Resolving BCP puzzles in QCD factorization An overview

Resolving B-CP puzzles in QCD factorization - An overview of charmless Hai-Yang Cheng hadronic B decays Academia Sinica Based on 3 papers with Chun-Khiang Chua Ø B-CP puzzles in QCDF Ø Bu, d decays Ø B (K, K*, K 0*, K 2*)( , ’) May 6, 2010 at NTHU 1

Day in the life – The Emperor’s Tea : Murayama

Direct CP asymmetries K - + + - ACP(%) -9. 8+1. 2 -1. 1 38 6 S 8. 5 6. 3 K - K*0 -37 9 19 5 AK 14. 8 2. 8 5. 3 4. 1 3. 8 K- 0 37 11 -13 4 3. 4 3. 3 AK =ACP(K- 0) – ACP(K- +) K*- + +K- K- 0 ACP(%) -18 7 15 6 5. 0 2. 5 S 2. 6 2. 5 2. 0 - 0 0 - + -13 7 43+25 -24 11 6 1. 9 1. 8 CDF: ACP(Bs K+ -)=0. 39 0. 17 (2. 3 ) 3

In heavy quark limit, decay amplitude is factorizable, expressed in terms of form factors and decay constants. Encounter several difficulties: Rate deficit puzzle: BFs are too small for penguin-dominated PP, VV modes and for tree-dominated decays 0 0, 0 0 CP puzzle: CP asymmetries for K- +, K*- +, K- 0, + - are wrong in signs Polarization puzzle: f. T in penguin-dominated B VV decays is too small 1/mb power corrections ! 4

A(B 0 K- +) ua 1+ c(a 4 c+r a 6 Theory c) Br 13. 1 x 10 -6 ACP 0. 04 Expt (19. 4 0. 6)x 10 -6 -0. 098+0. 012 -0. 011 Im 4 c 0. 013 wrong sign for ACP 4 c charming penguin, FSI penguin annihilation 1/mb corrections penguin annihilation 5

has endpoint divergence: XA and XA 2 with XA 10 dy/y Beneke, Buchalla, Neubert, Sachrajda Adjust A and A to fit BRs and ACP A 1. 10, A -50 o Im( 4 c+ 3 c) -0. 039 6

New CP puzzles ACP(%) S K - + + - AK -9. 8+1. 2 -1. 1 38 6 8. 5 K - K*0 K- 0 14. 8 2. 8 -37 9 19 5 37 11 -13 4 6. 3 5. 3 4. 1 3. 8 3. 4 3. 3 mb 3. 3 PA ( 1. 9) K*- + +K- K- 0 ACP(%) -18 7 15 6 5. 0 2. 5 S 2. 6 2. 5 2. 0 - 0 0 - + -13 7 43+25 -24 11 6 1. 9 1. 8 mb PA Penguin annihilation solves CP puzzles for K- +, + -, …, but in the meantime introduces new CP puzzles for K- , K*0 , … Also true in SCET with penguin annihilation replaced by charming penguin 7

All “problematic” modes receive contributions from c’=C’+P’EW. AK 0 if c’ is negligible T’ a 1, C’ a 2, P’EW (-a 7+a 9), P’c. EW (a 10+r a 8), t’=T+P’c. EW AK puzzle is resolved, provided c’/t’ ~ 1. 3 -1. 4 with a large negative phase (naively, |c’/t’| 0. 9) a large complex C’ or P’EW Large complex C’: Charng, Li, Mishima; Kim, Oh, Yu; Gronau, Rosner; … Large complex P’EW needs New Physics for new strong & weak phases Yoshikawa; Buras et al. ; Baek, London; G. Hou et al. ; Soni et al. ; Khalil et al.

The two distinct scenarios can be tested in tree-dominated modes where PEW<<C. CP puzzles of - , 0 0 & large rates of 0 0, 0 0 cannot be explained by a large PEW Power corrections have been systematically studied by n Beneke, Neubert: S 2, S 4 n Ciuchini et al. , 0801. 0341 n Duraisamy & Kagan, 0812. 3162 n Li & Mishima, 0901. 1272 9

a 2 a 2[1+ Cexp(i C)] C 1. 3, C -70 o for PP modes a 2(K ) 0. 51 exp(-i 58 o) Two possible sources: n spectator interactions NNLO calculations of V 2 & H 2 are now available [Bell, Pilipp] Real part of a 2 comes from H and imaginary part from vertex a 2( ) 0. 33 - 0. 09 i =0. 34 exp(-i 15 o) for B = 250 Me. V a 2(K ) 0. 51 exp(-i 58 o) H = 4. 9 & H -77 o n final-state rescattering [C. K. Chua] Neubert: In the presence of soft FSIs, there is no color suppression 10 of C w. r. t. T

