Analytic Tadpole Coefficients of One-loop Integrals

One remaining problem of unitarity cut method for one-loop integral reduction is that tadpole coefficients can not be straightforward obtained through this way. In this paper, we reconsider the problem by applying differential operators over an auxiliary vector $R$. Using differential operators, we establish the corresponding differential equations for tadpole coefficients at the first step. Then using the tensor structure of tadpole coefficients, we transform the differential equations to the recurrence relations for undetermined tensor coefficients. These recurrence relations can be solved easily by iteration, and we can obtain analytic expressions of tadpole coefficients for arbitrary one-loop integrals.


Introduction
Recent decades have witnessed much progress in the calculation of higher loop scattering amplitudes. Among them, one of the most remarkable achievements is the almost completely solution of the calculation of the one loop amplitude by many tools, such as PV reduction [1], OPP reduction [2], Unitarity cut [3][4][5] etc..
It's well known that [1,2,4] under dimensional regularization, an one-loop scattering amplitude or a general one-loop Feynman integral in D = d 0 − 2ǫ-dimensional spacetime can be rewritten as a linear combination of some scalar master integrals (MIs) as where the coefficient C is s (s = 1, · · · , d 0 + 1) is a rational function of external momenta, and I is s is the s-gon scalar integral. If d 0 = 4, then the corresponding master integrals are traditionally referred as tadpole, bubble, triangle, box and pentagon integrals (the tadpole integrals vanish if the corresponding propagators are massless). So the task of computing an one-loop amplitude is reduced to determining the coefficients of these master integrals. These coefficients can be derived by either integrand reduction [2] algebraically, or the unitarity method [3][4][5]. The idea underlying the unitarity method is to constrain amplitudes by their branch cuts, more explicitly, comparing two sides of (1.1) after cutting several propagators. By using the usual unitarity cuts (cut two propagators corresponding to a physical channel), the explicitly analytical expressions of these coefficients have been given in [6][7][8]. However, the analytic results of the tadpole coefficients are missing.
To obtain the coefficients of tadpole integrals by the usual unitary cut method, an idea was proposed to add an auxiliary, unphysical propagator in the integrand [9]. Since we need to compare the reduction coefficients of the physical integrand and auxiliary integrand in this new frame, the method isn't very efficient to calculate tadpole coefficients. Another idea is that although the tadpole integrals vanish under the unitarity cuts in physical channels, it survives under the single cut. Then naturally the single cut method was used to calculate the tadpole coefficients [10]. Because of the divergence of integrals after single cut and the dependence on the tensor reduction of the integrand, it's not easy to compute the tadpole coefficients for a general case.
In this paper, we will reconsider the computation of tadpole coefficients by using differential operators. Differential operators have played an important role in the area of scattering amplitude, for example, deriving the IBP relations and differential equations of Feynman integrals [11], relating tree-level amplitudes of different theories [12] and the expansion of Einstein-Yang-Mills amplitude [13,14].
Roughly speaking, for a general tensor one-loop integral, we will first introduce an auxiliary vector R µ and assume its reduction to scalar master integrals 1 , then consider applying some differential operators with respect to R to the integral, so we obtain the differential equations of tadpole coefficients after comparing two sides of the equation. Actually, the action of these differential operators of R is a little similar to the traditional PV reduction [1]. Instead of trying to solve these differential equations directly, we transform the differential equations into recurrence relations with the help of the general tensor form of tadpole coefficients. With some known initial conditions, we can solve these expansion coefficients iteratively. So the problem of calculating tadpole coefficients is reduced into solving these recurrence relations, we will provide a general algorithm.
Our plan of this paper is following. In section 2, we discuss the integral reduction of a general tensor 1-loop Feynman integral. First, we consider the action of differential operators and obtain the differential equations of reduction coefficients. Second, we transform these differential equations into recurrence relations. In section 3, we derive the recurrence relations of tadpole coefficients for four tensor integrals, namely bubbles, triangles, boxes, pentagons, and provide a general algorithm for calculating tadpole coefficients with some examples. Appendix A briefly reviews the traditional PV-reduction method to deal with the tadpole coefficient of a tensor tadpole, which is consistent with our results.

