The Proton Radius Puzzle
Gil Paz
Department of Physics and Astronomy, Wayne State University Introduction: The proton radius puzzle
Gil Paz (Wayne State University) The Proton Radius Puzzle 2 Form Factors Matrix element of EM current between nucleon states give rise to two form factors (q = pf pi ) − X iσµν N(p ) e q¯γµq N(p ) =u ¯(p ) γµF (q2)+ F (q2)qν u(p ) f q i f 1 2m 2 i h | q | i Sachs electric and magnetic form factors 2 2 2 q 2 2 2 2 GE (q ) = F1(q ) + 2 F2(q ) GM (q ) = F1(q ) + F2(q ) 4mp p p G (0) = 1 G (0) = µp 2.793 E M ≈ p The slope of GE
dG p r 2 p = 6 E h iE dq2 q2=0 q determines the charge radius r p r 2 p E ≡ h iE The proton magnetic radius p 2 2 p 6 dGM (q ) hr i = p M 2 2 GM (0) dq q =0
Gil Paz (Wayne State University) The Proton Radius Puzzle 3 Charge radius
p The slope of GE
dG p r 2 p = 6 E h iE dq2 q2=0 q determines the charge radius r p r 2 p E ≡ h iE
The proton charge radius is a fundamental physics constant listed by Committee on Data for Science and Technology (CODATA)
Yesterday’s Shanshan Cao’s talk: Charge radii as input to the hadronization of heavy quarks model
Gil Paz (Wayne State University) The Proton Radius Puzzle 4 Charge radius from atomic physics X iσµν p(p ) e q¯γµq p(p ) =u ¯(p ) γµF p(q2)+ F p(q2)qν u(p ) f q i f 1 2m 2 i h | q | i For a point particle amplitude for p + ` p + ` → 1 Zα U(r) = M ∝ q2 ⇒ − r Including q2 corrections from proton structure 1 4πZα q2 = 1 U(r) = δ3(r)(r p)2 M ∝ q2 ⇒ 6 E Proton structure corrections mr = m`mp/(m` + mp) m` ≈ 4 2(Zα) 3 p 2 ∆Er p = mr (r ) δ` 0 E 3n3 E p Muonic hydrogen can give the best measurement of rE!
Gil Paz (Wayne State University) The Proton Radius Puzzle 5 Charge radius from atomic physics
Lamb shift in muonic hydrogen [Pohl et al. Nature 466, 213 (2010)] p rE =0 .84184(67) fm p more recently rE = 0.84087(39) fm [Antognini et al. Science 339, 417 (2013)] CODATA value [Mohr et al. RMP 80, 633 (2008)] p rE = 0.87680(690) fm p more recently rE = 0.87750(510) fm [Mohr et al. RMP 84, 1527 (2012)] extracted mainly from (electronic) hydrogen 5σ discrepancy!
This is the proton radius puzzle
Gil Paz (Wayne State University) The Proton Radius Puzzle 6 1) Problem with the electronic extraction? (Part 1 of this talk)
2) Hadronic Uncertainty? (Part 2 of this talk)
3) New Physics?
What could be the reason for the discrepancy?
What could the reason for the discrepancy?
Gil Paz (Wayne State University) The Proton Radius Puzzle 7 2) Hadronic Uncertainty? (Part 2 of this talk)
3) New Physics?
What could be the reason for the discrepancy?
What could the reason for the discrepancy?
1) Problem with the electronic extraction? (Part 1 of this talk)
Gil Paz (Wayne State University) The Proton Radius Puzzle 7 3) New Physics?
What could be the reason for the discrepancy?
What could the reason for the discrepancy?
1) Problem with the electronic extraction? (Part 1 of this talk)
2) Hadronic Uncertainty? (Part 2 of this talk)
Gil Paz (Wayne State University) The Proton Radius Puzzle 7 What could be the reason for the discrepancy?
What could the reason for the discrepancy?
1) Problem with the electronic extraction? (Part 1 of this talk)
2) Hadronic Uncertainty? (Part 2 of this talk)
3) New Physics?
Gil Paz (Wayne State University) The Proton Radius Puzzle 7 Outline
Introduction: The proton radius puzzle
Part 1: Proton radii from scattering
Part 2: Hadronic Uncertainty?
Backup slides: Connecting µ p scattering and muonic hydrogen −
Conclusions and outlook
Gil Paz (Wayne State University) The Proton Radius Puzzle 8 Part 1: Proton radii from scattering
Gil Paz (Wayne State University) The Proton Radius Puzzle 9 Problem with the electronic extraction?
Recent development: use of the z expansion based on known analytic properties of form factors
The method for meson form factors [Flavor Lattice Averaging Group, EPJ C 74, 2890 (2014)]
p p n Applied successfully for to extract rE , rM , rM , mA...
Gil Paz (Wayne State University) The Proton Radius Puzzle 10 p PDG 2016: rE
[Hill, GP PRD 82 113005 (2010)] [Lee, Arrington, Hill, PRD 92, 013013 (2015)]
Gil Paz (Wayne State University) The Proton Radius Puzzle 11 p PDG 2016: rM
[Epstein, GP, Roy PRD 90, 074027 (2014)] [Lee, Arrington, Hill, PRD 92, 013013 (2015)]
Gil Paz (Wayne State University) The Proton Radius Puzzle 12 n PDG 2016: rM
[Epstein, GP, Roy PRD 90, 074027 (2014)]
Gil Paz (Wayne State University) The Proton Radius Puzzle 13 Neutron and proton magnetic radii
Comparing the proton and neutron magnetic radii r p = 0.87 0.02 fm M ± r n = 0.89 0.03 fm M ±
They are equal within errors...
Proton Magnetic moment/ Neutron Magnetic moment ratio can be explained by SU(6) symmetry or quark model
Is there a reason to the relation between the normalized slopes?
Gil Paz (Wayne State University) The Proton Radius Puzzle 14 Part 2: Hadronic Uncertainty?
[Hill, GP arXiv:1611.09917]
Gil Paz (Wayne State University) The Proton Radius Puzzle 15 The bottom line Scattering: - World e – p data [Lee, Arrington, Hill ’15] p rE = 0.918 0.024 fm - Mainz e – p±data [Lee, Arrington, Hill ’15] r p = 0.895 0.020 fm E ± - Proton, neutron and π data [Hill , GP ’10] r p = 0.871 0.009 0.002 0.002 fm E ± ± ± Muonic hydrogen - [Pohl et al. Nature 466, 213 (2010)] p rE = 0.84184(67) fm - [Antognini et al. Science 339, 417 (2013)] p rE = 0.84087(39) fm
The bottom line: using z expansion scattering disfavors muonic hydrogen
Is there a problem with muonic hydrogen theory? Gil Paz (Wayne State University) The Proton Radius Puzzle 16 Muonic hydrogen theory Is there a problem with muonic hydrogen theory?
