Table Of ContentB ρπ decays and final state phases
→
Fayyazuddin
7
0
National Centre for Physics and Department of Physics
0
2 Quaid-i-Azam University
n
Islamabad, Pakistan.
a
J
4 February 2, 2008
3
v
8
Abstract
0
0
4 Using the isospin analysis, the fact that penguin is pure ∆I = 1/2
0 transition, the unitarity for tree graph and C-invariance of strong
6 interactions, it is shown that δ = 0 =˜δ , r = tr ,δ δ = π,
0 t t −0 0− −0 0−
− ±
h/ 2trfrf¯cos(δf−δf¯) =(rf2+t2rf2¯)−r−20,whereδ’s are finalstate phases
p and r’s are penguin to tree ratios defined in the text.
- Using the factorization for tree graph as input and the experimen-
p
e tal data, we have obtained the following bounds on rf,rf¯,δf and δf¯:
h 0.11 r 0.21,0.18 r 0.30;11◦ δ 57◦,23◦ δ 90◦ for
: ≤ f ≤ ≤ f¯≤ ≤ f ≤ ≤ f¯≤
v thecasez = cosαcosδ < 0. Forz < 0andz > 0,weobtainthe
f,f¯ f,f¯ f f¯
i
X following bounds for rf¯ and δf¯ : 0.14 ≤ rf¯ ≤ 0.46; 90◦ ≤ δf¯ ≤ 170◦.
r From experimental data, for the decays B− ρ−π0(ρ0π−) we get
a ǫ = 0.28 0.10,ǫ = 0.51 0.10, A−CP0 = →0.8 0.1 i.e. A−0 and
−0 ± 0− ± A0− − ± CP
CP
A0C−P have opposite sign, where 1 + ǫ0−,0−= |T−0|,T0−−+0,C0−0−|,−0|. In the
naive quark model the above values imply a /a = 0.39 0.14 and
2 1
±
a /a = 0.37 0.07 consistent with each other.
2 1
±
1 Introduction
The decays of B into two light mesons which belong to an octet or a nonet
representation of SU(3) have been extensively investigated experimentally
and analysed theoretically to obtain information about CP-violating phases
1
α,β,γ and to test theoretical models [1] . As is well known, CP asymmetries
involve not only weak phases but also strong phases. It is not easy to reliably
estimate the final state strong phases.
In the conventional frame work, the decays are analysed in terms of the
tree amplitude (T); the color-supressed amplitude (C) and the penguin am-
plitude (P) (loop supressed). The supression factor is expressed in terms of
WilsoncoefficentsintheeffectiveLagrangianforB decaysviz |C| a2 0.18
|T| ∼ a1 ≈
, |P| a4 VtbVt∗d 0.03 VtbVt∗d .
|T| ≈ (cid:12)a1(cid:12)(cid:12)VubVu∗d(cid:12) ≈ (cid:12)VubVu∗d(cid:12)
How(cid:12)eve(cid:12)r(cid:12) for B(cid:12) ππ d(cid:12)ecays,(cid:12) to fit the data, rather large values of C
(cid:12) (cid:12)(cid:12) (cid:12) → (cid:12) (cid:12) T
and P are required; against the spirit of the model. The pions are identi(cid:12)ca(cid:12)l
T (cid:12) (cid:12)
boso(cid:12)ns(cid:12)and their wave functions in the final state must be symmetric. This
(cid:12) (cid:12)
is explict in the isospin analysis of these decays. It is not clear how Bose
statistics is satisfied in terms of T,C and P amplitudes. However ρ and
π being not identical bosons, there is no problem with Bose statistics in
terms of analysing these decays in terms of amplitudes T,C and P. The
