Large- n f Contributions to the Four-Loop Splitting Functions in QCD - PowerPoint PPT Presentation

Large- n f Contributions to the Four-Loop Splitting Functions in QCD Nucl. Phys. B915 (2017) 335-362, arXiv:1610.07477 Joshua Davies Department of Mathematical Sciences University of Liverpool Collaborators: A. Vogt (University of Liverpool), B. Ruijl, T. Ueda, J. Vermaseren (Nikhef) 25th International Workshop on Deep Inelastic Scattering and Related Topics 6th April

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS I NTRODUCTION Deep Inelastic Scattering : a lepton scatters from a proton Boson Lepton Q C q,g Quark or Gluon p xp f q , f g Proton Boson: γ, H , Z 0 (Neutral Current) or W ± (Charged Current) � � Cross-section: σ ∼ � a F a ( x , Q 2 ) = � C a , q ⊗ f q + C a , g ⊗ f g a F a – “Structure Function” C a , j – “Coefficient Function” ⊗ – “Mellin Convolution” f j – “Parton Distribution Function” 1/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS I NCLUSIVE DIS To compute C a , q , C a , g , we use the optical theorem . Compute forward scattering amplitudes : 2 ∼ Im Use Dim. Reg. ( D = 4 − 2 ε ). Divergences appear as poles in ε . Renormalization of a s removes UV poles. “Collinear” poles remain, � � C a , j = ˜ ˜ x , a s , Q 2 / µ 2 C a , j r , ε . 2/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS C OLLINEAR F ACTORIZATION We need to deal with these collinear poles: renormalize the PDF. � � C a , j ⊗ ˜ ⊗ ˜ F a = ˜ x , a s , µ 2 f j = C a , j ⊗ Z ji r / µ 2 f , ε f i = C a , j ⊗ f j . C a , j is finite. Z ji contains only poles in ε . Factorization at scale µ 2 f , implies f j has scale dependence: d d d Z ji ⊗ ˜ Z jk ⊗ Z − 1 f j = f i = ⊗ f i . ki d ln µ 2 d ln µ 2 d ln µ 2 f f f � �� � P ji ◮ this is the DGLAP evolution equation ◮ P ji are the Splitting Functions 3/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS S PLITTING F UNCTIONS Know Z ji from calculation of ˜ C a , j , so we can extract P ji . PDFs are universal to all hadron interactions; Splitting Functions are also. DGLAP evolution: system of 2 n f +1 coupled equations. By defining the distributions n f n f � � ( f i + ¯ q ± ns , ij = ( f i ± ¯ f i ) − ( f j ± ¯ ( f i − ¯ q s = f i ) , f j ) , q V = f i ) , i = 1 i = 1 we have the evolution equations, (setting µ 2 f = Q 2 ): � q s � P qq � q s � � � d P qg = ⊗ , g P gq P gg g d ln Q 2 d d ln Q 2 q ± ns , ij = P ± ns q ± ns , ij , d d ln Q 2 q V = P V q V . 4/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS I N M ELLIN SPACE ... Take the Mellin transform , � 1 F a ( N , Q 2 ) = d x x N − 1 ˆ F a ( x , Q 2 ) . 0 Now all convolutions ( ⊗ ) are simple products. We compute Mellin moments of ˜ C a , j , N = 2 , 4 , 6 , ... , not an analytic expression for arbitrary N (which gives x -space expression via IMT). ◮ Mellin moments of Splitting Functions P ij . Q: Given some fixed number of Mellin moments of P ij , can we derive an analytic expression for general N ? ◮ this is the goal of this project. 