Functional equations for Feynman integrals

Transcription

1 Functional equations for Feynman integrals O.V. Tarasov JINR, Dubna, Russia October 9, 016, Hayama, Japan O.V. Tarasov (JINR) Functional equations for Feynman integrals 1 / 34

2 Contents 1 Functional equations (FE) from recurrence relations Derivation of FE from dimensional invariance Algorithm One loop propagator integral One-loop box type integral 3 Derivation of FE by using algebraic relations for propagators for products of two propagators for products of 3 propagators Abel s addition theorem and FE for Feynman integrals 4 Concluding remarks O.V. Tarasov (JINR) Functional equations for Feynman integrals / 34

3 Derivation of FE from recurrence relations FE from recurrence relations Functional equations (FE) for Feynman integrals were proposed in O.V.T. Phys.Lett. B670 (008) 67. Feynman integrals satisfy recurrence relations which can be written as Q j I j,n = R k,r I k,r j k,r<n where Q j, R k are polynomials in masses, scalar products of external momenta, d, and powers of propagators. I k,r - are integrals with r external lines. In recurrence relations some integrals are more complicated than the others: they have more arguments than the others. General method for deriving functional equations: By choosing kinematic variables, masses, indices of propagators remove most complicated integrals, i.e. impose conditions : Q j = 0 keeping at least some other coefficients R k 0. O.V. Tarasov (JINR) Functional equations for Feynman integrals 3 / 34

4 Introduction Derivation of FE from recurrence relations Example: one-loop n-point integrals Integrals I n (d) satisfy generalized recurrence relations O.T. in Phys.Rev.D54 (1996) p.6479 G n 1 ν j j + I (d+) n ( j n )I (d) n = n ( j k n )k I (d) where j ± shifts indices ν j ν j ± 1, j p 1 p 1 p 1 p... p 1 p n 1 G n 1 = n......, p 1 p n 1 p p n 1... p n 1 p n 1 Y 11 Y 1... Y 1n n =......, Y ij = mi +mj p ij, p ij = (p i p j ), Y 1n Y n... Y nn k=1, mj n, O.V. Tarasov (JINR) Functional equations for Feynman integrals 4 / 34

5 Derivation of FE from recurrence relations One-loop propagator type integral At n = 3, j = 1, m 3 = 0, imposing conditions G = 0, 3 = 0 we get I (d) (m 1, m, p 1 ) = p 1 + m 1 m α 1 p 1 I (d) (m 1, 0, s 13 ) where + p 1 m 1 + m + α 1 p 1 I (d) (0, m, s 3 ) s 13 = 1 + p 1 m 1 (p 1 + m 1 m )α 1 p 1, s 3 = 1 + p 1 m + (p 1 m 1 + m )α 1 p 1, α 1 = ± 1. ij = p ij + m 4 i + m 4 j p ij m i p ij m j m i m j. Integral with arbitrary masses and momentum can be expressed in terms of integrals with one propagator massless!!! O.V. Tarasov (JINR) Functional equations for Feynman integrals 5 / 34

6 Derivation of FE from recurrence relations One-loop vertex type integral At n = 4, j = 1, m 4 = 0, imposing conditions G 3 = 0, 4 = 0 we get I (d) 3 (m1, m, m3, s 3, s 13, s 1 ) = s 13 + m3 m 1 + α 13 s 13 I (d) 3 (m, m3, 0, s (13) 34, s 4(m1, m3, s 3, s 13, s 1 ), s 3 ) + s 13 m 3 + m 1 α 13 s 13 I (d) 3 (m 1, m, 0, s 4 (m 1, m 3, s 3, s 13, s 1 ), s (13) 14, s 1) Again as it was for integral I (d) integral I (d) 3 with arbitrary arguments can be expressed in terms of integrals with at least one propagator massless!!! O.V. Tarasov (JINR) Functional equations for Feynman integrals 6 / 34