Integral reduction by differential operators
In this section, we will consider the integral reduction of a general 1-loop tensor integral using differential operators with respect to an auxiliary vector R. First, in subsection 2.1 we will derive the differential equations for the coefficients of the master integrals. Secondly, using the general tensor structure of reduction coefficients, we transform the differential equations of tadpole coefficients into recurrence relations in subsection 2.2.

Differential equations of reduction coefficients
Let us start with the following general one-loop m-rank tensor integral with n + 1 propagators where the i-th propagator is given by P i = (ℓ − K i ) 2 − M 2 i and K 0 = 0 (i.e., we have chosen to translate the loop momentum ℓ → ℓ + K 0 to simplify the 0-th propagator P 0 ). To simplify the manipulation of the tensor structure and utilize the tool of differential operators, we introduce an auxiliary vector R µ and contract I µ 1 ···µm n+1 with m R µ 's to arrive Note that the auxiliary vector R µ is in the D = (4 − 2ǫ)-dimensional space as the D-dimensional loop momenta ℓ. By setting R = m i=1 α i R i into (2.2) and expanding the result to find the coefficients of α 1 ...α m , it is easy to see that we will get the reduction of (2.1) up to a numerical factor. The simple but useful transformation from the form (2.1) to the form (2.2) is, in fact, our first crucial step.
It's well known that in dimensional regularization scheme, the integral I [R] can be reduced into the linear combination of master integrals (including pentagon, box, triangle, bubble and tadpole scalar integrals) as where the reduction coefficients C is s (m), s = 1, · · · , 5 are rational functions of external momenta, masses and R. An important point of the reduction is that R can only appear in the numerator of the reduction coefficients. The Lorentz invariance means that it can only have following types of contractions: To find these reduction coefficients C is s (m), we will try to establish some differential equations by virtue of the following differential operators: n+1; i means that the propagator P i is removed from I (m−1) n+1 . On the right-hand side, since D i acts only on the coefficients C is s (m), we have 5 s=1 is (D i C is s (m)) I is s . Identifying both sides, we get the equation At this point, the differential operator D i doesn't help us much for calculating the reduction coefficients. However, if we make the inductive assumption that the reduction of tensor integral I (m ′ ) n ′ +1 's are already known for m ′ < m, n ′ < n, or m ′ < m, n ′ = n and m ′ = m, n ′ < n, for example where 0 of C is s (m − 1; 0) is to remind us C is s (m − 1; 0) is the reduction coefficient of an integral with propagator P 0 being canceled out, the (2.7) can be written as By comparing the master integrals at the two sides of the above formula, we can get the differential equation for each particular reduction coefficient C is s (m). It is found that the differential equations given by D i 's are not enough to uniquely determine the reduction coefficients, so we need to consider the action of T on (2.3). Similar to D i 's, on the left-hand side, we get while on the right-hand side, we have 5 s=1 is (T C is s (m)) I is s . So after the reduction of integrals on the left-hand side, we can get another group of differential equations for unknown reduction coefficients

