one step beyond multiple polylogarithms stefan weinzierl
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One step beyond multiple polylogarithms Stefan Weinzierl ur Physik, - PowerPoint PPT Presentation

One step beyond multiple polylogarithms Stefan Weinzierl ur Physik, Universit Institut f at Mainz I : Periodic functions and periods II : Differential equations III : The two-loop sun-rise diagramm in collaboration with L. Adams, Ch.


  1. One step beyond multiple polylogarithms Stefan Weinzierl ur Physik, Universit¨ Institut f¨ at Mainz I : Periodic functions and periods II : Differential equations III : The two-loop sun-rise diagramm in collaboration with L. Adams, Ch. Bogner, S. M¨ uller-Stach and R. Zayadeh

  2. 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 = { n ω | n ∈ Z } , • a double lattice, Λ = { n 1 ω 1 + n 2 ω 2 | n 1 , n 2 ∈ Z } .

  3. Examples of periodic functions • Singly periodic function: Exponential function exp ( z ) . exp ( z ) is periodic with peridod ω = 2 π i . • Doubly periodic function: Weierstrass’s ℘ -function � � ℘ ( z ) = 1 ( z + ω ) 2 − 1 1 z 2 + ∑ , Λ = { n 1 ω 1 + n 2 ω 2 | n 1 , n 2 ∈ Z } , ω 2 ω ∈ Λ \{ 0 } Im( ω 2 / ω 1 ) � = 0 . ℘ ( z ) is periodic with periods ω 1 and ω 2 .

  4. Inverse functions The corresponding inverse functions are in general multivalued functions. • For the exponential function x = exp ( z ) the inverse function is the logarithm = ln ( x ) . z • For Weierstrass’s elliptic function x = ℘ ( z ) the inverse function is an elliptic integral ∞ dt 1 1 � g 2 = 60 ∑ ω 4 , g 3 = 140 ∑ = , ω 6 . z � 4 t 3 − g 2 t − g 3 ω ∈ Λ \{ 0 } ω ∈ Λ \{ 0 } x

  5. 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: 1 dt � 2 π i √ = 1 − t 2 . 2 i − 1 • Periods of Weierstrass’s ℘ -function: Assume that g 2 and g 3 are two given algebraic numbers. Then t 2 t 2 dt dt � � ω 1 = 2 ω 2 = 2 , , � � 4 t 3 − g 2 t − g 3 4 t 3 − g 2 t − g 3 t 1 t 3 where t 1 , t 2 and t 3 are the roots of the cubic equation 4 t 3 − g 2 t − g 3 = 0 .

  6. 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 in R 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. 2 dt � = t . ln2 1

  7. Feynman integrals A Feynman graph with m external lines, n internal lines and l loops corresponds (up to prefactors) in D space-time dimensions to the Feynman integral � µ 2 � n − lD / 2 l d D k r n 1 � ∏ ∏ = I G D ( − q 2 j + m 2 Γ ( n − lD / 2 ) j ) i π 2 r = 1 j = 1 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 m as l m ∑ ∑ = λ ij k j + σ ij p j , λ ij , σ ij ∈ {− 1 , 0 , 1 } . q i j = 1 j = 1

  8. Feynman parametrisation The Feynman trick: n n 1 1 � d n x δ ( 1 − ∏ ∑ = Γ ( n ) x j ) � n � n P j j = 1 j = 1 ∑ x j P j x j ≥ 0 j = 1 We use this formula with P j = − q 2 j + m 2 j . We can write n l l l ∑ x j ( − q 2 j + m 2 ∑ ∑ ∑ j ) = − k r M rs k s + 2 k r · Q r + J , j = 1 r = 1 s = 1 r = 1 where M is a l × l matrix with scalar entries and Q is a l -vector with momenta vectors as entries.