Potentially yes! [Hill, GP PRL 107 160402 (2011)]
p Muonic hydrogen measures ∆E and translates it to rE
- [Pohl et al. Nature 466, 213 (2010) Supplementary information] ∆E = 206.0573(45) 5.2262(r p)2 +0 .0347(r p)3 meV − E E - [Antognini et al. Science 339, 417 (2013), Ann. of Phy. 331, 127] ∆E = 206.0336(15) 5.2275(10)(r p)2 +0 .0332(20) meV − E p In both cases apart from rE need two-photon exchange
l l ¡ p p
Gil Paz (Wayne State University) The Proton Radius Puzzle 17 1 X Z W µν = i d 4x eiq·x k, s T Jµ (x)Jν (0) k, s 2 e.m. e.m. s h | { }| i µ ν µ ν µν q q µ k q q ν k q q = g + W1 + k · k · W2 − q2 − q2 − q2
Dispersion relations (ν = 2k q, Q2 = q2) · − 2 Z ∞ 0 2 2 2 ν 02 ImW1(ν , Q ) W1(ν, Q ) = W1(0, Q ) + dν 02 02 2 π 2 2 ν (ν ν ) νcut(Q ) −
Z ∞ 0 2 2 1 02 ImW2(ν , Q ) W2(ν, Q ) = dν 02 2 π 2 2 ν ν νcut(Q ) − W1 requires subtraction...
Two photon exchange p In both cases apart from rE we have two-photon exchange
l l ¡ p p
Gil Paz (Wayne State University) The Proton Radius Puzzle 18 Dispersion relations (ν = 2k q, Q2 = q2) · − 2 Z ∞ 0 2 2 2 ν 02 ImW1(ν , Q ) W1(ν, Q ) = W1(0, Q ) + dν 02 02 2 π 2 2 ν (ν ν ) νcut(Q ) −
Z ∞ 0 2 2 1 02 ImW2(ν , Q ) W2(ν, Q ) = dν 02 2 π 2 2 ν ν νcut(Q ) − W1 requires subtraction...
Two photon exchange p In both cases apart from rE we have two-photon exchange
l l ¡ p p 1 X Z W µν = i d 4x eiq·x k, s T Jµ (x)Jν (0) k, s 2 e.m. e.m. s h | { }| i µ ν µ ν µν q q µ k q q ν k q q = g + W1 + k · k · W2 − q2 − q2 − q2
Gil Paz (Wayne State University) The Proton Radius Puzzle 18 Two photon exchange p In both cases apart from rE we have two-photon exchange
l l ¡ p p 1 X Z W µν = i d 4x eiq·x k, s T Jµ (x)Jν (0) k, s 2 e.m. e.m. s h | { }| i µ ν µ ν µν q q µ k q q ν k q q = g + W1 + k · k · W2 − q2 − q2 − q2
Dispersion relations (ν = 2k q, Q2 = q2) · − 2 Z ∞ 0 2 2 2 ν 02 ImW1(ν , Q ) W1(ν, Q ) = W1(0, Q ) + dν 02 02 2 π 2 2 ν (ν ν ) νcut(Q ) −
Z ∞ 0 2 2 1 02 ImW2(ν , Q ) W2(ν, Q ) = dν 02 2 π 2 2 ν ν νcut(Q ) − W1 requires subtraction... Gil Paz (Wayne State University) The Proton Radius Puzzle 18 Cannot reproduce it from its imaginary part: Dispersion relation requires subtraction
2 Need poorly constrained non-perturbative function W1(0, Q )
Calculable in small Q2 limit using NRQED [Hill, GP, PRL 107 160402 (2011)]
Two photon exchange p In both cases apart from rE we have two-photon exchange
l l ¡ p p
Imaginary part of TPE related to data: form factors, structure functions
Gil Paz (Wayne State University) The Proton Radius Puzzle 19 Two photon exchange p In both cases apart from rE we have two-photon exchange
l l ¡ p p
Imaginary part of TPE related to data: form factors, structure functions
Cannot reproduce it from its imaginary part: Dispersion relation requires subtraction
2 Need poorly constrained non-perturbative function W1(0, Q )
Calculable in small Q2 limit using NRQED [Hill, GP, PRL 107 160402 (2011)]
Gil Paz (Wayne State University) The Proton Radius Puzzle 19 2 3 ¯ 2 Q 2mpβ 2 2 2 p 2 2 p 2 4 W1(0, Q ) = 2ap(2+ap)+ 2 ap (1+ap) mp(rM ) mp(rE ) + Q mp α − −3 − O 4 3 - ap = 1.793, β¯ = 2.5(4) 10 fm × − - rM = 0.776(34)(17) fm, H µH - rE = 0.8751(61) fm or rE = 0.84087(26)(29) fm 2 2 Q 4 W1(0, Q ) = 13.6 + 2 ( 54 7) + Q mp − ± O 15
10
5 )
2 0 ( 0, Q 1 -5 W
-10
-15
-20 0.0 0.2 0.4 0.6 0.8 1.0 Q2(GeV2)
Two Photon exchange: small Q2 limit Small Q2 limit using NRQED [Hill, GP, PRL 107 160402 (2011)] The photon sees the proton “almost” like an elementary particle
Gil Paz (Wayne State University) The Proton Radius Puzzle 20 4 3 - ap = 1.793, β¯ = 2.5(4) 10 fm × − - rM = 0.776(34)(17) fm, H µH - rE = 0.8751(61) fm or rE = 0.84087(26)(29) fm 2 2 Q 4 W1(0, Q ) = 13.6 + 2 ( 54 7) + Q mp − ± O 15
10
5 )
2 0 ( 0, Q 1 -5 W
-10
-15
-20 0.0 0.2 0.4 0.6 0.8 1.0 Q2(GeV2)
Two Photon exchange: small Q2 limit Small Q2 limit using NRQED [Hill, GP, PRL 107 160402 (2011)] The photon sees the proton “almost” like an elementary particle 2 3 ¯ 2 Q 2mpβ 2 2 2 p 2 2 p 2 4 W1(0, Q ) = 2ap(2+ap)+ 2 ap (1+ap) mp(rM ) mp(rE ) + Q mp α − −3 − O
Gil Paz (Wayne State University) The Proton Radius Puzzle 20 15
10
5 )
2 0 ( 0, Q 1 -5 W
-10
-15
-20 0.0 0.2 0.4 0.6 0.8 1.0 Q2(GeV2)
Two Photon exchange: small Q2 limit Small Q2 limit using NRQED [Hill, GP, PRL 107 160402 (2011)] The photon sees the proton “almost” like an elementary particle 2 3 ¯ 2 Q 2mpβ 2 2 2 p 2 2 p 2 4 W1(0, Q ) = 2ap(2+ap)+ 2 ap (1+ap) mp(rM ) mp(rE ) + Q mp α − −3 − O 4 3 - ap = 1.793, β¯ = 2.5(4) 10 fm × − - rM = 0.776(34)(17) fm, H µH - rE = 0.8751(61) fm or rE = 0.84087(26)(29) fm 2 2 Q 4 W1(0, Q ) = 13.6 + 2 ( 54 7) + Q mp − ± O
