Final state interactions in heavy mesons decays.

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Final state Final state interactions in heavy interactions in heavy mesons decays. mesons decays. A.B.Kaidalov and A.B.Kaidalov and M.I. Vysotsky M.I. Vysotsky ITEP, Moscow ITEP, Moscow

description

Final state interactions in heavy mesons decays. A.B.Kaidalov and M.I. Vysotsky ITEP, Moscow. Contents:. Introduction. Method of calculations. Applications to B  ππ and B ρρ decays. CPV in B  ππ and B ρρ decays. Conclusions. Introduction. - PowerPoint PPT Presentation

Transcript of Final state interactions in heavy mesons decays.

Page 1: Final state interactions in heavy mesons decays.

Final state interactions in Final state interactions in heavy mesons decays.heavy mesons decays.

A.B.Kaidalov andA.B.Kaidalov and

M.I. VysotskyM.I. Vysotsky

ITEP, MoscowITEP, Moscow

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Contents:Contents:

Introduction.Introduction. Method of calculations.Method of calculations. Applications to BApplications to Bππππ and B and Bρρρρ

decays.decays. CPV in CPV in BBππππ and B and Bρρρρ decays.decays. Conclusions.Conclusions.

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Introduction.Introduction.

Importance of theoretical understanding Importance of theoretical understanding of phases due to final state interactions in of phases due to final state interactions in hadronic B-decays.hadronic B-decays.

a) For extraction of CKM parameters from a) For extraction of CKM parameters from B-decays.B-decays.

b) A good laboratory to study large b) A good laboratory to study large distance aspects of QCD.distance aspects of QCD.

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Unitarity triangleUnitarity triangle

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The diagrams for B decaysThe diagrams for B decays

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The The ρρρρ – – ππππ puzzle. puzzle.

A large difference in +-/00 branching A large difference in +-/00 branching ratios for ratios for ππππ and and ρρρρ –decays.–decays.

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Matrix elements for BMatrix elements for Bππππ ( (ρρρρ) ) decaysdecays..

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Structure of the matrix elementsStructure of the matrix elements

are the phases due to final state are the phases due to final state interactions for tree (I=0,2) and interactions for tree (I=0,2) and penguin diagrams correspondingly.penguin diagrams correspondingly.

andand are CKM phases.are CKM phases.

P-contribution to these decays isP-contribution to these decays is

rather small (P/T ~ 0.1), but it israther small (P/T ~ 0.1), but it is

important for CP-violation.important for CP-violation.

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Determination of penguin Determination of penguin contribution.contribution.

The values of P can be determined, The values of P can be determined,

using SU(3)-symmetry,using SU(3)-symmetry,

from , -decays, wherefrom , -decays, where

penguin diagrams give dominant penguin diagrams give dominant contributions. contributions. M.Gronau, J.RosnerM.Gronau, J.Rosner

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Isospin analysisIsospin analysis

Neglecting by P-termNeglecting by P-term

From data on BFrom data on Bππππ decays we get decays we get

With account of P-termWith account of P-term

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Large deviation from factorizationLarge deviation from factorization

in phases.in phases.

For For BBρρρρ decays FSI-phases are decays FSI-phases are smaller:smaller:

Below the model will be presented, Below the model will be presented, which explains the pattern of FSI-which explains the pattern of FSI-

phases in Bphases in Bππππ , , BBρρρρ decays. decays.

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How to calculate FSI?How to calculate FSI?

For single channel case from unitarity For single channel case from unitarity follows Migdal-Watson theorem:follows Migdal-Watson theorem:

the phase of the matrix element forthe phase of the matrix element for

the decay Xthe decay Xab is equal to the phaseab is equal to the phase

of the elastic scattering amplitude of the elastic scattering amplitude δδabab..

Generalization to isospin.Generalization to isospin.

δδII(K(Kππππ)= )= δδII((ππππ))

At At MMK K δδ00((ππππ)=(35±3)º , )=(35±3)º , δδ22((ππππ)=(-7±2)º )=(-7±2)º

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Application to heavy (D,B)-mesons.Application to heavy (D,B)-mesons.

