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 j q j 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 j q j 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 j f gj 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 1 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 1 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 1 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 1 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 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 j f gj i µ ν µ ν µν q q µ k q q ν k q q = g + W1 + k · k · W2 − q2 − q2 − q2 l l ¡ p p 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 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 j f gj 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 1 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 1 l l 0 2 2 1 02 ImW2(ν ; Q ) W2(ν; Q ) = dν 02 2 2 2 π νcut(Q ) ν ν ¡ − p p 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 Imaginary part of TPE related to data: form factors, structure functions l l ¡ p p 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 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 ) l l Calculable in small Q2 limit using NRQED ¡ [Hill, GP, PRL 107 160402 (2011)]p p 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 Q 0, ( 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 Q 0, ( 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