Gil Paz (Wayne State University) The Proton Radius Puzzle 20 Two Photon exchange: small Q2 limit Small Q2 limit using NRQED [Hill, GP, PRL 107 160402 (2011)] The photon sees the proton “almost” like an elementary particle 2 3 ¯ 2 Q 2mpβ 2 2 2 p 2 2 p 2 4 W1(0, Q ) = 2ap(2+ap)+ 2 ap (1+ap) mp(rM ) mp(rE ) + Q mp α − −3 − O 4 3 - ap = 1.793, β¯ = 2.5(4) 10 fm × − - rM = 0.776(34)(17) fm, H µH - rE = 0.8751(61) fm or rE = 0.84087(26)(29) fm 2 2 Q 4 W1(0, Q ) = 13.6 + 2 ( 54 7) + Q mp − ± O 15
10
5 )
2 0 ( 0, Q 1 -5 W
-10
-15
-20 0.0 0.2 0.4 0.6 0.8 1.0 Q2(GeV2) Gil Paz (Wayne State University) The Proton Radius Puzzle 20 Was it?
Two Photon Exchange: large Q2 limit
l l
p p 2 Need poorly constrained non-perturbative function W1(0, Q ) Calculable in large Q2 limit using Operator Product Expansion (OPE) [J. C. Collins, NPB 149, 90 (1979)] The photon “sees” the quarks and gluons inside the proton 2 2 4 W1(0, Q ) = c/Q + 1/Q O Result was used to estimate two photon exchange effects c calculated in [J. C. Collins, NPB 149, 90 (1979)]
Gil Paz (Wayne State University) The Proton Radius Puzzle 21 Two Photon Exchange: large Q2 limit
l l
p p 2 Need poorly constrained non-perturbative function W1(0, Q ) Calculable in large Q2 limit using Operator Product Expansion (OPE) [J. C. Collins, NPB 149, 90 (1979)] The photon “sees” the quarks and gluons inside the proton 2 2 4 W1(0, Q ) = c/Q + 1/Q O Result was used to estimate two photon exchange effects c calculated in [J. C. Collins, NPB 149, 90 (1979)]
Was it?
Gil Paz (Wayne State University) The Proton Radius Puzzle 21 Two Photon Exchange: large Q2 limit
l l
p p
Z µν 1 X 4 iq·x µ ν W = i d x e hk, s|T {J (x)J (0)}|k, si 2 e.m. e.m. s qµqν k · q qµ k · q qν = −g µν + W + kµ − kν − W q2 1 q2 q2 2 2 W1(0, Q ) is dimensionless Proton O Proton 1 W1 h | | i + ∼ Q2 O Q4 O is a dimension 4 operator: µ ν ν µ 1 µν - Quarks: Spin 0: mqqq¯ Spin 2:q ¯ iD γ + iD γ iD/ g q − 4 αβ ρσ - Gluons: must be color singlet: Ga Ga - What gluon operators can we have?
Gil Paz (Wayne State University) The Proton Radius Puzzle 22 µν 2 2 - 1 scalar: G Gµν = 2(B~ E~ ) αβ− ρσ - 1 pseudo scalar: αβρσG G = E B: ruled out by parity · µα ν 1 αβ µν - 9 components of traceless symmetric tensor: G G G Gαβg α − 4 chromomagnetic stress-energy tensor - What else? 10 components of
µανβ 1 µαρσ νβκλ µβρσ νακλ O = + GρκGσλ all possible traces −4 − For example O0123 = G 01G 23 + G 03G 21 = E 1B1 E 3B3 For protons: Proton Oµανβ Proton = 0 − What about hMedium| Oµανβ| Mediumi ? Solution lookingh for a| problem...| i
Gluon operators
αβ ρσ Gluons: must be color singlet Ga Ga A product of (E i , Bi ) and (E j , Bj ) has 7 6/2 = 21 components: ×
Gil Paz (Wayne State University) The Proton Radius Puzzle 23 αβ ρσ - 1 pseudo scalar: αβρσG G = E B: ruled out by parity · µα ν 1 αβ µν - 9 components of traceless symmetric tensor: G G G Gαβg α − 4 chromomagnetic stress-energy tensor - What else? 10 components of
µανβ 1 µαρσ νβκλ µβρσ νακλ O = + GρκGσλ all possible traces −4 − For example O0123 = G 01G 23 + G 03G 21 = E 1B1 E 3B3 For protons: Proton Oµανβ Proton = 0 − What about hMedium| Oµανβ| Mediumi ? Solution lookingh for a| problem...| i
Gluon operators
αβ ρσ Gluons: must be color singlet Ga Ga A product of (E i , Bi ) and (E j , Bj ) has 7 6/2 = 21 components: µν 2 2 × - 1 scalar: G Gµν = 2(B~ E~ ) −
Gil Paz (Wayne State University) The Proton Radius Puzzle 23 µα ν 1 αβ µν - 9 components of traceless symmetric tensor: G G G Gαβg α − 4 chromomagnetic stress-energy tensor - What else? 10 components of
µανβ 1 µαρσ νβκλ µβρσ νακλ O = + GρκGσλ all possible traces −4 − For example O0123 = G 01G 23 + G 03G 21 = E 1B1 E 3B3 For protons: Proton Oµανβ Proton = 0 − What about hMedium| Oµανβ| Mediumi ? Solution lookingh for a| problem...| i
Gluon operators
αβ ρσ Gluons: must be color singlet Ga Ga A product of (E i , Bi ) and (E j , Bj ) has 7 6/2 = 21 components: µν 2 2 × - 1 scalar: G Gµν = 2(B~ E~ ) αβ− ρσ - 1 pseudo scalar: αβρσG G = E B: ruled out by parity ·
Gil Paz (Wayne State University) The Proton Radius Puzzle 23 10 components of
µανβ 1 µαρσ νβκλ µβρσ νακλ O = + GρκGσλ all possible traces −4 − For example O0123 = G 01G 23 + G 03G 21 = E 1B1 E 3B3 For protons: Proton Oµανβ Proton = 0 − What about hMedium| Oµανβ| Mediumi ? Solution lookingh for a| problem...| i
Gluon operators
αβ ρσ Gluons: must be color singlet Ga Ga A product of (E i , Bi ) and (E j , Bj ) has 7 6/2 = 21 components: µν 2 2 × - 1 scalar: G Gµν = 2(B~ E~ ) αβ− ρσ - 1 pseudo scalar: αβρσG G = E B: ruled out by parity · µα ν 1 αβ µν - 9 components of traceless symmetric tensor: G G G Gαβg α − 4 chromomagnetic stress-energy tensor - What else?
