The Charge Radius of the Proton Gil Paz Enrico Fermi Institute, The - PowerPoint PPT Presentation

The Charge Radius of the Proton Gil Paz Enrico Fermi Institute, The University of Chicago & Department of Physics and Astronomy, Wayne State University Richard J. Hill, GP PRD 82 113005 (2010) Richard J. Hill, GP [arXiv:1103.4617]

Form Factors Matrix element of EM current between nucleon states give rise to two form factors ( q = p f − p i ) � � γ µ F 1 ( q 2 ) + i σ µν � q γ µ q | p ( p i ) � = ¯ 2 m F 2 ( q 2 ) q ν � p ( p f ) | e q ¯ u ( p f ) u ( p i ) q Sachs electric and magnetic form factors G E ( q 2 ) = F 1 ( q 2 ) + q 2 F 2 ( q 2 ) G M ( q 2 ) = F 1 ( q 2 ) + F 2 ( q 2 ) 4 m 2 p G p E (0) = 1 G M (0) = µ p ≈ 2 . 793 The slope of G p E � E = 6 dG p � � r 2 � p E � dq 2 � � q 2 =0 � determines the charge radius r p � r 2 � p E ≡ E Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 2

Charge radius from atomic physics � 1 ( q 2 ) + i σ µν � � q γ µ q | p ( p i ) � = ¯ γ µ F p 2 m F p 2 ( q 2 ) q ν � p ( p f ) | e q ¯ u ( p f ) u ( p i ) q For a point particle amplitude for p + ℓ → p + ℓ M ∝ 1 U ( r ) = − Z α ⇒ q 2 r Including q 2 corrections from proton structure M ∝ 1 U ( r ) = 4 π Z α q 2 q 2 = 1 δ 3 ( r )( r p E ) 2 ⇒ 6 � � Proton structure corrections m r = m ℓ m p / ( m ℓ + m p ) ≈ m ℓ 2( Z α ) 4 r ( r p m 3 E ) 2 δ ℓ 0 ∆ E r p = 3 n 3 E Muonic hydrogen can give the best measurement of r p E ! Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 3

Charge radius from atomic physics Lamb shift in muonic hydrogen [Pohl et al. Nature 466 , 213 (2010)] r p E = 0 . 84184(67) fm CODATA value [Mohr et al. RMP 80 , 633 (2008)] r p E = 0 . 8768(69) fm extracted mainly from (electronic) hydrogen 5 σ discrepancy! We can also extract it from electron-proton scattering data Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 4

Charge radius from scattering data Problem: r p E in literature depends on functional form of G p E r p E not stable when we include more parameters E from e − p scattering data ( Q 2 ≤ 0 . 04 GeV 2 ) tabulated r p in Rosenfelder [arXiv:nucl-th/9912031] ( r p E in 10 − 18 m ) k max = 1 2 3 4 5 836 +8 867 +23 866 +52 959 +85 1122 +122 polynomial − 9 − 24 − 56 − 93 − 137 882 +10 869 +26 continued fraction − − − − 10 − 25 918 +9 868 +28 879 +64 1022 +102 1193 +152 z expansion (no bound) − 9 − 29 − 69 − 114 − 174 879 +38 880 +39 880 +39 918 +9 868 +28 z expansion ( | a k | ≤ 10) − 9 − 29 − 59 − 61 − 62 Only constrained z expansion is stable and model independent Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 5

The recent discrepancy [Hill, GP PRD 82 113005 (2010)] showed previous extractions are model dependent underestimated the error by a factor of 2 or more Based on a model independent approach using scattering data from proton, neutron and π π [Hill, GP PRD 82 113005 (2010)] r p E = 0 . 871(11) fm CODATA value (extracted mainly from electronic hydrogen) [Mohr et al. RMP 80 , 633 (2008)] r p E = 0 . 8768(69) fm Lamb shift in muonic hydrogen [Pohl et al. Nature 466 , 213 (2010)] r p E = 0 . 84184(67) fm Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 6

Lamb shift in muonic hydrogen CREMA measured [Pohl et al. Nature 466 , 213 (2010)] ∆ E = 206 . 2949 ± 0 . 0032 meV Comparing to the theoretical expression [Pachucki PRA 60 , 3593 (1999), Borie PRA 71 (3), 032508 (2005)] E ) 2 + 0 . 0347( r p E ) 3 meV ∆ E = 209 . 9779(49) − 5 . 2262( r p They got r p E = 0 . 84184(67) fm Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 7

The Theoretical Prediction Is there a problem with the theoretical prediction? [Pachucki PRA 60 , 3593 (1999), Borie PRA 71 (3), 032508 (2005)] E ) 3 meV 5 . 2262( r p 0 . 0347( r p E ) 2 ∆ E = 209 . 9779(49) − + ↑ ↑ ↑ mostly already where does µ QED discussed this term come from? Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 8

� Two-photon amplitude: “standard” calculation l l p p “standard” calculation: separate to proton and non-proton - non-proton ↔ DIS For proton - Insert form factors into vertices � ∞ dq 2 f ( G E , G M ) M = 0 - Using a “dipole form factor” G i ( q 2 ) ≈ G i ( q 2 ) / G i (0) ≈ [1 − q 2 / Λ 2 ] − 2 E ) 3 term - M is a function of Λ ⇒ ( r p Using, Λ 2 = 0 . 71 GeV 2 ⇒ ∆ E ≈ 0 . 018 meV [Pachucki, PRA 53 , 2092 (1996)] Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 9