Recurrence relations for tadpole coefficients
Since our main goal is to compute the reduction coefficients of tadpole integral, we will concentrate on the calculation of tadpole coefficients here. Without loss of generality, let's consider the tadpole integral with a propagator P 0 2 . Comparing the two sides of equations (2.9) and (2.11), since the 0-th propagator P 0 has been removed from I is s; 0 , it won't contribute to the tadpole coefficient of propagator P 0 , so we get the following differential equations where the superscript of C (0) 1 (m) reminds us it's the tadpole coefficient with a propagator P 0 .
Note that the above two differential equations (2.12) and (2.13) relate tadpole coefficient C (0) 1 (m) with rank m to tadpole coefficients with lower ranks (m − 1) or (m − 2), which are already known according to the inductive assumption. Since directly solving the differential equations (2.12) and (2.13) are complicated, we will try to transform the differential equations into much simpler recurrence relations by noticing the tensor structure of the tadpole coefficient, i.e., it can be expanded as with unknown coefficients c (m) i 0 ,i 1 ,i 2 ,i 3 ,...in (called expansion coefficient) being rational functions of (K i · K j ), M 2 i , and the summing indices satisfying 2i 0 + n k=1 i k = m (so the summing is written as ′ to emphasize the constraint and if we choose i 1 , i 2 , · · · , i n as free indices, we can just write c (m) i 1 ,i 2 ,··· ,in ). The above formula (2.14) always exists because of the previous mentioned contractions (2.4). For simplicity, we will also adopt the following conventions: • First, we define the notations s 00 = R · R, s 0i = R · K i , for i = 1, ..., n, and s ij = K i · K j , for i, j = 1, 2, ..., n.
• Second, the mass dimension of C • Fourthly, due to any a permutation σ : n+1 and the tadpole integral I 1 with propagator P 0 invariant, we have c The permutation symmetry of the expansion coefficients can be used to check our results.
After taking above notations, the tadpole coefficient in (2.14) becomes and our goal is to determine the unknown expansion coefficients c (m) i 0 ,i 1 ,··· ,in . When using above notation to express C (0) 1 (m − 1; i), one need to notice that it doesn't contain the external momenta K i for the propagator P i has been canceled in integral I is s; i , so its expansion is where in the second equation, we have added δ 0j i and the summation over j i for later convenience.
To get the recurrence relations, we need to consider the action of D i , T on the (2.15). Using the chain rule, it's easy to get where D is the space-time dimension. First, we consider the action of differential operator D j in (2.12) where in the second equation we have used the fact 2i 0 + n k=1 i k = m and redefined the indices i j and i l , andĩ ≡ n k=1 i k . Using the expansion (2.15), the right-hand side is Comparing the two sides we have with α j ≡ f j /M 2 0 , β jl ≡ s jl /M 2 0 . Secondly considering the operator T , we have at the left hand side and at the right-hand side. Comparing two sides, we get Up to now, we have successfully transformed the differential equations to algebraic recurrence relations (2.21) and (2.24) for tadpole coefficients, then the next step is to solve these relations explicitly.

Tadpole coefficients of tensor integrals
In the previous section, we have derived two types of recurrence relations (we will call them D-type and T -type) of unknown expansion coefficients appearing in (2.15).
In this section, we will show that by the inductive assumption the expansion coefficients with larger indices i k are related to those expansion coefficients with lower indices through the recurrence relations. Then with the boundary conditions 3 , these recurrence relations are enough to recursively determine the tadpole coefficients for a general tensor 1-loop integral with any rank. To illustrate this, we will take the tensor 1-loop integrals with up to 5 propagators as examples in next subsections, and an explicitly recursive algorithm is provided in each subsection. Before moving on complicated examples, let us first apply our method to the simplest example, i.e., the reduction of tensor tadpole integrals. With n = 0 in (2.2), it is easy to see that the tadpole coefficients are nonzero only when m is even, because the only existing Lorentz contraction is R 2 . When m is even, according to (2.15), the tadpole coefficient of a tensor tadpole integral is given by Since the tadpole coefficient only contains an unknown function c (m) , only one recurrence relation is necessary. For a tensor tadpole integral, there is no external momenta K i , so we can only consider the T -type recurrence relation (2.24) With the boundary condition c (0) = 1, we immediately get so when m is even, the tadpole coefficient is .

(3.5)
This result (3.5) is consistent with that given by the traditional PV-reduction method in the Appendix A.

Tadpole coefficients of tensor bubble integral
Now we consider the first nontrivial case, n = 1 of (2.2), i.e., tensor bubble integral. For now, the general form of tadpole coefficients (2.15) becomes where in the second equation we have used the constraint 2i 0 + i 1 = m to solve i 0 . Note that when m is an even integer, i must be even, while if m is an odd integer, i must be odd. We consider the recurrence relations resulted by the differential operator D i . Since n = 1, (2.21) can just give one recurrence relation We should note that c (m−1) in (3.7) is the known expansion coefficients appearing in the reduction of tensor tadpole integrals (see (3.4)). Replacing i by i + 1, we get Note that c . Now the task becomes the determination of the initial expansion coefficient c which means that we can obtain c with m = 2r. When n = 1, the recurrence relation (2.24) of the differential operator T with m = 2r, i = 0 becomes (3.10) with With the obvious boundary condition c (0) 0 = 0, this recurrence relation (3.12) can be solved as i . The points represented by black squares are the zero expansion coefficients, while the points represented by blue or red circles are the unknown expansion coefficients we need to calculate. The red thick arrow represents the recurrence relation (3.12), the orange thick arrows represents the recurrence relation (3.9), and the cyan thick arrow represents the recurrence relation (3.8).
After getting the analytic expression of c (2r) 0 , we will show how to obtain other expansion coefficients c 1 is necessary to calculate. Using (3.9), we easily get 0 is given directly by (3.12) (3.17) 1 and c 3 . First, we use (3.9) to calculate c The above three examples are enough to illustrate the procedure of calculating expansion coefficients c successively by using (3.14) and (3.8).
• Final step: Combine all expansion coefficients to get the tadpole coefficient by (3.6).
It's obvious that with the help of Mathematica, one can easily implement the three key relations (3.14), and (3.8), thus the analytic expression of tadpole coefficients can be nicely generated.

