Recursion Relation for Boundary Contribution

It is well known that under a BCFW-deformation, there is a boundary contribution when the amplitude scales as O(1) or worse. We show that boundary contributions have a similar recursion relation as scattering amplitude. Just like the BCFW recursion relation, where scattering amplitudes are expressed as the products of two on-shell sub-amplitudes (plus possible boundary contributions), our new recursion relation expresses boundary contributions as products of sub-amplitudes and boundary contributions with less legs, plus yet another possible boundary contribution. In other words, the complete scattering amplitude, including boundary contributions, can be obtained by multiple steps of recursions, unless the boundary contributions are still non-zero when all possible deformations are exploited. We demonstrate this algorithm by several examples. Especially, we show that for standard model like renormalizable theory in 4D, i.e., the theory including only gauge boson, fermions and scalars, the complete amplitude can always be computed by at most four recursive steps using our algorithm.


The introduction
In recent years BCFW recursion relation [1,2] has become a standard method to compute tree-level scattering amplitudes. In its original form, BCFW recursion relation was presented for 4d Yang-Mills theory in the language of spinors, but soon the method were applied to various other theories 1 . Despite its successes, BCFW recursion relation has met difficulties applying to certain theories 2 whose amplitudes do not have the desired vanishing scaling in the large limit of deformation parameter. A naive application of BCFW recursion relation fails to capture a piece of amplitude (usually called boundary contributions), which corresponds to the residue at infinity.
Several proposals have been made to find boundary contributions. The first [6,7] is to introduce auxiliary fields so that in the enlarged theory, there are no boundary contributions. The second [8,9,10] is to carefully analyze Feynman diagrams and then isolate their boundary contributions, which can be evaluated directly or recursively afterwards. The third [11,12,13] is to express boundary contributions in terms of roots of amplitudes. These three methods are, however, effective only for limited types of theories. Recently a systematical algorithm, based on carefully analysis of pole structure of boundary contributions, has been proposed in [14]. Though in principle the method is applicable to any quantum field theory, in practice it suffers from high computational complexity.
In this paper, we present a new method to compute boundary contributions. The key observation is that with properly chosen deformations, boundary contributions satisfy similar recursion relations as scattering amplitudes. Just like the BCFW recursion relation, where scattering amplitudes are expressed as (a sum of) the products of two on-shell sub-amplitudes (plus possible boundary contribution), our new recursion relation expresses boundary contributions as (a sum of) products of sub-amplitudes and boundary contributions with less legs, plus yet another possible boundary contribution. The new boundary contribution is subsequently computed by a new shift, and the recursion ends whenever the remaining boundary contribution vanishes. This multi-step recursion is (almost) as efficient as BCFW recursion, but applicable to more general models.
The paper is organized as follows. In section 2 after a short discussion of pole structure, we present our main result: the recursion relations for boundary contributions. In section 3 we show that a pure scalar φ m theory amplitude can be computed via a (m − 1) step recursion. In section 4 we analyze all possible boundary contributions of Standard Model like theories, and show that any amplitude in this theory can be computed via a (at most) 4 step recursion. In section 5, we present two explicit examples using our method. In Appendix A, we discuss some mathematical aspects of multi-variable integrations. In Appendix B, we present recursion relation for boundary contributions under other choices of deformations. In Appendix C, propagator in light-cone gauge has been discussed.

The recursion relation for boundary contribution
The key idea of BCFW recursion is determining scattering amplitudes by their poles. In order to find a recursion relation of boundary contributions, we also need to be very clear about the poles of the boundary contribution. First consider the primary deformation (BCFW-deformation) 1|n], Let us use indices I, J to denote subsets of remaining particles T ≡ {2, 3, ..., n − 1}. For later convenience, we also define q µ i = 1 2 [i|γ µ |n , then (2.1) can be written as Under the deformation, the expression of tree-level amplitudes coming from Feynman diagrams will be where C J 's are the residues of corresponding poles and B 1|n] is the boundary contribution we want to find. To read out B 1|n] , a good way is to do the large z expansion in the first line of (2.3). Using and B 1|n] can be read out by selecting same power of z in numerator f (z) and denominators. In other words, poles of B can be It is worth to notice that, in principle, B 1|n] can have terms which are pure polynomials in momentum (i.e. they do not have any pole), and our method is not applicable. This can happen in many effective theories with higher dimension operators.

