Hadronic Parity Violation in Effective Field Theory Matthias R. - PowerPoint PPT Presentation

Hadronic Parity Violation in Effective Field Theory Matthias R. Schindler University of South Carolina The 19th Particles and Nuclei International Conference July 24–29, 2011 Collaborators: H. W. Grießhammer, D. R. Phillips, R. P . Springer 1 / 18

Hadronic parity violation and effective field theory Two-nucleon sector Three-nucleon sector Conclusion & Outlook 2 / 18

Hadronic parity violation Weak interaction between quarks induces parity-violating component in nucleon-nucleon ( NN ) interaction Effects highly suppressed ∼ 10 − 6 − 10 − 7 Parity violation to isolate weak component Range of weak quark-quark interactions ∼ 0 . 002 fm Probe of nonperturbative QCD 3 / 18

Few-nucleon experiments Complex nuclei: enhancement up to ∼ 10 % effect Relation to NN interaction? Theoretically difficult Two-nucleon system � pp scattering (Bonn, PSI, TRIUMF , LANL) � np → d γ (SNS, LANSCE, Grenoble) � np spin rotation? Few-nucleon systems � n α spin rotation (NIST) � p α scattering (PSI) 3 He ( � n , p ) 3 H (SNS) � nd → t γ (SNS?) � nd spin rotation? 4 / 18

Theory goals Variety of experiments Unified framework Model-independent Check consistency of results Defendable theoretical errors Two- and few-body systems Energies � 10s of MeV Ideally suited for pionless effective field theory [EFT( π / )] 5 / 18

Pionless EFT Model-independent framework Lagrangian with all terms allowed by symmetries At very low energy cannot resolve pion exchange NN contact terms with increasing number of derivatives Leading-order parity-conserving Lagrangian � ∇ 2 2 M ) N − 1 8 C ( 1 S 0 ) L = N † ( i ∂ 0 + ( N T τ 2 τ a σ 2 N ) † ( N T τ 2 τ a σ 2 N ) 0 − 1 8 C ( 3 S 1 ) ( N T τ 2 σ 2 σ i N ) † ( N T τ 2 σ 2 σ i N ) + . . . , 0 6 / 18

Parity violation in EFT( π / ) At leading order: 5 independent PV NN contributions 1 Experimental input to determine 5 coefficients � � � � † ↔ C ( 3 S 1 − 1 P 1 ) � N T σ 2 � N T σ 2 τ 2 i L PV = − στ 2 N · ∇ N � � � † ↔ + C ( 1 S 0 − 3 P 0 ) � N T σ 2 τ 2 � N T σ 2 � τ N · σ · τ 2 � τ i ∇ N (∆ I = 0 ) � † � σ · τ 2 τ b ↔ � + C ( 1 S 0 − 3 P 0 ) ǫ 3 ab � N T σ 2 τ 2 τ a N N T σ 2 � ∇ N (∆ I = 1 ) � † � � ↔ + C ( 1 S 0 − 3 P 0 ) I ab � N T σ 2 τ 2 τ a N N T σ 2 � σ · τ 2 τ b i ∇ N (∆ I = 2 ) j � † � ↔ �� + C ( 3 S 1 − 3 P 1 ) ǫ ijk � N T σ 2 σ i τ 2 N N T σ 2 σ k τ 2 τ 3 ∇ N + h . c . 1 Savage, Springer (1998); Zhu et. al. (2005); Girlanda (2008); Phillips, MRS, Springer (2009) 7 / 18

Nucleon-nucleon scattering Simplest process � NN cross section Strong contribution does not depend on helicity Weak contribution does depend on helicity Consider asymmetry in � N + N A L = σ + − σ − σ + + σ − Interference between strong and weak P S + P S 8 / 18