K - + + - AK K - K*0 K- 0 ACP(%) -9. 8+1. 2 -1. 1 38 6 14. 8 2. 8 -37 9 19 5 37 11 -13 4 S 8. 5 6. 3 5. 3 4. 1 3. 8 3. 4 3. 3 mb 3. 3 PA ( 1. 9) large complex a 2 K*- + +K- K- 0 ACP(%) -18 7 15 6 5. 0 2. 5 S 2. 6 2. 5 2. 0 - 0 0 - + -13 7 43+25 -24 11 6 1. 9 1. 8 mb PA large complex a 2 All new CP puzzles are resolved ! 11

B - K - 0 1= a 1, 2= a 2 A(B 0 K- +) = A K( pu 1+ 4 p+ 3 p)= t’+p’ 2 A(B- K- 0) = A K( pu 1+ 4 p+ 3 p)+AK ( pu 2+3/2 3, EWp)= t’+p’+c’ In absence of C’ and P’EW, K- 0 and K- + have similar CP violation arg(a 2)=-58 o mb penguin ann large complex a 2 Expt ACP(K- 0)(%) 7. 3 -5. 5 4. 9+5. 9 -5. 8 5. 0 2. 5 AK (%) 3. 3 1. 9 12. 3+3. 0 -4. 8 14. 8 2. 8 12

(B PP) A=1. 10 A= -50 o C=1. 3 C= -70 o Large K ’ rates are naturally accounted for in QCDF 13 partial NLO

B K(*) , K(*) ’ BRs in units of 10 -6 In q & s flavor basis ( q=(uu+dd)/√ 2, s=ss) =39. 3 Interference between (b) & (c) K ’ K For K*, (b) is governed by a 4 -r a 6, (c) by a 4; a 4, a 6 are negative and |a 4|< |a 6|; chiral factor r is of order unity additional sign difference between (b) & (c) for K* (‘) 14

P(B PP)(%) C Several SCET predictions are in conflict with experiment 15

B 0 K 0 0 A(B- K 0 -) = A K( 4 p+ 3 p) = p’ 2 A(B 0 K 0 0) = A K(- 4 p- 3 p) + AK ( pu 2+ pc 3/2 3, EWc) = -p’+c’ In absence of C’ and P’EW, K 0 + and K 0 0 have similar CP violation of both K 0 - & K 0 0 is naively expected to be very small A’K =ACP(K 0 0) – ACP(K 0 -) = 2 sin Imr. C+… - AK mb penguin ann large complex a 2 Expt -1 10 ACP(K 0 0)(%) -4. 0 0. 75 -10. 6+6. 2 -5. 7 A’K (%) -4. 7 0. 57 -11. 0+6. 1 -5. 7 -- Ba. Bar: -0. 13 0. 03, Belle: 0. 14 0. 13 0. 06 for ACP(K 0 0) ACP (K 0 0)= -0. 15 0. 04 ACP (K 0 0)=-0. 073 0. 041 Toplogical quark diagram approach An observation of ACP(K 0 0) ACP (K 0 0)= -0. 08 -0. 12 16 - (0. 10 0. 15) power corrections to c’

B - K - n Destructive interference penguin amp is comparable to tree amp more sizable CP asymmetry in K than K ’ n Although f c=-2 Me. V is very small compared to f q = 107 Me. V, f s = -112 Me. V , it is CKM enhanced by Vcb. Vcs*/(Vub. Vus*) mb penguin ann ACP(K- )(%) -23. 3 12. 7 -2. 0 -14. 5 -37 9 ACP( - )(%) -11. 4 -5. 0 -13 7 large complex a 2 (w/o a 2 (with charm) Expt n Charm content of plays a crucial role for ACP(K- ), but not for ACP( - ) n Prediction of ACP(K- ) still falls short of data 17

p. QCD prediction is very sensitive to mqq, mass of q ACP(K- ) = 0. 0562, 0. 0588, -0. 3064 for mqq= 0. 14, 0. 18, 0. 22 Ge. V Akeroyd, Chen, Geng Two issues: (i) with anomaly: (ii) stability w. r. t. mqq Xiao et al. (0807. 4265) reply on NLO corrections to get a correct sign: ACP(K- )= 0. 093 to LO, (-11. 7+8. 4 -11. 4)% at NLO 1). If NLO effects flip the sign of ACP, p. QCD calculations should be done consistently to NLO 2). Missing parts of NLO: hard spectator & weak annihilation 18