5/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS S OFTWARE qgraf : generate diagrams (1.2 million!) [Nogueira ‘93] TFORM : physics, project Mellin moments. [Kuipers,Ueda,Vermaseren,Vollinga ‘13] Produces 2-point tensor integrals, which must be reduced to masters. To 3 loops, we can use MINCER . [Larin,Tkachov,Vermaseren ‘91] At 4 loops, FORCER . State of the art . [Ruijl,Ueda,Vermaseren] 6/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS W HAT DO P ij “ LOOK LIKE ”? To a 3 s , written in terms of harmonic sums , N N ( − 1 ) i 1 � � S m ( N ) = i m , S − m ( N ) = , i m i = 1 i = 1 N [( − 1 ) i ] � S [ − ] m 1 , m 2 ,..., m l ( N ) = S m 2 ,..., m l ( i ) , i m i = 1 � � p and denominators , D p 1 i = . N + i Define ◮ harmonic weight : � l i = 1 | m i | , ◮ overall weight : harmonic weight + p . ∞ � P ( n ) a n + 1 P ij = ij . s n = 0 s , P ( n ) To a 3 written as terms of overall weight up to ( 2 n + 1 ) . ij 7/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS 2-L OOP E XAMPLE � � P ( 1 ) � = − 8 ( 2 D 2 − 2 D 1 + D 0 ) S − 2 + 8 ( 2 D 2 − 2 D 1 + D 0 ) S 1 , 1 � qg C A n f � + 16 ( D 2 2 − D 2 1 ) S 1 + 8 ( 4 D 3 2 + 2 D 3 1 + D 3 0 ) OW 3 � 4 � 3 ( 44 D 2 2 + 12 D 2 1 + 3 D 2 − 0 ) OW 2 � 4 � + 9 ( 20 D − 1 − 146 D 2 + 153 D 1 − 18 D 0 ) OW 1 ◮ At overall weight i , up to factor ( 1 / 3 ) ( 3 − i ) , coefficients are integers . Possible basis: { S − 2 , S 1 , 1 , S 2 } · { D 0 , D 1 , D 2 } { S 1 } · { D 1 , 2 0 , D 1 , 2 1 , D 1 , 2 2 } { 1 } · { D 1 , 2 , 3 , D 1 , 2 , 3 , D 1 , 2 , 3 , D − 1 } 0 1 2 Assuming ( 1 / 3 ) ( 3 − i ) , need to determine 25 integer coefficients . 8/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS 2-L OOP E XAMPLE Compute Mellin moments: � P ( 1 ) 35 / ( 3 3 ) � ( 2 ) = � qg C A n f � ( 4 ) = − 16387 / ( 2 3 3 2 5 3 ) P ( 1 ) � � qg C A n f � ( 6 ) = − 867311 / ( 2 3 3 3 5 1 7 3 ) P ( 1 ) � � qg C A n f � ( 8 ) = − 100911011 / ( 2 6 3 6 5 3 7 1 ) P ( 1 ) � � qg C A n f . . . With moments N = 2 , 4 , . . . , 50 we can solve for 25 basis coefficients. Can we do better? ◮ Use that the coefficients are integer . ◮ It is a system of Diophantine equations . 9/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS L ATTICE B ASIS R EDUCTION Lenstra-Lenstra-Lov´ asz Lattice Basis Reduction: [Lenstra,Lenstra,Lov´ asz ‘82] ◮ find a short lattice basis in polynomial time ◮ can be used to find integer solutions to equations axb : ◮ part of calc [ www.numbertheory.org ] ◮ LLL-based solver for systems of Diophantine equations See also, Mathematica , Maple , fpLLL , ... , many more. To solve: �   P ( 1 ) C A n f ( 2 )     � b 1 ( 2 ) , . . . , b 25 ( 2 ) c 1 qg . . .       . .  = .   . . .      � b 1 ( m ) , . . . , b 25 ( m ) P ( 1 ) c 25 C A n f ( m ) � qg b i ( N ) , c i : basis elements, coefficients. c i ∈ Z . 