For heavy mesons there are many open For heavy mesons there are many open channels and application of unitarity channels and application of unitarity is not straightforward.is not straightforward.

Different ideas about FSI for heavy Different ideas about FSI for heavy quark decays.quark decays.

a) Effects of FSI should decrease with a) Effects of FSI should decrease with the mass of heavy quark Mthe mass of heavy quark MQQ..

Arguments Arguments (J.D.Bjorken).(J.D.Bjorken).

b) FSI do not decrease with Mb) FSI do not decrease with MQQ.. (at least for two-body final states)(at least for two-body final states)

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Classification of FSI in 1/N-expansion.Classification of FSI in 1/N-expansion.

In 1/N-expansion the In 1/N-expansion the following diagrams with following diagrams with FSI are possible:FSI are possible:

The diagram a)~ 1/N² and The diagram a)~ 1/N² and does not decrease with MQ does not decrease with MQ (pomeron), while the diagram(pomeron), while the diagram b)~1/N and decreases as b)~1/N and decreases as 1/MQ (reggeon).1/MQ (reggeon).

A.Kaidalov(1989)A.Kaidalov(1989)

Similar conclusions:Similar conclusions:

J.P.Donoghue et al.(1996)J.P.Donoghue et al.(1996)

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Experimental results on FSI phases in Experimental results on FSI phases in D, B-decaysD, B-decays

Data on DData on Dππππ branching ratios lead to: branching ratios lead to:

IIδδ2 2 – – δδ0 0 I= (86º±4º)I= (86º±4º)In B-decays:In B-decays: From BFrom BDDππ decays FSI difference between decays FSI difference between

I=1/2 and I=3/2 amplitudes I=1/2 and I=3/2 amplitudes δδDDππ= 30º±7º= 30º±7º From analysis of BFrom analysis of Bππππ decays: decays:

IIδδ2 2 – – δδ00II = (37º±10º) = (37º±10º) Large FSI phases!Large FSI phases!

However small phases in BHowever small phases in Bρρρρ ((ππππ,,ρρρρ-puzzle).-puzzle).

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Method of calculations.Method of calculations.We use Feynman diagrams approach,We use Feynman diagrams approach,which is often applied to high-energywhich is often applied to high-energyhadronic interactions.hadronic interactions.

Amplitudes for the transitions abAmplitudes for the transitions abik with ik with large masses of the states i,k should belarge masses of the states i,k should be strongly suppressed (as powers of 1/M²strongly suppressed (as powers of 1/M²i(k)i(k) ). ). It is possible to prove that M²It is possible to prove that M²i(k) i(k) ~~The states with MThe states with Mi ~i ~ 1 GeV are taken into account. 1 GeV are taken into account.

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Method of calculations (cont).Method of calculations (cont).

TransformingTransforming ∫dk ∫dk ∫d²k ∫d²kttdMdMii²dM²dMkk²²We obtainWe obtain

MMII(B(Bab)=∑ Mab)=∑ MIIºº(B(Bik)(ik)(δδiaiaδδkbkb + + i Ti TII(ik(ikab))ab))

TTII(ik(ikab) ab) isis J=0 J=0 projection of the projection of the

corressponding scattering amplitude.corressponding scattering amplitude.

Note that for real Note that for real TTII(ik(ikab) ab) this formula givesthis formula gives

the same result as unitarity condition. Howeverthe same result as unitarity condition. However

at high energies amplitudes have substantialat high energies amplitudes have substantial

imaginary parts.imaginary parts.

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Method of calculations (cont).Method of calculations (cont).

There are many papers There are many papers on this subject,on this subject,

which use two-body which use two-body intermediate statesintermediate states

for calculations of effects for calculations of effects due to FSI.due to FSI.

For example: For example: H-Y. Cheng, H-Y. Cheng, C-K. Chua and A.SoniC-K. Chua and A.Soni

The diagrams of the The diagrams of the following type are following type are used:used:

Triangle diagramTriangle diagram

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Reggeization of t-channel exchanges.Reggeization of t-channel exchanges.