Leading-order results: pp / nn 2 = 8 k A pp / nn A pp / nn L 1 S 0 C 0 � C ( 1 S 0 − 3 P 0 ) + C ( 1 S 0 − 3 P 0 ) + C ( 1 S 0 − 3 P 0 ) � A pp = 4 (∆ I = 0 ) (∆ I = 1 ) (∆ I = 2 ) � C ( 1 S 0 − 3 P 0 ) − C ( 1 S 0 − 3 P 0 ) + C ( 1 S 0 − 3 P 0 ) � A nn = 4 (∆ I = 0 ) (∆ I = 1 ) (∆ I = 2 ) No Coulomb interaction for pp ( ∼ 3% at 15 MeV) Depends on ratio of PV and PC constant 1 S 0 ⇒ Renormalization point-dependence of A pp / nn dictated by C 0 2 Phillips, MRS, Springer (2009) 9 / 18

Leading-order results: np  1 S 0 3 S 1  1 S 0 3 S 1 d σ d σ A A A np np np d Ω d Ω L = 8 k +   d Ω + 3 d σ 3 S 1 1 S 0 1 S 0 d Ω + 3 d σ 3 S 1 1 S 0 3 S 1 d σ d σ C C 0 0 d Ω d Ω 1 S 0 � C ( 1 S 0 − 3 P 0 ) − 2 C ( 1 S 0 − 3 P 0 ) � A np = 4 (∆ I = 0 ) (∆ I = 2 ) 3 S 1 � C ( 3 S 1 − 1 P 1 ) − 2 C ( 3 S 1 − 3 P 1 ) � A np = 4 � − 1 �� 1 � 2 d σ + k 2 d Ω = a 1 S 0 3 S 1 Measure at 2 different energies: disentangle A np and A np 10 / 18

� np spin rotation � np scattering amplitude related to spin rotation angle 3 1 d φ PV = M 1 � Re [ M + ( m p ) − M − ( m p )] ρ d l 2 k 2 m p = ± 1 2 � Rotation angle at NLO with g ( X − Y ) = C ( X − Y ) 32 π : M C X 0 d φ np 1 � ( 18 . 1 ± 1 . 8 ) g ( 3 S 1 − 3 P 1 ) + ( 9 . 0 ± 0 . 9 ) g ( 3 S 1 − 1 P 1 ) PV = ρ d l rad + ( − 37 . 0 ± 3 . 7 ) g ( 1 S 0 − 3 P 0 ) + ( 74 . 4 ± 7 . 4 ) g ( 1 S 0 − 3 P 0 ) � (∆ I = 0 ) (∆ I = 2 ) 1 MeV 2 � d φ np � 10 − 6 − 10 − 7 � rad � � � PV � ≈ � � d l m � � � 3 Grießhammer, MRS, Springer 11 / 18

Electromagnetic processes Asymmetry in � np → d γ C ( 3 S 1 − 3 P 1 ) A γ = 32 M 1 S 0 ) 3 S 1 3 κ 1 ( 1 − γ a C 0 NPDGamma @ SNS Related to deuteron anapole moment through C ( 3 S 1 − 3 P 1 ) 4 Circular polarization in np → d � γ C ( 1 S 0 − 3 P 0 ) − 2 C ( 1 S 0 − 3 P 0 ) P γ ∼ a C ( 3 S 1 − 1 P 1 ) (∆ I = 0 ) (∆ I = 2 ) + b 3 S 1 1 S 0 C C 0 0 Experimental result consistent with P γ = 0 Use high-intensity free electron lasers for � γ d → np ? 4 Savage (2001); MRS, Springer (2009); Knyazkov (1983) 12 / 18

Three-nucleon systems Parity-conserving sector Describe 3 N systems with NN interactions only? Na¨ ıve power counting: | 2 N | > | 3 N | > | 4 N | > . . . Analysis of nd scattering in 2 S 1 2 wave 5 3 N interaction at leading order for renormalization Additional experimental input (e.g. scattering length) Parity-violating sector Na¨ ıve power counting: 3 N interaction higher order If not: even more experiments needed 5 Bedaque, Hammer, van Kolck (1999) 13 / 18

Parity violation in 3 N sector Analyze high-momentum behavior of loop integrals, e.g., Parity-conserving amplitude known Include leading-order PV 2 N interaction 6 No PV 3 N interaction at LO and NLO 6 Grießhammer, MRS (2011) 14 / 18

