Computing Nucleon Electric Dipole Moments in Lattice QCD Hiroshi - PowerPoint PPT Presentation

ロゴマーク組み合わせ一覧 食物栄養学科 生活環境学部 生活環境学部 Computing Nucleon Electric Dipole Moments in Lattice QCD Hiroshi Ohki Nara Women’s University (RBC/UKQCD collaboration) RIKEN BNL Research Center References: M. Abramczyk, S. Aoki, T . Blum, T . Izubuchi, H. Ohki, S. Syritsyn, Phys.Rev. D96 (2017) no.1, 014501 N. Yamanaka, S. Hashimoto, T . Kaneko, H. Ohki (JLQCD), PRD 98, 054516 S. Syritsyn, T . Izubuchi, H. Ohki, 1901.05455, and work in progress Frontiers in Lattice QCD and related topics, April 26, 2019

outline • Introduction • Parity mixing problem on lattice EDM calculation • Lattice Study — old formula v.s. new formula (on lattice) numerical check using chromo-EDM •Implication to the θ -EDM •Noise reduction for θ -EDM • quark EDM • Summary

Introduction ■ Electric Dipole Moment d 
 Energy shift of a spin particle in an electric field ■ Non-zero EDM : P&T (CP through CPT) violation T + 
 + 
 - - P → H EDM is CP-odd ! - 
 + → H EDM is P-odd

■ Origin of EDM: CP-violating (CP-odd) interactions CKM: CP violating interaction in SM But, electron and quark EDM’s are zero at 1 and 2 loop level. at least three loops to get non-zero EDM’s. EDM’s are very small in the standard model. t g W d,s,b u c nucleon EDM from CKM : ~ 10 -32 [e cm] SM contribution (3-loop diagram) t Ref: [A. Czarnecki and B. Krause ’97] CP violation (CPV) in SM is not sufficient to reproduce matter/antimatter asymmetry. Large CPV beyond SM is required. (Sakharov’s three conditions) SM prediction 10 20 : 1 photon: matter •http://www.esa.int/ESA Observation 10 10 : 1

Origin of EDM: CP-violating (CP-odd) BSM physics BSM particles CPV int. CP-odd four-quark Weinberg op. BSM may induce EDM in lower loop level: a good probe of new physics EDM is usually measured using composite particles (neutron, atoms, etc) S nucleon EDM effects may be enhanced in the composite system. [N. Yamanaka, et al. Eur. Phys. J. A53 (2017) 54, Ginges and Flambaum Phys. Rep. 397, 63, 2004]

• Nucleon EDM e, µ EDM BSM physics: Higgs doublets Paramagnetic Supersymmetry Atom EDM / Molecules e-N int e-q int Left-Right Leptoquark Diamagnetic Extradimension q EDM Atom MQM N EDM EDM Composite q cEDM Schiff models (RGE) moment 4-q int N-N int Standard Model ggg (PQM) Nuclear θ -term EDM ( θ -term) ( P Q M ) Energy scale Atomic Nuclear Hadron QCD TeV : Observable available at experiment observable Important bottleneck : Sizable dependence of the EDM calculation! : Weak dependence : Matching [N. Yamanaka, et al. Eur. Phys. J. A53 (2017) 54, Ginges and Flambaum Phys. Rep. 397, 63, 2004] Role of (lattice) QCD : connect quark/gluon-level (effective) operators to hadron/nuclei matrix elements and interactions (Form factor, dn) Non-perturbative determination is important → Lattice QCD calculation! � 6