For exchange by an elementary For exchange by an elementary ρρ-meson in-meson in

the t-channel the partial wave amplitudes do not the t-channel the partial wave amplitudes do not

decrease as energy increases. decrease as energy increases.

However it is well known from phenomenologyHowever it is well known from phenomenology

of high-energy binary reactions that of high-energy binary reactions that ρρ-exchange -exchange

should be reggeized.should be reggeized.

In this case its contribution to the FSI decreasesIn this case its contribution to the FSI decreases

as as exp(-(1-exp(-(1-ααρρ(0))ln(s))~1/s½~1/M(0))ln(s))~1/s½~1/MQQ

for for ααρρ(0)=0.5(0)=0.5

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Reggeization of t-channel exchanges.Reggeization of t-channel exchanges.

Situation is even more Situation is even more drastically changeddrastically changed

for D*-trajectory withfor D*-trajectory with

ααD*D*(0) ≈ -0.8(0) ≈ -0.8 . .

We approximate high-We approximate high-

energy scattering energy scattering

amplitudes byamplitudes by

exchanges of Regge exchanges of Regge poles.poles.

Amplitudes at highAmplitudes at high

energies in Reggeenergies in Regge

model.model.

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Applications to BApplications to Bππππ and B and Bρρρρ decays.decays.

For For BBππππ andand B Bρρρρ decays the decays the ππππ, , ρρρρ and and ππAA1 1 intermediate states were used.intermediate states were used.

In the amplitudes of In the amplitudes of ππππ ππππ P, f P, f and and ρρ - -exchanges have been taken into account.exchanges have been taken into account.

In the amplitudes of In the amplitudes of ππππ ρρρρ ππ--exchangeexchange

gives the main contribution to the gives the main contribution to the

longitudinally polarized rho.longitudinally polarized rho.

In the amplitudes of In the amplitudes of ππππ ππAA1 1 ρρ - -exchange exchange

contributes.contributes.

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Applications to BApplications to Bππππ and B and Bρρρρ decays.decays.

The pion exchange in contribution ofThe pion exchange in contribution of ρρρρ intermediate state (neglected by other authorsintermediate state (neglected by other authors~1/M²~1/M²QQ) plays an important role in the ) plays an important role in the resolution of resolution of ππππ--ρρρρ puzzle. puzzle.

The pomeron and f-exchanges do notThe pomeron and f-exchanges do not contribute to the phase difference of contribute to the phase difference of amplitudes with I=0 and I=2 and it decreasesamplitudes with I=0 and I=2 and it decreases as ~1/Mas ~1/MQQ for M for MQQ∞.∞.

Vertices of reggeons with pions were takenVertices of reggeons with pions were taken from analysis of from analysis of ππN, NN-scattering and ReggeN, NN-scattering and Regge factorization.factorization.

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Results.Results.

Branching Branching BBρρ++ρρ- - ≈ ≈ 55 times larger than thetimes larger than the

one for one for BBππ++ππ- and - and ρρρρ – –intermediate state isintermediate state is

very important in very important in BBππππ decays, while decays, while ππππ – –intermediate state plays a minor role in Bintermediate state plays a minor role in Bρρρρ decays.decays.Final result is: Final result is:

BBππππ: : δδ00= 30º ; = 30º ; δδ00- - δδ22=40º (±15º)=40º (±15º) δδ22=-10º=-10º BBρρρρ: : δδ00= 11º ; = 11º ; δδ00- - δδ22=15º (±5º)=15º (±5º) δδ22=-4º=-4º

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CP-violation in B-decaysCP-violation in B-decays

Time-dependent CP assymetryTime-dependent CP assymetry

AACPCPtt)=)= SSsin(sin(mmt) – t) – CCcos(cos(mmt)t)

↑ ↑↑ ↑ Mixing induced CPV Direct CPVMixing induced CPV Direct CPV

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Direct CPV in BDirect CPV in Bππππ

CPV-parameter CPV-parameter C C is sensitive to a is sensitive to a magnitude of penguin contribution magnitude of penguin contribution and phases.and phases.

