SLIDE 1
Resolving New Physics Puzzle in B to Kπ
Yoon Yeo Woong (Yonsei Univ.) Collaborated with CS.Kim, SC.Oh In Progressing
Resolving New Physics Puzzle in B to K Yoon Yeo Woong (Yonsei - - PowerPoint PPT Presentation
Resolving New Physics Puzzle in B to K Yoon Yeo Woong (Yonsei - - PowerPoint PPT Presentation
Resolving New Physics Puzzle in B to K Yoon Yeo Woong (Yonsei Univ.) Collaborated with CS.Kim, SC.Oh In Progressing OUT Line B K Puzzle Model Independent Analysis Numerical Analysis(In progressing) Summary Experimental
SLIDE 2
SLIDE 3
Experimental Data of B Kπ
HFAG since ICHEP06
6
(10 ) Br
−
K π+ K π
+
K π
+ −
K π Blue color: Preliminary
SLIDE 4
Experimental Data of B Kπ
HFAG since ICHEP06
CP
A
K π+ K π
+
K π
+ −
,
S S
K K
C S
π π
0. 0.12 12±0.11 0.11 0. 0.33 33±0.21 0.21
PDG 2006 Avg. 0.08±0.14 PDG 2006 Avg. 0.34±0.28
SLIDE 5
Model Independent Analysis
SLIDE 6
Diagram approach B Kπ
Amplitude parameterization Adjust Parameter
( )
+ + =
⎡ ⎤ ⎛ ⎞ ⎢ ⎥ ⎟ ⎜ → = + + − + ⎟ ⎜ ⎟ ⎜ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦
∑
* * , ,
1 2 3 3
i i
C C ub us ib is i i EW EW i u c t
B K AV V V V P EP P EP π A
( )
( )
+ + =
⎡ ⎤ ⎛ ⎞ ⎢ ⎥ ⎟ ⎜ → = − + + + + + + + ⎟ ⎜ ⎟ ⎜ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦
∑
* * , ,
1 2 2 2 3 3
i i i
C C ub us ib is i i EW EW EW i u c t
B K V V T C A V V P EP P P EP π A
( )
+ − =
⎡ ⎤ ⎛ ⎞ ⎢ ⎥ ⎟ ⎜ → = − + + + + − ⎟ ⎜ ⎟ ⎜ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦
∑
* * , ,
2 1 3 3
i i i
C C ub us ib is i i EW EW EW i u c t
B K V V T V V P EP P P EP π A
( )
=
⎡ ⎤ ⎛ ⎞ ⎢ ⎥ ⎟ ⎜ → = − − + − − − ⎟ ⎜ ⎟ ⎜ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦
∑
* * , ,
1 1 1 2 3 3
i i i
C C ub us ib is i i EW EW EW i u c t
B K V V C V V P EP P P EP π A
*
1 2 | | 3 3 ⎛ ⎞ ⎟ ⎜ ≡ + − − − + ⎟ ⎜ ⎟ ⎜ ⎝ ⎠
- C
C tb ts t t c c EW EW
P V V P EP P EP P EP
( )
*
≡ + + − −
- ub
us u u c c
T V V T P EP P EP
( )
*
≡ − − + +
- ub
us u u c c
C V V C P EP P EP
( )
* ub us u u c c
A V V A P EP P EP ≡ + + − −
- (
)
* C EW tb ts EW EW
P V V P EP ≡ +
- (
)
* C C C EW tb ts EW EW
P V V P EP ≡ −
SLIDE 7
Diagram approach B Kπ
Final form
( )
+ + +
→ ≡ = − B K P π A A
( )
( )
+
+ + +
− → ≡ = − + 1 2
C EW T
i i i EW i i i
B K e P Ce e P e Te e
γ δ γ δ α δ
π A A
( )
+−
+ − +−
→ ≡ = −
T
i i i
B K e P Te e
α γ δ
π A A
( )
( )
→ ≡ = − − +
00
00
1 2
C EW
i i i EW i
B K Ce e P e P e
γ δ δ α
π A A
- We Neglect
- We set the the strong phase of P as 0 all phase is relative to
- There are 7 unknown parameters
- we consider is given by other analysis
- are real, are phases of their amplitude
,
C EW
P A
, , , , , ,
EW T C EW
P T C P δ δ δ
γ
P
δ
P
δ
ij
A
ij
α
SLIDE 8
Botella’s arguments We assume NP comes into PEW part(or C part) only
New Physics comes in
sin( ) sin( ) sin( ) sin( )
i i i
e e e
φ θ η
φ η φ θ θ η θ η − − = − − −
- For weak phase , always
- We can choose arbitrary at will, for any given
Botella, Silva 2005
− = − sin sin( ) 2 2 sin 2 sin
N N N N EW EW EW EW
N N N N N i i i i i EW EW EW EW EW
P P P e e e e e
φ δ δ γ δ
φ φ γ γ γ
Absorbed into C Absorbed into EW
φ , θ η φ