Recursion relation for boundary contribution
As discussed in [14], in order to determine B 1|n] , we must use a different deformation. Without loss of generality, we will choose the deformation 2|n]. A crucial merit of this deformation is that spurious poles n|P J⊂T |1] in (2.6) as well as others n|P J⊂T |i] generated in middle steps are invariant under the deformation. In other words, under this deformation only physical single poles P 2 I⊂T in (2.6) are detected. Following the proof of BCFW recursion relations, we evaluate the contour integration where B 12|n] is the possible remaining boundary contribution and the residue part is given by recursion relation To prove (2.8), first we notice that .., p n−1 , p 1 + wq 1 + zq 2 ) (2.10) where we have used (2.9) at the second line. Above integration can be parameterized by z = z I + ǫe iα and w = Re iβ , thus the contour integration becomes following double integrations − ǫ (2π) 2 2π 0 dαdβ z I + ǫe iα A(p 1 − Re iβ q 1 , p 2 − (z I + ǫe iα )q 2 , p 3 , ..., p n−1 , p 1 + Re iβ q 1 + (z I + ǫe iα )q 2 ) (2.11) Now for R big enough but finite and ǫ small enough but finite, A is finite (i.e., there is no pole along the integral path). Using the Fubini-Tonelli theorem reviewed in Appendix A, we can exchange the ordering of two integrations, thus (2.10) becomes = h A L ( p 2 (z I ), I, −P h (z I )) 1 (p 2 + P I ) 2 B 1|n] (p 1 , p n (z I ), I, P −h (z I )) (2.12) 3 Here 1 / ∈ I I, because according to (2.6), B 1|n] does not have poles at (p1 + p2 + PJ ) 2 .
Thus we have proved (2.8). If B 12|n] = 0, we can take the third deformation, for example 3|n]. First we write Using the contour integration, we obtain 14) It is important to emphasize that since above two integrations are around infinity, in general we can not change the ordering (see the discussion in the Appendix A), i.e., B 12|n] = B 21|n] . Nevertheless, we can still change the order of z and w i integration, and we find Before ending this section, let us give some remarks. For the application of above result, it seems crucial that there is a choice such that after finite steps, we should have B 1···k|n] = 0. In later part of the paper, we will present various theories where such a choice exists. But there are theories the boundary contributions do not vanish after exploiting all shifts. If we define the recursion part of the i-th deformation as A 1···i|n] , then we have (2. 16) In order for our algorithm to be complete, an efficient method to determine the last boundary contribution, B 1···n−2|n] , is desirable. At the same time, it is equally important to explore whether the later terms in (2.16) are suppressed(for example by some large energy scale). In this case one can use the first several terms as a good approximation of the complete amplitude.

Scalar Theory
Starting from this section, we will demonstrate our algorithm by several examples. The first simple example is the real scalar theory with φ m interaction term, i.e., the Lagrangian is given by The vertex φ m will contribute possible boundary terms for n-point amplitude when n ≥ m. For n = m, the contribution is just σ and we could not detect it using pole. Thus we will consider the case n > m. In fact, to get nontrivial Feynman diagrams, we need to have n = 2 + (m − 2)V where V is the number of vertices. It is easy to see that under the primary deformation 1|n], the boundary contribution comes from Feynman diagrams where 1, n attach to same vertex (see Figure 2). If we define a non-overlapping (m − 2)splitting K of the set {2, 3, · · · , n − 1} as with each K i having at least one element (the ordering does not matter in the splitting), the boundary contribution is given by where Λ is the set of all allowed splitting and P i=1,...,m−2 are, in fact, inner particles in Figure 2. From (3.3), it is easy to see that since n > m, there are at least two vertices. Thus for worst diagrams, where there is only one inner particle connecting to the vertex attached by 1, n, at most (m − 2) new deformations of i|n] type besides 1|n] will be enough to completely determine the boundary contribution B 1|n] . For example, for φ 4 theory, under the 1|n] (n > 4), boundary part are given by Feynman diagrams where 1, n attach to same vertex. Under the second deformation 2|n], only these Feynman diagrams where 1, 2, n attach to same vertex are undetected, but they will be detected by the third deformation, for example, 3|n]. Thus by total three steps we can determine the full amplitude.
It is worth to mention that our above discussion of pure scalar theory does not depend on the detail if the theory contains lower point vertex φ p with p < m in the Lagrangian.

