Exploring the function space of Feynman integrals. Stefan Weinzierl

Transcription

1 Exploring the function space of Feynman integrals Stefan Weinzierl Institut für Physik, Universität Mainz Mathematics intro: Physics intro: Part I: Part II: Periodic functions and periods Precision calculations Multiple Polylogarithms Elliptic generalisations

2 Mathematics intro Periodic functions and periods

3 Periodic functions Let us consider a non-constant meromorphic function f of a complex variable z. A period ω of the function f is a constant such that for all z: f (z+ω) = f (z) The set of all periods of f forms a lattice, which is either trivial (i.e. the lattice consists of ω=0 only), a simple lattice, Λ={nω n Z}, a double lattice, Λ={n 1 ω 1 + n 2 ω 2 n 1,n 2 Z}.

4 Examples of periodic functions Singly periodic function: Exponential function exp(z). exp(z) is periodic with peridod ω=2πi. Doubly periodic function: Weierstrass s -function ( ) (z)= 1 1 z 2+ 1 ω Λ\{0} (z+ω) 2, Λ={n ω 2 1 ω 1 + n 2 ω 2 n 1,n 2 Z}, Im(ω 2 /ω 1 ) 0. (z) is periodic with periods ω 1 and ω 2.

5 Inverse functions The corresponding inverse functions are in general multivalued functions. For the exponential function x=exp(z) the inverse function is the logarithm z = ln(x). For Weierstrass s elliptic function x= (z) the inverse function is an elliptic integral z = x dt 4t3 g 2 t g 3, g 2 = 60 ω Λ\{0} 1 ω 4, g 3=140 ω Λ\{0} 1 ω 6.

6 Periods as integrals over algebraic functions In both examples the periods can be expressed as integrals involving only algebraic functions. Period of the exponential function: 2πi 1 = 2i 1 dt 1 t 2. Periods of Weierstrass s -function: Assume that g 2 and g 3 are two given algebraic numbers. Then t 2 ω 1 = 2 dt t 2, ω 2 = 2 4t3 g 2 t g 3 dt 4t3 g 2 t g 3, t 1 t 3 where t 1, t 2 and t 3 are the roots of the cubic equation 4t 3 g 2 t g 3 = 0.

7 Numerical periods Kontsevich and Zagier suggested the following generalisation: A numerical period is a complex number whose real and imaginary parts are values of absolutely convergent integrals of rational functions with rational coefficients, over domains inr n given by polynomial inequalities with rational coefficients. Remarks: One can replace rational with algebraic. The set of all periods is countable. Example: ln2 is a numerical period. ln2 = 2 1 dt t.

8 Physics intro Precision calculations

9 Precision calculations Due to the smallness of all coupling constants g, we may compute an observable at high energies reliable in perturbation theory, σ = ( g ) 4 ( g ) 6 ( g ) 8 σlo + σnlo + σnnlo π 4π 4π Cross section related to the square of the scattering amplitude: σ A 2. Perturbative expansion of the amplitude: where A (l) contains l loops. A = g 2 A (0) + g 4 A (1) + g 6 A (2) +...,

10 Loop amplitudes The computation of the tree amplitude A (0) poses no conceptional problem. For loop amplitudes we have to calculate Feynman integrals. Let us write A (l) = c j I j, j c j : coefficients, computation tree-like, I j : Feynman integrals We may take the set of Feynman integrals {I 1,I 2,...} to consist of scalar integrals. (Tarasov, 96, 97)

11 Feynman integrals A Feynman graph with n external lines, r internal lines and l loops corresponds (up to prefactors) in D space-time dimensions to the family of Feynman integrals, indexed by the powers of the propagators ν j I ν1 ν 2...ν r = l d D k s s=1 iπ D 2 r j=1 1 ( q 2 j + m2 j )ν j The momenta flowing through the internal lines can be expressed through the independent loop momenta k 1,..., k l and the external momenta p 1,..., p n as q i = l λ i j k j + j=1 n σ i j p j, j=1 λ i j,σ i j { 1,0,1}.