⎛ ⎞ = ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ ⎝ ⎠ θ γ η
SLIDE 9
New Physics comes in
NP term is absorbed
( )
+
⊃ − + + ⎛ ⎞ − ⎟ ⎜ = − + + − ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ ⎠
00
1 , 2 1 ) 2 sin sin sin sin(
C EW C E N N EW EW N N W EW W E
i i i N EW EW i i i i i i N N N i N i EW EW EW EW EW
Ce e P e P Ce e P e e P e P e e e
γ δ δ φ δ δ γ δ γ δ δ
γ γ φ φ γ A A
′ =
′ − + sin sin
N E C C W
N N i EW i E i W
C P e e C e δ
δ δ
φ γ
′
− = ′ − + sin sin( )
N W W E E E W
N N i EW EW i i EW EW
P e e P e P
δ δ δ
γ γ φ
( )
′ ′
= + ′ − ′ 1 2
E C W
i E i W i
P e e C e δ
δ γ
SLIDE 10
Original Form does not change
( )
+ + +
→ ≡ = − B K P π A A
( )
( )
+
+ ′ + ′ +
→ ≡ = − − ′ + ′ 1 2
T C EW
i i i i EW i i
B K e P Te e e P e e C
α γ δ δ γ δ
π A A
( )
+−
+ − +−
→ ≡ = −
T
i i i
B K e P Te e
α γ δ
π A A
( )
( )
00
00
1 2
′ ′
→ ≡ = − − ′ + ′
C EW
i EW i i i
B K C e e P e P e
α δ γ δ
π A A
The analytic solution
- If there Is NP
( ) ( )
,
, , ≠ ≠ ≠ ′ ′ ′ ′ ≠
E SM C SM EW SM S W EM M SM C
C C P P δ δ δ δ
SLIDE 11
The analytic solution
P Br + =
2
sin2 1 cot 1 1 1 cos
+ +− +− +− +
⎡ ⎤ ⎛ ⎞ ⎢ ⎥ ⎟ ⎜ = ± + − ⎟ ⎜ ⎢ ⎥ ⎟ ⎜ ⎝ ⎠ ⎢ ⎥ ⎣ ⎦
T CP
Br Br A Br Br γ δ γ cot cot
CP T
T Br Br A Br γ δ
+− + +− +−
= − +
Step 1 -
T
P T δ , ,
2
0.0008 2sin
+− +− +
⎛ ⎞ ⎟ ⎜ ⎟ ≈ ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ ⎠
CP
A Br Br γ
+−
+−
= −
T
i i i
e P Te e
α γ δ
A
+
= − P A We neglect
SLIDE 12
The analytic solution
Step 2 -
2 2 00 2 00 00
2 2 cos( ) 2 2 X X α ζ
+
− − − = A A A
2 2 00 2 00 00
2 2 cos( ) 2 2 X X α ζ
+
− − − = A A A
00 00
, α α
( )
00
00
2 2
+
+
− = − ≡
T
i i i i i
e e P Te e Xe
α α γ δ ζ
A A
( )
00
00
2 2
+
+ −
− = − ≡
T
i i i i i
e e P Te e Xe
α α γ δ ζ
A A
SLIDE 13
The analytic solution
Step 3 -
′ ′ ′ ′ , , ,
EW C
C δ EW δ
( )
( )
⎡ ⎤ ⎡ ⎤ = + − = − − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ′ ⎦
2 2 * 00 00 00 00 00 2 2 2
1 1 2Re cos 4 sin sin C y y y y Br α α γ γ A A
( )
⎡ ⎤ = + − ⎢ ⎥ ⎣ ⎦ ′
2 2 2 * 2 2
1 2Re 4sin
i EW
y y y y P e
γ
γ − = − ′ − R tan e( ) Im( )
C
y y y y δ
− −
− = − ′ − Re( ) Im( ) tan
EW i i i i
ye ye ye ye
γ γ γ γ
δ
′ ′ ′ − ′
− + = + ≡ − ≡ ′ + + ′ ′ ′ =
00 00
00 00
2 2
EW EW C C
i i EW i E i i W i i i
C e C e e P e P e P e y e e P y
γ α γ α δ δ δ δ
A A
SLIDE 14
New Physics Test
Hierarchy Test
( )
′ ′ > ⇒ >
2 2 *
Im
i
EW C y y e γ
Strong phase Test
= ⇒ =
00 00 00 00
cos cos
C P
δ δ α α A A
( )
2 2 2 2 * 2 2
1 cot cot 2Re 4 sin
+− + +− +−
′ > ⎡ ⎤ − + > + − ⎢ ⎥ ⎣ ⎦
i CP T
T EW Br Br A Br y y y ye
γ
γ δ γ
2
sin2 1 Im( ) 1 1 1 cos Re( )
+ +− − +− +− + −
= ⎡ ⎤ ⎛ ⎞ − ⎢ ⎥ ⎟ ⎜ ± + − = − ⎟ ⎜ ⎢ ⎥ ⎟ ⎜ ⎝ ⎠ − ⎢ ⎥ ⎣ ⎦
T EW i i i i CP
Br Br ye ye A Br Br ye ye
γ γ γ γ
δ δ γ γ
SLIDE 15
SLIDE 16
Numerical Analysis
Isospin quadrangle 0.8 σ violation
A + A+− A+
00
A
The solution within 0.8 σ variation
4.79 0.96 P T = = 0.22
T
δ = −
1.05, 1.54, 2.89, 1.52 2.10, 0.63, 3.01 , 2.12 1.22, 1.10, 0.06, 0.49 0.14, 1.62, 0.19, 1.22
EW C EW EW C EW EW C EW EW C EW
C P δ δ C P δ δ C P δ δ C P δ δ ⎡ ′ ′ ′ ′ = = = = ⎢ ⎢ ′ ′ ′ ′ = = = = ⎢ ⎢ ′ ′ ′ ′ = = = = − ⎢ ⎢ ′ ′ ′ ′ = = = − = − ⎢ ⎣
SLIDE 17