StatePasquini and Vanderhaeghen (PRC 70, 2004) showed that inelastichadronic intermediate states have a larger contribution to theasymmetry than the elastic case.Consider the case of near forward limit.

Approach

Elastic Scattering ContributionInterference of one- and two-photon exchange.Interference of one-photon and Z exchange.

Approach

Hadronic Inelastic Intermediate State

Pasquini and Vanderhaeghen (PRC 70, 2004) showed that inelastichadronic intermediate states have a larger contribution to theasymmetry than the elastic case.Consider the case of near forward limit.

Approach

Elastic Scattering

Elastic scattering for electron-proton scattering,

e(p1) + N(p2)→ e(p3) + N(p4).

�p2p1

p2 + q1

Elastic Scattering

Kinematic Variables

Elastic scattering process:

e(p1) + N(p2)→ e(p3) + N(p4)

Using the following notation for kinematic variables:

P = p2 + p42 , K = p1 + p3

2 , q = p1 − p3, Q2 = −q2;

s = (p1 + p2)2, τ = Q2

4M2 , ν = P · K , ε = ν2 −M4τ(1 + τ)ν2 + M4τ(1 + τ) ;

W 2 = (p2 + q1)2, Q21 = −q2

1 , Q22 = −q2

for electron mass me and hadron mass M.

Elastic Scattering

Kinematic Variables

Elastic scattering process:

e(p1) + N(p2)→ e(p3) + N(p4)

Using the following notation for kinematic variables:

P = p2 + p42 , K = p1 + p3

2 , q = p1 − p3, Q2 = −q2;

s = (p1 + p2)2, τ = Q2

4M2 , ν = P · K , ε = ν2 −M4τ(1 + τ)ν2 + M4τ(1 + τ) ;

W 2 = (p2 + q1)2, Q21 = −q2

1 , Q22 = −q2

for electron mass me and hadron mass M.

Elastic Scattering Beam Normal SSA

Beam Normal Single Spin Asymmetry: SetupUsing six invariant amplitudes given by Goldberger et al. (Ann. Phys. 2,1957), we can construct a general elastic lepton-nucleon scatteringamplitude:

T = T non-flip + T flip,

T non-flip = e2

Q2 ue(p3)γµue(p1) · uN(p4)(

GMγµ − F2

M + F3/KPµ

)uN(p2)

T flip = e2

[ue(p3)ue(p1) · uN(p4)

(F4 + F5

)uN(p2)

+ F6ue(p3)γ5ue(p1) · uN(p4)γ5uN(p2)].

T non-flip = e2

GMγµ − F2

M + F3/KPµ

)uN(p2)

T flip = e2

(F4 + F5

)uN(p2)

T non-flip = e2

GMγµ − F2

M + F3/KPµ

)uN(p2)

T flip = e2

(F4 + F5

)uN(p2)

Beam Normal Single Spin Asymmetry: Setup

In the two previous equations,

GM , F2, F3, F4, F5, F6

are complex functions of ν and Q2. In the Born approximation, thesereduce to

GBornM (ν,Q2) = GM(Q2)

F Born2 (ν,Q2) = F2(Q2)

F Born3,4,5,6(ν,Q2) = 0.

We can also writeGE = GM − (1 + τ)F2.

GM , F2, F3, F4, F5, F6

F Born2 (ν,Q2) = F2(Q2)

F Born3,4,5,6(ν,Q2) = 0.

GM , F2, F3, F4, F5, F6

F Born2 (ν,Q2) = F2(Q2)

F Born3,4,5,6(ν,Q2) = 0.

Beam Normal SSA

Gorchtein et al. found that for a spin parallel (anti-parallel) to the normalpolarization vector

Sµ = (0, ~Sn), ~Sn = (~p1 × ~p3) / |~p1 × ~p3|

there is a beam normal SSA Bn due to M ∗γMγγ . It is equal to

Bn = 2meQ

√2ε(1− ε)

√1 + 1

M + G2E

×{−τGM Im

(F3 + 1

1 + τ

)− GE Im

(F4 + 1

1 + τ

Beam Normal SSA

Gorchtein et al. found that for a spin parallel (anti-parallel) to the normalpolarization vector

Sµ = (0, ~Sn), ~Sn = (~p1 × ~p3) / |~p1 × ~p3|

there is a beam normal SSA Bn due to M ∗γMγγ . It is equal to

Bn = 2meQ

√2ε(1− ε)

√1 + 1

M + G2E

×{−τGM Im

(F3 + 1

1 + τ

)− GE Im

(F4 + 1

1 + τ

Beam Normal SSA

Figure 3. Transverse Spin Polarization

If we consider a general transverse spin

~s = cosφs x + sinφs y ,

the general beam normal SSA can be written

Bn, gen = Bn sinφs .

Beam Normal SSA

Elastic Scattering Beam Transverse Single Spin Asymmetry

Beam Transverse SSA

We found that a beam transverse SSA exists at the Born level due toM ∗

γMZ . It is equal to

Bt = GF me2πα

(s −M2)

√ε(1− ε)τ2(τ + 1)

M + ε

geV GMGZ

Mτ(τ + 1) + geA

A τ(1 + τ − ν) + GE GZE (1− τ − ν)

where GZM and GZ

E are the weak form factors and GF is Fermi’s couplingconstant.

