The Golden Modes Bd J Ks and Bs · Reminder: Why are these modes “golden”? Precise predictions...

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Reminder: Why are these modes “golden”? Precise predictions I: Theoretical Attempts Precise predictions II: Using Data Controlling SU(3) Breaking The Golden Modes B d J K s and B s J /ψφ How can we know Penguin/Tree? Thomas Mannel Theoretische Physik I, Universität Siegen based on work with H. Boos, J. Reuter, S. Faller, R. Fleischer and M. Jung, hep-ph/0403085, arXiv:0809.0842 [hep-ph], arXiv:0810.4248 [hep-ph], SI-HEP-2009-01 5.2.2009, Neckarzimmern-Workshop Thomas Mannel, Uni. Siegen The golden modes ...

Transcript of The Golden Modes Bd J Ks and Bs · Reminder: Why are these modes “golden”? Precise predictions...

Page 1: The Golden Modes Bd J Ks and Bs · Reminder: Why are these modes “golden”? Precise predictions I: Theoretical Attempts Precise predictions II: Using Data Controlling SU(3) Breaking

Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

The Golden ModesBd → J/ψKs and Bs → J/ψφ

How can we know Penguin/Tree?

Thomas Mannel

Theoretische Physik I, Universität Siegen

based on work with H. Boos, J. Reuter, S. Faller, R. Fleischer and M. Jung,hep-ph/0403085, arXiv:0809.0842 [hep-ph], arXiv:0810.4248 [hep-ph], SI-HEP-2009-01

5.2.2009, Neckarzimmern-Workshop

Thomas Mannel, Uni. Siegen The golden modes ...

Page 2: The Golden Modes Bd J Ks and Bs · Reminder: Why are these modes “golden”? Precise predictions I: Theoretical Attempts Precise predictions II: Using Data Controlling SU(3) Breaking

Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Contents

1 Reminder: Why are these modes “golden”?

2 Precise predictions I: Theoretical Attempts

3 Precise predictions II: Using DataBd → J/ψKs

Bs → J/ψφ

4 Controlling SU(3) Breaking

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Introduction

Key Observable: Time-Dependent CP Asymmetries

ACP(t ; f ) ≡ Γ(B0(t)→ f )− Γ(B̄0(t)→ f )

Γ(B0(t)→ f ) + Γ(B̄0(t)→ f )

In case we can neglect the lifetime difference

ACP(t ; f ) = C(f ) cos(∆Md t)− S(f ) sin(∆Md t),

Theory Prediction of C(f ) and S(f ) in general difficult

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Some “Theory” ...

The decay amplitude for B0 → J/ψK 0

A(B0 → J/ψK 0) = (VcbV ∗cs)A(c)T

+(VubV ∗us)A(u)P + (VcbV ∗cs)A(c)

P + (VtbV ∗ts)A(t)P

CKM Unitarity yields

VtbV ∗ts = −VcbV ∗cs − VubV ∗us

Eliminate VtbV ∗ts

A(B0 → J/ψK 0) = (VcbV ∗cs)A(c)T

+(VubV ∗us)[A(u)P − A(t)

P ] + (VcbV ∗cs)[A(c)P − A(t)

P ]

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

A(B0 → J/ψK 0) = (VcbV ∗cs)[A(c)T +A(c)

P −A(t)P ]+(VubV ∗us)[A(u)

P −A(t)P ]

Identify the contributions with different weak phases

A(B0 → J/ψK 0) = A[1 + εeiγa eiθ]

with

A = (VcbV ∗cs)[A(c)T + A(c)

P − A(t)P ] and

a eiθ =

[A(u)

P − A(t)P

A(c)T + A(c)

P − A(t)P

]εeiγ =

VubV ∗us

VcbV ∗cs

ε ∼ 5%: This is why the mode is golden!