7 Dimensional invariance Derivation of FE from dimensional invariance It will be highly desirable to derive functional equations from some invariance principle. Feynman integrals depend on two rather different sets of variables: kinematical variables and masses as well as on parameter of the space time dimension d and powers of propagators. The general form of functional equations for a given integral I (d) ({x j }: x j (d, {s ij }, {m k }) I (d) ({s ij }, {mk}) = 0. j where x j, m j are different sets of kinematical variables, and d is arbitrary. Functional equations must be valid for arbitrary value of d. We can take this property as a basic principle for finding functional equations among Feynman integrals. O.V. Tarasov (JINR) Functional equations for Feynman integrals 7 / 34

8 Dimensional invariance Namely, we will try to find some combination of integrals in the form Φ(d) = j x j (d, {s ij }, {m}) I (d) ({s ij }, {m j }). which is invariant with respect to shift d d +, i.e. Φ(d + ) = Φ(d), Any integral with space time dimension d + can be expressed in terms finite set of basis integrals with space time dimension d (O.V.T, Phys.Rev. D54 (1996) 6479 ). This dimensional recurrence relations can be used for constructing dimensionally invariant combinations of integrals with different arguments and discovering functional equations. O.V. Tarasov (JINR) Functional equations for Feynman integrals 8 / 34

9 Dimensional invariance New principle Algorithm for deriving functional equations Algorithm for deriving functional equations : For a given integral write combination Φ(d) of this integral with undetermined arguments {s ij }, mk and undetermined coefficients x k Shift space time dimension by : d d + and express integrals I (d+) in terms of integrals with dimension d. By equating arguments of integrals and coefficients in front of them derive equations for undetermined arguments and coefficients. Find additional equations for x j by considering limiting cases like d, d d 0 (d 0 = 0, 1,, 3...) or by taking imaginary part of integrals on some cut. Check that the discovered dimensionally invariant combination is zero by considering particular cases s ij = s (0) ij and/or by differentiating it with respect to kinematical variables and masses. O.V. Tarasov (JINR) Functional equations for Feynman integrals 9 / 34

10 One-loop example Dimensional invariance Example Let s consider simple example: I (d) (q, m ) = 1 iπ d/ According to our proposal we write: d d k 1 (k 1 + iη)((k 1 q) m + iη) Φ(d) = I (d) (q, m ) + x 1 I (d) (q 1, m 1) + x I (d) (q, m ) = 0. We can try to find functional equation with x j independent of d. Then Φ(d + ) = I (d+) (q, m ) + x 1 I (d+) (q 1, m 1) + x I (d+) (q, m ) = 0. O.V. Tarasov (JINR) Functional equations for Feynman integrals 10 / 34

11 Dimensional invariance One-loop propagator integral Example Integral I (d) (q, m ) satisfies the following dimensional recurrence relation: I (d+) (q, m ) = (m q ) q (d 1) I (d) (q, m ) (m + q ) q (d 1) T (d) 1 (m ), where T (d) 1 (m ) = 1 iπ d/ d d ( k 1 k1 m + iη = Γ 1 d ) m d Substituting this relation into Φ(d + ) gives: z 0 I (d) (q, m ) + z 1 I (d) (q 1, m 1) + z I (d) (q, m ) + z 3 T (d) 1 (m ) + z 4 T (d) 1 (m 1) + z 5 T (d) 1 (m ) = 0. O.V. Tarasov (JINR) Functional equations for Feynman integrals 11 / 34

12 Dimensional invariance One-loop propagator integral Example Solving equation Φ(d) = 0 with respect to I (d) (q, m ) and substituting solution into the new equation leads to the following relation: where Y 1 I (d) (q 1, m 1) + Y I (d) (q, m ) (m + q ) q (d 1) T (d) 1 (m ) (m 1 + q 1 ) (d 1)q1 x 1 T (d) 1 (m1) (m + q ) (d 1)q x T (d) 1 (m) = 0, Y 1 = ( q m1 4 + q m1 q 1 q (q1 ) + q1 m4 q1 m q + q1 (q ) ) (d 1)q q1 x 1, Y = (q m4 q m q + q (q ) q m 4 + q m q q (q ) ) (d 1)q q x. O.V. Tarasov (JINR) Functional equations for Feynman integrals 1 / 34