For a general transverse spin,

Bt, gen = Bt cosφs .

Beam Transverse SSA

Bt = GF me2πα

(s −M2)

√ε(1− ε)τ2(τ + 1)

M + ε

geV GMGZ

Mτ(τ + 1) + geA

A τ(1 + τ − ν) + GE GZE (1− τ − ν)

where GZM and GZ

Beam Transverse SSA

Bt = GF me2πα

(s −M2)

√ε(1− ε)τ2(τ + 1)

M + ε

geV GMGZ

Mτ(τ + 1) + geA

A τ(1 + τ − ν) + GE GZE (1− τ − ν)

where GZM and GZ

Elastic Scattering Combination of Asymmetries

Combination of Asymmetries

The interference between one- and two-photon exchange onlyproduces a normal BSSA.The interference between one-photon and Z exchange only producesa transverse BSSA.

The combination of these two asymmetries gives

B = Bt cosφs + Bn sinφs

B2t + B2

n sin (φs + δ)

where δ = tan−1(

B2t + B2

n sin (φs + δ)

where δ = tan−1(

B2t + B2

n sin (φs + δ)

where δ = tan−1(

Combination of Asymmetries: Q-weak Kinematics

In Q-weak, Bn was measured as

Bn ≈ −5.350 ppm.

Using Q-weak kinematics, Q2 = 0.025 GeV, E1 = 1.125 GeV (electronbeam energy), we find

|δ| =∣∣∣∣tan−1

)∣∣∣∣ ≈ ∣∣∣∣BtBn

∣∣∣∣ ≈ 1.116× 10−11

5.350× 10−6 = 2.086× 10−6.

→ This is too small to affect the Q-weak measurements.

Bn ≈ −5.350 ppm.

|δ| =∣∣∣∣tan−1

)∣∣∣∣ ≈ ∣∣∣∣BtBn

∣∣∣∣ ≈ 1.116× 10−11

5.350× 10−6 = 2.086× 10−6.

Bn ≈ −5.350 ppm.

|δ| =∣∣∣∣tan−1

)∣∣∣∣ ≈ ∣∣∣∣BtBn

∣∣∣∣ ≈ 1.116× 10−11

5.350× 10−6 = 2.086× 10−6.

Bn ≈ −5.350 ppm.

|δ| =∣∣∣∣tan−1

)∣∣∣∣ ≈ ∣∣∣∣BtBn

∣∣∣∣ ≈ 1.116× 10−11

5.350× 10−6 = 2.086× 10−6.

Order of Magnitude Estimates

Beam Normal SSA:

Bn ∼αmeM ∼ 5× 10−6 → 5 ppm

APV (from longitudinal polarization):

APV ∼Q2

M2Z∼ Q2 × 10−4

Beam Transverse SSA:

Bt ∼Q2

meM ∼ Q2 ×

(5× 10−8

What factors contribute to Bn?

Beam Normal SSA:

Bn ∼αmeM ∼ 5× 10−6 → 5 ppm

APV ∼Q2

M2Z∼ Q2 × 10−4

Bt ∼Q2

meM ∼ Q2 ×

(5× 10−8

Beam Normal SSA:

Bn ∼αmeM ∼ 5× 10−6 → 5 ppm

APV ∼Q2

M2Z∼ Q2 × 10−4

Bt ∼Q2

meM ∼ Q2 ×

(5× 10−8

Beam Normal SSA:

Bn ∼αmeM ∼ 5× 10−6 → 5 ppm

APV ∼Q2

M2Z∼ Q2 × 10−4

Bt ∼Q2

meM ∼ Q2 ×

(5× 10−8

What factors contribute to Bn?Pratik Sachdeva Beam Single Spin Asymmetries in Electron-Proton Scattering 17 / 26

Inelastic Hadronic Intermediate State

Inelastic Hadronic Intermediate StateScattering process:

e(p1) + N(p2)→ e(p3) + N(p4)

with intermediate leptonic momentum k and intermediate hadronicmomentum W = p2 + q1.

Inelastic Hadronic Intermediate State M∗γ Mγγ Interference

One- and Two- Photon Interference

We consider the contribution to the normal BSSA from M ∗γMγγ .

For polarized beam, spinor relation becomes

ue(k)ue(k) = (1 + γ5/s)(/k + me).

The BSSA comes from the absorptive part of the two-photonexchange amplitude (Pasquini & Vanderhaeghen PRC 70, 2004):

∑spins

M ∗γ · Abs Mγγ

∑spins|Mγ |2

ue(k)ue(k) = (1 + γ5/s)(/k + me).

∑spins

∑spins|Mγ |2

ue(k)ue(k) = (1 + γ5/s)(/k + me).