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Impact of Penguins

In general we have (r = Penguin over Tree ratio)

A(B0 → f ) = Af[1 + eiγr eiθ]

A(B0 → f ) = Af

[1 + e−iγr eiθ]

Insert the time dependent B0 state and f = f

Γ[f , t ] = |Af (t)|2 + |Af (t)|2 = RfL e−Γ

(s)L t + Rf

H e−Γ(s)H t

|Af (t)|2 − |Af (t)|2 = 2 e−Γt [AfD cos(∆Mst) + Af

M sin(∆Mst)]

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

In terms of the params of the amplitude

Γ[f , t = 0] = RfL + Rf

H = 2|Af |2[1 + 2rf cos θf cos γ + r 2

f

]Af

D = −2|Af |2rf sin θf sin γ

AfM = |Af |2

[sinφs + 2rf cos θf sin(φs + γ) + r 2

f sin(φs + 2γ)]

|Af (t)|2 − |Af (t)|2

|Af (t)|2 + |Af (t)|2=

AfD cos(∆Mst) + Af

M sin(∆Mst)

cosh(∆Γst/2)−Af∆Γ sinh(∆Γst/2)

AfD = C(f ) and Af

M = S(f ) ∆Γ ∼ 0

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Back to J/ΨK : In the Standard Model rJ/ΨK ≤ 5%:

C(J/ψKS,L) ≈ 0, S(J/ψKS,L) ≈ −ηS,L sin 2β

Penguin contamination small, suppressed by εIs it really small ?

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

If there is new physics in B0–B̄0 mixing:

φd = 2β + φNPd

“True value” of β from |Vub/Vcb| and γ

(sin 2β)true = 0.76+0.02−0.04

+0.04−0.05 and

(φd )J/ψK 0 − 2βtrue = −(8.7+2.6−3.6 ± 3.8)◦

Reliable Calculation needed:Is this “new physics” or is it ony “oversized penguins”?

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Precise predictions I: Theoretical Attempts

Try to calculate the relevant matrix elements for rf

Corrections to the mixing phase from charm loops

∆=[

M12

|M12|

]≈ −2

m2c

m2t

ln(

m2c

M2W

)≈ −4× 10−4

These are calculable and safely small

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Up quark penguin corrections to the decay rateAlmost impossible to compute!Perturbative (un)reasoning:

HPeng.eff (b → cc̄s) = −GF√

2

3π(sb)V−A (cc)V ·

[1 +O

(M2

Ψ

M2Z

)]

+αs(k2)

3π(sT ab)V−A (cT ac)V

}·(

53− ln

(k2

µ2

)+ iπ

)Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Use µ = mb and k2 = M2J/ψ:

This yields a tiny numberAs a rule of thumb:Perturbative Estimates tend to underestimate ....

S(J/Ψ KS) = (sin 2β)0 − (2.16± 2.23)× 10−4

C(J/Ψ KS) = (5.0± 3.8)× 10−4

This is far beyond the current experimental accuracyHard to asess the uncertainties of this estimate

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

Precise predictions II: Using Data

Use of data: Using Flavour SymmetriesProblem: Flavour SU(3) is severely brokenTwo Strategies:

Assume SU(3) relations,but leave (generous) uncertaintiesTry to get a hand on SU(3) breaking (see below)

In the case at hand:Compare b → sc̄c with its SU(3) friend b → dc̄c

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

Penguins in Bd → J/ψKs

Remember

A(B0 → J/ψK 0) = A[1 + εeiγa eiθ]

Parametrize (φd = B − B Mixing phase)

S(J/Ψ KS) = sin(φd + ∆φd )

tan ∆φd =2εa cos θ sin γ + ε2a2 sin 2γ

1 + 2εa cos θ cos γ + ε2a2 cos 2γ

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

“Control Channel” for B0 → J/ψK 0: B0 → J/ψπ0

√2A(B0 → J/ψπ0) = A′

[1− a′eiθ′eiγ

]Measurements (HFAG Uncertainties!):

C(J/ψπ0) = −0.10±0.13, S(J/ψπ0) = −0.93±0.15

SU(3) limit: Identify the hadronic amplitudes

A′ =Vcd

VcsA a′ = a θ′ = θ

Of course, this is debatable ....

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

Aside from the CP Observables we have(Φ: Phase Space Corrections)

H ≡ 2ε

[BR(Bd → J/ψπ0)

BR(Bd → J/ψK 0)

] ∣∣∣∣VcdAVcsA′

∣∣∣∣2 ΦJ/ψK 0

ΦJ/ψπ0

=1− 2a′ cos θ′ cos γ + a′2

1 + 2εa cos θ cos γ + ε2a2 ,

Note that in the SU(3) Limit:∣∣∣ VcdA

VcsA′

∣∣∣2 = 1

Include “some SU(3) breaking effects” by assuming∣∣∣∣VcdAVcsA′

∣∣∣∣ =f +B→K (M2

J/ψ)

f +B→π(M2

Jψ)= 1.34± 0.12.