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Direct CPV in BDirect CPV in Bππππ

The ratios of amplitudes was determinedThe ratios of amplitudes was determined

above:above:

AA00//AA22 = 0.8 = 0.8±0.09, ±0.09, PP//AA2 2 = 0.092±0.02= 0.092±0.02

Phases Phases δδ0 0 andand δδ2 2 were calculated.were calculated.

What about phases of penguin contribution? What about phases of penguin contribution?

In PQCD the phase In PQCD the phase δδP P isis ~ 10º ~ 10º and positive.and positive.

The sign of the phase for contribution ofThe sign of the phase for contribution of

intermediate state in Regge model depends onintermediate state in Regge model depends on

the intercept of the intercept of D*D*-trajectory.-trajectory.

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For linear D*-trajectory with For linear D*-trajectory with

αα=0.5 GeV- : =0.5 GeV- : ααD*D*(0)= -0.8(0)= -0.8

and and δδP P is negative. is negative. δδP P ~ - 10º~ - 10º

In leading log PQCD calculation In leading log PQCD calculation

ααD*D*(0)≥ 0 and (0)≥ 0 and δδP P is positive.is positive.

Thus the sign of Thus the sign of δδP P gives an gives an important information on dynamicsimportant information on dynamics

of high-energy interactions.of high-energy interactions.

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If If δδP P is positive it is possible to obtainis positive it is possible to obtain

the lower bound for the lower bound for CC+-: +-: CC+- +- > - 0.18> - 0.18Belle and BABAR give different resultsBelle and BABAR give different results

for this quantity: for this quantity: CC+-(Belle)=-0.55(0.09),+-(Belle)=-0.55(0.09),

CC+-(BaBar)=-0.21(0.09)+-(BaBar)=-0.21(0.09)

Using d s symmetry and Using d s symmetry and CC from from

BBKKππ decay one obtains decay one obtains

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If negative If negative δδP P is allowed, then it isis allowed, then it is

possible to obtain possible to obtain CC+-+- closer to Belle closer to Belle

result.result.

For For BBππººππº º decay we havedecay we have

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Very large Very large llCC0000l l is predicted. It is not is predicted. It is not

sensitive to sensitive to δδP.P. Present experimental Present experimental error is too big:error is too big:

CC0000= = -0.36 -0.36 ± 0.32± 0.32

Belle and BABAR agree on the Belle and BABAR agree on the SS+-+-

SS+-+- == -0.59 -0.59 ± 0.09± 0.09

Without P-contributionWithout P-contribution

sin 2sin 2ααT T = = SS+-+- , , ααTT= (108= (108 ± 3)º± 3)º

With P: With P: αα= (88= (88 ± 4(exp) ± 5(th))º± 4(exp) ± 5(th))º

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Relative smallness of P-contribution inRelative smallness of P-contribution in

BBρρρρ –decays allow us to determine –decays allow us to determine

αα with better theoretical accuracy.with better theoretical accuracy.

From experimental resultFrom experimental result

((SS+-+-))ρρρρ == -0.06 -0.06 ± 0.18± 0.18

we obtainwe obtain αα= (87= (87 ± 5(exp) ± 1(th))º± 5(exp) ± 1(th))º

These values of These values of αα are in a good are in a good agreement with results of a global fit agreement with results of a global fit of unitarity triangleof unitarity triangle

ααUTfitUTfit= (94= (94 ± 4)º± 4)º

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Conclusions.Conclusions. FSI play an important role in two-body FSI play an important role in two-body

hadronic decays of heavy mesons.hadronic decays of heavy mesons. Theoretical estimates with account of Theoretical estimates with account of

the lowest intermediate states give a the lowest intermediate states give a

satisfactory agreement with experimentsatisfactory agreement with experiment

and provide an explanation of aand provide an explanation of a

difference between the properties ofdifference between the properties of

ππππ and and ρρρρ – –final states in B decays.final states in B decays.

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ConclusionsConclusions

CC+- +- gives an important information gives an important information on FSI and, in particular, on the on FSI and, in particular, on the phase of the penguin diagram. phase of the penguin diagram.

It is important to resolve the problem It is important to resolve the problem with difference in experimental with difference in experimental

measurments of measurments of CC+- +-

in Bin Bππππ decays.decays.