Learning(0) from B decaysChuan-Hung Chen

Department of Physics, National Cheng-Kung University, Tainan, Taiwan

Introduction & Our question

-0 mixing

B! K(*) (0) decays

Discussion

Introduction: what questions can we ask in B Physics?

• Determine the CP violating phases:

23

22

3 2

1 ( )2

1 , 0.819, 0.222

(1 ) 1

CKM

A i

V A A

A i A

A3Rbe-i3

A3Rte-i1

1

0.410.44

0.733 0.057(stat) 0.028(syst) (Belle)

sin 2 ( / ) 0.741 0.067( ) 0.034( ) (Babar)

0.79 ( ) (CDF)SJ K stat sys

stat sys

Precision measurement

Find new CP violating sources

B

b

d

t tb

Bd

122 itdV e

S

S

/ K

K ( eigenstates)

J

CP

• Test standard model & search for the new effects:• Extra dimensions • Noncommutative spaces …• Supersymmetric models …• Grand unified theories (GUTs) …

• Test QCD approach

• QCD factorization approach, perturbative QCD approach…• Do final state interactions play important role in B decays?

• Clarify and find new states or new decay modes, such as

• DsJ(2317) , DsJ(2457), B! X(3872) K,…•Why PL» PT in B! K* decays?

Belle Collaboration

Our question:

Br(B! 0 K0 )» 35 £ 10-6

Theoretical estimation

C.W. Chiang etal., hep-ph/0404037

How to understand?

(in units of 10-6 )

0 mixing:

According to quark model with SU(3) flavor symmetry, the mesonicstates could be obtained by 3 3 8 1

1

1

3uu dd ss

8

12

6uu dd ss

0 1, isotriplet

2uu dd

ud

usds

du

sdsu

0

8

1

If UA(1) is a good symmetry and quarks are massless, there are nine Goldstone bosons

Since ms>>mu,d, 8 and 1 will mix.

However, UA(1) is broken by anomaly,1 q0 while m 0m

1 cannot be a Goldstone boson, m0=958 MeV; m=547 MeV

8

1

cos sinThe mixing could be described

' sin cos

8 18 1 ...

P P P P

g cP gg cc

With a parton Fock state decomposition, 1

0

( )2 2

ii PP

c

fdx x

N

8 18 1

8 1' 8 ' 1

cos , sin

sin , cos

1

0

in which ( )2 2

ii

c

fdx x

N

One-angle scheme:

885 8

115 1

00 0

00 0

fJi p

fJ

8 1 8 15 5

8 1 8 11' '

8

5 5

0cos sin

0sin co0 ' s

0 0

0 '

fi p

f

J J f fi p

J J f f

The decay constants, defined as

will have the relation

85 5 5 5

15 5 5 5

12

61

3

J u u d d s s

J u u d d s s

Combining P! J/ ! P decays and P transition form factors, it has been shown that one-angle parametrization cannot match withthe results of ’Pt and experiments

RJ/=5.0 ± 0.8

Therefore, two-angle scheme is introduced.

8 1 8

8 1 1

cos sin

sin cos'

8 18 8 1 1

8 1' 8 8 ' 1 1

cos , sin

sin , cos

8 1

8 1'

1

8 1'

8 8

1

cos sin 0

sin cos 0

f f

f

f

f f

Leutwyler, NPPS64, 223(‘98)

Another quark-flavor scheme is introduced, T. Feldmann, P. Kroll, B. Stech, PRD58,114006(98); PLB449,339 (99)

q s

cos sin 1, ,

' sin cos 2qq

ss

uu dd ss

5

5

0 0 0;

00 0

qq q

sss

J fi p

fJ

5 5

5 5 ' '

0 0,

0 ' 0 '

q s q s

q s q s

J J f fi p

J J f f

' '

0cos sin

0sin cos

q sq

sq s

f f f

f ff

8 1

8 18 1

8 1' ' ' '

8

8 1 1

How to related to the convention and

cos sin 0

sin cos 0

1 2

3 3

2 1

3 3

0cos sin

0sin

c

o

s

q

s

s

s

q

q

f

f

f f f f

f f f

f

f

f

1 2

3 3

2 1

3 3

2 2 18 8

2 2 11 1

12 , ( 2 / )

31

2 , ( 2 / )3

q s s q

q s q s

f f f Tan f f

f f f Tan f f

85 5 5 5

15 5 5 5

12

61

3

J u u d d s s

J u u d d s s

Why do we need another scheme?