The standard model like theory
In this section, we discuss standard model like theory, for which the Lagrangian is given by 4 For the case (a), as has been proved in [15,16], if external particles contain at least one gluon, there is always a good deformation without boundary contribution, so we will not consider case (a) further.
For case (b), there is also a one-step deformation to completely determine amplitude. If there is at least three positive fermions, for example, 1, 2, 3, we can make following Risager deformation [17] Now we count the power of z. First the wave functions of three particles are |1] , |2] , |3], so they scale as z 0 . Secondly, each fermionic propagator scales as z 0 while each bosonic propagator scales as z −1 . Thirdly, for the vertex, only A 3 , Aφ∂φ contribute z 1 factors. But since these vertices are attached to at least two 5 bosonic propagators, thus the number of vertices is always less than the number of bosonic propagator (see, Figure 3), thus under the large z limit, the integrand vanishes. If there are two positive fermion ψ 1 , ψ 2 and two negative fermion ψ 3 , ψ 4 , we can do following deformation Since under this deformation, the wave functions of ψ i , i = 1, 2, 3 are not changed, the power counting of z is similar to the case where all ψ i , i = 1, 2, 3 are positive. The deformation (4.3) is kindly of the union of two BCFW-deformation using same z variable. In the next several subsections we will show that for cases (c) and (d), the amplitude can be computed by a (at most) 4 step recursion.

The case (c) with only two fermions
Now we consider n-point amplitude with 2 fermions and (n − 2) scalars. With out loss of generality we assume the particles 1, 2 are scalars while n is the negative fermion, so its wave function is given by |n . We will work in light cone gauge, which is most convenient for the analysis of boundary behavior.
Since we are mainly interested in gauge coupling, we will neglect Yukawa and quartic scalar coupling terms in (4.1) for now. The Light-cone gauge Lagrangian is given by All inner products of basis vanish except q·q = −1, h·h = 1. The advantage of Light-cone gauge Lagrangian (4.4) is that under the deformation i|n], many z-factor coming from vertices will be canceled out. For example, for L (4) part, only ∂ − operator appears, but it is equal to p − = q · p, so under the deformation i|n], the z-dependent part will be p − (z) ∼ q · z λ i λ n = 0. In other words, four point vertex in Light-cone gauge will never contribute z factor. Similar observation can be made for L (3) part. Now we will have Under the deformation, we will have (z λ i λ n ) · h = 0 although (z λ i λ n ) · h = 0 when i = 1. In other words, under the i|n]-deformation, only A h A h A h vertex, φφA h vertex and A h ψ + ψ − vertex contribute factor z. We will use this important observation to discuss the large z behavior. Now we consider the boundary contribution under the primary deformation 1|n]. First by our above analysis, vertices of L (3) , L (4) will not contribute z factors. The z-dependence coming from bosonic and fermionic propagators. Since bosonic propagator is 1 z and fermionic propagator is 1 z 0 , the worst Feynman diagrams are these without bosonic propagators and scale as 1 z 0 . Now we consider the vertex n attached. If 1 is not attached to same vertex, there is one fermionic propagator depending on z connected to n. If it is Yukawa coupling we will have |P +zq|n (P +zq) 2 = |P |n (P +zq) 2 . If it is gauge coupling we will have |P +zq|γ µ |n where we have used |qγ µ |n = |n [1|γ µ |n ∼ |n q µ . However, under the Light-cone gauge, z |n q µ · A µ = 0. Thus there is an overall 1 z contribution from the vertex n attached and these Feynman diagrams do not give boundary contribution.