12 Pinching of propagators If for some exponent we have ν j = 0, the corresponding propagator is absent and the topology simplifies: ν 4 =0 =

13 Integration by parts Within dimensional regularisation we have for any loop momentum k i {p 1,..., p n,k 1,...,k l } and v l d D k s s=1 iπ D 2 k µ i v µ r j=1 1 ( q 2 j + m2 j )ν j = 0. Working out the derivatives leads to relations among integrals with different sets of indices (ν 1,...,ν r ). This allows us to express most of the integrals in terms of a few master integrals. Tkachov 81, Chetyrkin 81

14 Laporta s algorithm Expressing all integrals in terms of the master integrals requires to solve a rather large linear system of equations. This system has a block-triangular structure, originating from subtopologies. Order the integrals by complexity (more propagators more difficult) Solve the system bottom-up, re-using the results for the already solved sectors. Laporta 01

15 Differential equations Let t be an external invariant (e.g. t = (p i + p j ) 2 ) or an internal mass. Let I i {I 1,...,I N } be a master integral. Carrying out the derivative t I i under the integral sign and using integration-by-parts identities allows us to express the derivative as a linear combination of the master integrals. t I i = N a i j I j j=1 (Kotikov 90, Remiddi 97, Gehrmann and Remiddi 99)

16 Differential equations Let us formalise this: I =(I 1,...,I N ), x=(x 1,...,x n ), set of master integrals, set of kinematic variables the master integrals depend on. We obtain a system of differential equations of Fuchsian type where A is a matrix-valued one-form d I A = = A I, n A i dx i. i=1 The matrix-valued one-form A satisfies the integrability condition da A A = 0. Computation of Feynman integrals reduced to solving differential equations!

17 Part I Multiple Polylogarithms

18 Feynman integrals and periods Laurent expansion in ε =(4 D)/2: I G = c j ε j. j= 2l Question: What can be said about the coefficients c j? Theorem: For rational input data in the euclidean region the coefficients c j of the Laurent expansion are numerical periods. (Bogner, S.W., 07) Next question: Which periods?

19 One-loop amplitudes All one-loop amplitudes can be expressed as a sum of algebraic functions of the spinor products and masses times two transcendental functions, whose arguments are again algebraic functions of the spinor products and the masses. The two transcendental functions are the logarithm and the dilogarithm: Li 1 (x) = ln(1 x)= Li 2 (x) = x n n=1 n x n n=1 n 2

20 Generalisations of the logarithm Beyond one-loop, at least the following generalisations occur: Polylogarithms: Li m (x) = n=1 x n n m Multiple polylogarithms: This is a nested sum: Li m1,m 2,...,m k (x 1,x 2,...,x k ) = n 1 >n 2 >...>n k >0 x n 1 1 n m 1 1 xn 2 2 n m xn k k n m k k n j n j =1 x n j j n j m j n j 1... n j+1 =1

21 Multiple polylogarithms Definition based on nested sums: Li m1,m 2,...,m k (x 1,x 2,...,x k ) = Definition based on iterated integrals: n 1 >n 2 >...>n k >0 x n 1 1 n m 1 1 xn 2 2 n m xn k k n m k k G(z 1,...,z k ;y) = y 0 dt 1 t 1 z 1 t 1 0 dt 2 t 2 z 2... t k 1 0 dt k t k z k Conversion: Li m1,...,m k (x 1,...,x k ) = ( 1) k G m1,...,m k ( 1 x 1, ) 1 1,..., ;1 x 1 x 2 x 1...x k Short hand notation: G m1,...,m k (z 1,...,z k ;y) = G(0,...,0 }{{},z 1,...,z k 1,0...,0 }{{},z k ;y) m 1 1 m k 1

22 The ε-form of the differential equation If we change the basis of the master integrals J=U I, the differential equation becomes d J = A J, A = UAU 1 UdU 1 Suppose one finds a transformation matrix U, such that A = ε j C j d ln p j ( x), where - ε appears only as prefactor, - C j are matrices with constant entries, - p j ( x) are polynomials in the external variables, then the system of differential equations is easily solved in terms of multiple polylogarithms. Henn 13

23 Transformation to the ε-form We may perform a rational / algebraic transformation on the kinematic variables often done to absorb square roots. change the basis of the master integrals (x 1,...,x n ) (x 1,...,x n), I U I, where U is rational in the kinematic variables Henn 13; Gehrmann, von Manteuffel, Tancredi, Weihs 14; Argeri et al. 14; Lee 14; Meyer 16; Prausa 17; Gituliar, Magerya 17; Lee, Pomeransky 17;

