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

Transcription

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

2 1/16 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS INTRODUCTION Deep Inelastic Scattering: a lepton scatters rom a proton Lepton Quark or Gluon Proton p q, g Q xp Boson C q,g Boson: γ, H, Z 0 (Neutral Current) or W ± (Charged Current) Cross-section: σ a F a(x, Q 2 ) = [ ] a Ca,q q + C a,g g F a Structure Function C a,j Coeicient Function Mellin Convolution j Parton Distribution Function

3 2/16 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS INCLUSIVE DIS To compute C a,q, C a,g, we use the optical theorem. Compute orward scattering amplitudes: 2 Im Use Dim. Reg. (D = 4 2ε). Divergences appear as poles in ε. Renormalization o a s removes UV poles. Collinear poles remain, C a,j = C a,j ( x, as, Q2 /µ 2 r, ε ).

4 3/16 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS COLLINEAR FACTORIZATION We need to deal with these collinear poles: renormalize the PDF. F a = C a,j ( j = C a,j Z ji x, as, µ2 r/µ 2, ε ) i = C a,j j. C a,j is inite. Z ji contains only poles in ε. Factorization at scale µ 2, implies j has scale dependence: d d ln µ 2 j = d d ln µ 2 Z ji i = d d ln µ 2 Z jk Z 1 ki } {{ } P ji i. this is the DGLAP evolution equation P ji are the Splitting Functions

5 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS SPLITTING FUNCTIONS Know Z ji rom calculation o C a,j, so we can extract P ji. PDFs are universal to all hadron interactions; Splitting Functions are also. DGLAP evolution: system o 2n +1 coupled equations. By deining the distributions n q s = ( i + i ), q ± ns,ij = ( i ± i ) ( j ± j ), q V = ( i i ), i=1 we have the evolution equations, (setting µ 2 = Q2 ): ( ) ( ) ( d qs Pqq P = qg qs d ln Q 2 g P gq P gg g d d ln Q 2 q± ns,ij = P± ns q± ns,ij, d d ln Q q 2 V = P V q V. ), n i=1 4/16

6 5/16 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS IN MELLIN SPACE... Take the Mellin transorm, F a (N, Q 2 ) = 1 0 dx x N 1ˆF a (x, Q 2 ). Now all convolutions ( ) are simple products. We compute Mellin moments o C a,j, N = 2, 4, 6,..., not an analytic expression or arbitrary N (which gives x-space expression via IMT). Mellin moments o Splitting Functions P ij. Q: Given some ixed number o Mellin moments o P ij, can we derive an analytic expression or general N? this is the goal o this project.

7 6/16 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS SOFTWARE qgra: 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 o the art. [Ruijl,Ueda,Vermaseren]

8 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS WHAT DO P ij LOOK LIKE? To a 3 s, written in terms o harmonic sums, S m (N) = N i=1 S [ ]m1,m 2,...,m l (N) = and denominators, D p i = Deine 1 i m, S m(n) = ( 1 N+i) p. N i=1 ( 1) i i m, N [( 1) i ] S m2,...,m l (i), i=1 harmonic weight: l i=1 m i, overall weight: harmonic weight + p. i m P ij = n=0 a n+1 s P (n) ij. To a 3 s, P(n) ij written as terms o overall weight up to (2n + 1). 7/16

9 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS 2-LOOP EXAMPLE P (1) qg CAn = + [ 8(2D 2 2D 1 + D 0 )S 2 + 8(2D 2 2D 1 + D 0 )S ] 1,1 + 16(D 2 2 D2 1 )S 1 + 8(4D D3 1 + D3 0 [ ) OW3 4 3 (44D D D2 0 [ ]OW2 ) 4 9 (20D 1 146D D 1 18D 0 )] OW1 At overall weight i, up to actor (1/3) (3 i), coeicients are integers. Possible basis: {S 2, S 1,1, S 2 } {D 0, D 1, D 2 } {S 1 } {D 1,2 0, D1,2 1, D1,2 2 } {1} {D 1,2,3 0, D 1,2,3 1, D 1,2,3 2, D 1 } Assuming (1/3) (3 i), need to determine 25 integer coeicients. 8/16

10 9/16 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS 2-LOOP EXAMPLE Compute Mellin moments: P (1) qg (2) = 35/(3 3 ) CAn P (1) qg (4) = 16387/( ) CAn P (1) qg (6) = /( ) CAn P (1) (8) = /( ) CAn qg. With moments N = 2, 4,..., 50 we can solve or 25 basis coeicients. Can we do better? Use that the coeicients are integer. It is a system o Diophantine equations.

