/T interactions and chiral symmetry

Transcript

1 /T interactions and chiral symmetry Emanuele Mereghetti Los Alamos National Lab January 3th, 015 ACFI Workshop, Amherst

2 Introduction M T BSM probing CP with EDM entails physics at very different scales M /T v ew M m π... understanding nature of CP requires v EW several, orthogonal probes robust theoretical tools M nonptb Nucleon d n, d p Light nuclei d d, d t, d h atoms d 199 Hg, d05 Tl,... molecules d ThO,... m

3 Introduction M T BSM probing CP with EDM entails physics at very different scales M /T v ew M m π... understanding nature of CP requires v EW several, orthogonal probes robust theoretical tools M nonptb Nucleon d n, d p Light nuclei d d, d t, d h treated in same theory framework Chiral EFT m

4 Introduction M T BSM v EW M nonptb m 10 % nuclear uncertainty... when expressed in hadronic couplings summary of recent calculations of light nuclei EDM, from U. van Kolck, EM, in preparation

5 Introduction M T BSM probing CP with EDM entails physics at very different scales M /T v ew M m π... understanding nature of CP requires v EW M several, orthogonal probes Nucleon Light nuclei d n, d p d d, d 3 H, d3 He robust theoretical tools treated in same theory framework Chiral EFT nonptb m missing link: LECs from What do we learn from symmetry?

6 /T at the quark-gluon level θ term L 4 = θ g s 3π εµναβ TrG µνg αβ q L Me iρ q R q R Me iρ q L, dimension six L 6 = 1 q iσµν γ 5 d 0 + d 3 τ 3 q F µν 1 q iσµν γ 5 d0 + d 3 τ 3 G µνq + d W 6 f abc ε µναβ G a αβ Gb µρ Gc ν ρ ImΣ 18 qq qiγ 5 q qτ q qτ iγ 5 q + 1 qγ 4 Im Ξ 18ε 3ij µ τ i q qiγ µγ 5 τ j q qγ µ τ i q qiγ µγ 5 τ j q quark electric qedm and chromo-electric dipole moment qcedm see Jordy s talk d 0,3 = mδ 0,3 M/T, d0,3 = m δ 0,3 M/T chiral breaking, assume m u + m d δ 0,3, δ 0,3 are O1

7 /T at the quark-gluon level dimension four: θ term L 4 = θ g s 3π εµναβ TrG µνg αβ q L Me iρ q R q R Me iρ q L, dimension six L 6 = 1 q iσµν γ 5 d 0 + d 3 τ 3 q F µν 1 q iσµν γ 5 d0 + d 3 τ 3 G µνq + d W 6 f abc ε µναβ G a αβ Gb µρ Gc ν ρ ImΣ 18 qq qiγ 5 q qτ q qτ iγ 5 q + 1 qγ 4 Im Ξ 18ε 3ij µ τ i q qiγ µγ 5 τ j q qγ µ τ i q qiγ µγ 5 τ j q gluon chromo-electric dipole moment gcedm & χi four-quark d W, Σ 1,8 = {w, σ 1, σ 8 } 1 M /T chiral invariant see Jordy s talk

8 /T at the quark-gluon level dimension four: θ term L 4 = θ g s 3π εµναβ TrG µνg αβ q L Me iρ q R q R Me iρ q L, dimension six L 6 = 1 q iσµν γ 5 d 0 + d 3 τ 3 q F µν 1 q iσµν γ 5 d0 + d 3 τ 3 G µνq + d W 6 f abc ε µναβ G a αβ Gb µρ Gc ν ρ ImΣ 18 qq qiγ 5 q qτ q qτ iγ 5 q + 1 qγ 4 Im Ξ 18ε 3ij µ τ i q qiγ µγ 5 τ j q qγ µ τ i q qiγ µγ 5 τ j q left-right four-quark FQLR operators Ξ 1,8 = ξ 1 M /T isospin breaking, not v ew see Jordy s talk

9 /T at the hadronic level include dim-four and dim-six /T in χpt Lagrangian L /T = N d0 + d 1 τ 3 S µ v ν NF µν ḡ0 F π Nπ τ N ḡ1 F π π 3 NN F π π 3 π + C 1 NN µ NS µ N + C Nτ N µ NS µ τ N at LO, EDMs expressed in terms of a few couplings d0, d 1 neutron & proton EDM, one-body contribs. to A nuclei ḡ 0, ḡ 1, pion loop to nucleon & proton EDMs leading /T OPE potential C 1, C short-range /T potential relative size depends on /T source = different signals for one, two, three-nucleon EDMs Can we go beyond NDA?

