Hard spectator interactions in B\to\pi\pi at order \alpha_{s}^{2}

# Hard spectator interactions in B→ππ at order α2s

###### Abstract

I compute the hard spectator interaction amplitude in at NLO i.e. at . This special part of the amplitude, whose LO starts at , is defined in the framework of QCD factorization. QCD factorization allows to separate the short- and the long-distance physics in leading power in an expansion in , where the short-distance physics can be calculated in a perturbative expansion in .

In this calculation it is necessary to obtain an expansion of Feynman integrals in powers of . I will present a general method to obtain this expansion in a systematic way once the leading power is given as an input. This method is based on differential equation techniques and easy to implement in a computer algebra system.

The numerical impact on amplitudes and branching ratios is considered. The NLO contributions of the hard spectator interactions are important but small enough for perturbation theory to be valid.

LMU-ASC 68/07

Hard spectator interactions in at order

[4ex] Volker Pilipp111volker.pilipp@itp.unibe.ch

Arnold Sommerfeld Center, Department für Physik

Ludwig-Maximilians-Universität München

Theresienstrasse 37, 80333 München, Germany

[3ex] Institute of Theoretical Physics

Universität Bern

Sidlerstrasse 5, 3012 Bern, Switzerland

## 1 Introduction

In the last decades physics has proven to be a promising field to determine parameters of the flavour sector with high precision. On the theoretical side QCD factorization [1, 2] has turned out to be an appropriate tool to calculate decay modes from first principles. Though the decay of the -meson is caused by weak interactions, strong interactions play a dominant role. It is however not possible to handle the QCD effects completely perturbatively. This is due to the energy scales that are contained in the -meson: Whereas at the mass of the -quark is a small parameter, the bound state of the quarks leads to an energy scale of which spoils perturbation theory. The idea of QCD factorization is to separate these scales. At leading power in we obtain the amplitude for in the following form:

 ⟨ππ|H|B⟩ ∼ FB→π∫10dxTI(x)fπϕπ(x)+ (1) ∫10dxdydξTII(x,y,ξ)fBϕB1(ξ)fπϕπ(x)fπϕπ(y)

Two different types of quantities enter this formula. On the one hand the hadronic physics is contained in the form factor and the wave functions and , which will be defined more precisely in the next section. These quantities contain the information about the bound states of the mesons. They have to be determined by non-perturbative methods like QCD sum rules or lattice calculations. Alternatively, because they are at least partly process independent, they might be extracted in the future from experiment. On the other hand the hard scattering kernels and contain the physics of the hard scale and the hard collinear scale and can be calculated perturbatively.

The Feynman diagrams that contribute to can be distributed into two different classes. The class of diagrams where there is no gluon line connecting the spectator quark with the rest of the diagram (fig. 1) contributes to . We obtain by evaluating the hard spectator scattering diagrams, which are shown in LO in in fig. 2. The order corrections of are the topic of the present work. Through the soft momentum of the constituent quark of the -meson the hard collinear scale comes into play. This leads to the fact that in contrast to , which is completely governed by the scale , comes with formally large logarithms. These logarithms cannot be resummed in the present QCD calculation. It will be shown by numerical analysis that the scale dependence of the hard spectator scattering amplitude and the absolute size of its NLO corrections are small enough for pertubation theory to be valid.

My calculation of the hard spectator scattering amplitude is not the first one as it has been calculated recently by [3, 4]. It is however the first pure QCD calculation, whereas [3, 4] used the framework of soft-collinear effective theory (SCET) [5, 6, 7] an effective theory, where the expansion in is performed at the level of the Lagrangian rather than of Feynman integrals. It is the main result of this paper to confirm the results of [3, 4] and to show by explicit calculation that pure QCD and SCET lead to the same result in this case.

Hard spectator scattering corrections to the penguin diagram (third diagram of fig. 1) are beyond the scope of this publication. This class of diagrams does not influence the cancellation of the scale dependence of the “tree amplitude” i.e. the diagrams of fig. 2 and higher order corrections. For phenomenological applications, however, the hard spectator penguin amplitudes, which have been recently calculated in [8], should be taken into account. Also the order of is important for phenomenological applications. This calculation has been partly performed by [9, 10]. There the complete imaginary part and a preliminary result of the real part of the amplitude is given.