B ρπ decays have been extensively studied [2, 3]. In particular, the
dec→ays B¯0 ρ−π+(f), B¯0 ρ+π−(f¯) and B− ρ−π0(ρ−π0) are analysed
→ → → ˜
in terms of one weak phase α, six strong phases δ ,δ ,δ ,δ ,δ and δ and
f f¯ t −0 0− t
seven parameters:
Tf¯ Tf¯
= (cid:12) (cid:12)ei(δTf¯−δTf) teiδt
Tf (cid:12)Tf(cid:12) ≡
(cid:12) (cid:12)
| |
Pf¯ Pf¯
Tf¯ = (cid:12)(cid:12)Tf¯(cid:12)(cid:12)ei(δPf¯−δTf¯) ≡ rf¯eiδf¯
(cid:12) (cid:12)
Pf (cid:12)Pf(cid:12)
= (cid:12) (cid:12)ei(δPf−δTf) r eiδf
Tf (cid:12)Tf(cid:12) ≡ f
(cid:12) (cid:12)
| |
T˜0− = T0− +C0− = (cid:12)T˜0−(cid:12)ei(δT0˜−−δT−˜0) = 1+ǫ0− tei˜δt
T˜−0 T−0 +C−0 (cid:12)(cid:12)T˜−0(cid:12)(cid:12) (cid:18)1+ǫ−0(cid:19)
(cid:12) (cid:12)
P0− = r˜ ei(δP0−−δT0˜−) =(cid:12)(cid:12) r(cid:12)(cid:12)0− eiδ0−
T˜0− 0− 1+ǫ
0−
P−0 = r˜ ei(δP−0−δT−˜0) = r−0 eiδ−0 (1)
T˜−0 −0 1+ǫ
−0
2
where
T˜0− T˜−0
1+ǫ = (cid:12) (cid:12),1+ǫ = (cid:12) (cid:12),√2 T0− = Tf¯ ,√2 T−0 = Tf
0− (cid:12)T0−(cid:12) −0 (cid:12)T−0(cid:12)
(cid:12)| (cid:12)| (cid:12)| |(cid:12) (cid:12) (cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12)
Thus there are thirteen parameters besides the weak phase α. There are
seven independent observables viz the CP violating asymmetries A± and
CP
A∓ , the direct CP violation parameter C and dilution parameter ∆C, the
CP
mixing induced CP violation parameter S and dilution parameter ∆S, and
the total decay rate Γ for B¯0 decays and four observables for B− decays
ρ±π∓
viz two decay rates and two CP asymmetries.
Hence it is required to reduce number of parameters by some theoretical
input. In particular in this paper (see section 2); using the isospin analysis
and the fact that penguin is pure ∆I = 1/2 transition, we have obtained the
following relations
r = tr ;δP δP = π
−0 0− 0− − −0 ±
r2 (r tr )2
1 cos(δP δP) = −0 − f − f¯
− f − f¯ 2tr r
f f¯
tr sin(δP δP) = r sin(δP δP)
f¯ f − f¯ −0 −0 − f
Further, using theunitarity for thetreeamplitude andC-invarianceof strong
˜
interactions, it isshown thatonepossible solutiongives δ = 0 = δ .Selecting
t t
this solution we have
δP δP = δ δ
f − f¯ f − f¯
δP δP = δ δ = π
−0 − 0− −0 − 0− ±
These results are used to analyse the experimental data in section 3.
2 Isospin Constraints and Final State Phases
The decay amplitudes in terms of T,P and C for B¯0 f,f¯decays are given
→
by
S¯ = e−iβ f H B¯0 e−iβA¯ = e−iβ Tfe−iγ +Pfeiβ
f f
h | | i ≡
= Tf eiα r eiδf (cid:2) (cid:3) (2)
f
− −
S¯f¯ = e−iβA¯(cid:2)f¯= Tf¯ ei(cid:3)α rf¯eiδf¯ (3)
− −
(cid:2) (cid:3)
3
For B0 f¯and B0 f change α α in Eqs.(2) and (3) respectively.
→ → → −
For B− ρ−π0
→
S¯ = e−iβA¯ = T−0eiα +C−0eiα P−0
−0 −0
− −
(cid:2) (cid:3)
= T˜−0eiα P−0 = T˜−0 eiα r˜ eiδ−0 (4)
−0
− − − −
h i
(cid:2) (cid:3)
S¯ = e−iβA¯ = T˜0− eiα r˜ eiδ0− (5)
0− 0− 0−
− −
(cid:2) (cid:3)
For B+ ρ+π0,ρ0π+ change α to α in Eqs.(4) and (5) respectively.
