1. Learning  ( 0 ) from B decays Chuan-Hung Chen Department of Physics, National Cheng-Kung University, Tainan, Taiwan  Introduction & Our question   -  0 mixing  B! K (*)  (0) decays  Discussion

2.  Introduction: what questions can we ask in B Physics? Determine the CP violating phases: A 3 R b e - i  3 A 3 R t e -i  1 Precision measurement  Find new CP violating sources b d tt b d

3. Test standard model & search for the new effects: Extra dimensions Noncommutative spaces … Supersymmetric models … Grand unified theories (GUTs) … Test QCD approach QCD factorization approach, perturbative QCD approach… Do final state interactions play important role in B decays? Clarify and find new states or new decay modes, such as D sJ (2317), D sJ (2457), B! X(3872) K,… Why P L » P T in B!  K * decays? Belle Collaboration

4. Our question: Br(B !  0 K 0 ) » 35 £ 10 -6 Theoretical estimation C.W. Chiang etal., hep-ph/0404037 How to understand? (in units of 10 -6 )

5.  0 mixing: According to quark model with SU(3) flavor symmetry, the mesonic states could be obtained by 00 88 11 If U A (1) is a good symmetry and quarks are massless, there are nine Goldstone bosons Since m s >>m u,d,  8 and  1 will mix. However, U A (1) is broken by anomaly,  1 cannot be a Goldstone boson, m  0 =958 MeV; m  =547 MeV

6. With a parton Fock state decomposition, One-angle scheme: The decay constants, defined as will have the relation

7. Combining P !   J/  ! P  decays and P  transition form factors, it has been shown that one-angle parametrization cannot match with the results of  ’Pt and experiments R J/  =5.0 ± 0.8

8. Therefore, two-angle scheme is introduced. Leutwyler, NPPS64, 223(‘98)

9. Another quark-flavor scheme is introduced, T. Feldmann, P. Kroll, B. Stech, PRD58,114006(98); PLB449,339 (99)

10. Why do we need another scheme? In B decays, we need to deal with the matrix elements, for instance, If we know the matrix element for axial current, It seems can be obtained in terms of equation of motion But, m s ! 0, f and M  ’  0 For displaying the SU(3) limit explicitly, it is better to use bases  qq and  ss We can have mass matrix

11. Free parameters: The mass matrix can be diagonalized via  8 and  1 are not independent

12. B d ! K 0  (0) decays Effective interactions for b ! s qq u b u s Tree W V ub V us penguin b s qq t WV tb V ts g Effective operators Tree u b u s g penguin b s qq g Penguin Hence, b s qq V ± A V-A C 3-6 u C 1,2 b u s V-A

13. Topologies for B d ! K 0  ( 0 ) Since V ub V us <<V tb V ts, penguin dominates. Penguin emission b s d d B (0)(0) K V-AV± A (a) bs d,u B (0)(0) K V-A V± A (c) Tree’s contributions are similar to (c) except the CKM matrix elements b s ss K V-A V± A (0)(0) (d) B Penguin annihilation d b B K V-A V± A d s d d (0)(0) (e) b d B K V-A V± A d s s s (0)(0) (f) Usually, (e), (f) < (a), (b), (c), (d) b s s s B (0)(0) K V-AV± A (b)

14. Hadronic matrix elements: b s d d B (0)(0) K V-A (a) Only show the factorizable effects b s d d B (0)(0) K V-AV+A (a) (V-A)  (V+A)=-2(S-P)  (S+P) b s s s B (0)(0) K V-A (b) b s s s B (0)(0) K V-AV+A (b)

15. (0)(0) b s d,u B K V-A (c) b s d,u B (0)(0) K V-A V+A (c) bs ss K V-A (0)(0) (d) B bs ss K V-A V+A (0)(0) (d) B (cont’ed)

16.  In order to calculate hadronic matrix elements, such as we need to know the wave functions of B, K,  q and  s Numerical analyses:  The wave functions of B and K meson have been studied in the literature.  We have to assume that  q and  s have the same asymptotic behavior as those of  -meson.

17. (cont’ed) T. Feldmann, P. Kroll, B. Stech, PRD58,114006(98); PLB449,339 (99) Another way to understand above assumptions, we can use the mass matrix of octet-singlet, in which we know M 2 88 =(4m 2 K -m 2  )/3, Gell-Mann-Okubo relation By basis rotations, we obtain M 2 88 =(2m 2 ss +m 2 qq )/3, if we set m qq =m , we get m 2 ss =2m 2 K -m 2 . If one takes recourse to the first order of flavor symmetry breaking, one expects Angle  T. Feldmann and P. Kroll, hep-ph/0201044

18. (cont’ed) In the framework of perturbative QCD However, F 0 B ! K (0)=0.35 ± 0.05 ) Taking the conventional values of f q(s) and m 0 q(s) cannot enhance B d !  0 K 0

19. (cont’ed)  By M 2 88 =(2m 2 ss +m 2 qq )/3, why not m 2 ss =2(m 2 K -m 2  ) and m 2 qq =3m 2  ? T. Feldmann and P. Kroll, hep-ph/0201044  And also Maybe we should take f q >f  and m qq >m   Or M 2 88 =(4m 2 K -m 2  )/3 (1+ , why not m 2 qq > m 2  ? Anomaly

20. (cont’ed) Taking m qq » 1.65 m  GeV, f q =1.07f , Besides, we also calculate B d !  ( 0 ) K *0 F + B ! K =0.38

21. Other contributions: Intrinsic charm-quark T. Feldmann, P. Kroll, B. Stech, PRD58,114006(98) Two-gluon content

22. C.S. Kim et al., hep-ph/0305032 M. Beneke and M. Neubert, Nucl.Phys. B651 (2003) 225-248 F + B!  (0) » 0.21 (0.32) F + B !  0 (0) » 0.32 (0.27)

23. Discussion:  If taking m qq > m , the branching ratio of the decay B d !  0 K 0 could be enhanced efficiently.  The considering effects could be distinguished from BN’s two-gluon mechanism, in which With our consideration,  The possible exotic mechanisms can be also tested by B !  ( 0 ) ℓ ℓ decays, in which the original BR is order of 10 -8. With BN’s two-gluon content the BR could reach to 10 -7 that is the same order of magnitude as the decays B ! K ℓ ℓ, measured by Belle and Babar. B !  ( 0 ) decays could be the candidates

24.  The more serious one: Babar Collaboration, hep-ex/0403046 B Thero. << 10 -6  One phenomenon is worth noticing (mild), i.e. why is the branching ratio of B + !  0 K + so high? We expect B(B + !  0 K + )/B(B 0 !  0 K 0 ) »  (B + }/  (B 0 ) =1.08 1.48 1.19 Final state interactions ? or “New” effects ?

26. M. Beneke and M. Neubert, Nucl.Phys. B651 (2003) 225-248

27. Babar Collaboration, hep-ph/0308015

28. Two-angle scheme

30. As a result, the decay constants, defined as With a parton Fock state decomposition, For simplicity, we can redefine the wave functions as Hence,

31. b d tt b d

32. Babar Collaboration, hep-ex/0403046