The case (d) with only scalars
For the third deformation 3|n], by similar analysis, especially there is no gluon propagator along the hard line, only the type of diagrams in Figure (7) scales as O(1). It is worth to mention that factor (4.8) is not affected by the deformation 3|n] although other part of Feynman diagrams will be affected in general.
Finally for the fourth deformation 4|n], by similar analysis, especially when the hard line has gluon propagators scaling behavior will be suppressed by an extra 1 z factor, we found that no matter how we insert the particle 4 into Figure 7, we will always get at least 1 z scaling. Thus by three steps, we can completely determine boundary contributions of Figure (5

)(a).
Type (b): Now we consider the second deformation 2|n] for the type (b) in Figure 5. Unlike the type (a) where 1, n are attached to φ 4 vertex, here 1, n are attached to φ 2 A h A h vertex. If the particle 2 is along the line A h , it can be shown using Lagrangian (4.4) that the large z behavior is 1 z at least. But if particle 2 is along the line A h , it will contribute to boundary part. After this we will get diagrams like these given in Figure 8. Next we consider the third deformation 3|n]. There are two cases. For the first case 3 is not directly connected to 2 by scalar line, thus using the same analysis for the type (a), it is 1 z behavior at least. For the second case, 3 is directly connected to 2 by scalar line, thus like the Figure 7, it gives nonzero boundary contributions. Finally, like the case (a), the fourth deformation 4|n] will make diagrams in 8 vanishing at large z limit.

Type (c):
The type (c) of Figure 5 is most complicated one because under our light-cone gauge choice, p 1n · q = 0, thus we can not impose Light-cone gauge on the gauge field A. To solve this problem, we shift momentum basis to where |n ǫ = |n] + ǫ|y . Thus when we take ǫ → 0 after finishing calculations, we will come back to original light-cone gauge. The type (c) can grow to following four diagrams given in Figure 9. Now we discuss these four diagrams one by one. For diagram (c1), using Feynman rules given in (4.4), it is easy to find thus If the bottom vertex is M HV , the diagram vanishes. If the bottom vertex is M HV , the diagram reads Figure 9: Four different diagrams containing the Aφφ vertex with q · P = 0 for the gluon.
This quantity goes to infinity when ǫ → 0. However, for four scalars, term ( 1 ∂ − J − ) 2 in the L (4) part of Lagrangian will be singular too. Its contribution is When combining these two together, we arrive where the ǫ has been canceled out. It is worth to notice that (4.13) scales as O(z) for 1|n]-deformation, but under i|n]-deformation it scales as O(1) ( notice that p 5 or p 6 will be shifted). Thus after two steps of 1|n] and 2|n], we arrive to the similar structure as Figure 6. Thus by same argument, two further deformations will enable us to detect all contributions.
For the diagram (c2), the same bottom vertex must be M HV to give nonzero contribution. If the top vertex is M HV , it reads ( In fact, this calculation shows that the diagram (c2) is similar to the type (b) of Figure 5. Thus by same argument, we need at most four steps of deformations to determine its contributions.
For the last diagrams (c3), (c4), to have nonzero contribution, the bottom vertex must be M HV , while the top vertex is z 0 scaling. Thus under the second deformation 2|n], whole diagram scales as 1 z . In other words, by two steps of deformations, we can determine its contributions.

Conclusion:
After above careful analysis, we can see that for all external particles to be scalars, at most four steps of deformations are enough to completely determine amplitudes by our algorithm.

Examples
In this section, we will use two examples to demonstrate our method. These two examples correspond to the case (c) and (d) in previous section. We will use A and B to denote color ordered amplitudes and boundary contributions, and A and B to denote the complete amplitudes and boundary contributions dressed with color factors.