Tadpole coefficients of tensor triangle integral
Now we consider the second nontrivial case, n = 2 for (2.2), i.e., tensor triangle Feynman integral. In this case, the general form of the tadpole coefficients (2.15) can be written as As a result, the D-type recurrence relations coming from D i , i = 1, 2 with n = 2 of (2.21) are where we have added [ 1] or [ 2] behind c to stress that it corresponds to the tadpole coefficients of the integrals with 1st or 2nd propagator being canceled. The above formulas can be rewritten in a compact form as 4 where we use the notation Here we have defined the rescaled Gram matrix G(1, 2, · · · , n) with (i, j) element being And we will denote its corresponding determinant as ∆(1, 2, · · · , n) and its (i, j) cofactor as ∆ Note that two equations of (3.22) are symmetric for exchanging (1,2), so its result is also symmetric for (1,2). We can easily solve c (m) (3.26) More explicitly, they are For a certain rank m, by the inductive assumption given in the subsection 2.1, the expansion coefficients c (m−1) So according to rank m is even or odd, the expansion coefficient c   with larger indices, we need to reduce them to expansion coefficients with smaller indices. Using (3.27) we have ) are two vectors, and G −1 is the inverse of Gram matrix G(1, 2). Using the boundary condition c where we have used the boundary condition (c (0) 0 ) T = (0, 0), so the summation is from i = 2 to r.
After obtaining the analytic expression of c (2r) 0,0 recursively using (3.32), we will consider how to obtain all expansion coefficients c (m) i,j by using the recurrence relations (3.27), (3.28) and (3.32) step by step. First, we note that two equations of (3.28) can be rewritten as i,j and c (6) i,j . The points represented by black squares are the zero expansion coefficients (we don't draw those expansion coefficients with i + j = even for m = 5 and i + j = odd for m = 6, for they all vanish according to the definition) while the points represented by blue or green circles are the unknown expansion coefficients we need to calculate. The red thick arrow represents the first one of the recurrence relations (3.27) and the cyan thick arrow represents the second one of the recurrence relations (3.27).
1,0 , c 0,1 , c 1,2 , c 2,1 , c 0,3 , c 1,0 + 3β 12 c . (3.37) Other expansion coefficients can be got by using the permutation symmetry. i,j from lower rank to higher rank, and at a fixed rank m, we prefer to calculate the expansion coefficients c 1,0 are given by (3.34) as illustrated before.
With the help of Mathematica, one can easily implement the relations (3.32) and (3.42), thus analytic reduction coefficients can be obtained for any rank.

Tadpole coefficients of tensor box integral
Now we consider the third nontrivial case, n = 3 of (2.2), i.e., the tensor box Feynman integral, which is similar to the previous cases. In this case, the general form of the tadpole coefficients (2.15) can be written as As a result, the D-type recurrence relations are given by setting n = 3 in (2.21): where the notation [î] corresponds to the integral with i-th propagator being canceled. These equations can be rewritten in a more compact form as where we have used the notation G ≡ (β ij ) with i, j = 1, · · · , 3 and defined the vector Then we can easily solve c (m) (3.42) More explicitly, they are c (m) As in previous subsections, according to rank m is even or odd, the expansion coefficient c 1,1 + α 2 ∆ 1,2 + α 3 ∆ 1,3 c (2r−2) 0,0 where we have used the same notations as in (3.2). Using the boundary condition c where we have considered (c are easy to get. Second, we will take several examples to illustrate the procedure of calculation. Here for simplicity we denote ∆(i 1 , i 2 , · · · , i n ; j 1 , j 2 , · · · , j n ) as the determinant of a n × n matrix A with entry A ab = β iaj b . Specially, we denote ∆(i 1 , i 2 , · · · , i n ) ≡ ∆(i 1 , i 2 , · · · , i n ; i 1 , i 2 , · · · , i n ).
• m < 3: The essence of one-loop reduction is to expand the auxiliary vector R with external momenta to cancel the propagators. To reduce a tensor box integral to the scalar tadpole integral with propagator P 0 , we need to cancel other three propagators, ie., P 1 , P 2 , P 3 , which means three R's are required at least. So all the expansion coefficients vanish for rank m < 3. This result can also be obtained from our explicit recurrence relation (3.47) and (3.42).
• Step 2: Consider the rank m = 3, calculate the expansion coefficients c 3,0,0 by (3.43) as we illustrated before. • Final step: Combine all expansion coefficients to get the tadpole coefficient by (3.38).
With the help of Mathematica, it is easy to implement recurrence relation (3.47) and (3.42) to automatically produce analytic expression for reduction coefficients of any rank.