→ −
Isospin analysis of these decays gives useful constraints on the decay am-
plitudes. The effective weak Lagrangian contains both ∆I = 1 and ∆I = 3
2 2
parts. As is well known, the decay amplitudes can be written in terms of
four complex amplitudes corresponding to I = 0,I = 2 symmetric isospin
wave functions for ρπ states and two antisymmetric isospin wave functions
corresponding to I = 1. Thus we have
1 1 1 1
A¯ = A + A + A A′ (6)
f −2 1 √6 0 √12 2 − 2 1
1 1 1 1
A¯ = A + A + A + A′ (7)
f¯ 2 1 √6 0 √12 2 2 1
1 √3 1
A¯ = A + A A′ (8)
−0 −√2 1 2√2 2 − 2√2 1
1 √3 1
A¯ = A + A + A′ (9)
0− √2 1 2√2 2 2√2 1
1 1
A¯ = A A (10)
00 0 2
√6 − √3
The amplitudes A , A and A ,A′correspond to ∆I = 1 and ∆I = 3 parts
0 1 2 1 2 2
of the effective weak Lagrangian respectively. From Eqs (6-10), we get the
relations
A¯ +A¯ √2 A¯ +A¯ = 2A¯ (11)
f f¯ −0 0− 00
−
(cid:0) (cid:1) (cid:0) (cid:1) √3
A¯ +A¯ = A (12)
−0 0− 2
2
and
2 A¯ A¯ √2 A¯ A¯ = A′ (13)
f − f¯ − −0 − 0− − 1
(cid:0) (cid:1) (cid:0) (cid:1)
4
Since both A and A′ are pure ∆I = 3 and the penguin P is pure ∆I = 1,
2 1 2 2
we must have
P−0 +P0− = 0 (14)
2(Pf Pf¯) √2 P−0 P0− = 0 (15)
− − −
(cid:0) (cid:1)
From Eqs.(14) and (15), we get
δP δP = π
−0 − 0− ±
T0−
r = r | | (16)
−0 0− T−0
| |
= tr
−0
and
r2 r tr 2
1 cos δP δP = −0 − f − f¯ (17)
− f − f¯ 2(cid:0)tr r (cid:1)
(cid:16) (cid:17) f f¯
tr sin δP δP = r sin δP δP (18)
f¯ f − f¯ −0 −0 − f
(cid:16) (cid:17)
(cid:0) (cid:1)
One can get some information about the final state phases δT and δT as
f f¯
follows. The factorization ansatz for tree graph is on strong footing [4]. This
combined with a physical picture [5] that in the weak decay of B meson,
the b quark is converted into b u + q + q¯ and that for the tree graph
→
the configuration is such that q and q¯ essentially go together into the color
singlet states with the third quark recoiling, there is a significant probability
that the system will hadronize as a two body final state. The strong phase
shifts aregenerated after hadronization by rescattering. Thus it is reasonable
to use unitarity to get information about strong phases at least for the tree
graph.
Unitarity gives
ImAi = M∗ Ai (19)
f nf n
Xn
whereA isthedecay amplitudeandM isthescattering amplitudeforf
f nf
→
n.ThesuperscriptiindicatesthattheEq. (19)holdsforeachweakamplitude
andnot forthe whole amplitude A .Eq.(19) canbe used to estimate thefinal
f
state phase shifts [6]. Eq. (15) can be written as
ImAi M∗ Ai = M∗ A (20)
f − ff f nf n
Xn
5
Since the decays are P-wave decays, we can use M = S−1 where S is the
2i
S-matrix for l = 1 partial wave. In terms of S, Eq. (19) gives
ImAi 1+S∗ +iReAi 1 S∗ = i S∗ Ai (21)
f f f − f nf n
(cid:0) (cid:1) (cid:0) (cid:1) Xn
Now parametrizing S-matrix as ηe2i∆ [7] and noting ReAi = Ai cosδi,
f f f
ImA = A sinδ and taking the absolute square on both sides,(cid:12)we(cid:12)get
f f f
| | (cid:12) (cid:12)
Ai 2 1+η2 2ηcos2 δi ∆ = AiS∗ Ai∗S (22)
f − f − n nf n′ n′f
(cid:12) (cid:12) (cid:2)(cid:0) (cid:1) (cid:0) (cid:1)(cid:3) nX′,n6=f
(cid:12) (cid:12)
Note that in single channel description η the absorption coefficent take care
of all the inelastic channels [8]. This is an exact equation. In the random
phase approximation of [6], we can put
AiS∗ Ai∗S = Ai 2 S 2 = Ai 2 1 η2 (23)
n nf n′ n′f n | nf| | n| −
nX′,n6=f Xn6=f (cid:12) (cid:12) (cid:0) (cid:1)
(cid:12) (cid:12)
Then using Eq.(23), Eq.(22) can be written in the form [7]
1 η ρ2 1−η
tan2 δi ∆ = − − 1+η (24)
f − (cid:18)1+η(cid:19) 1 ρ21−η
(cid:0) (cid:1) − 1+η
where
Ai 2
ρ2 = | n| (25)
Ai 2
f
(cid:12) (cid:12)
1 η
− (cid:12)ρ2(cid:12) 1
1+η ≤ ≤
Using the C-invariance of S-matrix [9]
f S n = f C−1CSC−1C n = f¯ S n¯ (26)
h | | i h i h | | i
(cid:12) (cid:12)
we get (cid:12) (cid:12)
1 η ρ¯2 1−η
tan2 δi ∆ = − − 1+η (27)
(cid:16) f¯− (cid:17) (cid:18)1+η(cid:19) 1 ρ¯21−η
− 1+η
where
Ai 2
ρ¯2 = | n¯| (28)
2
Ai
f¯
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
6
In particular we use Eqs.(24) and (27) for the tree amplitudes i = T. First
we note that for the minimum value of ρ2 and ρ¯2, Eq.(24) and (27) gives
δT = ∆ or π +∆ (29)
f,f¯ ±
Thus either we have δ δT δT = 0 or δ = π. We select the solution
t ≡ f¯ − f t ±
δ = 0in agreement with the result of refrence [10]. This reinforces the point
t
of view that for the tree graph final state phases shift is generated by the
rescattering.