Example I: two fermions with three scalars
The first example we will consider is where we have defined the color factor For this example, we need two steps to determine the amplitude according to our discussion in case (c) of section four. We will start with deformation 1|5]. Under this shift, the recursive part gives F a 2 a 3 a 1 a 4 a 5 (5.3) Now we calculate the boundary part using the boundary recursion relation with deformation 2|5]. Naively, there will be following splitting diagrams: (1) A 4 (2, 3, 4, P )B (−P, 1, 5, 3). Among these three cases, only case (1) gives nonzero contribution. The reason is that the sum k of negative helicity should be four (i.e., we should have η A with A = 1, 2, 3, 4 appear four times), thus since one side is four-point amplitude with k = 2, another side of three-point amplitude must have k = 2. This can not be true for A 3 (2, 4, P ) with η 14 2 η 134 3 η i P or η 14 2 η 134 3 η ijk P no matter how we choose i = j = k from {1, 2, 3, 4}. For A 3 (2, 3, P ), we need to choose η 14 2 η 23 3 η 1234 P , but since now A 3 is MHV amplitude and the deformation of λ 2 makes λ 2 ∼ λ 3 ∼ λ P , we get zero.

Example II: six scalars
The theory we are considering is the scalar-Yang-Mills theory For this theory, a standard method is to consider the color-ordering amplitudes. For six-point amplitudes with three + scalars and three − scalars, there are following three primary color ordering amplitudes

Discussions
In this paper we showed that boundary contributions satisfy similar recursion relations as scattering amplitudes, and presented a new algorithm to compute boundary contributions. We analyzed large z scaling of amplitudes and boundary contributions in standard model like theories via light cone gauge, and gave two explicit calculations.
It is worth noting that although we only discussed on-shell amplitudes, our method can be applied to amplitudes with off shell currents. The recursion relations for amplitudes with off shell currents was discussed in [10], and one complication there was besides physical states g ± , longitudinal and time-like states also contributes. Fortunately, only physical states contributes in light cone gauge.
In section 2, we mentioned in general two deformations do not commute. It would be interesting to investigate what is the commutator of two deformations. And it might help us to determine the best choice of deformations.
In general, with more derivatives in vertexes the large z scaling of amplitudes get worse. Thus for these theories the boundary might not vanish when all possible deformations of the type i|n] are exploited. It would be interesting to decide under which conditions, our new recursion algorithm ends in finite steps. One particular theory is the one with matters coupling to gravity. It would be nice if we can give a similar analysis for the theory, like the example (i.e., the standard model like theory) studied in the paper.
Last, let us point out that boundary contributions serves as a bridge between on-shell and off-shell quantities. On one hand, boundary contributions stem from on shell scattering amplitude, and can be computed using on-shell methods. On the other hand, in many cases Feynman diagrams contributing to boundary part B 1|n] have the topology of the vertex V (1, n, P ) connected to off-shell current J(23 · · · n − 1, P ) (where P is the propagator). So using boundary contributions we can compute off-shell quantities like correlations function effectively.
In this part, we discuss some aspects of integration of multiple variables related to our study.

C. Light cone propagator
Although it is not used explicitly, we like to discuss one aspect of light-cone propagator given by Π µν = 1 p 2 η µν − q µ p µ + p µ q µ q · p (C.1) Using the basis q, q, h, h (see (4.5)), it is easy to rewrite it as η µν − q µ p ν + p µ q ν q · p = e µēν + e νēµ − p 2 (q · p) 2 q µ q ν (C.2) where we have defined e(p) = h − h · p q · p q,ē(p) =h −h · p q · p q, (C. 3) which have been used in main text and are proportional to the gluon polarization vectors ǫ + (p) and ǫ − (p), respectively. The p 2 of the term p 2 (q·p) 2 q µ q ν will cancel the denominator of Π µν , thus this term will give an effective 4-point vertex.