10 Theta Term L 4 = θ g s 3π εµναβ TrG µνg αβ q L Me iρ q R q R Me iρ q L, rotate θ away physics depends on θ = θ n F ρ perform vacuum alignment i.e. kill /T iso-breaking terms qiγ 5 τ 3 q L 4 = mr θ qq + r 1 θ mε qτ 3 q + m sin θ qiγ 5 q CP-even quark mass and mass difference CP-odd isoscalar mass term m = m u + m d, mε = m d m u

11 Theta Term L 4 = θ g s 3π εµναβ TrG µνg αβ q L Me iρ q R q R Me iρ q L, rotate θ away physics depends on θ = θ n F ρ perform vacuum alignment i.e. kill /T iso-breaking terms qiγ 5 τ 3 q L 4 = mr θ qq + r 1 θ mε qτ 3 q + m sin θ qiγ 5 q CP-even quark mass and mass difference CP-odd isoscalar mass term m = mum d = m 1 ε m u + m d r θ = 1 1 ε θ +...

12 The Theta Term. Chiral Lagrangian and NDA ḡ 0 ḡ 1 /Fπ d0,1 Q C 1, Fπ Q θ m π 1 ε m π M M ε Q Q Q M M M NDA Chiral properties of θ determine size of LECs breaks chiral symmetry but not isospin isoscalar ḡ 0 at LO isobreaking requires insertion of mε ḡ 1 and suppressed higher dimensionality of Nγ and NN operators costs Q/M

13 Theta Term. Symmetry L 4 = mr θ qq + r 1 θ mε qτ 3 q + m sin θ qiγ 5 q θ term and mass splitting are chiral partners qiγ5 q SUA qτ q qα τ q α qiγ 5 q nucleon matrix elements are related i.e. one spurion enough to construct iso- and T-breaking couplings T violation isospin breaking = 1 ε sin θ ρ θ ε powerful at LO breaks down at OQ /M ignorance of CP-even LECs too many operators when including EM

14 Theta Term. ḡ 0 L 1 = m N 1 π NN Fπ + 1 [ δm N N τ 3 π 3π τ Fπ N ρ θ N π τ ] N F π m N nucleon sigma term δm N = m n m p st, strong mass splitting 1 ε ḡ 0 = δm N sin θ ε

15 Theta Term. ḡ 0 L 1 = m N 1 π NN Fπ + 1 [ δm N N τ 3 π 3π τ Fπ N ρ θ N π τ ] N F π m N nucleon sigma term δm N = m n m p st, strong mass splitting 1 ε ḡ 0 = δm N sin θ ε δm N not directly accessible experimentally, δ emm N δm N accessible via existing lattice calculations δm N =.39 ± 0.1 MeV A. Walker-Loud, 14; Borsanyi et al, 14. ε = 0.37 ± 0.03 MeV Aoki 13, FLAG Working group. precise 10% determination of ḡ 0 ḡ 0 F π = 15 ± 10 3 sin θ errors from lattice only

16 Theta Term. ḡ 0 [ ḡ 0 = ḡ0 1 + m π F π F π πf π 3g A + 1 log µ m π m n m p st = δm N [ 1 + m π πf π + g A + 1 3g A + 1 log µ m + g A + 1 π ] + δ3 m N ] + δ 3 m N ρ θ + δḡ 0 F π F π same loop corrections to ḡ 0 and δm N finite LEC δḡ 0 only correct πn coupling T violation isospin breaking = 1 ε sin θ ρ θ... but... ε corrections appear at NNLO and are not log enhanced

17 Theta term. ḡ 0 what about strangeness? in SU3 χpt ḡ 0 δm N = ρ θ F π F π and ḡ 0 = m Ξ m Σ m1 ε = 10 3 sin θ F π m s m F π J. de Vries, EM, A. Walker-Loud, in progress large Om K /M corrections to m n m p m K + m K 0, η-π mixing and ḡ 0 πkk, ππη CP-odd vertex

18 Theta term. ḡ 0 what about strangeness? in SU3 χpt ḡ 0 δm N = ρ θ F π F π and ḡ 0 = m Ξ m Σ m1 ε = 10 3 sin θ F π m s m F π a b c d a b c d e f e f g h i l m n large... too large... Om K /M corrections to m n m p m K + m K 0, η-π mixing and ḡ 0 πkk, ππη CP-odd vertex under control NNLO corrections

