Hot Topics in decays?

Transcription

1 B Kπ Hot opics in decays? 1) Is there any puzzle of new physics in B Kπ decays? (arxiv: ) γ 2) Extraction of via iso-spin analysis of B Kπ decays (arxiv: ). S. Kim (Yonsei Univ.) ollaboration with Sechul Oh, Y.W. Yoon

2 B Kπ Puzzle Branching Ratios HFAG, September 2007 Fleischer Hep-ph/ At September 2007 Rc = 1.12 ± 0.07 Rn = 0.98 ± 0.08

3 B Kπ Puzzle P Asymmetries HFAG, September 2007 A ( B K π ) A ( B K π ) = 0.15 ± 0.03 P P (sin 2 β) (sin 2 β) = 0.30 ± 0.19 K S π 0 ccs

4 Quark Diagram Approach in B Kπ Amplitude parameterization ( Kπ ) AB AB 0 + ( ) AB ( ) ( Kπ ) 2AB with re-definition of = P + A K π = P P K π = P P P A = P P 1 1 P + EP P EP P 3 3 A + EP A

5 Quark Diagram Approach in B Kπ Hierarchy between the parameters 1 λ 2 λ 3 λ P = VV P + V V P P + P P tc, P, P A, P uc * * tb ts tc ub us uc tc uc λ 2 λ 4 P Gm (, k, μ) Gm (, k, μ) P ( ) ln Gm (, k, μ) uc u c 2 tc 2 μ Ext W 2 2 k / m b P cu / P tu P / P uc c tc 2 1 k 1 4 m 2 2 b Buras, Fleischer PLB (1995) P P uc tc ishima, Yoshikawa PRD (2004) > P

6 Quark Diagram Approach in B Kπ Final form + AB ( K 0 + π ) A 0 + = P iα iγ iδ AB ( K π ) A e = P re e ( ) i α ( i γ i δ i γ i δ i δ π = 1 + ) iα 1 iγ iδ iδ ( π ) = ( 1 + ) AB K A e P re e re e r e (1 ) AB K Ae P re e r e 2 We eglect P, P, A uc We set the strong phase of P to be zero all phase is relative to it We hold 7 unknown parameters P, r, r, r, δ, δ, δ We use γ value given by other analysis ij ij A are real and positive, α are phases of their amplitude P P = P, r =, r =, r = tc P tc P tc Ptc

7 Re-Parameterization Invariance Re-parametrization Invariance For any phase φ Botella, Silva 2005 sin( φ η) sin( φ θ) e = e e sin( θ η) sin( θ η) iφ iθ iη We can choose arbitrary θ, η at will, for any given We assume P comes into P part(or part) r r sin φ r sin( φ γ) e e = e e e 2 2 sin γ 2 sin γ iφ iδ iδ iγ iδ θ = γ η = 0 Absorbed into Absorbed into φ

8 Re-Parameterization Invariance P term is absorbed into S term iγ iδ iδ iφ iδ ( r e e ) + A, A P r e e + r e γ δ δ sin φ iδ iγ sin( φ γ) P r e e r e r e e r e 2 sinγ sinγ i i i iδ = = P e e + e 2 r r ( iγ iδ iδ r ) re W δ i δ i e δ i e = re r i E W i e = r e r sin φ sin γ δ δ sin( φ γ) i e δ sin γ

9 Analytic Solution Original Form does not change ( π ) AB K 0 A 0 = P iα iγ iδ ( π ) = AB K A e P(1 re e ) ( ) ( ) i α i γ i i i i e δ e γ δ π = 1 + e δ AB K A e P re r e r 2 If there is P iα iγ iδ ( ) ( ) iδ π = P 1 r e + re We AB K Ae e r r ( r ), δ ( δ ) S S ( r ), δ ( δ ) S S

10 Analytic Solution Step 1 -,, P r δ reject A = P 0+ + A e P (1 r e e ) + iα iγ iδ = R + B τ + B = 0.90 ± B τ B 0 R = 1+ r 2r cosδ cosγ + P 2 A R = 2r sinδ sinγ 0 P Br + γ + 2 sin 2 1 P R δ R A cot = ± AP R γ 1 2 ( ) cos γ 2sin r = R(1 A cot γcot δ ) 1 + P P = (49.9 ± 1.1)eV r = 0.14 ± 0.07 δ = 20 ± 12