(Values from QCD Sum rules and from pole extrapolation)

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

This yields H = 1.53± 0.16BR ± 0.27FF

Use C, S and H to extract a′ → a and θ′ → θ

Extract ∆φd

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

Results for B → J/ψK

Using SU(3) for a and θ: ∆φd ∈ [−3.9,−0.8]◦

Allowing 50% SU(3) breaking in a andθ, θ′ ∈ [90,270]◦ indepedently: ∆φd ∈ [−6.7,0.0]◦

Hints at negative ∆φd

Softens the tension with the SM fitHowever, still quite debatable SU(3) assumptionsThis is likely much larger then the perturbativeestimate! (Ala Boos, Reuter M.)

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

Future possibilites

Assme a future reduction of uncertainties on CPobservables by a factor of 2Assume a reduction of the uncertainty of γ and onthe BR’s by a factor of 5Scenario (a): “High S”:C(J/ψπ0) = −0.10± 0.03, H = 1.53± 0.03± 0.27, S = −0.98± 0.03

Scenario (b): “Low S”:C(J/ψπ0) = −0.10± 0.03, H = 1.53± 0.03± 0.27, S = −0.85± 0.03

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

Bs → J/ψφ

Theory is practically the same, except:Final State is Vector-Vector

→ Project out the various CP componentsφs = −2λ2η: Small

Experiment:

∆Ms =

{(18.56± 0.87)ps−1 (D0 coll.),(17.77± 0.10± 0.07)ps−1 (CDF coll.).

φs =(−44+17

−21

)◦∨(−135+21

−17

)◦(HFAG)

φs could be sizable?

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

“New Physics” in the mixing

Modify ∆B = 2 matrix element of the mass matrix:

Ms12 = Ms,SM

12

(1 + κseiσs

)

0 50 100 150 200 250 300 3500.0

0.5

1.0

1.5

2.0

2.5

!s !deg"

" s

#s$#%135%17&21$'

#s$#%44%21&17$'

However, What, if we have “oversized penguins”?Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

Controlling the Penguins

Pick a control channel: B0s → J/ψK̄ ∗0

b

c

c

s

J/ψ

WB0

s

φ

s

s

b

c

c

s

J/ψ

u, c, t

W

colour singletexchange

B0

s

s

s

φ

b

c

c

J/!

WB0

s

s

s

d

K̄!0

b

c

c

J/!

u, c, t

W

colour singletexchange

B0

s

s

sK̄

!0

d

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

MeasureB0

s → J/ψ[→ `+`−]φ[→ K +K−]B0

s → J/ψ[→ `+`−]K̄ ∗0[→ π+K−]

However, B0s → J/ψK̄ ∗0 is flavour specific!

→ No mixing induced CP Asymmetry→ less observables

One still can extract θ′ and a′ from data(using H and the direct CP Asymmetry)

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

Numercal example: H = 1.44 and AD = −0.33

180 200 220 240 260 280 300 320 340 3600

0.2

0.4

0.6

0.8

1

1.2

1.4

Θ f¢ @degD

af¢

H f =1.44

cosΘ f¢

>0

cosΘ f¢

<0

A`

D

f ¢

=-0.33

correctsolution

Still a twofold ambiguity ...Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Bd → J/ψKsBs → J/ψφ

More Control Channels: B0d → J/ψρ0

Very much the same as B0d → J/ψφ

→ The same set of observables

0 30 60 90 120 150 180 210 240 270 300 330 3600

0.2

0.4

0.6

0.8

1

1.2

1.4

Θ f¢ @degD

af¢

H f =1.44A`

D

f ¢

=-0.33

Η f A`

M

f ¢

= 0.90

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Controlling SU(3) Breaking

SU(3) has SU(2) Subgroup: Either I, U or V spin

Pattern of SU(3) breaking:I Spin is a very good symmetryU Spin and V Spin are equally bad

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

U Spin and U Spin breaking

U Spin: d and s form the fundamental doubletParametrizing U Spin breaking

Ls,dm = md d̄d + mss̄s

=12

(ms + md )(s̄s + d̄d) +12

(ms −md )(s̄s − d̄d)

=12

(ms + md ) q̄q +12

∆m q̄τ3q

Structure of the breaking: Triplett (j = 1, jz = 0),Hbreak = 1

2∆m q̄τ3q = εB(1)0

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Advantage: s and d have the same charge:Electroweak Penguins are singletHadronic B-decays: Heff doublet under U-spinRelatively simple goup theory... Currently under investgation ...