From a technical point of view the calculation in this paper consists of the evaluation of about 60 one-loop Feynman diagrams. The challenges of this task are due to the fact that these diagrams come with up to five external legs and three independent ratios of scales. In order to reduce the number of master integrals and to perform power expansions of the Feynman integrals, integration by parts methods and differential equation techniques will prove appropriate tools. They provide a general method to obtain higher powers of a Feynman integral once the leading power is given.

The paper is organized as follows: I define my notations in section 2. In section 3 I show how to get at LO. In section 4 I present a method which uses differential equation techniques and allows for the extraction of higher powers of Feynman integrals once the leading power is given. Sections 5 is dedicated to the technical details of the calculation. After some remarks how to evaluate the Feynman diagrams occuring at NLO I show how to deal with meson wave functions at NLO. Especially the correct treatment of evancescent structures will be explained in detail. In section 6 the analytic results of the calculation will be given, whereas section 7 will provide the numerical analysis. I end up with the conclusions.

## 2 Notation and basic formulas

### 2.1 Kinematics

For the process we will assign the momenta and to the pions (fig. 2) which fulfil the condition

 p2,q2=0. (2)

This is the leading power approximation in . Let us define two Lorentz vectors , by:

 nμ+≡(1,0,0,1),nμ−≡(1,0,0,−1). (3)

In the rest frame of the decaying meson can be defined to be in the direction of and to be in the direction of . Light cone coordinates for the Lorentz vector are defined by:

 z+≡z0+z3√2,z−≡z0−z3√2,z⊥≡(0,z1,z2,0) (4)

So one can decompose into:

 zμ=z⋅pp⋅qqμ+z⋅qp⋅qpμ+zμ⊥ (5)

such that

 z⊥⋅p=z⊥⋅q=0. (6)

### 2.2 Colour factors

In our calculations we will use the following three colour factors, which arise from the algebra:

 CN=12,CF=N2c−12NcandCG=Nc, (7)

where is the number of colours.

### 2.3 Meson wave functions

The pion light cone distribution amplitude is defined by

 ⟨π(p)|¯q(z)α[…]q′(0)β|0⟩z2=0=ifπ4(⧸pγ5)βα∫10dxeixp⋅zϕπ(x). (8)

The ellipsis stands for the Wilson line

 [z,0]=Pexp(∫10dtigsz⋅A(zt)), (9)

which makes (8) gauge invariant. For the definition of the -meson wave function we need the special kinematics of the process. Following [2] let us define

 ΨαβB(z,pB)=⟨0|¯qβ(z)[…]bα(0)|B(pB)⟩=∫d4l(2π)4e−il⋅zϕαβB(l,pB). (10)

In the calculation of matrix elements we get terms like:

 ∫d4l(2π)4tr(A(l)ϕB(l))=∫d4l(2π)4∫d4zeil⋅ztr(A(l)ΨB(z)). (11)

We have to consider only the case that the amplitude depends on only through :

 A=A(l⋅p) (12)

In this case we can use the -meson wave function on the light cone which is given by [2]:

 ⟨0|¯qα(z)[…]bβ(0)|B(pB)⟩∣∣∣z−,z⊥=0 =−ifB4[(⧸pB+mb)γ5]βγ∫10dξe−iξp−Bz+[ΦB1(ξ)+⧸n+ΦB2(ξ)]γα

where

 ∫10dξΦB1(ξ)=1and∫10dξΦB2(ξ)=0. (14)

It is now straightforward to write down the momentum projector of the -meson:

 ∫d4l(2π)4tr(A(2l⋅p)^Ψ(l)) (15) =−ifB4tr(⧸pB+mB)γ5∫10dξ(ΦB1(ξ)+⧸n+ΦB2(ξ))A(ξm2B)

At this point we give the following definitions

 mBλB ≡ ∫10dξξϕB1(ξ) (16) λn ≡ λBmB∫10dξξlnnξϕB1(ξ). (17)

## 3 Hard spectator interactions at LO

The effective weak Hamiltonian we deal with is given by [11]:

 Heff=GF√2V∗udVub[C1O1+C2O2]+h.c., (18)

where

 O1 = (¯dp)V−A(¯pb)V−A, O2 = (¯dipj)V−A(¯pjbi)V−A. (19)

Explicit expressions for the short-distance coefficients can be obtained from [11]. The decay amplitude of is given by

 A(B→ππ)≡⟨ππ|Heff|B⟩. (20)

For later convenience we define

 A(B→ππ)≡A(B→ππ)I+A(B→ππ)II (21)

where () belongs to the first (second) term of (1). Because and contain different hadronic quantities, the renormalisation scale dependence of both of them has to vanish separately. So we can set their scales to different values and . As in there occurs only the mass scale we can set . In there occurs also the hard-collinear scale . As we will see this scale is an appropriate choice for .

Because we only deal with the tree amplitude and do not consider penguin contractions only the matrix elements of the operators and are taken into account. These operators close under renormalisation such that the corresponding hard spectator amplitude is independent of the renormalisation scale.

The decay amplitudes of can be written in terms of as follows [12]:

 −A(¯B0→π+π−) = [λ′ua1+λ′p(ap4+rπχap6)]Aππ −√2A(B−→π−π0) = λ′u(a1+a2)Aππ A(¯B0→π0π0) = [−λ′ua2+λ′p(ap4+rπχap6)]Aππ (22)

where

 Aππ=iGF√2(m2B−m2π)fBπ+fπ

and

 rπχ(μ)=2m2π¯mb(μ)(¯mu(μ)+¯md(μ)). (23)

For the LO and NLO results of the I refer to [12]. We define analogously to (21)

 ai=ai,I+ai,II (24)

where the labels ‘I’ and ‘II’ refer to the contribution to and .

The leading order of the hard spectator interactions which start at is shown in fig. 2. The hard spectator scattering kernel , which does not depend on the wave functions, can be obtained by calculating the transition matrix element between free external quarks, to which we assign the momenta shown in fig. 2. The variables are the arguments of , which arise from the projection on the pion wave function (8). In the sense of power counting we count all components of of , while the components of and are or exactly zero. We define the following quantities

 ξ≡l⋅pp⋅q,θ≡l⋅qp⋅q. (25)

We will see that in the end the dependence on vanishes in leading power such that we can use (15).

We consider the three cases , and . In the case, that the external quarks come with the flavour content of , the LO hard spectator amplitude for the effective operator reads:

 A(1)spect.(¯B0→π+π−)≡ (26) ⟨¯d(¯xp)u(xp)¯u(¯yq)d(yq)|O2|¯d(l)b(p+q−l)⟩spect.= 4παsCFNc1¯xξm2B¯d(l)γμd(¯xp)¯u(xp)γν(1−γ5)b(p+q−l) ¯d(yq)(2⧸pgμν¯y−⧸py¯yγμγν)(1−γ5)u(¯yq),

where the quark antiquark states in the input and output channels of the matrix element form colour singlets. The subscript “spect.” means that only diagrams with a hard spectator interaction are taken into account. The amplitude of vanishes to this order in . In the case of we get the tree amplitude from the matrix element of . The case does not need to be considered separately, because from isospin symmetry follows [13, 12]:

 √2A(B−→π−π0)=A(¯B0→π+π−)+A(¯B0→π0π0). (27)

On the other hand the full amplitude is the convolution of with the wave functions, given by (1). To extract from (26) we need the wave functions with the same external states we have used in (26), i.e. we have to calculate the matrix elements (8) and (10), where the pion or -meson states are replaced by free external quark states. To the order we get

 ϕ(0)π−αβ(y′) ≡ ∫d(z⋅q)e−iz⋅qy′⟨¯u(¯yq)d(yq)|¯diβ(z)uiα(0)|0⟩z−,z⊥=0 = 2πNcδ(y′−y)¯dβ(yq)uα(¯yq) ϕ(0)π+αβ(x′) = 2πNcδ(x′−x)¯uβ(xp)dα(¯xp) (28) ϕ(0)Bαβ(l′−) ≡ ∫dz+eil′−z+⟨0|¯dβ(z)ibα(0)i|¯d(l)b(p+q−l)⟩z−,z⊥=0 = 2πNcδ(l′−−l−)¯dβ(l)bα(p+q−l)