11 Analytic Solution Step 2 - α α α, α ( ) ( ) 2 = (2 ) A e A e P r e e xe iα iα iγ iδ iζ 2 = (2 ) + A e A e P r e e xe iα iα iγ iδ iζ A A P x = ζ ± 2 2 Arcos A P x A A P x = ζ ± 2 2 Arcos A P x A +0 A +0 x 00 A 00 A wo-fold ambiguity occurs. 2 X 2 = 4 fold ambiguities in tatal.

12 Analytic Solution One Isospin quadrangle two fold discrete ambiguity

13 Analytic Solution Step 3 - r, r, δ, δ A r e + = e P + ye A r e + = e P + y r r δ δ iγ iδ iδ iα iη e re iγ iδ iδ iα iη e re We 2 1 e y + y = yy cos( η η) sin γ y + y = yy cos(2 γ + η η) sin γ 2 y η y η cos cos = Arcan ysin η y sin η y η γ y η γ cos( ) cos( + ) = Arcan y η γ y η + γ sin( ) sin( ) y r e r iγ γ γ y o discrete ambiguity

14 Analytic Solution 4 different solutions for r, r, δ, δ We reject ase 3 due to large prediction he S estimate ase 2: Large ase 4: Large S K π s 0 S π 0 = 0.38 ± 0.19(data) r = 0.12 > r = 0.039, δ 61, δ = 22 K S

15 Analytic Solution Step 4 - solutions for P term r r δ i δ i e δ i e = re r i E W i e = r e r sin φ sin γ δ δ sin( φ γ) i e δ sin γ 4 Equations VS 7 unknowns - r, r, δ, δ, r, φ, δ eed at least 3 additional inputs to fix P terms

16 Additional Inputs a) Additional inputs from Flavour SU(3) Sym. From B ππ decays Assuming no P in B ππ HFAG, September 2007

17 Additional Inputs Additional inputs from Flavour SU(3) Sym. B ππ parametrization hisq-fitting with 5 measurements iγ iδ iγ iδ ( π π ) = + 2 AB ( e e e e ) 0 AB π π = e e + Pe + iγ iδ iβ ( ) ( ) iγ iδ iβ 2 AB ( ππ) = ( e e Pe ) + 3 Br, P( ), with 5 parameters A ππ S + π π δ δ P Vus = = (3.8 ± 0.4)eV Vud iδ 3 c9 + c10 1 iδ iδ re = ( r e + r e ) 2 2 c + c λ R 1 2 Gronau, Pirjol, Yan (1999) b A ( ππ)=0.48 (data) P 0.31 ( r, δ ) = (0.076 ± 0.008, 14 ± 15 ) ( r, δ ) = ( 0.14 ± 0.04, 8 ± 10 )

18 Additional Inputs b) Additional inputs from PQD result Li, ishima, Sanda, PRD72, (2005) ( r, δ ) = (0.039, 61 ) ( r, δ ) = (0.12, 22 )

19 Determining P parameters Solution for P term with additional inputs Defining iδδ iδ iδ Δre r e re iδδ iδ iδ Δr e r e r e sin φ Δ re = r e sin γ iδδ iδ sin( φ γ) Δ re = r e sin γ iδδ iδ Δr Δr δ δ or δ π r = Δ Δ = sin φ sin( φ γ) sin γ = Δr sin φ With inputs from SU(3) sym. With inputs from PQD results ases 2&4 are suitable and consistent each other between two methods.

20 Discussions Dependance on γ Dependance on S K S π 0

21 Summary Due to the Re-parametrization Invariance(RI) the P terms absorbed into the S terms and P in pair. In order to extract P parameters we need at least 3 additional inputs. We could pin down each hadronic parameter under four-fold discrete ambiguity using analytic method. And also P parameter for given additional inputs Results shows that there should be quite large P contribution with maximal weak phase π /2

22 K EXRAIO OF γ ISOSPI AALYSIS OF DEAYS VIA B Kπ

23 HISORY OF EXRAIG γ B ± 0 ± D K decays Gronau, London, Wyler (1991) Weak phase γ u Amp * ~ V cb V us Amp * ~ V ub V cs olor suppressed olor allowed But the triangles are quite squashed!