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

To first order in U-spin breaking:

〈f̃ |O(0)|̃i〉 = 〈f |O(0)|i〉 + (−i)∫

d4x 〈f |T [O(0)Hbreak(x)]|i〉

O(x): (Irreducible U Spin tensor) operator|i〉, |f 〉: U-spin symmetric states|̃i〉, |f̃ 〉: States including U-spin breaking

Classify the statesUse Wigner Eckart Theorem ...

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Simple Example: B− → J/ψ(π/K )−

Interesting because:Related to the golden modesFactorization does not work (neither naive nor QCDF)All observables measured

Decay BR/10−4 ACP

B− → J/ψK− 10.26± 0.37 0.017± 0.016(∗)B− → J/ψπ− 0.48± 0.04(∗) 0.09± 0.08

Table 1: Data taken from the PDG. (*): Inconsistentmeasurements, error enhanced by the PDG.

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

U-spin limit:Only one amplitude, with two CKM structuresPredicts

ACP(J/ψK−)BR(J/ψK−) +

ACP(J/ψπ−)BR(J/ψπ−)!

= 0 exp= 0.22± 0.17

Not conclusive at the moment, due to uncertaintiesNaive factorization does not describe the breakingwell:

BR(B− → J/ψK−)

BR(B− → J/ψπ−)

∣∣∣∣λcd

λcs

∣∣∣∣2 ∼(

F B→K (M2J/ψ)

F B→π(M2J/ψ)

)2

⇐⇒ 1.1± 0.1 ∼ 1.8± 0.3

Thomas Mannel, Uni. Siegen The golden modes ...

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Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

U-spin breaking:|J/ψ π−

K− 〉 = |1/2,±1/2〉 ⇒ no ∆U = 3/2 breakingResults in⟨

B+|Heff |J/ψK +

π+

⟩=∑

q=u,c

λqs(Aq,1/2 ± Aε

q,1/2

)Doubling of amplitudes, 7 hadronic parametersPossible strategy: Au,1/2 < Ac,1/2 expected

consider Aεc,1/2 only⇒ 5 parameters, 4 observablesnot fully determined, study correlations in 2D-plots

Thomas Mannel, Uni. Siegen The golden modes ...

Page 35: The Golden Modes Bd J Ks and Bs · Reminder: Why are these modes “golden”? Precise predictions I: Theoretical Attempts Precise predictions II: Using Data Controlling SU(3) Breaking

Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

PRELIMINARY fit results. Plotted: ε = Aεc,1/2/Ac,1/2.

Observable missing⇒ only correlation, no best fit sReal part finite, Imaginary part ∼ 100% correlatedwith |Au,1/2|

Thomas Mannel, Uni. Siegen The golden modes ...

Page 36: The Golden Modes Bd J Ks and Bs · Reminder: Why are these modes “golden”? Precise predictions I: Theoretical Attempts Precise predictions II: Using Data Controlling SU(3) Breaking

Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

PRELIMINARY: A simple future scenario (σexp/5).Plotted: ε = Aε

c,1/2/Ac,1/2.

Thomas Mannel, Uni. Siegen The golden modes ...

Page 37: The Golden Modes Bd J Ks and Bs · Reminder: Why are these modes “golden”? Precise predictions I: Theoretical Attempts Precise predictions II: Using Data Controlling SU(3) Breaking

Reminder: Why are these modes “golden”?Precise predictions I: Theoretical Attempts

Precise predictions II: Using DataControlling SU(3) Breaking

Conclusion

Perturbative Estimates may be misleading!(typically underestimate the effects)If there are large non-perturbative contributions,SCET/PQCD/QCDF Ansätze will not yield preciseresults for an SM test→ may be an interesting lab for QCD studies.With sufficient ammount of data (LHC-b and SFF):(Approximate) Flavour Symmetries will be the way totest the SM... and possibly identify “new physics”

Thomas Mannel, Uni. Siegen The golden modes ...