24 Numerical evaluations of multiple polylogarithms Multiple polylogarithms have branch cuts. Numerical evaluation of multiple polylogarithms Li m1,m 2,...,,m k (x 1,x 2,...,x k ) as a function of k complex variables x 1, x 2,..., x k : Use truncated sum representation within its region of convergence. Use integral representation to map arguments into this region. Acceleration techniques to speed up the computation. Implementation in GiNaC, using arbitrary precision arithmetic in C++. J. Vollinga, S.W. 04

25 Part II Elliptic generalisations

26 Differential equations again If it is not feasible to compute the integral directly: Pick one variable t from the set s jk and m 2 i. 1. Find a differential equation for the Feynman integral. r p j (t) d j j=0 dt ji G(t) = i q i (t)i Gi (t) Inhomogeneous term on the rhs consists of simpler integrals I Gi. p j (t), q i (t) polynomials in t. 2. Solve the differential equation. Remark: A single differential equation of order r is equivalent to a system of r first-order differential equations.

27 Differential equations: The case of multiple polylogarithms Suppose the differential operator factorises into linear factors: r p j (t) d j = j=0 dt j ( a r (t) d ) dt + b r(t) Iterated first-order differential equation. Denote homogeneous solution of the j-th factor by t ψ j (t) = exp ds b j(s). a j (s) Full solution given by iterated integrals t I G (t) = C 1 ψ 1 (t)+c 2 ψ 1 (t) Multiple polylogarithms are of this form. 0 ψ 2 (t 1 ) t dt 1 a 1 (t 1 )ψ 1 (t 1 ) +C 3ψ 1 (t) (... a 2 (t) d )( dt + b 2(t) a 1 (t) d ) dt + b 1(t) 0 0 ψ 2 (t 1 ) t 1 ψ 3 (t 2 ) dt 1 dt 2 a 1 (t 1 )ψ 1 (t 1 ) a 2 (t 2 )ψ 2 (t 2 )

28 Differential equations: Beyond linear factors Suppose the differential operator does not factor into linear factors. r p j (t) d j j=0 dt j The next more complicate case: The differential operator contains one irreducible second-order differential operator a j (t) d2 dt 2+ b j(t) d dt + c j(t)

29 An example from mathematics: Elliptic integral The differential operator of the second-order differential equation [t ( 1 t 2) d 2 dt 2+( 1 3t 2) ] d dt t f(t) = 0 is irreducible. The solutions of the differential equation are K(t) and K( 1 t 2 ), where K(t) is the complete elliptic integral of the first kind: K(t) = 1 0 dx (1 x2 )(1 t 2 x 2 ).

30 An example from physics: The two-loop sunrise integral S ( p 2,m ) = m m m p Two-loop contribution to the self-energy of massive particles. Sub-topology for more complicated diagrams.

31 Single scale integrals beyond multiple polylogarithms Starting from two-loops, there are integrals which cannot be expressed in terms of multiple polylogarithms. Simplest example: Two-loop sunrise integral with equal masses. Slightly more complicated: Two-loop kite integral. Both integrals depend on a single scale t/m 2. Change variable from t/m 2 with q=e iπτ. to the nome q or the parameter τ Sabry, Broadhurst, Fleischer, Tarasov, Bauberger, Berends, Buza, Böhm, Scharf, Weiglein, Caffo, Czyz, Laporta, Remiddi, Groote, Körner, Pivovarov, Bailey, Borwein, Glasser, Adams, Bogner, Müller-Stach, Schweitzer, S.W, Zayadeh, Bloch, Vanhove, Tancredi, Pozzorini, Gunia,...

32 The elliptic curve How to get the elliptic curve? From the Feynman graph polynomial: x 1 x 2 x 3 t+ m 2 (x 1 + x 2 + x 3 )(x 1 x 2 + x 2 x 3 + x 3 x 1 ) = 0 From the maximal cut: y 2 ( x t m 2 )( x ) t 4m2 ( x 2 + 2x+1 4 t ) m 2 m 2 = 0 Baikov 96; Lee 10; Frellesvig, Papadopoulos, 17; Bosma, Sogaard, Zhang, 17; Harley, Moriello, Schabinger, 17 The periods ψ 1, ψ 2 of the elliptic curve are solutions of the homogeneous differential equation. Adams, Bogner, S.W., 13; Primo, Tancredi, 16 Set τ = ψ 2 ψ 1, q = e iπτ.