∑spins

∑spins|Mγ |2

Inelastic Hadronic Intermediate State Leptonic and Hadronic Tensors

Leptonic and Hadronic TensorsWe can write this as

Bn = e6

(2π)3Q2

(∑spins|Mγ |2

)−1 ∫ d3~kEk

Im {LαµνHαµν} ,

Lαµν = Tr[1

2(1 + γ5/s)( /p1 + me)γα( /p1 − /q + me)γµ( /p1 − /q1 + me)γν].

and we take the forward limit on the hadronic tensor:

Hαµν = 2pα2 W µν

where (de Rujula et al. Nucl. Phys. B35, 1971)

W µν =(−gµν + qµ1 qν1

)W1+ 1

(pµ2 −

p2 · q1q2

)(pν2 −

p2 · q1q2

Bn = e6

(2π)3Q2

(∑spins|Mγ |2

)−1 ∫ d3~kEk

Lαµν = Tr[1

)W1+ 1

(pµ2 −

p2 · q1q2

)(pν2 −

p2 · q1q2

Bn = e6

(2π)3Q2

(∑spins|Mγ |2

)−1 ∫ d3~kEk

Lαµν = Tr[1

)W1+ 1

(pµ2 −

p2 · q1q2

)(pν2 −

p2 · q1q2

Bn = e6

(2π)3Q2

(∑spins|Mγ |2

)−1 ∫ d3~kEk

Lαµν = Tr[1

)W1+ 1

(pµ2 −

p2 · q1q2

)(pν2 −

p2 · q1q2

Inelastic Hadronic Intermediate State Normal BSSA

Finally, we can write the asymmetry as

Bn = − 1(2π)3

D(s,Q2)1

8E1E3M

∫ 2π

M2dW 2

∫ Q21,max

× Im {LαµνHαµν}Q2

1 |~k| (1− cos θ cos θk − sin θ sin θk cosφk)

where E1 and E3 are the energies of the incoming and outgoing leptons, θkis the angle between ~p1 and ~k, φk is the azimuthal angle, and

D(s,Q2) = 8Q4

1− ε

M + ε

Bn = − 1(2π)3

D(s,Q2)1

8E1E3M

∫ 2π

M2dW 2

∫ Q21,max

D(s,Q2) = 8Q4

1− ε

M + ε

Bn = − 1(2π)3

D(s,Q2)1

8E1E3M

∫ 2π

M2dW 2

∫ Q21,max

D(s,Q2) = 8Q4

1− ε

M + ε

Once again,

Bn = − 1(2π)3

D(s,Q2)1

8E1E3M

∫ 2π

M2dW 2

∫ Q21,max

The tensor contraction is

Im LαµνHαµν = 8meM2Q2

[2W1M2Q2

1 (εp1p2qs + εp2qq1s)

+W2εp2qq1s

((M2 − Q2

1 −W 2)(p1 · p2)−M2Q21

→ Currently evaluating for Q-weak kinematics.

Once again,

Bn = − 1(2π)3

D(s,Q2)1

8E1E3M

∫ 2π

M2dW 2

∫ Q21,max

[2W1M2Q2

+W2εp2qq1s

((M2 − Q2

1 −W 2)(p1 · p2)−M2Q21

Once again,

Bn = − 1(2π)3

D(s,Q2)1

8E1E3M

∫ 2π

M2dW 2

∫ Q21,max

[2W1M2Q2

+W2εp2qq1s

((M2 − Q2

1 −W 2)(p1 · p2)−M2Q21

Conclusion

1 Beam single spin asymmetries result from the difference in crosssections produced by the beam polarization in electron-protonscattering experiments.

2 The interference between one- and two-photon exchange amplitudesproduces a beam normal single spin asymmetry.

3 The interference between one-photon and Z exchange amplitudesproduces a beam transverse single spin asymmetry.

4 The combination of these two asymmetries produces a small phaseshift that may be detectable in the future.

5 The inelastic hadronic intermediate state produces an importantcontribution to the BSSA (soon to be quantified).

6 The framework developed here could potentially be used to computeother beam or target SSA at near-forward angles.

Conclusion

References

1 Waidyawansa, B. P. (2013). A 3% Measurement of the Beam NormalSingle Spin Asymmetry in Forward Angle Elastic Electron-ProtonScattering using the Qweak Setup. Ohio University: U.S.A.

2 B. Pasquini and M. Vanderhaeghen, Physical Review C 70, 045206(2004).

3 M. Gorchtein et al., Nucl. Phys. A741, 234 (2004).4 M.L. Goldberger, Y. Nambu, R. Oehme, Ann. Phys. 2 (1957) 226.5 A. De Rujula, J. M. Kaplan, and E. de Rafael, Nucl. Phys. B35, 365

(1971).

Conclusion

Acknowledgements

I would like to thank my mentor, Dr. Wally Melnitchouk, for his guidance,instruction, and patience. I would also like to thank Dr. Peter Blunden forhis aid in completing the calculations for this project.

Conclusion

Questions?

Beam Single Spin Asymmetries in Electron-Proton Scattering · p 2 p 1 p 2+q 1 q 1 p 4 q 2 k p 3...

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