50 's s In B decays, we need to deal with the matrix elements, for instance,

If we know the matrix element for axial current, 50 's s ip f

It seems 50 's s can be obtained in terms of equation of motion

2 25 5 '0 ' 0 2 'ss s m si s p f M f

2

'50 '

2 s

Msi s f

m But, ms ! 0, f and M’ 0

For displaying the SU(3) limit explicitly, it is better to use bases qq and ss

5 524

s ssJ GG m si s

5 5 5

22

4q s

u dJ GG m ui u m di d

2 25 5

2 2

5 5

0 0

0 0

q sq q q qq s qs

q sq sq s sss s

J J f M f M

f M f MJ J

We can have mass matrix5 52 2

2 2

5 5

1 10 0

1 10 0

q sq q

q sqq qs

q ssq sss s

q s

J Jf fM M

M MJ J

f f

2

2 2

2 2

2

2 10 0

4 4

2 10 0

4 4

s sqq q q

q sqq qs

sq ss s ss ss s

q s

m GG GGf fM M

M MGG m GG

f f

25 5

25

20

20

qq u d qq

ss s ss

m m ui u m di df

m m si sf

Free parameters:

2 2q s

2 1, , f , f , , 0 = 0

4 4s s

qq ss q sq s

m m GG GGf f

2 2 21

2 2 2'

2 21 88 81

8 1 8 12 218 11

0( ) ( )

0

( , ) ( , )

qq qs

sq ss

M M MU U

M M M

M MU U

M M

The mass matrix can be diagonalized via

8 and 1 are not independent

cos sin( )

sin cosU

Bd! K0 (0) decaysEffective interactions for b! s qq

u

b

u

s

Tree

W

Vub Vus

penguinb s

q q

t

WVtb Vts

g

Effective operators

1 5 5

2 5 5

ˆ 1 1

ˆ 1 1

O b uu s

O b u u s

Tree

u

b

u

s

g

penguinb s

q qg

4,6 5 5

3,5 5 5

ˆ 1 1

ˆ 1 1

O b s q q

O b sq q

Penguin

Hence, b s

q qV± A

V-A

C3-6

uC1,2b

u

sV-A V-A

2 0.043tsV A

0.2233.6 10

Topologies for Bd! K0(0)

Since VubVus<<VtbVts, penguin dominates.

Penguinemission

b

s d

d

B (0)

K

V-A V± A

(a)

b s

d,ud,u

B

(0)

KV-A

V± A

(c)

Tree’s contributions are similar to (c) except the CKM matrix elements

b s

ss

KV-A

V± A

(0)

(d)

B

Penguinannihilation d

b

B

KV-A

V± A

d

s

dd

(0)

(e)b

d

B

K

V-A

V± A

d

s

ss

(0)(f)

Usually, (e), (f) < (a), (b), (c), (d)

b

s s

s

B

(0)

KV-A V± A

(b)

Hadronic matrix elements:

bs d

d

B (0)

K

V-A V-A

(a)

Only show the factorizable effects

4 4 5 4 5

4

ˆ~ ~ (1 ) 0 (1 )

q q

K q K

K a O B K d s a b d B

f a bp d B

6 6 5 6 5

2

6

ˆ~ ~ 2 (1 ) 0 (1 )

2

q q

KK q

s d

K a O B K d s a b d B

mf a bd B

m m

bs d

d

B (0)

K

V-A V+A

(a)

(V-A)(V+A)=-2(S-P)(S+P)

bs s

s

B

(0)

KV-A V-A

(b)

bs s

s

B

(0)

KV-A V+A

(b)

6 6 5 6 5

2

6

ˆ~ ~ 2 (1 ) 0 (1 )

2

s s s

sss

s s

K a O B s s a b s B

mf K a bs B

m m

4 4 5 4 5

4 B

ˆ~ ~ (1 ) 0 (1 )

, q=p

s s

s K

K a O B s s K a b s B

f K a bqs B p

24 ( ) 4 10ba m

26 ( ) 6 10ba m

(0)

b s

d,ud,u

B KV-A

V-A

(c)

3 3 5 3 5

3 B

ˆ~ ~ (1 ) 0 (1 )

, q=p

q q

q K

K a O B q q K a b s B

f K a bqs B p

5 5 5 5 5

5

ˆ~ ~ (1 ) 0 (1 )

q q q

q

K a O B q q a b s B

f K a bqs B

b s

d,ud,u

B

(0)

KV-A

V+A

(c)

b s

ss

KV-A

V-A

(0)

(d)

B

b s

ss

KV-A

V+A

(0)

(d)

B

(cont’ed)

3 3 5 3 5

3 B

ˆ~ ~ (1 ) 0 (1 )

, q=p

s s

s K

K a O B s s K a b s B

f K a bqs B p

5 5 5 5 5

5

ˆ~ ~ (1 ) 0 (1 )

q s q

s

K a O B s s a b s B

f K a bqs B

23( ) 1 10ba m

25 ( ) 1 10ba m

In order to calculate hadronic matrix elements, such as

we need to know the wave functions of B, K, q and s

Numerical analyses:

4 4, q Ka bp d B K a bqs B

The wave functions of B and K meson have been studied in the literature.