Neutron EDM S • Nucleon EDM Experiments 10 -18 • Previous Expts Neutron EDM Upper Limit ( e cm) Future Expts 10 -20 10 -22 Current nEDM limits: 10 -24 199Hg spin precession (UW) [Graner et al, 2016] • Ultracold Neutrons in a trap (ILL) [Baker 2006] 10 -26 Supersymmetry � | d Hg | < 7 . 4 × 10 − 30 e · cm Predictions 10 -28 | d n | < 2 . 6 × 10 − 26 e · cm 10 -30 SM nucleon EDMs expectation is Standard Model much smaller than the current bound. Predictions 10 -32 1950 1970 1990 2010 Year of Publication [B. Yoon, talk at Lattice 2017] Several experimental projects are on going. ■ nucleon, charged hadrons, lepton, PSI EDM, Munich FRMII, SNS nEDM, RCNP/TRIUMF , J-PARC etc…

• Effective CPV operators dim=4, θ QCD dim=5, chromo EDM dim=5, e, quark EDM dim=6, Weinberg three gluon C (4 q ) O (4 q ) X dim=6, Four-quark operators + i i ¯ : Strong CP problem θ ≤ O (10 − 10 ) m q / Λ 2 Dim=5 operators suppressed by -> effectively dim=6, quark EDM … the most accurate lattice data for EDM (~5% for u,d) Others are not well determined. cEDM, Weinberg ops just started.

θ QCD induced Nucleon EDMs Phenomenological estimates Lattice calculations Neutron 0 method value ChPT/NDA ∼ 0 . 002 e fm QCD sum rules [1,2] 0 . 0025 ± 0 . 0013 e fm -0.05 n (e fm) 0 . 0004 +0 . 0003 QCD sum rules [3] − 0 . 0002 e fm N f =2+1 DWF, F 3 ( θ ), DSDR 32c N f =2+1 DWF, F 3 ( θ ), Iwasaki 24c d N N f =2 DWF, F 3 ( θ ) -0.1 N f =2 clover, ∆ E( θ ) [1] M. Pospelov, A. Ritz, Nuclear Phys. B 573 (2000) 177, N f =2 clover, F 3 ( θ ) [2] M. Pospelov, A. Ritz, Phys. Rev. Lett. 83 (1999) 2526, N f =2 clover, F 3 (i θ ) -0.15 N f =3 clover, F 3 (i θ ) [3] J. Hisano, J.Y . Lee, N. Nagata, Y . Shimizu, Phys. Rev. D N f =2+1+1 TM, F 3 ( θ ) 85 (2012) 114044. 0 0.2 0.4 2 ) 2 (GeV m π [E. Shintani, T . Blum, T . Izubuchi, A. Soni, PRD93, 094503(2015)] Phenomenology: |dn| ~ θ QCD 10^{-3} e fm -> | θ QCD | < 10^{-10} Lattice : |dn| ~ θ QCD 10^-2 e fm -> severer constraint on | θ QCD | Problem: a spurious mixing between EDM and magnetic moments in all previous lattice computations of nucleon form factor.

Parity mixing problem on the CP-violating nucleon form factors M. Abramczyk, S. Aoki, T . Blum, T . Izubuchi, H. Ohki, and S. Syritsyn, Lattice calculation of electric dipole moments and form factors of the nucleon Phys.Rev. D96 (2017) no.1, 014501, selected editor’s suggestions

Definition of nucleon form factors Nucleon form factor in C, P-symmetric world (CP-even) F 1 ( Q 2 ) γ µ + F 2 ( Q 2 ) i σ µ ν q ν  � h p 0 , σ 0 | J µ | p, σ i = ¯ u p 0 , σ 0 u p, σ 2 m N Q 2 = − q 2 ) ( q = p 0 − p, u p : spinor wave function for the nucleon ground state |p, σ > (/ p − m N ) u p = 0 J : electromagnetic current N N

Definition of nucleon form factors Nucleon form factor in CP-broken world F 1 ( Q 2 ) γ µ + F 2 ( Q 2 ) i σ µ ν q ν � F 3 ( Q 2 ) γ 5 σ µ ν q ν  � h p 0 , σ 0 | J µ | p, σ i = ¯ u p 0 , σ 0 u p, σ 2 m N 2 m N P , T even P , T odd Refs. [many textbooks, e.g. Itzykson, Zuber, “Quantum Field Theory“] The same spinor u p (F1, F2 are same as CP-even case.) Non-zero F3 is a signature of the CP violation (F3= 0 -> CP-even) permanent EDM: All previous lattice studies (prior to 2017) use a different spin structure for the form factors.

