Resolving New Physics Puzzle in B to K Yoon Yeo Woong (Yonsei - PowerPoint PPT Presentation

Resolving New Physics Puzzle in B to Kπ Yoon Yeo Woong (Yonsei Univ.) Collaborated with CS.Kim, SC.Oh In Progressing

OUT Line � B � K π Puzzle � Model Independent Analysis � Numerical Analysis(In progressing) � Summary

Experimental Data of B � Kπ � HFAG since ICHEP06 − 6 Br (10 ) K π + 0 K π + 0 + − K π K π 0 0 Blue color: Preliminary

Experimental Data of B � Kπ � HFAG since ICHEP06 A CP K π + 0 + 0 K π + − K π C , S 0 0 π π K K S S 12 ± 0.11 0. 0.12 0.11 33 ± 0.21 0. 0.33 0.21 PDG 2006 Avg. 0.34 ± 0.28 PDG 2006 Avg. 0.08 ± 0.14

Model Independent Analysis

Diagram approach B � Kπ � Amplitude parameterization ⎡ ⎤ ⎛ 1 2 ⎞ ( ) ⎢ ∑ ⎥ ⎟ + 0 + * * ⎜ C C B → K π = AV V + V V P + EP − P + EP ⎟ A ⎜ ⎜ ⎟ ⎢ ⎥ ub us ib is ⎝ i i EW EW ⎠ 3 3 i i ⎣ ⎦ = i u c t , , ⎡ ⎤ ⎛ ⎞ 2 1 ∑ ( ) ⎢ ⎟ ⎥ ⎜ 0 + − * * C C B → K π = − V V T + V V P + EP + P + P − EP ⎟ A ⎜ ⎟ ⎜ ⎢ ⎥ ⎝ ⎠ ub us ib is i i EW EW EW 3 3 i i i ⎣ ⎦ i = u c t , , ⎡ ⎤ 1 ⎛ 2 2 ⎞ ( ) ∑ ⎢ ⎟ ⎥ ⎜ + → + π 0 = − * ( + + ) + * + + + C + C B K V V T C A V V P EP P P EP ⎟ A ⎜ ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ub us ib is i i EW EW EW 2 3 3 i i i ⎣ ⎦ = i u c t , , ⎡ ⎤ ⎛ ⎞ 1 1 1 ( ) ⎢ ∑ ⎥ ⎟ ⎜ 0 → 0 π 0 = − * − * + − − C − C B K V V C V V P EP P P EP ⎟ A ⎜ ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ub us ib is i i EW EW EW 2 3 3 i i i ⎣ ⎦ i = u c t , , � Adjust Parameter ⎛ 1 2 ⎞ � ( ) � ≡ * + + − − ⎟ T V V T P EP P EP ⎜ ≡ * + − − − C + C P | V V | P EP P EP P EP ⎟ ⎜ ⎜ ⎟ ub us u u c c ⎝ ⎠ tb ts t t c c EW EW 3 3 � ( ) C ≡ V V * C − P − EP + P + EP ( ) � * C P ≡ V V P + EP ub us u u c c EW tb ts EW EW � ( ) * A ≡ V V A + P + EP − P − EP ub us u u c c ( ) � C * C C P ≡ V V P − EP EW tb ts EW EW

Diagram approach B � Kπ � Final form 0 ( ) + + + → 0 π ≡ = − B K P A A ( ) + − 0 + − + − i α i γ i δ → π ≡ = − B K e P Te e A A T 1 ( ) ( ) + 0 + + + α γ δ γ δ δ → π 0 ≡ 0 i = − i i − i i + i B K e P Te e Ce e P e A A T C EW EW 2 1 ( ) ( ) 00 α γ δ δ 0 → 0 π 0 ≡ 00 i = − − i i + i B K e P Ce e P e A A C EW EW 2 C � We Neglect P , A EW � We set the the strong phase of P as 0 � all phase is relative to δ δ P P � There are 7 unknown parameters δ δ δ P T C P , , , , , , EW T C EW � we consider is given by other analysis γ α ij � ij are real, are phases of their amplitude A

New Physics comes in � Botella’s arguments Botella, Silva 2005 � For weak phase , always φ φ − η φ − θ sin( ) sin( ) φ θ η i = i − i e e e θ − η θ − η sin( ) sin( ) � We can choose arbitrary at will, for any given θ η φ , � We assume NP comes into P EW part(or C part) only N N N N N φ φ − γ P P sin P sin( ) N N N N i φ i δ i δ i γ i δ = − EW e e EW EW e e EW EW e EW EW EW EW γ γ 2 2 sin 2 sin ⎛ θ = γ ⎞ ⎟ ⎜ ⎟ ⎜ ⎟ Absorbed into EW ⎜ Absorbed into C ⎟ η = ⎜ 0 ⎟ ⎝ ⎠

New Physics comes in � NP term is absorbed 1 ( ) N N + 0 00 i γ i δ i δ N i φ i δ ⊃ − + + , Ce e P e P e e A A C EW EW EW EW EW 2 ⎛ ⎞ φ N φ N − γ 1 sin sin( ) ⎟ ⎜ N N i γ i δ i δ N i δ i γ N i δ = − Ce e + P e + P EW e e − P EW e ⎟ ⎜ C E W E W EW ⎟ ⎜ ⎟ ⎜ EW EW EW γ γ 2 ⎝ sin sin ⎠ 1 ( ) ′ ′ ′ ′ i γ e δ i i δ = − + C e P e C E W E W 2 φ N sin ′ = N ′ C e δ i − N i δ + i δ P EW e C e C E W C E W γ sin φ N − γ sin( ) ′ N ′ δ δ δ i = − N i + i P e P EW e P e E W E W E W EW EW EW γ sin