However if we assume ρ2 = ρ¯2,then from Eqs.(24) and (27), we get
tan δT ∆ = tan δT ∆ (30)
f − ± f¯ −
(cid:16) (cid:17)
(cid:0) (cid:1)
Then from Eq. (30), it follows that
(i) δT = δT; δT δT δ = 0
f f¯ f¯ − f ≡ t
(ii) δT ∆ = π + δT ∆
f¯ − f −
(iii) δT ∆ = δ(cid:0)T ∆ ,(cid:1)
f¯ − − f −
(iv) δT ∆ = π(cid:0) δT (cid:1) ∆ ,
f¯ − ± − f −
(cid:0) (cid:1)
Thus besides other solutions for which δ is arbitrary, δ = 0 is also a possible
t t
solution.
We now discuss the final state phases for the amplitudes T˜−0 and T˜0−.
Since the tree and color supressed amplitudes have same weak phase, we
can use similar analysis as above to obtain the solution δT˜ = δT˜ or δT˜ =
0− −0 0−
π +˜δT . Now using δT = δT and δT˜ = δT˜ we get from Eqs.(16) and (17)
∓ −0 f f¯ 0− −0
δ δ = π
−0 0−
− ±
r2 (r tr )2
1 cos(δ δ ) = −0 − f − f¯
− f − f¯ 2tr r
f f¯
We note that δ = δ if r = (r tr ) which is experimentally testable.
f f¯ −0 f f¯
± −
3 Observables
In this section, we define the observables which areexperimentally measured.
1) The average decay rate
7
Define
R = Sf¯ 2 + S¯f 2 = 1 Γ +Γ¯
f (cid:12) (cid:12) 2 (cid:12) (cid:12) 2 f¯ f
(cid:12) (cid:12) (cid:12) (cid:12) (cid:0) (cid:1)
R = |Sf|2 + S¯f¯ 2 = 1 Γ +Γ¯ (31)
f¯ 2 (cid:12) (cid:12) 2 f f¯
(cid:12) (cid:12) (cid:0) (cid:1)
S 2 + S¯ 2
+0 −0
R = | |
−0 2 (cid:12) (cid:12)
(cid:12) (cid:12)
S 2 + S¯ 2
0+ 0−
R = | |
0− 2 (cid:12) (cid:12)
(cid:12) (cid:12)
where
R = Tf 2 1 2r cosαcosδ +r2 Tf 2B
f − f f f ≡ f
(cid:12) (cid:12)2(cid:2) (cid:3) (cid:12) (cid:12)2
R = (cid:12)Tf¯(cid:12) 1 2r cosαcosδ +r2 (cid:12)Tf¯(cid:12) B (32)
f¯ − f¯ f¯ f¯ ≡ f¯
(cid:12) (cid:12) h i (cid:12) (cid:12)
R−0 = (cid:12)(cid:12)T−(cid:12)(cid:12)0 2 (1+ǫ−0)2 −2(1+ǫ−0)r−0(cid:12)(cid:12)cos(cid:12)(cid:12)αcosδ−0 +r−20 ≡ T−0 2B−0
(cid:12) (cid:12) (cid:2) (cid:3) (cid:12) (cid:12) (33)
(cid:12) (cid:12) (cid:12) (cid:12)
R = T−0 2 (1+ǫ )2 2(1+ǫ )r cosαcosδ +r2 T0− 2B
0− 0− − 0− 0− 0− 0− ≡ 0−
(cid:12) (cid:12) (cid:2) (cid:3) (cid:12) (cid:12)
It is conve(cid:12)nien(cid:12)t to define, the average rates for B-decays to ρ+π− a(cid:12)nd ρ(cid:12)−π+
S 2 + S¯ 2
Γ± = f¯ f¯
(cid:12) (cid:12) 2 (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
S 2 + S¯ 2
Γ∓ = | f| f (34)
2 (cid:12) (cid:12)
(cid:12) (cid:12)
R +R = Γρπ = Γ± +Γ∓
f f¯
Γ± +(cid:0)Γ∓ R(cid:1)
Tf 2 = = f (35)
B +t2B B
(cid:12) (cid:12) f f¯ f