Gil Paz (Wayne State University) The Proton Radius Puzzle 23 µανβ 1 µαρσ νβκλ µβρσ νακλ O = + GρκGσλ all possible traces −4 − For example O0123 = G 01G 23 + G 03G 21 = E 1B1 E 3B3 For protons: Proton Oµανβ Proton = 0 − What about hMedium| Oµανβ| Mediumi ? Solution lookingh for a| problem...| i
Gluon operators
αβ ρσ Gluons: must be color singlet Ga Ga A product of (E i , Bi ) and (E j , Bj ) has 7 6/2 = 21 components: µν 2 2 × - 1 scalar: G Gµν = 2(B~ E~ ) αβ− ρσ - 1 pseudo scalar: αβρσG G = E B: ruled out by parity · µα ν 1 αβ µν - 9 components of traceless symmetric tensor: G G G Gαβg α − 4 chromomagnetic stress-energy tensor - What else? 10 components of
Gil Paz (Wayne State University) The Proton Radius Puzzle 23 For protons: Proton Oµανβ Proton = 0 What about hMedium| Oµανβ| Mediumi ? Solution lookingh for a| problem...| i
Gluon operators
αβ ρσ Gluons: must be color singlet Ga Ga A product of (E i , Bi ) and (E j , Bj ) has 7 6/2 = 21 components: µν 2 2 × - 1 scalar: G Gµν = 2(B~ E~ ) αβ− ρσ - 1 pseudo scalar: αβρσG G = E B: ruled out by parity · µα ν 1 αβ µν - 9 components of traceless symmetric tensor: G G G Gαβg α − 4 chromomagnetic stress-energy tensor - What else? 10 components of
µανβ 1 µαρσ νβκλ µβρσ νακλ O = + GρκGσλ all possible traces −4 − For example O0123 = G 01G 23 + G 03G 21 = E 1B1 E 3B3 −
Gil Paz (Wayne State University) The Proton Radius Puzzle 23 Gluon operators
αβ ρσ Gluons: must be color singlet Ga Ga A product of (E i , Bi ) and (E j , Bj ) has 7 6/2 = 21 components: µν 2 2 × - 1 scalar: G Gµν = 2(B~ E~ ) αβ− ρσ - 1 pseudo scalar: αβρσG G = E B: ruled out by parity · µα ν 1 αβ µν - 9 components of traceless symmetric tensor: G G G Gαβg α − 4 chromomagnetic stress-energy tensor - What else? 10 components of
µανβ 1 µαρσ νβκλ µβρσ νακλ O = + GρκGσλ all possible traces −4 − For example O0123 = G 01G 23 + G 03G 21 = E 1B1 E 3B3 For protons: Proton Oµανβ Proton = 0 − What about hMedium| Oµανβ| Mediumi ? Solution lookingh for a| problem...| i
Gil Paz (Wayne State University) The Proton Radius Puzzle 23 Summary: Possible operators In total we have four operators with non-zero proton matrix elements.
Quarks:
- Spin 0: mqqq¯ 1 - Spin 2:q ¯ iDµγν + iDνγµ iD/ g µνq − 4
Gluons: µν - Spin 0: G Gµν µα ν 1 αβ µν - Spin 2: G G G Gαβg α − 4
Gil Paz (Wayne State University) The Proton Radius Puzzle 24 Summary: Possible operators In total we have four operators with non-zero proton matrix elements.
Quarks:
- Spin 0: mqqq¯ 1 - Spin 2:q ¯ iDµγν + iDνγµ iD/ g µνq − 4
Gluons: µν - Spin 0: G Gµν µα ν 1 αβ µν - Spin 2: G G G Gαβg α − 4
Gil Paz (Wayne State University) The Proton Radius Puzzle 24 2 For W1(0, Q ) you need also spin-2 operators
µ ν ν µ 1 µν µα ν 1 αβ µν P q¯ iD γ +iD γ iD/ g q P , P G G G Gαβg P h | −4 | i h | α −4 | i
Need to calculate the spin-2 contribution [Hill, GP arXiv:1611.09917]
Quark Gluon Spin-0 Collins ’78 Collins ’78 Spin-2 Hill, GP ’16 Hill, GP ’16
Collins’s result is not enough for muonic hydrogen!
Large Q2 behavior In 1978 Collins calculated EM corrections to the nucleon mass with an emphasis on mn mp The mass only depends on− spin-0 operators (q quark, G µν gluon) µν P mqqq¯ P , P G Gµν P h | | i h | | i Quark Gluon Spin-0 Collins ’78 Collins ’78
Gil Paz (Wayne State University) The Proton Radius Puzzle 25 Need to calculate the spin-2 contribution [Hill, GP arXiv:1611.09917]
Quark Gluon Spin-0 Collins ’78 Collins ’78 Spin-2 Hill, GP ’16 Hill, GP ’16
Collins’s result is not enough for muonic hydrogen!