11 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS LATTICE BASIS REDUCTION Lenstra-Lenstra-Lovász Lattice Basis Reduction: [Lenstra,Lenstra,Lovász 82] ind a short lattice basis in polynomial time can be used to ind integer solutions to equations axb: part o calc LLL-based solver or systems o Diophantine equations See also, Mathematica, Maple, plll,..., many more. To solve: b 1 (2),..., b 25 (2). b 1 (m),..., b 25 (m) c 1. c 25 = P qg (1) P qg (1) [ CAn (2). CAn (m) b i (N), c i : basis elements, coeicients. c i Z. 10/16

12 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS 2-LOOP EXAMPLE: RECONSTRUCTION Determines P qg (1) (25 integer coeicients) with just 9 Mellin moments. CAn 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) }{{}}{{}}{{} SW0 SW1 SW2 What i 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

13 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS FOUR-LOOP SPLITTING FUNCTIONS Large-n contributions: subset o diagrams, much easier or FORCER to compute smaller reconstruction bases (terms o lower overall weight) Singlet Splitting Functions, colour actors at n 3, P (3) qq {C Fn 3 } P (3) gq {C Fn 3 } P(3) qg {C An 3, C Fn 3 } P(3) gg {C An 3, C Fn 3 } Guess bases using lower order inormation. Number o coeicients: P (3) qq {69} P (3) gq {38} P(3) qg {125, 101} P(3) gg {34, 54} Moments used or reconstruction, (check), N = 2,... P (3) qq {30(44)} P (3) gq {18(28)} P(3) qg { ( ), 40(54)} P(3) gg {20(28), 26(32)} 12/16

14 13/16 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS HARDEST SINGLET CASE P (3) : Basis with 125 unknown integer coeicients. qg CAn 3 N = 2,..., 46 insuicient to determine a good solution. Moment calculations become very computationally demanding. Hardest diagram computed at N = 46, 2 weeks wall-time 13TB peak disk usage by TFORM no more moments! [16 cores, 192GB RAM] We need to somehow make the basis smaller. Use additional constraints: large-x limit: no irrational constants other than ζ i #S 1,2 = #S 2,1-1 coe. -7 coe. 117 unknowns. Solution with N = 2,..., 44, N = 46 checks.

15 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS NON-SINGLET SPLITTING FUNCTIONS n 3 terms o P ns (3),± are already known. [Gracey 94] We determine the n 2 terms o P ns (3),+ (even N) and P ns (3), (odd N). Colour actors to determine at n 2: CF 2n 2 C A C F n 2 diagrams are very hard to compute! Method: decompose in two ways, P (3),± ns {n 2 {C 2 F, C AC F }} = n 2 = n 2 ( 2C 2 F A + C F(C A 2C F )B ±) ( 2C 2 F (A B± ) + C F C A B ±). A should be common to both P ± ns ; use both odd and even N. Large n c. Compute (easier) C 2 F n 2 diagrams to higher N to determine (A B ± ). From these, determine B + and B and hence P (3),+ ns and P ns (3),. 14/16

16 15/16 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS VERIFICATION Check against existing results: Linear combinations o n 3 terms o P (3) qq, P(3) gq, and P(3) gq, P(3) gg [Gracey 96, 98] Large-N prediction o P qq (3), P(3) gg Small-x Double Log Resummations [Dokshitzer,Marchesini,Salam 06] [Davies,Kom,Vogt] Large-x Double Log Resummations [Soar,Moch,Vermaseren,Vogt 10] Cusp Anomalous Dimension at a 4 s : Given by A in large-n limit [Henn,Smirnov,Smirnov,Steinhauser 16] [Grozin 16] Everything is in agreement.

17 INTRODUCTION SPLITTING FUNCTIONS LLL RECONSTRUCTION CONCLUSIONS OUTLOOK Using FORCER, we determine moments o 4-loop Splitting Functions. We have used these moments to derive analytic all-n expressions or n 3 terms o P qq (3), P(3) qg, P(3) gq, P(3) gg n 2 terms o P (3),± ns and P (3) V. 0.8 γ (3)± ns (N) Using the OPE, n 1 and n 0 terms o A. large-n c P ns (3),± complete. To come: [Moch, RUVV, to appear] Numerical approx. to (rest o), using 8 moments. P (3),± ns Suitable or N 3 LO analysis, at least at x n = 3 n = 4 large n c points: ± at even/odd N N 16/16

18 1/2 BACKUP: NON-SINGLET SPLITTING FUNCTIONS To determine the basis coeicients, A: basis with 54 unknown coeicients reconstruct with N = 2, 3,..., 17. N = 18, 19,..., 22 check. (A B + ) and (A B ) are harder: bases with 139 unknown coeicients additional constraints reduce to 115, like P (3) qg approach reconstruct (A B + ) with N = 2,..., 40, N = 42 checks reconstruct (A B ) with N = 3,..., 37, N = 39 checks.

19 BACKUP: SIMPLE LLL EXAMPLE Suppose r = is a (rounded) solution to a quadratic equation with integer coeicients. Form the matrix r r A new basis consists o vectors o the orm (a, b, c, 10000(ar 2 + br + c)). Apply LatticeReduce[] (Mathematica): x 2 + x + 1 = 0 = x = (6 s..) 7x x 48 = 0 = x = (6 s..) 11x x 78 = 0 = x = (6 s..). 2/2