revisit of the nucleon CP-odd (EDM) form

Nucleon 2 point function in CP-conserving theory N = u [ u T C γ 5 d ] Lattice nucleon operator for sink and source Nucleon ground state in CP-even vacuum h 0 | N | p, σ i CP − even = Zu p, σ u p is a solution spinor of the free Dirac equation: (/ p − m N ) u p = 0 p ; t ) | ¯ C 2 pt ( ~ p ; t ) CP − even = h N ( ~ N ( ~ p ; 0) i CP − even 2 3 | k, � ih k, � | 4X 5 ¯ = h N ( ~ p, t ) N ( ~ p ; 0) i CP − even + (excited states) 2 E k k, σ t →∞ | Z | 2 e − E p t X ! ( u p, σ ¯ u p, σ ) 2 E p Completeness condition for free Dirac σ spinor = | Z | 2 e − E p t m N � i / p 2 E p (From now on excited states are omitted.)

Nucleon 2 point function in CP-broken theory h 0 | N | p, σ i � Nucleon ground state in CP-broken vacuum CP = Z ˜ u p, σ � ˜ p − m N e − 2 i αγ 5 )˜ is a solution spinor of the free Dirac equation: (/ u p = 0 u p Asymptotic state is modified: (CP-violating) γ 5 mass is allowed in general. p ; t ) | ¯ C 2 pt ( ~ p ; t ) � CP = h N ( ~ N ( ~ p ; 0) i � � � CP = | Z | 2 e − E p t u p, σ ¯ X ( ˜ u p, σ ) ˜ 2 E p Completeness condition for free Dirac spinor σ = | Z | 2 e − E p t m N e 2 i αγ 5 � i / p 2 E p u p = e i αγ 5 u p is a solution to the above Dirac equation. ˜ u p, σ ) e i αγ 5 = m N e 2 i αγ 5 − i / X u p, σ ¯ X u p, σ = e i αγ 5 ( ˜ ˜ u p, σ ¯ p σ σ [Completeness condition for free Dirac spinor with γ 5 mass]

① ② ③ Calculation of 3 point function in CP-broken theory e � i ~ z, ⌧ ) ¯ C 3 pt ( ~ X p 0 · ~ y + i ~ p · ~ z h N ( ~ y, t ) J µ ( ~ p 0 , t ; ~ p, ⌧ ) � CP = N (0) i � � � CP y, ~ ~ z = | Z | 2 e � E p 0 ( t � ⌧ ) � E p ( ⌧ ) X CP h p 0 , � | J µ | p, � 0 i � h N ( p 0 ) | p 0 , � i � CP h p, � 0 | N ( p ) i � � � � CP 4 E p 0 E p � , � 0 ① & ③ : h 0 | N | p, σ i � CP = Z ˜ u p, σ � F 2 ( Q 2 ) i σ µ ν q ν F 3 ( Q 2 ) γ 5 σ µ ν q ν  � F 1 ( Q 2 ) γ µ + ˜ ② : CP = ¯ ˜ � ˜ h p 0 , σ 0 | J µ | p, σ i � ˜ ˜ u p 0 , σ 0 u p, σ � 2 m N 2 m N Refs: original works since 2005 “All” previous (prior 2017) lattice studies: F 1 , ˜ ˜ F 2 , ˜ (˜ u ) : defined in the rotated spinor basis F 3 ( F 2 ( Q 2 ) 6 = ˜ F 2 ( Q 2 ) Two form factors are different! F 3 ( Q 2 ) 6 = ˜ F 3 ( Q 2 ) ( u ) (˜ u )