Large Q2 behavior In 1978 Collins calculated EM corrections to the nucleon mass with an emphasis on mn mp The mass only depends on− spin-0 operators (q quark, G µν gluon) µν P mqqq¯ P , P G Gµν P h | | i h | | i Quark Gluon Spin-0 Collins ’78 Collins ’78 2 For W1(0, Q ) you need also spin-2 operators
µ ν ν µ 1 µν µα ν 1 αβ µν P q¯ iD γ +iD γ iD/ g q P , P G G G Gαβg P h | −4 | i h | α −4 | i
Gil Paz (Wayne State University) The Proton Radius Puzzle 25 Large Q2 behavior In 1978 Collins calculated EM corrections to the nucleon mass with an emphasis on mn mp The mass only depends on− spin-0 operators (q quark, G µν gluon) µν P mqqq¯ P , P G Gµν P h | | i h | | i Quark Gluon Spin-0 Collins ’78 Collins ’78 2 For W1(0, Q ) you need also spin-2 operators
µ ν ν µ 1 µν µα ν 1 αβ µν P q¯ iD γ +iD γ iD/ g q P , P G G G Gαβg P h | −4 | i h | α −4 | i
Need to calculate the spin-2 contribution [Hill, GP arXiv:1611.09917]
Quark Gluon Spin-0 Collins ’78 Collins ’78 Spin-2 Hill, GP ’16 Hill, GP ’16
Collins’s result is not enough for muonic hydrogen! Gil Paz (Wayne State University) The Proton Radius Puzzle 25 Doing that, we found a mistake in Collins spin-0 calculation from 1978...
Collins didn’t calculate the spin-0 gluon contribution directly He extracted it from another calculation
For three light quark u, d, s P 2 2 2 1 2 1 2 2 Correct result: q eq = ( 3 ) + ( 3 ) + ( 3 ) = 3 P Collins: q = 3 Too large by a factor of 4.5...
Large Q2 behavior Requires 1-loop calculation
Gil Paz (Wayne State University) The Proton Radius Puzzle 26 Collins didn’t calculate the spin-0 gluon contribution directly He extracted it from another calculation
For three light quark u, d, s P 2 2 2 1 2 1 2 2 Correct result: q eq = ( 3 ) + ( 3 ) + ( 3 ) = 3 P Collins: q = 3 Too large by a factor of 4.5...
Large Q2 behavior Requires 1-loop calculation
Doing that, we found a mistake in Collins spin-0 calculation from 1978...
Gil Paz (Wayne State University) The Proton Radius Puzzle 26 For three light quark u, d, s P 2 2 2 1 2 1 2 2 Correct result: q eq = ( 3 ) + ( 3 ) + ( 3 ) = 3 P Collins: q = 3 Too large by a factor of 4.5...
Large Q2 behavior Requires 1-loop calculation
Doing that, we found a mistake in Collins spin-0 calculation from 1978...
Collins didn’t calculate the spin-0 gluon contribution directly He extracted it from another calculation
Gil Paz (Wayne State University) The Proton Radius Puzzle 26 Large Q2 behavior Requires 1-loop calculation
Doing that, we found a mistake in Collins spin-0 calculation from 1978...
Collins didn’t calculate the spin-0 gluon contribution directly He extracted it from another calculation
For three light quark u, d, s P 2 2 2 1 2 1 2 2 Correct result: q eq = ( 3 ) + ( 3 ) + ( 3 ) = 3 P Collins: q = 3 Too large by a factor of 4.5...
Gil Paz (Wayne State University) The Proton Radius Puzzle 26 Even worse, quark spin-0 and gluon spin-0 come with opposite signs After correcting the mistake they largely cancel 2 W1(0, Q ) is dominated by spin-2 contribution
Lesson: It is important to do a full calculation
Some good news: The mistake has no effect on mn mp − since gluon contribution is the same at lowest order in isospin breaking
Flip side: You cannot use mn mp to constrain muonic hydrogen −
Large Q2 behavior Quark Gluon Spin-0 Collins ’78 ————–Collins’78 Hill, GP ’16 Spin-2 Hill, GP ’16 Hill, GP ’16
Gil Paz (Wayne State University) The Proton Radius Puzzle 27 Some good news: The mistake has no effect on mn mp − since gluon contribution is the same at lowest order in isospin breaking
Flip side: You cannot use mn mp to constrain muonic hydrogen −
Large Q2 behavior Quark Gluon Spin-0 Collins ’78 ————–Collins’78 Hill, GP ’16 Spin-2 Hill, GP ’16 Hill, GP ’16
Even worse, quark spin-0 and gluon spin-0 come with opposite signs After correcting the mistake they largely cancel 2 W1(0, Q ) is dominated by spin-2 contribution
Lesson: It is important to do a full calculation
Gil Paz (Wayne State University) The Proton Radius Puzzle 27 Flip side: You cannot use mn mp to constrain muonic hydrogen −
Large Q2 behavior Quark Gluon Spin-0 Collins ’78 ————–Collins’78 Hill, GP ’16 Spin-2 Hill, GP ’16 Hill, GP ’16
Even worse, quark spin-0 and gluon spin-0 come with opposite signs After correcting the mistake they largely cancel 2 W1(0, Q ) is dominated by spin-2 contribution
Lesson: It is important to do a full calculation
Some good news: The mistake has no effect on mn mp − since gluon contribution is the same at lowest order in isospin breaking
Gil Paz (Wayne State University) The Proton Radius Puzzle 27 Large Q2 behavior Quark Gluon Spin-0 Collins ’78 ————–Collins’78 Hill, GP ’16 Spin-2 Hill, GP ’16 Hill, GP ’16
Even worse, quark spin-0 and gluon spin-0 come with opposite signs After correcting the mistake they largely cancel 2 W1(0, Q ) is dominated by spin-2 contribution
Lesson: It is important to do a full calculation
Some good news: The mistake has no effect on mn mp − since gluon contribution is the same at lowest order in isospin breaking
Flip side: You cannot use mn mp to constrain muonic hydrogen −
Gil Paz (Wayne State University) The Proton Radius Puzzle 27 Large Q2 behavior: Spin 0 contribution The correct spin 0 result 2 ! Q (spin 0) 2 X 2 (0) X 2 αs (0) W − (0, Q ) = 2 e f + e f˜ 2m2 1 q q q 12 g p − q q π quark and gluon matrix elements related by
X (0) β (0) 1 = (1 γ ) f f˜ m q 2g g − q − 0.1 2 p - Dashed blue: quark 0.05 )/ 2 m
2 - Dash-dotted red: gluon 0
(0, Q - Vertical stripes: perturbative −0.05 uncertainty Q/2 < µ < 2Q (spin-0) 1 −0.1 - Solid bands: hadronic uncertainties W 2
Q on matrix elements −0.15 0 2 4 6 8 10 Q2 (GeV2)
Gil Paz (Wayne State University) The Proton Radius Puzzle 28 Large Q2 behavior: Spin 2 contribution The new spin 2 result 2 ! 2 Q (spin−2) X X αs 5 4 Q W (0, Q2) = 2 e2f (2)(µ) + e2 + log f (2)(µ) 2m2 1 q q q 4 3 3 2 g p q q π − µ quark and gluon matrix elements related by X (2) (2) fq (µ) + fg (µ) = 1 q 0.8 2 p - Dashed blue: down quark 0.6 - Dash-dotted blue: up quark )/ 2 m 2
0.4 - Red: Gluon contribution (0, Q - Vertical stripes: perturbative 0.2 uncertainty Q/2 < µ < 2Q (spin-2) 1 W 2 0 - Solid bands: hadronic uncertainty Q 0 2 4 6 8 10 Q2 (GeV2)
Gil Paz (Wayne State University) The Proton Radius Puzzle 29 Large Q2 behavior: Total contribution The total contribution
0.8 2 p 0.6 )/ 2 m 2 0.4 (0, Q 1 0.2 W 2
Q 0
0 2 4 6 8 10 Q2 (GeV2)
- Dashed red: spin 0 - Dashed blue: spin 2 - Vertical stripes: total contribution with perturbative and hadronic errors added in quadrature
Gil Paz (Wayne State University) The Proton Radius Puzzle 30 Two Photon exchange: small Q2 and large Q2 Using NRQED we have control over low Q2
2 3 ¯ 2 Q 2mpβ 2 2 2 p 2 2 p 2 4 W1(0, Q ) = 2ap(2+ap)+ 2 ap (1+ap) mp(rM ) mp(rE ) + Q mp α − −3 − O Using OPE we have control over the high Q2
2 ! Q (spin−0) X X αs W (0, Q2) = 2 e2f (0) + e2 f˜(0) 2m2 1 q q q 12 g p − q q π 2 ! 2 Q (spin−2) X X αs 5 4 Q W (0, Q2) = 2 e2f (2)(µ) + e2 + log f (2)(µ) 2m2 1 q q q 4 3 3 2 g p q q π − µ
The problem, like the joke, is how to make a whole fish from a head and a tail...