Tadpole coefficients of tensor pentagon integral
At last, we consider the fourth nontrivial case, n = 4 for (2.2), i.e., tensor pentagon integral. In this case, the general form of tadpole coefficients (2.15) can be written as As a result, the D-type recurrence relations are given by setting n = 4 in (2.21): (3.58) These equations can be rewritten in a more compact form as where [î] means the i-th propagator is removed from the propagators P 0 , P 1 , P 2 , P 3 , P 4 . These equations can be rewritten in a more compact form as Then we can solve c (m) 3,4 c (2r−2) 0,0,0,0 where we have used the denotations in (3.2). Using the boundary condition c where we have considered (c • m < 4: The essence of one-loop reduction is to expand the auxiliary vector R with external momenta to cancel the propagators. To reduce a tensor pentagon integral to the scalar tadpole integral with propagator P 0 , we need to cancel other four propagators, ie., P 1 , P 2 , P 3 , P 4 , which means four R's are required at least. So all the expansion coefficients vanish for rank m < 4.  i,j,k,l from lower rank to higher rank, and at a fixed rank m, we prefer to calculate the expansion coefficients c (m) i,j,k,l with smaller index i + j + k + l first. For general rank m 0 , we need to calculate all expansion coefficients with rank m ≤ m 0 to calculate the tadpole applied to calculating the reduction coefficients of other master integrals. We will show how to do this in the further research.
As for the reduction of tensor higher-loop integral, by constructing differential operators and expanding the coefficients in a general form, our method can also give the recurrence relations. But differing from one-loop case, these relations are in general not enough to uniquely determine coefficients. A further complexity is that for the higher-loop integrals, the master basis are more complicated. In despite of these difficulties, it is still an interesting question to apply our method to higher-loop cases.

Acknowledgments
It is a pleasure to thank Yang Zhang, Chang Hu and Yaobo Zhang for inspiring discussions. This work is supported by Qiu-Shi Funding and Chinese NSF funding under Grant No.11935013, No.11947301, No.12047502 (Peng Huanwu Center).

A The reduction of tadpole by PV-method
In this appendix, we use the traditional PV-reduction method to study following tadpole integrals: The key of PV-reduction is that by Lorentz symmetry the tensor structure at the both sides of (A.1) must be the same. For the tadpole (A.1), there is no external momentum to provide the tensor index, thus the only available one is the metric g µν , which must be considered in the general D = 4 − 2ǫ-dimension. Furthermore, because all µ i 's are symmetric, the tensor structure must be symmetric under index permutation. Thus we have when s = 2r + 1, (A.1) is zero and when s = 2r we have To determine the constant A(2r), we contract both sides of (A.1) with, for example, g µ 1 µ 2 . Using (A.2) the RHS gives A(2r)(D + 2(r − 1))[g µ 3 µ 4 ...g µ 2r−1 µ 2r + symmetrization], r ≥ 2. (A.3) When doing the contraction, there are two types of tensor structures in (A.2), one is with g µ 1 µ 2 and another one, with g µ 1 µ i g µ 2 µ j . For the former, g µ 1 µ 2 g µ 1 µ 2 = D, while for the later g µ 1 µ 2 g µ 1 µ i g µ 2 µ j = g µ i µ j . Furthermore, for the later one, since for each i, j pair, which are different tensor structures before contraction, we get the same tensor structure after the contraction, thus when considering the remaining particular tensor structure, for example, g µ 3 µ 4 ...g µ 2r−1 µ 2r with (r − 1)'s g µν , we get the overall factor 2(r − 1). For the LHS, we have One can check that coefficients in (A.7) is the same as the one given in (3.5).