33 The elliptic dilogarithm Recall the definition of the classical polylogarithms: Li n (x) = j=1 x j j n. Generalisation, the two sums are coupled through the variable q: ELi n;m (x;y;q) = j=1 k=1 x j j n y k k mqjk. Elliptic dilogarithm: E 2;0 (x;y;q) = 1 i [ 1 2 Li 2(x) 1 2 Li ( ) ( 2 x 1 +ELi 2;0 (x;y;q) ELi 2;0 x 1 ;y 1 ;q )]. Various definitions of elliptic polylogarithms can be found in the literature Beilinson 94, Levin 97, Wildeshaus 97, Brown, Levin 11, Bloch, Vanhove 13, Adams, Bogner, S.W. 14, Remiddi, Tancredi 17

34 Elliptic generalisations In order to express the sunrise/kite integral to all orders in ε introduce ELi n1,...,n l ;m 1,...,m l ;2o 1,...,2o l 1 (x 1,...,x l ;y 1,...,y l ;q)= =... j 1 =1 j l =1 k 1 =1... k l =1 x j 1 1 j n x j l l j n l l y k 1 1 k m yk l l k m l l q j 1k j l k l l 1 ( j i k i j l k l ) o i i=1. Numerical evaluation: G. Passarino 16

35 The all-order in ε result (ELi-representation) Taylor expansion of the sunrise integral around D=2 2ε: S = ψ 1 π Each term in the ε-series is of the form ε j E ( j) j=0 E ( j) linear combination ofeli n1,...,n l ;m 1,...,m l ;2o 1,...,2o l 1 and Li n1,...,n l Using dimensional-shift relations this translates to the expansion around 4 2ε. The multiple polylogarithms extended byeli n1,...,n l ;m 1,...,m l ;2o 1,...,2o l 1 are the class of functions to express the equal mass sunrise graph to all orders in ε. Adams, Bogner, S.W., 15

36 Modular forms Denote by H the complex upper half plane. A meromorphic function f : H C is a modular form of modular weight k for SL 2 (Z) if (i) f transforms under Möbius transformations as f ( ) aτ+b cτ+d =(cτ+d) k f(τ) for ( a b c d ) SL 2 (Z) (ii) f is holomorphic onh, (iii) f is holomorphic at. Iterated integrals of modular forms: I(f 1, f 2,..., f n ;q) = (2πi) n τ τ 0 dτ 1 f 1 (τ 1 ) τ 1 τ 0 dτ 2 f 2 (τ 2 )... τ n 1 τ 0 dτ n f n (τ n )

37 The all-order in ε result (iterated integrals) S = ψ 1 π e εi( f 2;q)+2 {[ j=0( + j=0 ε j ε 2 j I j 2 k=0 I n=2 ( 1) n n ζ n ε n ( ) {1, f 4 } j ;q 1 ( ) )] 2 ε2 j+1 I {1, f 4 } j,1;q ε k B (k) k=0 ( ) {1, f 4 } k,1, f 3,{ f 2 } j 2k ;q Uniform weight: At order ε j one has exactly ( j+ 2) integrations. Alphabet given by modular forms 1, f 2, f 3, f 4. Adams, S.W., 17

38 The letters Example: The modular form f 3 is given by f 3 = 1 24 ( ψ1 π ) 3 t ( t m 2)( t 9m 2) m 6 = 3 i [ ELi0; 2 (r 3 ; 1; q) ELi 0; 2 ( r 1 3 ; 1; q)] = 3 η(2τ 2 ) 11 η(6τ 2 ) 7 3 η(τ 2 ) 5 η(4τ 2 ) 5 η(3τ 2 )η(12τ 2 ) = 3 3[E 3 (τ 2 ;χ 1,χ 0 )+2E 3 (2τ 2 ;χ 1,χ 0 ) 8E 3 (4τ 2 ;χ 1,χ 0 )] with τ 2 = τ/2, r 3 = exp(2πi/3), Dedekind s eta function η, Dirichlet characters χ 0 = ( 1 n ), χ 1=( 3 n ) and Eisenstein series E 3.

39 The ε-form of the differential equation for the sunrise/kite It is not possible to obtain an ε-form by a rational/algebraic change of variables and/or a rational/algebraic transformation of the basis of master integrals. However by the (non-algebraic) change of variables from t to τ and by factoring off the (non-algebraic) expression ψ 1 /π from the master integrals in the sunrise sector one obtains an ε-form for the kite/sunrise family: d dτ I = ε A(τ) I, where A(τ) is an ε-independent 8 8-matrix whose entries are modular forms.