Before this work we had only the low Q2 knowing the large Q2 allows to connect the dots
Gil Paz (Wayne State University) The Proton Radius Puzzle 31 How to connect the curves? 15
10
5 )
2 0 ( 0, Q 1 -5 W
-10
-15
-20 0.0 0.5 1.0 1.5 2.0 Q2(GeV2)
Two Photon Exchange: Modeling “Aggressive” modeling: use OPE for Q2 1 GeV2 4 2 2≥ 2 - Model unknown Q : add ∆L(Q ) = Q /ΛL with ΛL 500 MeV 4 2 ± 2 2 ≈ - Model unknown 1/Q : add ∆H (Q ) = Λ /Q with ΛH 500 MeV ± H ≈
Gil Paz (Wayne State University) The Proton Radius Puzzle 32 Two Photon Exchange: Modeling “Aggressive” modeling: use OPE for Q2 1 GeV2 4 2 2≥ 2 - Model unknown Q : add ∆L(Q ) = Q /ΛL with ΛL 500 MeV 4 2 ± 2 2 ≈ - Model unknown 1/Q : add ∆H (Q ) = ΛH /Q with ΛH 500 MeV How to connect the curves? ± ≈ 15
10
5 )
2 0 ( 0, Q 1 -5 W
-10
-15
-20 0.0 0.5 1.0 1.5 2.0 Q2(GeV2)
Gil Paz (Wayne State University) The Proton Radius Puzzle 32 10 ) 2 0 (0, Q 1
W −10
−20 0 0.5 1 1.5 2 Q2 (GeV2)
2 Energy contribution: δE(2S)W1(0,Q ) [ 0.046 meV, 0.021 meV] ∈ − − To explain the puzzle need this to be 0.3 meV Caveats: OPE might be only valid for∼ larger Q2 2 W1(0, Q ) might be different than the interpolated lines
Two Photon Exchange: Modeling “Aggressive” modeling: use OPE for Q2 1 GeV2 4 2 2≤ 2 - Model unknown Q : add ∆L(Q ) = Q /ΛL with ΛL 500 MeV 4 2 ± 2 2 ≈ - Model unknown 1/Q : add ∆H (Q ) = ΛH /Q with ΛH 500 MeV Interpolating: ± ≈
Gil Paz (Wayne State University) The Proton Radius Puzzle 33 2 Energy contribution: δE(2S)W1(0,Q ) [ 0.046 meV, 0.021 meV] ∈ − − To explain the puzzle need this to be 0.3 meV Caveats: OPE might be only valid for∼ larger Q2 2 W1(0, Q ) might be different than the interpolated lines
Two Photon Exchange: Modeling “Aggressive” modeling: use OPE for Q2 1 GeV2 4 2 2≤ 2 - Model unknown Q : add ∆L(Q ) = Q /ΛL with ΛL 500 MeV 4 2 ± 2 2 ≈ - Model unknown 1/Q : add ∆H (Q ) = ΛH /Q with ΛH 500 MeV Interpolating: ± ≈
10 ) 2 0 (0, Q 1
W −10
−20 0 0.5 1 1.5 2 Q2 (GeV2)
Gil Paz (Wayne State University) The Proton Radius Puzzle 33 Two Photon Exchange: Modeling “Aggressive” modeling: use OPE for Q2 1 GeV2 4 2 2≤ 2 - Model unknown Q : add ∆L(Q ) = Q /ΛL with ΛL 500 MeV 4 2 ± 2 2 ≈ - Model unknown 1/Q : add ∆H (Q ) = ΛH /Q with ΛH 500 MeV Interpolating: ± ≈
10 ) 2 0 (0, Q 1
W −10
−20 0 0.5 1 1.5 2 Q2 (GeV2)
2 Energy contribution: δE(2S)W1(0,Q ) [ 0.046 meV, 0.021 meV] ∈ − − To explain the puzzle need this to be 0.3 meV Caveats: OPE might be only valid for∼ larger Q2 2 W1(0, Q ) might be different than the interpolated lines Gil Paz (Wayne State University) The Proton Radius Puzzle 33 Experimental test How to test? New experiment: µ p scattering MUSE (MUon proton− Scattering Experiment) at PSI [R. Gilman et al. (MUSE Collaboration), arXiv:1303.2160]
Need to connect muon-proton scattering and muonic hydrogen can use a new effective field theory: QED-NRQED [Hill, Lee, GP, Mikhail P. Solon, PRD 87 053017 (2013)] [Steven P. Dye, Matthew Gonderinger, GP, PRD 94 013006 (2016)]
Gil Paz (Wayne State University) The Proton Radius Puzzle 34 Conclusions
Gil Paz (Wayne State University) The Proton Radius Puzzle 35 Conclusions Proton radius puzzle: > 5σ discrepancy between p - rE from muonic hydrogen - r p from hydrogen and e p scattering E −
Recent muonic deuterium results find similar discrepancies [Pohl et al. Science 353, 669 (2016)]
After 6 years the origin is still not clear 1) Is it a problem with the electronic extraction? 2) Is it a hadronic uncertainty? 3) is it new physics?