13 Dimensional invariance One-loop propagator integral Example We assume that the sets {q1, m 1 }, {q, m } are different and integrals depending on these sets are independent and nontrivial. Therefore it must be Y 1 = 0, Y = 0. These equations can be solved for q1 and q remain (m + q ) q (d 1) T (d) 1 (m ) After that three terms will (m 1 + q 1 ) (d 1)q1 x 1 T (d) 1 (m1) (m + q ) (d 1)q x T (d) 1 (m) = 0, In order to cancel the term with T (d) 1 (m ) and get rid off all T (d) 1 terms one should assume either m 1 = m, m = 0, or m 1 = 0, m = m, or m 1 = m = m. O.V. Tarasov (JINR) Functional equations for Feynman integrals 13 / 34

14 Dimensional invariance One-loop propagator integral Example At m 1 = m, m = 0 x 1 = m q, q 1 = m4 q, q = (q m ) q At m 1 = 0, m = m x = m q, q 1 = (q m ) q, q = m4 q In order to determine both x 1 and x for both solutions additional equation is required. It can be easily found by taking the limit (4 d)/ = ε 0 for Φ(d) : 1 + x 1 + x = 0. O.V. Tarasov (JINR) Functional equations for Feynman integrals 14 / 34

15 Dimensional invariance One-loop propagator integral Example At m 1 = m, m = m as follows from equations Y 1 = Y = 0: q 1 = q = m4 q, that from one side contradicts to our assumption about different sets of arguments and from the other side it leads to the dimensional invariant Φ(d) 0. In both considered cases we obtained the same functional equation: ( ) I (d) (q, m ) = m q I (d) m 4 q, m + (q m ( ) q I (d) (q m ) ) q, 0. where (q, 0) = Γ ( d ) Γ ( d 1) ( q ) d. Γ(d ) I (d) O.V. Tarasov (JINR) Functional equations for Feynman integrals 15 / 34

16 Dimensional invariance One-loop box type integral One-loop box type integral The same algorithm can be applied to more complicated integrals. We considered the box-type integral I (d) 4 ({s kr }) where I (d) 4 (s nj, s jk, s kl, s nl ; s jl, s nk ) = d d q iπ d/ 1 [(q p n ) m ][(q p j ) m ][(q p k ) m ][(q p l ) m ], s ij = (p i p j ), m = µ iη. taken at s 1 = s 3 = s 34 = 0. O.V. Tarasov (JINR) Functional equations for Feynman integrals 16 / 34

17 Dimensional invariance One-loop box type integral One-loop box type integral We derived a new functional equation for the integral with this kinematics: ( ) I (d) 4 (s 14, s 4, s 13, m s 13 ) = I (d) s 13 s 4 4 0,, s 13, m s 13 s 14 s 13 s 14 s ( ) 14 I (d) s 13 s 4 4 s 4,, s 14, m, s 13 s 14 s 13 s 14 where I 4 (s 14, s 4, s 13, m ) I 4 (0, 0, 0, s 14, s 4, s 13, m ). The most complicated part in derivation of this equation was analysis of all possible solutions of equations for unknown parameters and arguments of integrals. The procedure of finding all possible solutions was implemented in Maple and also in Mathematica. Sometimes Maple discovered not all solutions. O.V. Tarasov (JINR) Functional equations for Feynman integrals 17 / 34

18 Question Dimensional invariance One-loop box type integral We used recurrence relations to derive FE. All recurrence relations follow from the equation: d d k k µ f (k, {s ij }, {m r }) = 0. One can raise the question: Functional equations hold for integrals or they can be obtained as a consequence of a relation between integrands? O.V. Tarasov (JINR) Functional equations for Feynman integrals 18 / 34