19 Theta term. ḡ 0 what about strangeness? in SU3 χpt ḡ 0 δm N = ρ θ F π F π and ḡ 0 = m Ξ m Σ m1 ε = 10 3 sin θ F π m s m F π a b c d a b c d e f e f g h i l m n same divergent loop corrections to m n m p and ḡ 0 different loop corrections to m Ξ m Σ and ḡ 0, starting at NLO ḡ 0 m Ξ m Σ at NLO ḡ 0 δm N violated by NNLO finite LECs, but δḡ 0 m, not m s m

20 Theta Term. ḡ 1 and 1 L = δ stm π π 3 π 3 π ρ θ F π δ stm π strong contrib. to m π + m π 0 δ stm π = 87 ± 55 MeV L 3 = m N δm st m π fit to meson data G. Amoros, J. Bijnens, P. Talavera, 01. π 3 ρ θ NN c 3 ππn F π tadpole induced, related to nucleon sigma term c 3 ππn tiny contrib. to π-n scattering beyond accuracy of current analysis [ 1 ] π3 π 3 Fπ + ρ θ NN F π

21 Theta Term. ḡ 1 and 1 L = δ stm π π 3 π 3 π ρ θ F π δ stm π strong contrib. to m π + m π 0 δ stm π = 87 ± 55 MeV L 3 = 3 ± 10 3 sin θ π 3 NN c 3 ππn tadpole induced, related to nucleon sigma term c 3 ππn tiny contrib. to π-n scattering beyond accuracy of current analysis fit to meson data G. Amoros, J. Bijnens, P. Talavera, 01. [ 1 ] π3 π 3 Fπ + ρ θ NN F π ḡ 1 poorly determined but somewhat larger than expected, & extremely important for deuteron

22 Theta Term. d 0 and d 1 symmetry relation breaks down [ L 3 Nγ = c 1γ + c γ π 3 F π too much isospin violation from EM & quark masses ] NS µ v ν N ef µν + c 1γ ρ θ [ N c 3γ + c 4γ π τ ] + c 4γ F ρ θτ 3 S µ v ν N ef µν π no info on d 0,1 from CP-even pion photoproduction needs genuine CP-odd non-ptb information: fit to data or fit to lattice

23 Theta Term. d 0 and d 1 a b c d At NLO F 1 Q = d 1 + eg Aḡ 0 πf π F 0 Q = d 0 + eg Aḡ 0 πf π [L + log µ m + 5π π 4 [ ] 3π m π 4 m N ] m π g A ḡ 0 Q + e m N πf π 6m π 1 5π 4m N + h Q m π EDFF at various m π and Q allows to simultaneously fit ḡ 0, d 1,0 when more precision extract d 0,1 check ḡ 0 from symmetry = Tom & Taku s talk

24 Theta Term. Summary θ ḡ 0 ḡ 1 /Fπ d0,1 Q C 1, F πq m π 1 ε m π M M ε Q M NDA θ 10 3 F π symm. M Q Q M symmetry consideration very powerful at low order ḡ 0 well known and ḡ1 known with large errors no evident way to improve on ḡ 1 need experiment/lattice to determine d 0,1 four-nucleon C 1, are harder,... but power counting relegates them to subleading role Improving of lattice will allow to determine d 0,1 & check ḡ 0 in the near future

25 Quark CEDM L 6 = 1 qσµν g sg µν c 0 + iγ 5 τ 3 d3 q 1 qσµν g sg µν c 3 τ 3 + iγ 5 d0 q +r qiγ 5 d3 τ 3 ε q qcedm has CP-even chiral partner 1 qσ µν g sg µνq qσ µν iγ 5 τ g sg µνq isovector qcedm & isoscalar qcmdm 1 qσ µν iγ 5 g sg µνq qσ µν τ g sg µνq isoscalar qcedm & isovector qcmdm

26 Quark CEDM L 6 = 1 qσµν g sg µν c 0 + iγ 5 τ 3 d3 q 1 qσµν g sg µν c 3 τ 3 + iγ 5 d0 q +r qiγ 5 d3 τ 3 ε q qcedm has CP-even chiral partner 1 qσ µν g sg µνq qσ µν iγ 5 τ g sg µνq isovector qcedm & isoscalar qcmdm 1 qσ µν iγ 5 g sg µνq qσ µν τ g sg µνq isoscalar qcedm & isovector qcmdm d 3 causes vacuum misalignment re-alignment causes the appearance of a mass term r = 1 qσ µν g sg µνq = m π m qq m π c 0 need to know matrix elements of qσ µν iγ 5 G µνq and qiγ 5 q!