Motivates a reevaluation of our understanding of the proton
Gil Paz (Wayne State University) The Proton Radius Puzzle 36 1) Extraction of proton radii from scattering: Use an established tool of the z expansion Studies disfavor the muonic hydrogen value
2) The first full and correct evaluation of large Q2 behavior of forward virtual Compton tensor Can improve the modeling of two photon exchange effects
3) Direct connection between muon-proton scattering and muonic hydrogen using a new effective field theory: QED-NRQED
Much more work to do!
Thank you
Conclusions Discussed three topics:
Gil Paz (Wayne State University) The Proton Radius Puzzle 37 2) The first full and correct evaluation of large Q2 behavior of forward virtual Compton tensor Can improve the modeling of two photon exchange effects
3) Direct connection between muon-proton scattering and muonic hydrogen using a new effective field theory: QED-NRQED
Much more work to do!
Thank you
Conclusions Discussed three topics:
1) Extraction of proton radii from scattering: Use an established tool of the z expansion Studies disfavor the muonic hydrogen value
Gil Paz (Wayne State University) The Proton Radius Puzzle 37 3) Direct connection between muon-proton scattering and muonic hydrogen using a new effective field theory: QED-NRQED
Much more work to do!
Thank you
Conclusions Discussed three topics:
1) Extraction of proton radii from scattering: Use an established tool of the z expansion Studies disfavor the muonic hydrogen value
2) The first full and correct evaluation of large Q2 behavior of forward virtual Compton tensor Can improve the modeling of two photon exchange effects
Gil Paz (Wayne State University) The Proton Radius Puzzle 37 Much more work to do!
Thank you
Conclusions Discussed three topics:
1) Extraction of proton radii from scattering: Use an established tool of the z expansion Studies disfavor the muonic hydrogen value
2) The first full and correct evaluation of large Q2 behavior of forward virtual Compton tensor Can improve the modeling of two photon exchange effects
3) Direct connection between muon-proton scattering and muonic hydrogen using a new effective field theory: QED-NRQED
Gil Paz (Wayne State University) The Proton Radius Puzzle 37 Conclusions Discussed three topics:
1) Extraction of proton radii from scattering: Use an established tool of the z expansion Studies disfavor the muonic hydrogen value
2) The first full and correct evaluation of large Q2 behavior of forward virtual Compton tensor Can improve the modeling of two photon exchange effects
3) Direct connection between muon-proton scattering and muonic hydrogen using a new effective field theory: QED-NRQED
Much more work to do!
Thank you
Gil Paz (Wayne State University) The Proton Radius Puzzle 37 Gluon operators
αβ ρσ Gluons: must be color singlet Ga Ga A product of (E i , Bi ) and (E j , Bj ) has 7 6/2 = 21 components: µν 2 2 × - 1 scalar: G Gµν = 2(B~ E~ ) αβ− ρσ - 1 pseudo scalar: αβρσG G = E B: ruled out by parity · µα ν 1 αβ µν - 9 components of traceless symmetric tensor: G G G Gαβg α − 4 chromomagnetic stress-energy tensor - What else? 10 components of
µανβ 1 µαρσ νβκλ µβρσ νακλ O = + GρκGσλ all possible traces −4 − For example O0123 = G 01G 23 + G 03G 21 = E 1B1 E 3B3 For protons: Proton Oµανβ Proton = 0 − What about hMedium| Oµανβ| Mediumi ? Solution lookingh for a| problem...| i
Gil Paz (Wayne State University) The Proton Radius Puzzle 38 Backup: Connecting muon-proton scattering and muonic hydrogen
Gil Paz (Wayne State University) The Proton Radius Puzzle 39 MUSE Muonic hydrogen:
Muon momentum mµcα 1 MeV ∼ ∼ Both proton and muon non-relativistic
MUSE:
Muon momentum mµ 100 MeV ∼ ∼ Muon is relativistic, proton is still non-relativistic
QED-NRQED effective theory: - Use QED for muon - Use NRQED for proton
mµ/mp 0.1 as expansion parameter ∼
A new effective field theory suggested in [Hill, Lee, GP, Mikhail P. Solon, PRD 87 053017 (2013)]
Gil Paz (Wayne State University) The Proton Radius Puzzle 40 q q!
p p!
QED-NRQED Effective Theoryq p! q q! q q!
= p p! + p p! + Example: TPE at the lowest order in 1/mp p q! [Steven P. Dye, Matthew Gonderinger, GP, PRD 94 013006 (2016)]
Consider muon-protonFigure 2: Diagrammatic scattering representationµ( ofp the) matching + p of( thek amplitu) de forµ Comp-(p ) + p(k ) ton scattering obtained in the full theory and NRQED, to leading order0 0 in e.Theblackverticesinthediagramsontheright-handsidere0 0 0 →0 present - At lowest order in 1/minsertionsp: p of NRQED= p 1-photon0 operators.δ(p p0 ) ⇒ 0− - At the proton rest frame k = (mp,~0) k = 0 in NRQED 1 ⇒ NRQED propagator: l 0 l~2/2M + i − p L p! p L p!
L p L p! − −
k k L + p k kk+ L p! k − ! − ! Figure 3: Feynman diagrams for two-photon exchange with momentum labels. 1 1 + δ(L0 p0) p0 L0 + i L0 p0 + i ⇒ − − − 2 In total 0 0 0 0 0 0 0 0 δ(p p0 ) δ(L p ) = δ(L p ) δ(L p0 ) − − − − Gil Paz (Wayne State University) The Proton Radius Puzzle 41 q q!
p p!
q p! q q! q q!
p p! p p! = + + p q!
Figure 2: Diagrammatic representation of the matching of the amplitude for Comp- ton scattering obtained in the full theory and NRQED, to leading order in e.Theblackverticesinthediagramsontheright-handsiderepresent insertions of NRQED 1-photon operators.
QED-NRQED Effective Theory
p L p! p L p!