19 for propagators Analyzing one-loop FE one can see that integrands are rather similar and differ only by one propagator: Integrands for the one-loop propagator type integrals 1 D 1 D, 1 1,, D 0 D D 1 D 0 Integrands for the one-loop vertex type integrals 1 D 1 D D 3, 1 1 1,, D 0 D D 3 D 1 D 0 D 3 D 1 D D 0 where D 0 = (k 1 p 0 ) m 0 + iη, D = (k 1 p ) m + iη, D 1 = (k 1 p 1 ) m 1 + iη, D 3 = (k 1 p 3 ) m 3 + iη, Observation: since G n = 0 vectors p 1,p,...are linearly dependent O.V. Tarasov (JINR) Functional equations for Feynman integrals 19 / 34

20 for propagators Question: Would it be possible to find algebraic relations of the form: where 1 D 1 D = x 1 D 0 D + x D 1 D 0 p 0 = y 01 p 1 + y 0 p and x 1, x, y 01, y 0 being independent of k 1. The answer is - YES! Putting all terms over the common denominator and equating coefficients in front of different products of (k 1 )a (k 1 p 1 ) b (k 1 p ) c to zero we obtain system of equations: where s 1 = (p 1 p ) y 0 x = 0, y 01 x 1 = 0, x 1 + x = 1, p 1(x 1 y 01) + p (x y 0) + y 01 y 0 (s 1 p 1 p ) m 1x 1 m x + m 0 = 0. O.V. Tarasov (JINR) Functional equations for Feynman integrals 0 / 34

21 for propagators Solution of this system of equations is: and x 1 = y 01 = m m 1 + s 1 s 1 x = y 0 = m 1 m + s 1 s 1 + Λ + 4s 1 m0, s 1 Λ + 4s 1 m 0 s 1. Λ = s 1 + m m 4 s 1 (m 1 + m ) m 1m. O.V. Tarasov (JINR) Functional equations for Feynman integrals 1 / 34

22 for propagators Integrating obtained algebraic relation w.r.t. k 1 gives the following FE: I (d) (m 1, m, s 1 ) = s 1 + m 1 m λ s 1 where + s 1 m 1 + m + λ s 1 I (d) (m 1, m 0, s 13 (m 1, m, m 0, s 1 )) I (d) (m, m 0, s 3 (m 1, m, m 0, s 1 )). s 13 = Λ + s 1 (m1 + m 0 ) + m 1 m + s 1 λ s 1 s 1 s 3 = Λ + s 1 (m + m 0 ) + m 1 m s 1 λ. s 1 s 1 λ = Λ + 4s 1 m0 Parameter m 0 is arbitrary and can be taken at will. The same equation was obtained from recurrence relations by imposing conditions on Gram determinants. O.V. Tarasov (JINR) Functional equations for Feynman integrals / 34

23 for propagators Similar to the relation with two propagators one can find relation for three propagartors: 1 x 1 x x 3 = + +. D 1 D D 3 D D 3 D 0 D 1 D 3 D 0 D 1 D D 0 Here p 1, p and p 3 are independent external momenta, k 1 will be integration momentum and p 0 = y 01 p 1 + y 0 p + y 03 p 3. Multiplying both sides of equation by the product D 1 D D 3 D 0 and equating coefficients in front of k 1,k 1p 1,k 1 p,k 1 p 3 and term independent of k 1 we obtain system of equations y 01 x 1 = 0, y 0 x = 0, y 03 x 3 = 0, x 3 + x + x 1 1 = 0, [x 1 y 01 (y 03 + y 0 + y 01 )]p 1 + [x y 0 (y 03 + y 0 + y 01 )]p +[x 3 y 03 (y 03 + y 01 + y 0 )]p 3 +y 0 y 03 p 3 + y 01 y 03 p 13 + y 01 y 0 p 1 m 1x 1 m x m 3x 3 + m 0 = 0. O.V. Tarasov (JINR) Functional equations for Feynman integrals 3 / 34