10/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS 2-L OOP E XAMPLE : R ECONSTRUCTION � Determines P ( 1 ) C A n f ( 25 integer coefficients) with just 9 Mellin moments. � qg ◮ solution, ( c 1 , . . . , c 25 ) = ( 2 , 6 , 72 , 8 , 88 , 584 , 4 , 24 , − 612 , − 80 , 0 , 0 , 4 , 0 , − 4 , 0 , 2 , 4 , − 4 , 2 , 4 , − 4 , 0 , 0 , 0 ) � �� � � �� � � �� � SW 0 SW 1 SW 2 What if the basis were incorrect? For e.g., leave out D − 1 : ◮ solve with N = 2 , . . . , 18 , ( − 43 , 423 , 123 , 1492 , − 102 , 1332 , 4 , 24 , − 612 , − 15 , 437 , 102 , − 2399 , 80 , 1700 , − 146 , 180 , − 26 , − 1065 , 670 , 579 , − 919 , 490 , 605 ) ◮ solve with N = 2 , . . . , 20 , ( − 178 , 4391 , − 25712 , 412 , − 10348 , − 6476 , 4 , 24 , − 612 , − 572 , 25401 , − 2178 , − 5642 , − 3526 , − 20152 , − 3302 , − 3161 , 6474 , − 4011 , 5092 , 3775 , − 3283 , − 4617 , 11029 ) Claim: these solutions are “obviously bad” . 11/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS F OUR -L OOP S PLITTING F UNCTIONS Large- n f contributions: ◮ subset of diagrams, much easier for FORCER to compute ◮ smaller reconstruction bases (terms of lower overall weight) Singlet Splitting Functions , colour factors at n 3 f , P ( 3 ) qq { C F n 3 P ( 3 ) qg { C A n 3 f , C F n 3 f } f } P ( 3 ) gq { C F n 3 P ( 3 ) gg { C A n 3 f , C F n 3 f } f } Guess bases using lower order information. Number of coefficients: P ( 3 ) P ( 3 ) qq { 69 } qg { 125 , 101 } P ( 3 ) P ( 3 ) gq { 38 } gg { 34 , 54 } Moments used for reconstruction, (check), N = 2 , . . . P ( 3 ) P ( 3 ) qq { 30 ( 44 ) } qg {× ( × ) , 40 ( 54 ) } P ( 3 ) P ( 3 ) gq { 18 ( 28 ) } gg { 20 ( 28 ) , 26 ( 32 ) } 12/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS H ARDEST S INGLET C ASE � P ( 3 ) f : Basis with 125 unknown integer coefficients. � qg C A n 3 N = 2 , . . . , 46 insufficient to determine a good solution. Moment calculations become very computationally demanding. Hardest diagram computed at N = 46 , ◮ ∼ 2 weeks wall-time [16 cores, 192GB RAM] ◮ ∼ 13TB peak disk usage by TFORM → no more moments! We need to somehow make the basis smaller. Use additional constraints : ◮ large- x limit: no irrational constants other than ζ i -1 coeff. ◮ # S 1 , 2 = − # S 2 , 1 -7 coeff. 117 unknowns. Solution with N = 2 , . . . , 44 , N = 46 checks. 13/16

I NTRODUCTION S PLITTING F UNCTIONS LLL R ECONSTRUCTION C ONCLUSIONS N ON - SINGLET S PLITTING F UNCTIONS n 3 f terms of P ( 3 ) , ± are already known. [Gracey ‘94] ns f terms of P ( 3 ) , + (even N ) and P ( 3 ) , − We determine the n 2 (odd N ). ns ns Colour factors to determine at n 2 f : ◮ C 2 F n 2 f ◮ C A C F n 2 f – diagrams are very hard to compute! Method: decompose in two ways , � F A + C F ( C A − 2 C F ) B ± � P ( 3 ) , ± { n 2 f { C 2 F , C A C F }} = n 2 2 C 2 ns f � F ( A − B ± ) + C F C A B ± � = n 2 2 C 2 . f A should be common to both P ± ns ; use both odd and even N . Large n c . Compute (easier) C 2 F n 2 f diagrams to higher N to determine ( A − B ± ) . From these, determine B + and B − and hence P ( 3 ) , + and P ( 3 ) , − . � ns ns 14/16