� Two-photon amplitude: “standard” calculation l l p p Why is the insertion of form factors in vertices valid? Even if it was, result looks funny two-photon amplitude ⇔ the charge radius only for one parameter model for G E and G M In ”standard approach” two-photon ⇒ ∆ E ≈ 0 . 018 meV Need 0 . 258(90) meV (scattering) or 0 . 311(63) meV (spec.) to explain discrepancy Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 10

NRQED Model Independent approach: use NRQED [Caswell, Lepage PLB 167 , 437 (1986); Kinoshita Nio PRD 53 , 4909 (1996); Manohar PRD 56 , 230 (1997)] iD t + D 2 + D 4 � + c F e σ · B + c D e [ ∂ · E ] ψ † L e = e 8 m 3 8 m 2 2 m e 2 m e e e + c W 1 e { D 2 , σ · B } + ic S e σ · ( D × E − E × D ) 8 m 2 8 m 3 e e − c W 2 e D i σ · B D i + c p ′ p e σ · DB · D + D · B σ · D 4 m 3 8 m 3 e e + c A 1 e 2 B 2 − E 2 + ic M e { D i , [ ∂ × B ] i } − c A 2 e 2 E 2 � + ... ψ e 8 m 3 8 m 3 16 m 3 e e e Need also ψ † p σ ψ p · ψ † ψ † p ψ p ψ † e σ ψ e e ψ e L contact = d 1 + d 2 m e m p m e m p Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 11

NRQED From c i and d i determine proton structure correction, e.g. m 3 r ( Z α ) 3 � Z απ � d 2 c proton δ E ( n , ℓ ) = δ ℓ 0 − π n 3 2 m 2 D m e m p p Matching - Operators with one photon coupling: c i given by F ( n ) (0) i - Operators with only two photon couplings: c A i given by forward and backward Compton scattering - d i from two-photon amplitude Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 12

� Two-photon amplitude: matching l l p p 1 � � d 4 x e iq · x � k , s | T { J µ e . m . ( x ) J ν e . m . (0) }| k , s � i 2 s − g µν + q µ q ν k µ − k · q q µ k ν − k · q q ν � � � � � � = W 1 + W 2 q 2 q 2 q 2 Matching 4 π m r 2 m e m p λ − 2 π m r π m r � � F 2 (0)+4 m 2 p F ′ 1 (0) − λ 3 m 2 p λ + d 2 ( Z α ) − 2 � 2 � � 2 1 e log m p p log m e � m 2 λ − m 2 3 + − m 2 p − m 2 m e m p λ m e m p e � 1 � ∞ Q 3 = − m e � 1 − x 2 dx dQ ( Q 2 + λ 2 ) 2 ( Q 2 + 4 m 2 e x 2 ) m p − 1 0 � � (1 + 2 x 2 ) W 1 (2 im p Qx , Q 2 ) − (1 − x 2 ) m 2 p W 2 (2 im p Qx , Q 2 ) × Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 13

d 2 � In order to determine d 2 need to know W i l l Im ∼ Im W i p p can be extracted from on-shell quantities: Proton form factors and Inelastic structure functions To find W i from Im W i , need dispersion relations Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 14

Dispersion relation Dispersion relations ( ν = 2 k · q , Q 2 = − q 2 ) � ∞ W 1 ( ν, Q 2 ) = W 1 (0 , Q 2 ) + ν 2 ν cut ( Q 2 ) 2 d ν ′ 2 Im W 1 ( ν ′ , Q 2 ) ν ′ 2 ( ν ′ 2 − ν 2 ) π � ∞ ν cut ( Q 2 ) 2 d ν ′ 2 Im W 2 ( ν ′ , Q 2 ) W 2 ( ν, Q 2 ) = 1 ν ′ 2 − ν 2 π W 1 requires subtraction... - Im W p i from form factors - Im W c i from DIS - What about W 1 (0 , Q 2 )? Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 15

W 1 (0 , Q 2 ) Can calculate in two limits: [Hill, GP, arXiv:1103.4617] - Q 2 ≪ m 2 p The photon sees the proton “almost“ like an elementary particle Use NRQED to calculate W 1 (0 , Q 2 ) upto O ( Q 2 ) (including) F − 1) + 2 Q 2 W 1 (0 , Q 2 ) = 2( c 2 c A 1 + c 2 � � F − 2 c F c W 1 + 2 c M 4 m 2 p - Q 2 ≫ m 2 p The photon sees the quarks inside the proton Use OPE to find W 1 (0 , Q 2 ) ∼ 1 / Q 2 for large Q 2 In between you will have to model! Current calculation pretends there is no model dependence How big is the model dependence? Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 16

Bound State Energy 1) Proton: Im W p i using dipole form factor ∆ E = − 0 . 016 meV 2) Continuum: Im W c i [Carlson, Vanderhaeghen arXiv:1101.5965] ∆ E = 0 . 0127(5) meV 3) What about W 1 (0 , Q 2 )? “Sticking In Form Factors” (SIFF) model W SIFF (0 , Q 2 ) = 2 F 2 (2 F 1 + F 2 ) F i ≡ F i ( Q 2 ) 1 Gil Paz (The University of Chicago & Wayne State University) The Charge Radius of the Proton 17