24 for product of 3 propagators This system has the following solution x 1 = y 01 = 1 α y 0, x = y 0, x 3 = y 03 = α, where α is solution of the quadratic equation α p 13 + [m 3 m 1 p 13 + y 0 (p 13 + p 1 p 3 )]α +m 1 m 0 + (m m 1 p 1 + p 1 y 0 )y 0 = 0. Solution depends on arbitrary parameters: m 0, y 0. By integrating the obtained relation we get the same FE as it was given before. Functional relations for Feynman integrals with integrands being rational functions strongly remind Abel s addition theorem! O.V. Tarasov (JINR) Functional equations for Feynman integrals 4 / 34

25 Abelian integral is an integral in the complex plane of the form z z 0 R(x, y)dx, where R(x, y) is an arbitrary rational function of the two variables x and y. These variables are related by the equation F (x, y) = 0, where F (x, y) is an irreducible polynomial in y, F (x, y) φ n (x)y n φ 1 (x)y + φ 0 (x), whose coefficients φ j (x), j = 0, 1,...n are rational functions of x. Abelian integrals are natural generalizations of elliptic integrals, which arise when F (x, y) = y P(x), where P(x) is a polynomial of degree 3 and 4. If degree of the polynomial is greater than 4 then we have hyperelliptic integral. O.V. Tarasov (JINR) Functional equations for Feynman integrals 5 / 34

26 Abel s theorem Let C and C be plane curves given by the equations C : F (x, y) = 0, C : φ(x, y) = 0. These curves have n points of intersections (x 1, y 1 ),... (x n, y n ), where n is the product of degrees of C and C. Let R(x, y) be a rational function of x and y where y is defined as a function of x by the relation F (x, y) = 0. Consider the sum n xi,y i I = R(x, y)dx i=1 x 0,y 0 Integrals being taken from a fixed point to the n points of intersections. If some of the coefficients a 1,a,...,a k of φ(x, y) are regarded as continuous variables, the points (x i, y i ) will vary continuously and hence I will be a function, whose form is to be determined, of the variable coefficients a 1,a,...,a k. O.V. Tarasov (JINR) Functional equations for Feynman integrals 6 / 34

27 Abel s theorem Abel s theorem: The partial derivatives of the sum I, with respect to any of the coefficients of the variable curve φ(x, y) = 0, is a rational function of the coefficients and hence I is equal to a rational function of the coefficients of φ(x, y) = 0, plus a finite number of logarithms or arc tangents of such rational functions. Important: integrals themselves can be rather complicated transcendental functions but their sum can be simple. Example: Elliptic integral of the second type: E(k, x) = x 0 (1 k x)dx x(1 x)(1 k x). O.V. Tarasov (JINR) Functional equations for Feynman integrals 7 / 34

28 Abel s theorem We take as C and C C : y = x(1 x)(1 k x), C : y = ax + b. The elimination of y between two equations will give us as the abscissae x 1, x, x 3 of the points of intersection the three roots of the equation: φ(x) = k x 3 (1 + k + a )x + (1 ab)x b = 0. The corresponding sum will be I (a, b) = x1 0 R(x)dx + x Abel s theorem gives addition formula: x1 0 R(x)dx + x 0 0 R(x)dx + where κ is an arbitrary constant. x3 R(x)dx + R(x)dx 1/k x3 1/k R(x)dx = a + κ, O.V. Tarasov (JINR) Functional equations for Feynman integrals 8 / 34