� nd spin rotation nd scattering amplitude at NLO 7 � Verified that no PV 3NI required for renormalization d φ nd 1 � ( 15 . 9 ± 1 . 6 ) g ( 3 S 1 − 1 P 1 ) − ( 36 . 6 ± 3 . 7 ) g ( 3 S 1 − 3 P 1 ) PV = ρ d l rad +( 4 . 6 ± 1 . 0 ) ( 3 g ( 1 S 0 − 3 P 0 ) − 2 g ( 1 S 0 − 3 P 0 ) � ) (∆ I = 0 ) (∆ I = 1 ) 1 MeV 2 � � d φ nd � ≈ 10 − [ 6 ... 7 ] rad � � PV � � � d l � m � 7 Grießhammer, MRS, Springer 15 / 18

Cutoff dependence � � � � �� � � � � � � � ��� � � � � � � � � � � � � � 15 � � 1 � 2 � c � g � 3 S 1 � 1 P 1 � � � rad MeV � � � � � � � � � � � � � ��� � � � � � � 10 � 5 g � 3 S 1 � 1 P 1 � 0 � 200 500 1000 2000 5000 � � MeV � � � � � 5 � � � � � � � ��� � � � � �� � � � � 4 � 1 � 2 � � � � � c � � � � rad MeV � � � � � � � � 3 � � � � ��� � � � � � � � 2 � � � � 1 0 � 200 500 1000 2000 5000 � � MeV � 16 / 18

Conclusion & Outlook Hadronic parity violation Probe nonperturbative QCD phenomena: inside-out probe Current and proposed experiments Low-energy Few-nucleons Need consistent analysis and interpretation 17 / 18

Conclusion & Outlook EFT for parity-violating NN interactions 5 independent operators at LO in EFT( π / ) 2-body observables � pp scattering, � np spin rotation, np ↔ d γ Not enough information PV 3 body sector No PV 3-body operators at LO and NLO � nd spin rotation � nd → t γ Few-body observables Parity-conserving: EFT( π / ) up to A =6 No Core Shell Model, Resonating Group Method 8 Lattice 8 Stetcu, Barrett, van Kolck (2007), (2010); Kirscher, Grießhammer, Shukla, Hofmann (2010); Beane, Savage (2002) 18 / 18

Coulomb corrections / ) 9 Coulomb corrections can be included in EFT( π Coulomb parameter η = M α 2 p Integrals for cross section over finite range θ 1 ≤ θ ≤ θ 2 For T lab = 0 . 1 MeV: η ≈ 0 . 26 ⇒ expand in η L = 8 p A pp � � � � 1 − cos θ 1 � 1 1 A pp 1 + η ln 1 S 0 a S ( µ ) p cos θ 1 − cos θ 2 1 − cos θ 2 C 0 + O ( η ) 2 � 9 Kong and Ravndal (1999) 19 / 18

Comparison with experiment pp scattering experiments (23 o < θ lab < 52 o ) 10 A � pp L ( E = 13 . 6 MeV ) = ( − 0 . 93 ± 0 . 21 ) × 10 − 7 A � pp L ( E = 45 MeV ) = ( − 1 . 50 ± 0 . 22 ) × 10 − 7 From result at E = 13 . 6 MeV: A pp = ( − 1 . 5 ± 0 . 3 ) × 10 − 10 MeV − 1 1 S 0 C 0 Coulomb correction ∼ 3 percent Use to ‘predict’ asymmetry at 45 MeV A � pp L ( E = 45 MeV ) = ( − 1 . 69 ± 0 . 38 ) × 10 − 7 In agreement with experiment 10 Eversheim (1991); Kistryn (1987) 20 / 18

Higher-order corrections At E = 45 MeV center-of-mass momentum p > m π Resum higher-order corrections in PC sector Re-analyze low-energy pp measurement (no Coulomb) A pp ( µ = m π ) = ( − 1 . 1 ± 0 . 25 ) × 10 − 10 MeV − 1 1 S 0 C 0 ∼ 30 % difference “Prediction” for E = 45 MeV A � pp L ( E = 45MeV ) = ( − 2 . 6 ± 0 . 6 ) × 10 − 7 Compare to − 1 . 69 × 10 − 7 : > 50 % difference 21 / 18