40 Analytic continuation and numerical evaluations of the kite and sunrise integral Complete elliptic integrals efficiently computed from arithmetic-geometric mean. sunrise, real part q-series converges for all t R\{m 2,9m 2, } our result SecDec 6 path in q Re(I10101) t/m 2 ],0] t/m 2 [0 : 1] t/m 2 [1 : 9] t/m 2 [9, [ t/m 2 sunrise, imaginary part Im(q) our result SecDec Im(I10101) Re(q) t/m No need to distinguish the cases t < 0, 0< t < m 2, m 2 < t < 9m 2, 9m 2 < t! Bogner, Schweitzer, S.W., 17

41 Conclusions Differential equations are a powerful tool to compute Feynman integrals. If a system can be transformed to an ε-form, a solution in terms of multiple polylogarithm is easily obtained. There are system, where within rational transformations at order ε 0 two coupled equations remain. Kite/sunrise family: Sum representation in terms ofeli-functions. Iterated integral representation involving modular forms Analytic continuation / numerical evaluation easy.

42 Back-up Towards multi-scale integrals beyond multiple polylogarithms

43 A more complicated example Let s look at a two-loop example from t t-production. In the top topology we have 5 master integrals: Multi-scale problem (x 1 = s/m 2, x 2 = t/m 2 ), contains sunrise as sub-topology. Do we have to solve at order ε 0 a coupled system of 5 differential equations?

44 Reduction to a single-scale problem Let α=[α 1 :... : α n ] CP n 1, without loss of generality take α n = 1. Consider a path x i (λ) = α i λ, 1 i n. View the master integrals as functions of λ. For the derivative with respect to λ we have d dλ I = B I, B = n α i A i. i=1 Let us write B = B (0) + ε j B ( j). j>0

45 The Picard-Fuchs operator Consider the top sector and let us work modulo sub-topologies and ε-corrections. Let I be one of the master integrals {I 1,...,I N }. Determine the largest number r, such that the matrix which expresses I, (d/dλ)i,..., (d/dλ) r 1 I in terms of the original set {I 1,...,I N } has full rank. It follows that(d/dλ) r I can be written as a linear combination of I,...,(d/dλ) r 1 I. This defines the Picard-Fuchs operator L r for the master integral I with respect to λ: L r I = 0, L r = r d k R k k=0 dλ k. L r is easily found by transforming to a basis which contains I,...,(d/dλ) r 1 I.

46 Factorisation Consider as an example the differential operator L 2 = d2 dλ 2 ( 1 λ + 1 ) d λ 1 dλ + ( ) 1 λ(λ 1) + 1 (λ 1) 2. This operator factorises: L 2 = ( d dλ 1 ) ( d λ dλ 1 ) λ 1 Not every differential operator factorises into linear factors. We may decompose any differential operator into irreducible factors. van Hoeij 97

47 Factorisation Suppose L r factorises as a differential operator L r = L 1,r1 L 2,r2...L s,rs, where L i,ri denotes a differential operator of order r i. Then we may convert the system of differential equations at order ε 0 into block triangular form with blocks of size r 1, r 2,..., r s. A basis for block i is given by J i, j = d j 1 dλ j 1L i+1,r i+1...l s,rs I, 1 j r i. This decouples the original system into sub-systems of size r 1, r 2,..., r s. Adams, Chaubey, S.W. 17

48 Lifting Let us write the transformation to the new basis as J = V(α 1,...,α n 1,λ) I. Setting U = V,..., x ) n 1,x n x n x n ( x1 gives the transformation in terms of the original variables x 1,..., x n. Remark: Terms in the original A of the form d lnz(x 1,...,x n ), where Z(x 1,...,x n ) is a rational function in (x 1,...,x n ) and homogeneous of degree zero in (x 1,...,x n ), map to zero in B. These terms are in many cases easily removed by a subsequent transformation.

49 Example Let s return to the example of the double box integral for t t-production: Decoupling at ε 0 from the factorisation of the Picard-Fuchs operator: 5 = Need to solve only two coupled equations, not five!

50 Example Let us look at the following sector with two master integrals: The Picard-Fuchs operator factorises into two linear factors and we may transform to an ε-form: [( ) ( ) ( ) A = ε d ln(x ) d ln(x ) d ln(x ) ( ) ( ) ] d ln(x x 2 )+ d ln(x x 2 + 1), with s= m2 (1 x 1 ) 2 x 1, t = m 2 x 2.