1. What data show

2. Polarization analysis & Puzzle ?

3. Results by PQCD & speculation

4. Summary

Chuan-Hung Chen

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

Polarization Problem in B decays

What data show in B→K* decays

1.6 0.7 61.5 0.810.0 10BR

69.2 0.9 0.5 10BR

2.2 62.012.7 1.1 10BR

2.1 0.7 61.9 1.06.7 10BR

22 2

0 1A A A

What data show in B→K* decays

What data show in B→ ()() decays

Polarization analysis

u

b

u

s

Tree

W

Vub Vus

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 q

t

WVtb Vts

g

2 0.043tsV A

0.2233.6 10

Effective interactions for b sqq

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

1 1

2 3a aij kl il jk ij kl C3 (mW) ~ -C4 (mW) /Nc ;

C5 (mW) ~ -C6 (mW )/Nc

Vector mesons are composed of quark and anti-quark and spin=1

=anti-s s, K*0=anti-s d, +=anti-d u

moving

direction

+

+ 0

Meson helicity

For B → V1 V2 decays, V1,2 : vector mesons

Helicity basis:s s

b sVA VA

m

s s

b smK*

1 2

( ) * *2 1 1 2( ) ( )M ag bp p ic p p

2 2 200 1 2

1 2

2 2

1

2

+ 2

B

B

H m m m am m

m p b

BH a m pc

BH a m pc

s s

b sm

mB

*

00 : : 1: : K

B B B

mm mH H H

m m m

H00 >> H >> H++

If one sets the hadronic effects to be the same, i.e. F00 ~ F ~ F++

*

00 00 : : : : K

B B B

mm mH H H F F F

m m m

The resultants are

22 2

0 || A A A

Polarization fractions:

222 200

0 ||( )2 2

1,

2h hh h

HA A H H

H H

Estimation:

1. Set the QCD approach you like, naïve factorization, generalized factorization, QCDF, PQCD,… and so on

2. For simplicity, let’s concentrate on naïve factorization.

*3 3 4 4 5 5 6 6

*3 643 4 5 5

*3 4 5 5

0 (1 )

(1 )

c c c

K c O c O c O c O B

c ccc c c s s K b s B

N N N

a a a f m K b s B

By gluon penguin,

3 643 3 4 4 5 5, ,

c c c

c cca c a c a c

N N N

1 1

3 2a a

il jk ij kl ij kl

Fierz identities

Notation:

* *

*

* *

1 10 ,0,0, , 0 ,0,0, ,

1 11 0,1, ,0 , 1 0,1, ,0 ,

2 21 1

1 0,1, ,0 , 1 0,1, ,02 2

c cK KK

K K

p E p Em m

i i

i i

1 2

( ) * *2 1 1 2( ) ( )M ag bp p ic p p

* *

* *

*

*

2 22 2 2

00 1 2

1

4

2

2V

B cB BK K

BK K

B cB K

B K

f m pH a m m m m m A A

m m m

m pH am f m m A

m m

Helicity basis

3 4 5a a a a

By polarization basis

* *

* *

*

*

2 22 2 2

0 00 1 2

1

4

2

12

221

2 V2

B cB BK K

BK K

B K

B c

B K

f m pA H a m m m m m A A

m m m

A H H am f m m A

m pA H H am f

m m

By taking A1 ~ A2 ~ V and neglecting m2V/m2

B

2 222 2

0 2 2: : ~ 1: :

B B

m mA A A

m m

In SM, with naïve factorization, no way to solve the small longitudinal polarization. Unless we make some fine-tuning on the form factors; however, small BR will be the problem.

The results of PQCD in the SM

Inevitably, annihilation and nonfactorizable effects should be included, in which in general the nonfactorization includes final state interactions.

H.Y. Cheng, C.K. Chua, A. Soni, Phys. Rev. D71 (2005) 014030

Why annihilation is important ?

By Fierz transformation and equation of motion, (V-A)(V+A)→(S-P)(S+P)

Only O6 is important

, , B

Tree dominant processes: C1 >> C4,6

u

b

u

d

Tree

W

Vub=A3Rb

Vud ~O(1)

penguinb d

q q

t

WVtb Vtd

g

3td tV A R

Since tree only has (VA)(VA), for those color-allowed processes, annihilation contributions are negligible

The nonfactorizable effects are associated with C2/Nc, as known C2 << C1, we expect that they are also small.

Expectably, the results will be similar to the estimations of naïve factorization

The results by PQCD are summarized as follows:

* *; , B K ss K sd

No tree contributions in neutral B decay, purely penguin process

Since there involve (VA)(V+A) operators, annihilation contributions are important

Nonfactorizable effects are associated with C3()/Nc, C4 ()/Nc , C5()/Nc , C6()/Nc

=1.5 GeV 8.83 103, 1.67 102, 4.35 103, 2.45 102

=2.5 GeV 6.31 103, 1.27 102, 3.51 103, 1.70 102

and non-negligible

By conventional PQCD with including the transverse momentum, kT, and with the chosen condition for hard scale

1max ,1/ ,1/e Bt xm b b

b1 : the conjugate variable of k1T, light quark momentum of B b : the conjugate variable of kT , light quark momentum of meson

How do annihilation and nonfactorizable effects affect ?

Can we make the results be better ?

H.N. Li, Phys.Lett. B622, 63 (2005)

1max ,1/ ,1/ , , 1 GeVe Bt xm b b

By modifying the condition for hard scale to be

Nonperturbative wave functions as expansion of Gegenbauer polynomials:

3/ 2

1

2 2, 6 1 1 2 1n nn

x x x a C x

Usually is set to be 1 GeV

* *,B K K

penguin dominant processes

tree is important in Bd→ K*+

By conventional PQCD with including the transverse momentum, kT, and with the chosen condition for hard scale 1max ,1/ ,1/e Bt xm b b

1max ,1/ ,1/ , 1e Bt xm b b GeV

By modifying the condition for hard scale to be

Since the tree of Bu→ + K*0 is annihilation, it is negligible; that is, it is proper to take it as a pure penguin dominant decay

Compare to Bd→K*0 , LP of 65% with =1, it seems a little bit large; Why is it different ?

Are the annihilation and nonfactorizable effects important for polarization fractions ?

Comparisons: the decay amplitudes for B→VV

VtsV*tb

Factorizedparts

Nonfactorizedparts

annihilation

mode fV FLe103 MLe104 fB FLa 103 MLa 105

Bd → K*0 3.41 3.77 + i 5.11 0.15 i 1.35 0.26 i 9.52

Bu→0 K*+ 2.50 1.88 i 3.16 0.47 i 1.51 3.71 i 6.50

The real parts of FLe in Bd →K*0 and Bu→0 K*+ are opposite in sign

If we artificially change the sign of Bu→0 K*+ |A0| will change from 78% to 68%

Summary :

It is clear that only considering the factorized effects in the SM has no chance to solve the polarization problem of BK*

It is easy to understand the polarization fractions of B→()()

Is any anomaly between A|| and A ?

By modifying the chosen condition of PQCD for hard scale, the BRs of BK* are consistent with the data, and longitudinal polarizations could be around 60%

By PQCD, the annihilation contributions are strongly depend on the wave functions of involving mesons; consequently, the longitudinal polarizations are different in K* and K*

mode FTe103 MTe104 FTa 103 MTa 105

Bd → K*0 4.49 3.56 i 0.22 2.82 + i 5.82 0.17 + i 0.66

Bu→0 K*+ 3.21 1.80 i 0.36 0.03 + i 8.77 0.06 + i 1.19