  9. Feynman integrals After Feynman parametrisation the integrals over the loop momenta k 1 , ..., k l can be done: n x i ) U n − ( l + 1 ) D / 2 � d n x δ ( 1 − ∑ I G = F n − lD / 2 , U = det ( M ) , i = 1 x j ≥ 0 � � J + QM − 1 Q / µ 2 . F = det ( M ) The functions U and F are called the first and second graph polynomial. U is positive definite inside the integration region and positive semi-definite on the boundary. F depends on the masses m 2 i and the momenta ( p i 1 + ... + p i r ) 2 . In the euclidean region F is also positive definite inside the integration region and positive semi-definite on the boundary.

  10. Feynman integrals and periods Laurent expansion in ε = ( 4 − D ) / 2 : ∞ c j ε j . ∑ = I G j = − 2 l 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 ?

  11. 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: ∞ x n ∑ Li 1 ( x ) = − ln ( 1 − x ) = n n = 1 ∞ x n ∑ Li 2 ( x ) = n 2 n = 1

  12. Generalisations of the logarithm Beyond one-loop, at least the following generalisations occur: Polylogarithms: ∞ x n ∑ Li m ( x ) = n m n = 1 Multiple polylogarithms (Goncharov 1998) : x n 1 · x n 2 · ... · x n k ∞ ∑ 1 2 k Li m 1 , m 2 ,..., m k ( x 1 , x 2 ,..., x k ) = n m 1 n m 2 n m k n 1 > n 2 >...> n k > 0 1 2 k This is a nested sum: n j n j − 1 − 1 n j − 1 x j ∑ ∑ ... ... n jm j n j = 1 n j + 1 = 1

  13. Iterated integrals Define the functions G by t k − 1 y t 1 dt 1 dt 2 dt k � � � G ( z 1 ,..., z k ; y ) = ... . t 1 − z 1 t 2 − z 2 t k − z k 0 0 0 Scaling relation: G ( z 1 ,..., z k ; y ) = G ( xz 1 ,..., xz k ; xy ) Short hand notation: G m 1 ,..., 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 Conversion to multiple polylogarithms: � 1 � , 1 1 ( − 1 ) k G m 1 ,..., m k Li m 1 ,..., m k ( x 1 ,..., x k ) = ,..., . ;1 x 1 ... x k x 1 x 1 x 2

  14. Differential equations for Feynman integrals 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 = ∑ ∑ dt j I G ( t ) q i ( t ) I G i ( t ) j = 0 i Inhomogeneous term on the rhs consists of simpler integrals I G i . p j ( t ) , q i ( t ) polynomials in t . 2. Solve the differential equation. Kotikov, Remiddi, Gehrmann, Laporta

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

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

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

  18. An example from physics: The two-loop sunrise integral m 1 m 2 � � p p 2 , m 2 1 , m 2 2 , m 2 = S 3 m 3 • Two-loop contribution to the self-energy of massive particles. • Sub-topology for more complicated diagrams.

  19. The two-loop sunrise integral: Prior art Integration-by-parts identities allow to derive a coupled system of 4 first-order differential equations for S and S 1 , S 2 , S 3 , where ∂ = S i S ∂ m 2 i (Caffo, Czyz, Laporta, Remiddi, 1998) . This system reduces to a single second-order differential equation in the case of equal masses m 1 = m 2 = m 3 (Broadhurst, Fleischer, Tarasov, 1993) . Dimensional recurrence relations relate integrals in D = 4 dimensions and D = 2 dimensions (Tarasov, 1996, Baikov, 1997, Lee, 2010) . Analytic result in the equal mass case known up to quadrature, result involves elliptic integrals (Laporta, Remiddi, 2004) .

  20. The two-loop sunrise integral Is the system of 4 coupled first-order differential equations generic for the unequal mass case or can we do better ? Yes, we can ! Also in the unequal mass case there is a single second-order differential equation. The second-order differential equation follows from algebraic geometry. (S. M¨ uller-Stach, S.W., R. Zayadeh, arXiv:1112:4360)

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