By using

 A(1)spect.=∫dxdydl−ϕ(0)π+αα′(x)ϕ(0)π−ββ′(y)ϕ(0)Bγγ′(l−)TII(1)(x,y,l−)α′αβ′βγ′γ (29)

we finally obtain:

 TII(1)(x,y,l−)α′αβ′βγ′γ = 4παsCF(2π)3N2c1ξ¯xm2Bγμγ′α[γν(1−γ5)]α′γ (30) [(2⧸pgμν¯y−⧸py¯yγμγν)(1−γ5)]β′β.

It should be noted that only the first summand of the above equation contributes after performing the Dirac trace in four dimensions. The second summand is evanescent. This will be important, when we will calculate the NLO corrections of the wave functions (see section 5.2).

If we plug the hadronic wave functions defined by (8) and (15) into (29) i.e. we calculate the matrix element (26) between meson states instead of free quark states, we get for the LO amplitude 222 is used for the matrix elements of the operators between both free external quarks and hadronic meson states. It should become clear from the context what is actually meant.:

 A(1)spect.=−if2πfBCF4N2c4παs∫10dxdydξΦB1(ξ)ϕπ(x)ϕπ(y)1ξ¯x¯y. (31)

Following (1) and the conventions of [3] we write our amplitude in the form:

 Aspect.i=−im2B∫10dxdydξTIIi(x,y,ξ)fBΦB1(ξ)fπϕπ(x)fπϕπ(y). (32)

where in the case of we define

 Aspect.1 = ⟨O2⟩spect. Aspect.2 = ⟨O1⟩spect. (33)

and in the case we define

 Aspect.1 = ⟨O1⟩spect. Aspect.2 = ⟨O2⟩spect.. (34)

Because we use the NDR-scheme which preserves Fierz transformations for and , has the same form for both decay channels. From (31) and (32) we get:

 TII(1)1 = 4παsCF4N2c1ξ¯x¯ym2B TII(1)2 = 0. (35)

## 4 Calculation of Feynman diagrams with differential equations

In this section I will discuss the extraction of subleading powers of Feynman integrals with the method of differential equations [14, 15, 16]. This method will prove to be easy to implement in a computer algebra system. The idea to obtain the analytic expansion of Feynman integrals by tracing them back to differential equations has first been proposed in [14]. This method, which is demonstrated in [14] by the one-loop two-point integral and in [15] by the two-loop sunrise diagram, uses differential equations with respect to the small or large parameter, in which the integral has to be expanded.

In contrast to [14, 15] I will discuss the case where setting the small parameter to zero gives rise to new divergences. In this case the initial condition is not given by the differential equation itself and also cannot be obtained by calculation of the simpler integral that is defined by setting the expansion parameter to zero. It is not possible to give a general proof, but it seems to be a rule, that one needs the leading power as a “boundary condition”. An efficient way to calculate the leading power of Feynman integrals is provided by the method of regions [17, 18, 19, 20], whereas the subleading powers can be obtained from a differential equation. In the present section I will discuss which conditions the differential equation has to fulfil in order for this to work.

Although the examples I use below are taken from the present calculation, this method is very general and can be used in any case in which the expansion of Feynman integrals in small parameters is needed.

### 4.1 Description of the method

 I(p1,…,pn,m1,…,mn)=∫ddk(2π)d1D1…Dn (36)

where the propagators are of the form . We assume that there is only one mass hierarchy, i.e. there are two masses such that all of the momenta and masses and are of or of . We expand (36) in by replacing all small momenta and masses by and expand in . After the expansion the bookkeeping parameter can be set to .

We obtain a differential equation for by differentiating the integrand in (36) with respect to . This gives rise to new Feynman integrals with propagators of the form and scalar products in the numerator. Those Feynman integrals, however, can be reduced to the original integral and to simpler integrals (i.e. integrals that contain less propagators in the denominator) by using integration by parts identities.