Time-dependent CP asymmetries: SB PP Ø QCDF prediction for S( + -) agrees well with data Ø S( ’KS) is theoretically very clean in QCDF & SCET but not so in p. QCD Ø Around 2005, CCS and Beneke got S( ’KS) 0. 74 in QCDF. Why 0. 67 this time ? 19

sin 2 extracted from charmonium data is 0. 725 circ 2005, and 0. 672 0. 023 today. It is more sensible to consider the difference Sf = - f. Sf - sin 2 Sf = 2|rf|cos 2 sin cos f with rf=( u. Afu)/( c. Afc) small and could be + or – S Ks positive 20

B VV decays n Branching fractions tree-dominated decays: VV>PV>VP>PP (due to f. V > f. P) penguin-dominated decays: PP>PV~VV>VP (due to amplitudes a 4+r Pa 6, a 4+r Va 6, a 4 -r Pa 6, a 4+r Va 6 n Polarization puzzle in charmless B→VV decays In transversity basis Why is f. T so sizable ~ 0. 5 in B→ K*Á decays ? 21

A 00 >> A-- >> A++ 22

B→ K*Á ® 3=a 3+a 5, h=0 ® 4=a 4 -r Áa 6, ® 3, EW=a 9+a 7, ¯ 3= penguin ann h= - h=0 h= - Coefficients are helicity dependent ! with ¯ 3=0 constructive (destructive) interference in A- (A 0) ⇒ f. L¼ 0. 58 Yang, HYC NLO corrections alone can lower f. L and enhance f. T significantly ! 23

Although f. L is reduced to 60% level, polarization puzzle is not resolved as the predicted rate, BR» 4. 3£ 10 -6, is too small compared to the data, » 10£ 10 -6 for B →K*Á (S-P)(S+P) Kagan (S-P)(S+P) penguin annihilation contributes to A-- & A 00 with similar amount n Br & f. L are fitted by ½A=0. 60, ÁA= -50 o f || ¼ f? » 0. 25 24

• A=0. 78, A=-43 o for K* , A=0. 65, A=-53 o for K* • Rate for 0 is very small However, p. QCD prediction is larger than QCDF by a factor of 20 ! • Br(B 0 K*0 K*0)=1. 28+0. 35 -0. 30 0. 11 by Ba. Bar, 0. 3 0. 1 by Belle • Br(B 0 0 0)=0. 9+1. 5+2. 4 -2. 6 -1. 5 is obtained with C=0 soft corrections to a 2 are large for PP, moderate for VP and very small for VV r V<<r P doesn’t help! or due to Goldstone nature of the pion ? [Duraisamy, Kagan] 25

Conclusions u In QCDF one needs two 1/mb power corrections (one to penguin annihilation, one to color-suppressed tree amplitude) to explain decay rates and resolve CP puzzles. u CP asymmetries are the best places to discriminate between different models. 26

0 << - is expected due to near cancellation of a 2. Belle’s result 0 ’ > - ’ is the other way around 27

Spare slides 28

(B VP) A(VP)=1. 07 A(VP)= -700 A(PV)=0. 87 o A(PV)= -30 A(K )=0. 70 o A(K )= -40 C=0. 8 C= -70 o Belle: C. C. Chiang Br(B- -)=Br(B- -) sin 2 is an - mixing angle 3. 3 o 29

(B VP) A(VP)=1. 09 A(VP)= -700 A(PV)=0. 87 o A(PV)= -30 A(K )=0. 70 o A(K )= -40 C=0. 8 C= -70 o for K* K* ’ rates too small compared toby Ba. Bar but In QCDF heavypredictions quark limit, are too small (15 50)%, while **0 consistent with Belle: Br(K ’)<2. 9, Br(K ’)<2. 6 K are too small by a factor of 2 3 A( K*)> A( K*) 30

P(B VP)(%) C *- +)= -2 sin Imr (K* )+…. 0. 137 AK* =ACP (KA*- 0(K ) -*0 A (Kin CP c Data of ) is better agreement with QCDF than p. QCD CP *0 -, -K 0 have K A’ *0 0 small A *0 as they are pure * penguin processes =ACP(KThe )-SCET ACP(KCP )= 2 sin Imr ) +… -0. 111 K* SCET. & predictions arec(K ruled out by experiment. 31

SB VP Expt’l errors of S 0 0 are very large SB VP S K is negative and sensitive to soft corrections on a 2 32

• 0 is expected to have larger f. T as its tree contribution is small • b d penguin-dominated modes K*0 K*0, K*0 K*- are expected to have f. L 0. 5. Experimentally, f. L 0. 75 -0. 80 (why ? ) • For K*- 0, recent Ba. Bar measurement gives f. L=0. 9 0. 2 with 2. 5 significance • QCDF leads to 33