27 Quark CEDM L 6 = 1 qσµν g sg µν c 0 + iγ 5 τ 3 d3 q 1 qσµν g sg µν c 3 τ 3 + iγ 5 d0 q +r qiγ 5 d0 + d 3 τ 3 q qcedm has CP-even chiral partner 1 qσ µν g sg µνq qσ µν iγ 5 τ g sg µνq isovector qcedm & isoscalar qcmdm 1 qσ µν iγ 5 g sg µνq qσ µν τ g sg µνq isoscalar qcedm & isovector qcmdm if PQ solves strong CP problem θ d 0, d 3 isoscalar and isovector resume their original meaning r = 1 qσ µν g sg µνq = m π m qq m π c 0

28 Quark CEDM. Chiral Lagrangian and NDA ḡ 0 ḡ 1 /Fπ d0,1 Q C 1, Fπ Q θ m π 1 ε m π M M ε Q Q Q M M M M δ 0 M /T m π 1 ε m π M M ε Q Q Q M M M M δ 3 M /T m π Q Q ε 1 Q M M M M M δ 3 M /T m π ε m π Q Q M M 1 Q M M M assume d u m u, d d m d ; d m 0,3 = O δ 0,3 M /T Chiral Lagrangian very similar to θ but now iso-breaking! NDA NDA no PQ PQ if δ 0 δ 3, ḡ 0 ḡ 1, important for deuteron, N = Z nuclei

29 Quark CEDM. ḡ 0, and ḡ 1 no PQ mechanism ḡ 0 = δm d0 m N δm π d3 N c 3 m, π c 0 m ḡ 1 = m N m π d3 N m, π c 0 δm N correction to m n m p from c 3 m π, m N corrections to m π and sigma term from c 0

30 Quark CEDM. ḡ 0, and ḡ 1 PQ mechanism ḡ 0 = ḡ 1 = δm N correction to m n m p from c 3 m δm N + δm π N m π m N m N m π m π m π, m N corrections to m π and sigma term from c 0 ḡ 0 only depends on d 0 Do these hold beyond LO? c 3 d0, c 0 ε c 3 d3, c 0 ḡ 0 : yes for SU & SU3 loops violated by finite LECs ḡ 1 : yes for SU loops. SU3? violated by finite LECs ḡ 0,1 known if δm N, m π and m N

31 no symmetry relation; need lattice or experiment Quark CEDM. d 0 and d 1 a b c d F 1 Q = d 1 + eg Aḡ 0 [L πfπ + log µ m + 5π π 4 g A ḡ 0 Q +e πf π 1 5π + h 4m N F 0 Q = d 0 + eg Aḡ 0 πf π 6m π [ 3π 4 m π m N 1 + ḡ1 3ḡ 0 ] m π 1 + ḡ1 m N 5ḡ 0 Q m π ] ḡ 1 appears at NLO, only for d p EDFF at various m π and Q allows to simultaneously fit ḡ 0, d 1,0, & ḡ 1?

32 Four quark Left-Right Operators L 6 = ReΞ 1 qγ µ q qγ µq qγ µ γ 5 q qγ µγ 5 q τ τ τ 3 τ 3 +ImΞ 1 qγ µ q qγ µγ 5 q τ τ 3 more complicated transformation properties, 34 component of a symmetric tensor X = 1 4 τ i γ µ τ j γ µ τ i γ µ γ 5 τ j γ µγ 5 ɛ jkl τ k γ µ τ l γ µγ 5 ɛ ikl τ k γ µ τ l γ µγ 5 τ γ µ τ γ µ τ γ µ γ 5 τ γ µγ 5,

33 Four quark Left-Right Operators L 6 = ReΞ 1 X 44 X 33 ImΞ 1 X 34 +r LR qiγ 5 ImΞ 1 τ 3 εq more complicated transformation properties, 34 component of a symmetric tensor X = 1 4 τ i γ µ τ j γ µ τ i γ µ γ 5 τ j γ µγ 5 ɛ jkl τ k γ µ τ l γ µγ 5 ɛ ikl τ k γ µ τ l γ µγ 5 τ γ µ τ γ µ τ γ µ γ 5 τ γ µγ 5, ImΞ 1 causes vacuum misalignment re-alignment causes the appearance of a mass term r LR = LRm π m π m ReΞ 1