Finally we obtain for (36) a differential equation of the form

 ddλI(λ)=h(λ)I(λ)+g(λ) (37)

where contains only rational functions of and can be expressed by Feynman integrals with a reduced number of propagators. It is easy to see that and are unique if and only if and the integrals contained in are master integrals with respect to IBP-identities, i.e. they cannot be reduced to simpler integrals by IBP-identities. If is divergent in , , and have to be expanded in :

 I = ∑iIiϵi h = ∑ihiϵi g = ∑igiϵi. (38)

Plugging (38) into (37) gives a system of differential equations for , similar to (37). In the next paragraph we will consider an example for this case.

First let us assume that and have the following asymptotic behaviour in :

 h(λ) = h(0)+λh(1)+… g(λ) = ∑jλjg(j)(lnλ) (39)

i.e.  starts at , and we allow that starts at a negative power of . We count as so the may depend on . This dependence, however, has to be such that

 limλ→0λg(j)(lnλ)=0. (40)

The condition (40) is fulfilled, if the are of the form of a finite sum

 m∑n=n0anlnnλ. (41)

The limit however can spoil the expansion (39). E.g.  so the condition (40) is not fulfilled, which is due to the fact that we must not change the order of the limits and .

Further we assume that also starts at

 I(λ)=I(0)(lnλ)+λI(1)(lnλ)+… (42)

and plug this into (37) such that we obtain an equation which gives recursively:

 λiI(i)=∫λ0dλ′λ′i−1(i−1∑j=0h(j)I(i−1−j)(lnλ′)+g(i−1)(lnλ′)). (43)

I want to stress that, because starts at , (43) is a recurrence relation, i.e.  does not mix into if . As the integral is only well defined if , we need the leading power as “boundary condition” and (43) will give us all the higher powers in . It is easy to implement (43) in a computer algebra system, because we just need the integration of polynomials and finite powers of logarithms.

A modification is needed if starts at i.e.

 h=−nλ+h(0)+…. (44)

By replacing we obtain the differential equation

 ddλ¯I=(nλ+h)¯I+λng (45)

which is similar to (37) and leads to

 λi+nI(i)=∫λ0dλ′λ′i+n−1(i+n−1∑j=0h(j)I(i−1−j)(lnλ′)+g(i−1)(lnλ′)), (46)

which is valid for . So, if starts at , the subleading powers result from the leading power.

### 4.2 Examples

###### Example 4.1.
 I=∫ddk(2π)d1k2(k2−λ)(k2−1) (47)

where . The exact expression for this integral is given by:

 I=i(4π)2lnλ1−λ=i(4π)2lnλ(1+λ+λ2+…). (48)

We see that diverges for . As described e.g. in [20] we can obtain the leading power by expanding the integrand in the regions and . This leads in the first region to

 ∫ddk(2π)d−1k2(k2−λ)=−i(4π)2−ϵΓ(1+ϵ)(1ϵ+1−lnλ) (49)

and in the second region to

 ∫ddk(2π)d1k4(k2−1)=i(4π)2−ϵΓ(1+ϵ)(1ϵ+1) (50)

such that we finally obtain

 I(0)(lnλ)=i(4π)2lnλ. (51)

This is the result we obtain from the leading power of (48). We write the derivative of with respect to in the following form:

 ddλI=11−λ[I−∫ddk(2π)d1k2(k2−λ)2]. (52)

We obtained the right hand side of (52) by decomposing into partial fractions. Of course this decomposition is not unique which is due to the fact that itself is not a master integral but can be further simplified by partial fractioning. From (52) and (37) we get:

 h = 11−λ=1+λ+λ2+… g = i(4π)21λ(1−λ)=i(4π)2(λ−1+1+λ+…) (53)

such that the coefficients in the expansion in according to (39) do not depend on the power label :

 h(k)=1andg(k)=i(4π)2. (54)

We obtain for the recurrence relation (43):

 I(k)=1λk∫λ0dλ′λ′k−1(k−1∑j=0I(k−1−j)(lnλ′)+i(4π)2). (55)

Using the initial value (51) it is easy to prove by induction

 I(k)(lnλ)=i(4π)2lnλ∀k≥0. (56)

This result coincides with (48).