L p L p! − −
k k L + p k kk+ L p! k − ! − ! The amplitude Figure 3: Feynman diagrams for two-photon exchange with momentum labels. Z 4 4 4 2 4 d L 1 i (2π) δ (k + p k0 p0)2 = Z e M − − (2π)4 (L p)2L p )2 − − 0 0 i 0 0 0 0 0 u¯(p0)γ γ u(p)χ† χ(2π)δ(L p ) (2π)δ(L p0 ) × L/ m − − 3 4 − (2π) δ (~p ~p0 ~k0) × − −
The cross section 2 2 2 2 2 θ " θ θ # dσ Z α 4E 1 v sin Zαπv sin 1 sin = − 2 1 + 2 − 2 dΩ ~q 4 1 v 2 sin2 θ − Z = 1, E = muon energy, v = ~p /E, q = p p, θ scattering angle | | 0 − Gil Paz (Wayne State University) The Proton Radius Puzzle 42 QED-NRQED Effective Theory
QED-NRQED result
2 2 2 2 2 θ " θ θ # dσ Z α 4E 1 v sin Zαπv sin 1 sin = − 2 1 + 2 − 2 dΩ ~q 4 1 v 2 sin2 θ −
Same result as scattering relativistic lepton off static 1/r potential [Dalitz, Proc. Roy. Soc. Lond. 206, 509 (1951)] reproduced in [Itzykson, Zuber, “Quantum Field Theory”]
Same result as mp of “point particle proton” QED scattering → ∞ (For mp only proton charge is relevant) → ∞
Gil Paz (Wayne State University) The Proton Radius Puzzle 43 QED-NRQED Effective Theory beyond mp limit → ∞
QED-NRQED allows to calculate 1/mp corrections
Example: one photon exchange µ + p µ + p: → QED-NRQED = 1/mp expansion of form factors [Steven P. Dye, Matthew Gonderinger, GP, PRD 94 013006 (2016)]
Gil Paz (Wayne State University) The Proton Radius Puzzle 44 Connecting muon-proton scattering to muonic hydrogen
Matching
2 QED, QCD GE,M , Structure func., W1(0, Q )
Scale: mp 1 GeV ∼ ⇓ p 0 QED-NRQED: MUSE rE ,µγ ¯ µψp† ψp
Scale: mµ 0.1 GeV ∼ ⇓ p NRQED-NRQED: muonic H rE , ψµ† ψµψp† ψp
0 Need to match QED-NRQED contact interaction, e.g.µγ ¯ µψp† ψp
to NRQED-NRQED contact interaction, e.g. ψµ† ψµψp† ψp [Dye, Gonderinger, GP in progress]
Gil Paz (Wayne State University) The Proton Radius Puzzle 45 Connecting muon-proton scattering to muonic hydrogen
To do list:
1) Relate QED-NRQED contact interactions 2 to NRQED contact interactions and W1(0, Q )
p 2) Calculate dσ(µ + p µ + p) and asymmetry in terms of r and d2 → E
3) Direct relation between µ–p scattering and muonic H
Gil Paz (Wayne State University) The Proton Radius Puzzle 46 Backup: General
Gil Paz (Wayne State University) The Proton Radius Puzzle 47 Charge radius from Muonic Hydrogen
CREMA Collaboration measured for the first time 2SF =1 2PF =2 transition in Muonic Hydrogen 1/2 − 3/2 [Pohl et al. Nature 466, 213 (2010)]
Gil Paz (Wayne State University) The Proton Radius Puzzle 48 NRQED You already know NRQED!
∂ 0 Dt = + ieA , D = ∇ ieA ∂t − D2 Schr¨odingerequation: iDt + 2mp Hydrogen Fine Structure: - Spin-Orbit: σ B · - Relativistic correction: D4 - Darwin term: ∇ E ·
Organize operators in 1/mp, Lagrangian form:
D2 D4 σ B [∂ E] NRQED = ψ† iDt + + 3 + cF e · + cD e · 2 + ψ L 2mp 8mp 2mp 8mp ···
Gil Paz (Wayne State University) The Proton Radius Puzzle 49 NRQED (NRQCD) Lagrangian 2 The1 /mp were given in Caswell, Lepage ’86 D2 D4 σ B [∂ E] NRQED = ψ† iDt + + 3 + cF g · + cD g · 2 L 2mp 8mp 2mp 8mp σ (D E E D) D2, σ B +icS g · × −2 × + cW 1g { 3· } ψ 8mp 8mp
3 The1 /mp were given in Manohar ’97 Di σ BDi σ DB D + D Bσ D 0 NRQED = ψ† ... cW 2g · 3 + cp pg · · 3 · · L − 4mp 8mp i i 2 2 2 D , [∂ B] 2 B E 2 E +icM g { ×3 } + cA1g −3 cA2g 3 + 8mp 8mp − 16mp 2 σ · (B × B − E × E) 2 σ · (E × E) +icB1g 3 icB2g 3 ψ 8mp − 8mp Last line is non-zero only for Non-Relativistic Quantum Chromo Dynamics (NRQCD)
Gil Paz (Wayne State University) The Proton Radius Puzzle 50 4 NRQED Lagrangian 1/mp 4 For some applications need also 1/mp [Hill, Lee, GP, Solon, PRD 87 053017 (2013)] [D2, D E + E D] D2, [∂ E] NRQED = ψ† ... + cX 1g · 4 · + cX 2g { 4 · } L mp mp 2 i i [∂ ∂ E] 2 D , [E B] +cX 3g 4· + icX 4g { 4× } mp mp Di σ (D E E D)Di ijk σi Dj [∂ E]Dk +icX 5g · × 4− × + icX 6g 4 · mp mp
2 σ B[∂ E] 2 [E ∂σ B] 2 [B ∂σ E] +cX 7g · 4 · + cX 8g · 4 · + cX 9g · 4 · mp mp mp i i i i 2 [E σ ∂B ] 2 [B σ ∂E ] +cX 10g ·4 + cX 11g ·4 mp mp 2 σ E [∂t E ∂ B] +cX 12g · × 4 − × ψ mp Method explained in [GP MPLA 30, 1550128 (2015)]
Gil Paz (Wayne State University) The Proton Radius Puzzle 51 Two Photon exchange: small Q2 limit l l
p p
2 Need poorly constrained non-perturbative function W1(0, Q ) Calculable in small Q2 limit using NRQED [Hill, GP, PRL 107 160402 (2011)] The photon sees the proton “almost” like an elementary particle
2 2 2 Q 2 4 W1(0, Q ) = 2(cF 1)+2 2 cA1 + cF 2cF cW 1 + 2cM + Q − 4mp − O Matching - Operators with one photon coupling: (n) ci given by Fi (0) - Operators with only two photon couplings:
cAi given by forward and backward Compton scattering
Gil Paz (Wayne State University) The Proton Radius Puzzle 52