24 HISORY OF EXRAIG γ GLW From B + K + D, K + D, K + D and their P conjugates + he triangles are quite squashed ADS Gronau, Wyler (1991), Gronau, London (1991) Atwood, Dunietz, Soni (1997) Improvement of GLW method by seeing and 0 + B K D ( K π ). Enhancement of since AD ( 0 + K π )/ AD ( 0 + K π ) λ 2 GGSZGiri, Grossman, Soffer, Zupan (2003) 0 + Dalitz analysis from D K π π ombined all methods : γ = 88 ± 16 S Recent Ufit result B K D 0 ( K π ) + AB K D K π 0 + ( ( ))

25 HISORY OF EXRAIG γ B Kπ, ππ and SU(3) Flavour Symmetry Gronau, Rosner, London (1994); Buras Fleischer (1995); Deshpande, He (1995) Gronau, Hernandez, London, Rosner (1995); Fleischer (1996); eubert Rosner (1998); Buras, Fleischer (1999); Gronau, Rosner (2002) AB K = AB K = P 0 0 ( + π + ) ( π ) tc 0 i AB ( K + π ) = tc e γ P AB 0 ( K + π ) AB ( K π ) 2γ AB 0 ( K π + )

26 HISORY OF EXRAIG γ Fleischer annel Bound Fleischer, annel (1998) 0 < γ < γ, or 180 γ < γ < R γ = arccos 1 R 0 = τ τ B B + 0 Br B 0 + ( K ) ( ) Br B K π π = 72 = 0.90 ± 0.05

27 ISOSPI AALYSIS I B Kπ DEAYS Kπ Isospin Relation for Decays B AB K AB K AB K AB K ( π ) ( π ) = 2 ( π ) 2 ( π ) AB K AB K AB K AB K ( π ) ( π ) = 2 ( π ) 2 ( π ) One can make two quadrangle in complex plane from these two relations θ 2γ A A B K π ij ( +,0) i j ( )

28 ISOSPI AALYSIS I B Kπ DEAYS Determination of a quadrangle: 4 lengths + 1 angle All lengths can be obtained by 4 Br s and 4 Acp s. θ S from gives angle constrain on quadrangles K S π wo quadrangles are determined once is given γ θ We get function in terms of θ θ 2γ ixing induced PV S K S π 2 A A = A + A iθ 2iβ Im( e e ) A A B K π ij ( +,0) i j ( )

29 ISOSPI AALYSIS I B Kπ DEAYS ixing induced PV SK Sπ iddle value Error S ccs

30 DISREE ABIGUIY 0 i AB ( K + π ) = Ptc e γ 4 ambiguity 2γ 2γ 2γ 2γ 4ambiguity for two quadrangle θ 2γ

31 DISREE ABIGUIY otally Sixteen-fold Discrete Ambiguity occurs

32 EXRAIO OF γ S K S π = 2 A A A + A = 0.38 ± 0.19 iθ 2iβ Im( e e ) γ θ = 20 ± 12 or 115 ± 12 P Dashed line : > 1 (unreasonable) tc θ γ A1: A4: B2 and 2: B3: = γ = 71 γ = γ = 109 γ = or

33 EXRAIO OF γ Fleischer and annel Bound 0 < γ < γ, 180 γ < γ < γ R γ 0 = arccos 1 R = τ + Br( B K π ) B = = 0.90 ± τ 0Br( B K π ) B 0 < γ < 72, 108 < γ < 180 P Dashed line : > 1 (unreasonable) tc θ Our constraint on is much stronger γ

34 OLUSIO We adopt 1 Isospin relation in B Kπ decays. 2 Quark Diagram Approach in and it s charge conjugate modes B K π, K π All nine measurements of B Kπ is required to extract γ in this framework. ixing induced P asymmetry plays a crucial role. γ S K S π 0 We extract from only B Kπ decay system without using SU(3) flavor sym. as follows γ = 71 or