We have to assume that q and s have the same asymptotic behavior as those of -meson.

q(s)

0 0K

2

5 5

20

where m , and m are lated to

K 0 K 0 KK

s

KK K

qs q

mq s if p q s if

m m

mm

m m

2( )

q(s) 5 ( ) q(s) 5 ( )

2( )0

( )( )( (( ) ))

0 0 qq ssq s q s

qq s

q s q s

sq s

q s q s

mq q if p q q if

m

mm

m mm

(cont’ed)

25 5

2 22

2

5 2

20

20

qq u d qq

ss s ss

K

m m ui u m di df

m m m mf

m

si s

T. Feldmann, P. Kroll, B. Stech, PRD58,114006(98); PLB449,339 (99)

2 20.13 GeV, 2 0.18 GeVq s Kf f f f f

Another way to understand above assumptions, we can use the mass matrix ofoctet-singlet, in which we know M2

88=(4m2K-m2

)/3, Gell-Mann-Okubo relation

By basis rotations, we obtain M288=(2m2

ss+m2qq )/3, if we set mqq=m,

we get m2ss=2m2

K-m2.

If one takes recourse to the first order of flavor symmetry breaking, one expects

Angle T. Feldmann and P. Kroll, hep-ph/0201044

(cont’ed)

In the framework of perturbative QCD

However, F0B! K(0)=0.35± 0.05

) Taking the conventional values of fq(s) and m0q(s) cannot enhance Bd! 0 K0

(cont’ed)

By M288=(2m2

ss+m2qq )/3, why not m2

ss=2(m2K-m2

) and m2qq=3m2

?

T. Feldmann and P. Kroll, hep-ph/0201044And also

Maybe we should take fq>f and mqq>m

Or M288=(4m2

K-m2)/3 (1+, why not m2

qq > m2 ?

Anomaly

(cont’ed)

Taking mqq» 1.65 m GeV, fq=1.07f,2 2 2 ,2 0.18 GeV ss K sm m m f

0 6

0 6

' 58 10

3.6 10

d

d

Br B K

Br B K

Besides, we also calculate Bd! (0) K*0

*0 6

*0 6

' 7.7 10

25 10

d

d

Br B K

Br B K

F+B! K=0.38

Other contributions:

Intrinsic charm-quark

T. Feldmann, P. Kroll, B. Stech, PRD58,114006(98)

Two-gluon content

C.S. Kim et al., hep-ph/0305032

M. Beneke and M. Neubert, Nucl.Phys. B651 (2003) 225-248

F+B! (0) » 0.21 (0.32) F+

B! 0(0)» 0.32 (0.27)

Discussion:

If taking mqq > m, the branching ratio of the decay Bd! 0 K0 could be enhanced efficiently.

The considering effects could be distinguished from BN’s

two-gluon mechanism, in which 'B B v B B v

With our consideration, 'B B v B B v

The possible exotic mechanisms can be also tested by B! (0) ℓ ℓ decays,

in which the original BR is order of 10-8. With BN’s two-gluon content

the BR could reach to 10-7 that is the same order of magnitude

as the decays B! K ℓ ℓ , measured by Belle and Babar.

B! (0) decays could be the candidates

The more serious one: Babar Collaboration, hep-ex/0403046

BThero.<< 10-6

One phenomenon is worth noticing (mild), i.e. why is

the branching ratio of B+! 0 K+ so high?

We expect B(B+! 0 K+)/B(B0! 0 K0)» (B+}/(B0) =1.08

1.48

1.19

Final state interactions ? or “New” effects ?

M. Beneke and M. Neubert, Nucl.Phys. B651 (2003) 225-248

Babar Collaboration, hep-ph/0308015

Two-angle scheme

As a result, the decay constants, defined as

50 ( ) , i=8,1,P= , 'P

i iP p if pJ 15 5 5 5

1

3J u u d d s s

8 1 8 15 5

8 1 8 15 5 ' '

0 0,

0 ' 0 '

J J f fi p

J J f f

8 1

8 1 ...P P P P

g cP gg cc

With a parton Fock state decomposition, 1

0

( )2 2

ii PP

c

fdx x

N

885 8

115 1

00 0. . ;

00 0

fJi e i p

fJ

For simplicity, we can redefine the wave functions as

8 18 1

8 1' 8 ' 1

cos , sin

sin , cos

1

0

in which ( )2 2

ii

c

fdx x

N

8 1 8 15 5

8 1 8 11' '

8

5 5

0cos sin

0sin co0 ' s

0 0

0 '

fi p

f

J J f fi p

J J f f

Hence,

B

b

d

t tb

Bd

122 itdV e

S

S

/ K

K ( eigenstates)

J

CP

Babar Collaboration, hep-ex/0403046