34 Four quark Left-Right Operators L 6 = ReΞ 1 X 44 X 33 ImΞ 1 X 34 +r LR qiγ 5 ImΞ 1 τ 3 q more complicated transformation properties, 34 component of a symmetric tensor X = 1 4 τ i γ µ τ j γ µ τ i γ µ γ 5 τ j γ µγ 5 ɛ jkl τ k γ µ τ l γ µγ 5 ɛ ikl τ k γ µ τ l γ µγ 5 τ γ µ τ γ µ τ γ µ γ 5 τ γ µγ 5, ImΞ 1 causes vacuum misalignment re-alignment causes the appearance of a mass term if PQ, no isoscalar component r LR = LRm π m π m ReΞ 1

35 Four-quark LR Operators. Chiral Lagrangian and NDA ḡ 0 ḡ 1 /Fπ d0,1 Q C 1, Fπ Q θ m π 1 ε m π M M ε Q Q Q M M M ξ M M M /T M ε 1 Q Q Q M M ξ M M /T M ε m π M M 1 Q Q Q M M NDA no PQ PQ isobreaking couplings are more important large three-pion coupling if PQ, no ḡ 0 at LO vector qiγ 5τ 3q vs tensor X 34 incomplete cancellation of π matrix elements, important mainly for nuclei e.g three-body force, large correction to ḡ 1

36 Four-quark LR operators. ḡ 0, ḡ 1 and no PQ mechanism LR m π ḡ 0 = δm N m π ḡ 1 = ImΞ 1 ReΞ 1, LR m N m N LR m π m π = LR m ImΞ 1 π ReΞ 1 ImΞ 1 ReΞ 1 LR m π, LR m N corrections to m π and sigma term from Re Ξ 1 if PQ, no ḡ 0 at LO

37 Four-quark LR operators. ḡ 0, ḡ 1 and PQ mechanism ḡ 0 = 0, = LR m ImΞ 1 π ReΞ 1 LR m ḡ 1 = π LR m N m N m π ImΞ 1 ReΞ 1 LR m π, LR m N corrections to m π and sigma term from Re Ξ 1 if PQ, no ḡ 0 at LO

38 Four-quark LR operators. ḡ 0, ḡ 1 and PQ mechanism ḡ 0 = 0, = LR m ImΞ 1 π ReΞ 1 LR m ḡ 1 = π LR m N m N m π ImΞ 1 ReΞ 1 LR m π, LR m N corrections to m π and sigma term from Re Ξ 1 if PQ, no ḡ 0 at LO ḡ 0,1 and known if LR m N, LR m π need to evaluate CP even four-quark already done? affected by loop corrections?

39 Chiral Invariant /T sources θ m π 1 ε m π M M w M m M /T M π M ε m π M ḡ 0 ḡ 1 /Fπ d0,1 Q C 1, F πq ε Q Q M M ε Q3 π Q M 3 M Q M Q M NDA NDA L = dw 6 gsf abc ε µναβ G a αβ Gb µρ Gc ν ρ + g s 4 ImΣ 1,8 [ qq qiγ 5 q qτ q qiγ 5 τ q] 1 1, t a t a no CP-even partner π-n couplings suppressed by m π nucleon EDFF dominated by d 0,1, momentum independent F 1 Q = d 1 + O Q M, F 0 Q = d 0 + O Q M. ḡ 0,1, C 1, should be important for light nuclei but found small de Vries, et al, 11; Bsaisou, et al, 14

40 Conclusions Connection EDM to BSM physics several, orthogonal EDMs e.g. d n, d p, d d... robust theory at different physics scales first principle determination of d n, d p & /T pion-nucleon couplings chiral symmetry provides powerful constraints θ ḡ 0 from θ determined by m n m p st qcedm ḡ 0 and ḡ 1 determined by corrections to meson and baryon spectrum induced by CP-even qcmdm FQLR ḡ 0, ḡ 1 & determined by CP-even FQLR operators no info from symmetry on d n, d p, genuine non-ptb CP-odd info needed no info on four-nucleon couplings C 1,, little info on subleading couplings ḡ 1 θ Lattice is or is getting there!