29 One can find FE for Feynman integrals following closely derivation of relationships for usual algebraic integrals. Deriving relations for propagators we used orthogonality condition G n = 0. (1) In fact it is not needed to assume such a relation. For example, to fix parameters in algebraic relations for products of two propagators R (k 1, p 1, p, m1, m, m0) = 1 x 1 x = 0, () D 1 D D D 0 D 1 D 0 instead of (1) we can impose conditions Multiplying both sides of () by D 0 D 1 D we get D 0 x 1 D 1 x D x 1 k 1µ = x k 1µ = 0. (3) = (1 x 1 x )k 1 + x 1 k 1 p 1 + x k 1 p +x 1 m 1 + x m x 1 p 1 x p k 1 p 3 m 3 + p 3. (4) O.V. Tarasov (JINR) Functional equations for Feynman integrals 9 / 34

30 and similar expressions for other scalar products of p 3. Solution for x 1, x is the same as it was before and as a result we obtained the same relation between products of two propagators. O.V. Tarasov (JINR) Functional equations for Feynman integrals 30 / 34 Differentiating this relation, contracting with k 1, p 1, p, p 3 and taking into account (3) gives several equations: x 1 (k 1 k 1 p 1 ) x (k 1 k 1 p ) + k 1 k 1 p 3 = 0, (1 x 1 x )k 1 p 1 + x 1 p 1 + x p 1 p p 1 p 3 = 0, (1 x 1 x )k 1 p + p 1 p x 1 + x p p p 3 = 0, (1 x 1 x )k 1 p 3 + x 1 p 1 p 3 + x p p 3 p 3 = 0. (5) They can be used to express k 1 p 3, p 1 p 3, p p 3, x 1, x in terms of k1, k 1p 1, k 1 p, p1, p 1p, p considered to be independent variables. For example, we get: k 1 p 3 = x 1 k 1 p 1 x 1 k 1 p + k 1 p + x 1 (m 1 m p 1 + p ) + 1 (m p m 3 + p 3),

31 Solution of the above system of equations is rather similar to finding intersections of two plane curves considered in Abel s theorem. Similar to usual algebraic integrals for one variable we can construct various integrands out of our different relationships for products of propagator. These integrands will be rational functions in independent variables. Integrations should be done over d dimensional space. Rational function must resemble integrands for Feynman integrands. O.V. Tarasov (JINR) Functional equations for Feynman integrals 31 / 34

32 For example, integrating product of different relationships between two propagators multiplied by 1/[(k 1 k ) m 5 ]ν 5 with respect to k 1, k leads to the FE for the integral d d k 1 d d k [(k 1 k ) m 5 ]ν 5 R (k 1, p 1, p, m 1, m, m 0)R (k, p 3, p 4, m 3, m 4, m 0) = 0, corresponding to the following diagram k p k 1 k k 1 p 1 k p 4 k 1 p 3 O.V. Tarasov (JINR) Functional equations for Feynman integrals 3 / 34

33 By integrating product of the relationship R 3 (k 1, p 1, p, p 3, m1, m, m3, m0) 1 x 1 = D 1 D D 3 D D 3 D 0 and one loop-propagator integral x D 1 D 3 D 0 x 3 D 1 D D 0 = 0, d d k 1 d d k [k 1 m 1 ]ν 1 [(k1 k ) m 5 ]ν 5 R 3(k, p, p 3, p 4, m, m 3, m 4) = 0, we obtain the FE for the integral corresponding to the following diagram with arbitrary ν 1, ν 5, and arbitrary momenta and masses k p k 1 p 1 k1 k k p 4 k p 3 O.V. Tarasov (JINR) Functional equations for Feynman integrals 33 / 34

34 Concluding remarks Concluding remarks Three different methods for finding FE for Feynman integrals with any number of loops and external legs are proposed. FE reduce integrals with complicated kinematics to simpler integrals FE can be used for analytic continuation of Feynman integrals without knowing explicit analytic result. Application of these methods for some two- and three- loop integrals is in progress Systematic investigation of FE for Feynman integrals based on algebraic geometry and group theory is needed. Some improvements of these methods can be done by exploiting known methods for algebraic integrals. The methods can be extended for finding functional equations among hypergeometric as well as holonomic functions. O.V. Tarasov (JINR) Functional equations for Feynman integrals 34 / 34