The first nontrivial example, we want to consider, is the following three-point integral:

###### Example 4.2.
 I=∫ddk(2π)d1k2(k+un−+l)2(k+n++n−)2. (57)

Here and are collinear Lorentz vectors, which fulfil and , is a real number between and and is a Lorentz vector with and . Furthermore we define

 ξ=2l⋅n+andθ=2l⋅n−. (58)

We expand in , so we make the replacement and differentiate with respect to . The integral is not divergent in such that we obtain a differential equation of the form (37) where the Taylor series of starts at as in (39). In only two-point integrals occur, which are easy to calculate. I do not want to give the explicit expressions for and because they are complicated, their exact form is not needed to understand this example and they can be handled by a computer algebra system. Because the leading power of is of , (43) gives all of the subleading powers.

We obtain the leading power as follows: First we have to identify the regions, which contribute at leading power. If we decompose into

 kμ=2k⋅n+nμ−+2k⋅n−nμ++kμ⊥ (59)

we note that the only regions, which remain at leading power, are the hard region and the hard-collinear region

 k⋅n+ ∼ 1 k⋅n− ∼ λ kμ⊥ ∼ √λ. (60)

The soft region leads at leading power to a scaleless integral, which vanishes in dimensional regularisation. In the hard region we expand the integrand to

 1k2(k+un−)2(k+n++n−)2. (61)

By introducing a convenient Feynman parametrisation we obtain for the -dimensional integral over (61):

 i(4π)2−ϵΓ(1+ϵ)exp(iπϵ)1u(ln(1−u)ϵ−12ln2(1−u)). (62)

In the hard-collinear region we expand the integrand to

 1k2(k+un−+θn+)2(2k⋅n++1). (63)

The integral over (63) gives:

 i(4π)2−ϵΓ(1+ϵ)exp(iπϵ)1u(−ln(1−u)ϵ+2Li2(u)+12ln2(1−u)+lnuln(1−u)+ln(1−u)lnθ). (64)

Adding (62) and (64) together we get the leading power of (57):

 I(0)=i(4π)21u(2Li2(u)+lnuln(1−u)+ln(1−u)lnθ). (65)

By plugging (65) into (43) we obtain at :

 (66)

Now we want to consider the following four-point integral

###### Example 4.3.
 I=∫ddk(2π)d1k2(k+n−)2(k+l−n+)2(k+l−un+)2, (67)

where we used the same variables, which were introduced in (57). This example is very special, because in this case our method will allow us to obtain not only the subleading but also the leading power in . is divergent in such that we obtain after the expansion (38) a system of differential equations of the following form:

 ddλI−1 = h0I−1+g−1 ddλI0 = h0I0+h1I−1+g0. (68)

It turns out that in our example takes the simple form

 h=−2+2ϵλ (69)

such that analogously to (45) we can transform (68) into

 ddλ(λ2I−1) = λ2g−1 ddλ(λ2I0) = −2λI−1+λ2g0. (70)

This system of differential equations can easily be integrated to:

 I(i)−1 = 1λi+2∫λ0dλ′λ′i+1g(i−1)−1 I(i)0 = 1λi+2∫λ0dλ′λ′i+1(−2I(i)−1+g(i−1)0) (71)

where the superscript denotes the order in as in (39) and (42). Both and start at . Because (71) is valid for , it gives us the leading power expression, which reads:

 I(−1)=i(4π)2−ϵΓ(1+ϵ)2uξ(1ϵ−1−lnu1−u−lnξ) (72)

where as in the example above. The exact expression for (67) can be obtained from [21]. Thereby (72) can be tested.

In the last paragraph I want to return to Example 4.2. I will show how we can use differential equations to prove that the integral (57) depends in leading power only on the soft kinematical variable and not on . We need derivatives of the integral with respect to and , which we have to express through derivatives with respect to . These derivatives can be applied directly to the integrand, whose dependence on is obvious. We start from the following equations:

 nμ+∂∂lμI = ∂∂θI+ξ∂∂l2I nμ−∂∂lμI = ∂∂ξI+θ∂∂l2I (73) lμ∂∂lμI = ξ∂∂ξI+θ∂∂θI+2l2∂∂l2I