(cid:12) (cid:12)
8
2) CP-Violating Asymmetries:- Define direct CP asymmetries
A± = a = Sf¯ 2 − S¯f 2 = 2rf sinαsinδf
− CP f (cid:12)(cid:12)Sf¯(cid:12)(cid:12)2 +(cid:12)(cid:12)S¯f(cid:12)(cid:12) Bf
2(cid:12)r s(cid:12)inα(cid:12)sin(cid:12)δ
A∓ = a = (cid:12) f¯(cid:12) (cid:12) (cid:12) f¯ (36)
− CP f¯ B
f¯
2r (1+ǫ )sinαsinδ
A−0 = a = −0 −0 −0
− CP −0 B
−0
2r (1+ǫ )sinαsinδ
A0− = a = 0− 0− 0− (37)
− CP 0− B
0−
Thus it follows from Eqs.(32),(34) and (36)
Γ± = R +R (1+A )
f f¯ CP
Γ∓ = (cid:0)R +R (cid:1)(1 A ) (38)
f f¯ CP
−
(cid:0) (cid:1)
Rfaf Rf¯af¯ Γ± Γ∓
A = − = −
CP
R +R Γ± +Γ∓
f f¯
2sinα r sinδ t2r sinδ
f f f¯ f¯
= − (39)
(cid:2) B +t2B (cid:3)
f f¯
Inordertodiscuss themixinginducedCP asymmetries wefirstgiveageneral
expression for the time dependent decay rates forB0 f¯,f interms of these
→
asymmetries
e−Γt 1+(C ∆C)cos∆mt
Γ B0(t) → f¯,f = 2 Rf +Rf¯ (1±ACP)(cid:20) (S ±∆S)sin∆mt (cid:21)
(cid:0) (cid:1) (cid:0) (cid:1) − ±
(40)
where
S 2 S¯ 2 (R R )+(R a +R a )
f¯,f f¯,f f f¯ f f f¯ f¯
C ∆C = − = ± − (41)
± (cid:12)(cid:12)Sf¯,f(cid:12)(cid:12)2 +(cid:12)(cid:12)S¯f¯,f(cid:12)(cid:12)2 (Rf +Rf¯)(1±ACP)
(cid:12) (cid:12) (cid:12) (cid:12)
The decay rates Γ¯(cid:12) c(cid:12)an b(cid:12)e ob(cid:12)tained from Eq.(40) by changing cos∆mt
f¯,f
→
cos∆mt,sin∆mt sin∆mt
→ −
9
From Eqs.(36) ,(39)and (41), we obtain
1
R = R +R [(1+∆C)+CA ]
f 2 f f¯ CP
(cid:0) (cid:1)
1
R = R +R [(1 ∆C) CA ] (42)
f¯ 2 f f¯ − − CP
(cid:0) (cid:1)
C +A (1+∆C)
A± = a = CP (43)
CP − f −(1+∆C)+CA
CP
C A (1 ∆C)
A∓ = a = − CP − (44)
CP − f¯ −(1 ∆C) CA
CP
− −
The mixing induced CP asymmetry S and ∆S are given by
R +R (1+A )(S +∆S) = 2ImS∗S¯
f f¯ CP f¯ f¯
(cid:0) (cid:1)
sin(2α+δ ) r sin α+δ +δ
2t(Rf +Rf¯) t − f¯ f¯ t
= r sin(α δ(cid:0)+δ ) (cid:1)
f f t
B +tB − −
f f¯ +r r sin δ δ +δ
f f¯ f¯ f t
−
(cid:0) (cid:1) (45)
R +R (1 A )(S ∆S) = 2ImS∗S¯
f f¯ − CP − f f
(cid:0) (cid:1)
sin(2α δ ) r sin α δ δ
2t(Rf +Rf¯) − t − f¯ − f¯− t
= r sin(α+δ(cid:0) δ ) (cid:1)
f f t
B +tB − −
f f¯ +r r sin δ δ δ
f f¯ f f¯ t
− −
(cid:0) (cid:1) (46)
4 CP- Asymmetries and Bound on r ,r
f f¯
In this section, we try to extract information about r and r without as-
f f¯
suming δ = δ . From reference [11], we have
f f¯
C = 0.30 0.13 , ∆C = 0.33 0.13
± ±
a = 0.15 0.08 (47)
f
±
a = 0.53 0.30
f¯
±
Γρπ = R +R = (22.8 2.5) 10−6 (48)
f f¯
± ×
10