The analytic solution � Original Form does not change 0 ( ) + 0 + + → π ≡ = − B K P A A ( ) + − + − + − α γ δ 0 → π ≡ i = − i i B K e P Te e A A T 1 ( ) ( ) + 0 ′ ′ + + + α γ δ ′ γ δ ′ δ → π 0 ≡ 0 i = − i i − i i + i B K e P Te e C e e P e A A T C EW EW 2 1 ( ) ( ) 00 ′ ′ α ′ γ δ ′ δ 0 → 0 π 0 ≡ 00 i = − − i i + i B K e P C e e P e A A C EW EW 2 � If there Is NP ′ ′ ≠ δ ≠ δ C C , SM C C SM , ′ ( ) ′ ( ) ≠ δ ≠ δ P P , E W EW EM SM S M SM

The analytic solution � Step 1 - P T δ , , T 0 + = − P A + − + − α γ δ i = − i i e P Te e A T Br + = 0 P We neglect ⎡ ⎤ ⎛ ⎞ γ 0 + + − sin2 Br 1 Br ⎢ ⎥ ⎟ ⎜ δ = ± + − ⎟ cot 1 1 1 ⎜ 2 ⎛ ⎞ ⎢ ⎥ ⎟ + − + − A Br ⎜ T ⎟ + − + − + ⎜ 2 γ ⎝ 0 ⎠ A Br cos Br ⎟ ≈ ⎢ ⎥ CP 0.0008 ⎜ ⎣ ⎦ ⎟ CP ⎜ ⎜ ⎟ 0 + ⎝ γ Br ⎠ 2sin + − + + − + − = − 0 + γ δ T Br Br A Br cot cot CP T

The analytic solution � Step 2 - 00 00 α α , ( ) + 0 00 + α α γ δ ζ 0 i − 00 i = − i i ≡ i 2 e e 2 P Te e Xe A A T ( ) + 0 00 + α α − γ δ ζ 0 i − 00 i = − i i ≡ i 2 e e 2 P Te e Xe A A T 2 2 + 0 − 00 − 2 2 2 X A A 00 α − ζ = cos( ) 00 2 2 X A 2 2 + 0 00 2 2 − 2 − X A A 00 α − ζ = cos( ) 00 2 2 X A

The analytic solution ′ ′ ′ ′ � Step 3 - δ δ C , EW , , C EW 00 ′ ′ ′ ′ i γ i δ i δ i α − + = 00 + ≡ C e e P e 2 e P y C EW A EW ′ ′ 00 ′ − γ δ ′ δ α − i i + i = 00 i + ≡ C e e P e 2 e P y C EW A E W 1 1 ⎡ ( ) ⎤ ⎡ ( ) ⎤ ′ 2 2 2 = + − * = 00 − 00 00 α 00 − α 00 C y y 2Re y y Br cos A A ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 2 2 γ γ 4 sin sin 1 ⎡ ( ) ⎤ ′ 2 2 2 = + − * 2 i γ P y y 2Re y y e ⎢ ⎥ ⎣ ⎦ EW 2 γ 4sin − γ γ i − i − Re( ye ye ) − R e( y y ) ′ ′ δ = − δ = tan tan EW C − γ γ i − i − Im( ye ye ) Im( y y )

New Physics Test � Hierarchy Test ( ) ′ ′ y y e γ i 2 > 2 ⇒ * > EW C Im 0 ′ 2 > 2 T EW 1 ⎡ ( ) ⎤ 2 2 + − + + − + − γ − 0 + γ δ > + − * 2 i Br Br A Br cot cot y y 2Re y ye ⎢ ⎥ ⎣ ⎦ CP T 2 γ 4 sin � Strong phase Test δ = δ ⇒ 00 α 00 = 00 α 00 cos cos A A C P δ = δ T EW ⎡ ⎤ ⎛ ⎞ γ 0 + + − − i γ − i γ sin2 Br 1 Br Im( ye ye ) ⎢ ⎟ ⎥ ⎜ ± + − = − 1 1 1 ⎟ ⎜ ⎢ ⎥ ⎟ ⎜ + − + − + − γ γ 2 γ ⎝ 0 ⎠ i − i A Br cos Br Re( ye ye ) ⎢ ⎥ ⎣ ⎦ CP

Numerical Analysis Isospin quadrangle 0.8 σ violation 00 A A + 0 A + − A + 0 The solution within 0.8 σ variation δ = − = = 0.22 P 4.79 T 0.96 T ⎡ ′ ′ ′ ′ = = = = C 1.05, P 1.54, δ 2.89, δ 1.52 ⎢ EW C EW ⎢ ′ ′ ′ ′ = = δ = δ = C 2.10, P 0.63, 3.01 , 2.12 ⎢ EW C EW ⎢ ′ ′ ′ ′ = = δ = δ = − C 1.22, P 1.10, 0.06, 0.49 ⎢ EW C EW ⎢ ′ ′ ′ ′ = = = − = − ⎢ δ δ C 0.14, P 1.62, 0.19, 1.22 ⎣ EW C EW

Summary � Even there is New Physics, by using botella’s argument they changed into SM form. � We can get each topological amplitude analytically step by step. � For given Theoretical calculations, such as QCDF, pQCD we can resolve New physics part analytically – so we can resolve NP puzzle through hierarchical test and Strong phase comparison.