Sector Decomposition

Transcription

1 Sector Decomposition J. Carter Institute for Particle Physics Phenomenology University of Durham Student Seminar, 06/05/2009

2 Outline 1 What is Sector Decomposition? Why is Sector Decomposition Important? 2 3 Linear Divergences Phase Space t t Production

3 Outline What is Sector Decomposition? Why is Sector Decomposition Important? 1 What is Sector Decomposition? Why is Sector Decomposition Important? 2 3 Linear Divergences Phase Space t t Production

4 What is Sector Decomposition? Why is Sector Decomposition Important? A method of evaluating parameter integrals that occur in perturbative QFT Can be used to calculate virtual and real corrections to processes at higher orders

5 Outline What is Sector Decomposition? Why is Sector Decomposition Important? 1 What is Sector Decomposition? Why is Sector Decomposition Important? 2 3 Linear Divergences Phase Space t t Production

6 What is Sector Decomposition? Why is Sector Decomposition Important? Current experimental accuracy 1% Future precision experiments will require theoretical predictions at 0.1% Computation of higher order corrections is vital to achieve this level of accuracy

7 µ Dependence (I) What is Sector Decomposition? Why is Sector Decomposition Important? Evaluation of these high order corrections are formally infinite, so we use dimensional regularisation (D = 4 2ɛ) to describe these infinities. This introduces an energy scale µ R Processes with partonic initial states are factorized so that above a certain energy scale µ F, partonic interactions (gg, qg, q q...) are treated separately from the parton distribution function These µ R, µ F are put in by hand, and thus the true result should have no µ dependence (conventionally µ R = µ F = µ)

8 µ Dependence (II) What is Sector Decomposition? Why is Sector Decomposition Important? Anastasiou, Dixon, Melnikov and Petriello, hep-ph/

9 µ Dependence (III) What is Sector Decomposition? Why is Sector Decomposition Important? Anastasiou, Dixon, Melnikov and Petriello, hep-ph/

10 Outline 1 What is Sector Decomposition? Why is Sector Decomposition Important? 2 3 Linear Divergences Phase Space t t Production

11 Getting From This... Figure: A 9,1 Massless Three-Loop Form Factor

12 To This A 9,1 = iγ(3 + 3ɛ)( q 2 iη) 3 3ɛ ( /ɛ /ɛ /ɛ /ɛ /ɛ O(ɛ))

13 Outline 1 What is Sector Decomposition? Why is Sector Decomposition Important? 2 3 Linear Divergences Phase Space t t Production

14 Feynman Parameters Write down amplitude using Feynman rules Use Feynman parameters and integrate over loop momenta What is left is I = 1 0 ( N j=1 dx j)δ(1 N i=1 x i) U(x)a+bɛ F (x) c+dɛ U is a function of x, and F is a function of x and external invariants (s,m 2,...), and have zeroes when all or some of x i 0

15 Example As an example, I will consider the massless 1-loop box

16 Example N = 4 I d D k 1 k 2 (k+p 1 ) 2 (k+p 1 +p 3 ) 2 (k p 2 ) 2 d D k 1 0 d 4 x 1 0 d 4 x δ(1 P 4 δ(1 P 4 i=1 x i ) (k 2 (x 1 +x 2 +x 3 +x 4 )+2(x 2 p 1 +x 3 p 1 x 4 p 2 +x 3 p 3 ) k+2x 3 p 1 p 3 ) 4 i=1 x i )(x 1 +x 2 +x 3 +x 4 ) 2ɛ ( s 12 x 1 x 3 s 13 x 2 x 4 ) 2+ɛ

17 Primary Decomposition Split I = N k=1 I k, where I k is restricted to the region x k > x i i Rescaling, Relabelling and integrating out δ function wrt x k gives I k = 1 0 ( N 1 j=1 dt j)ũ(t)a+bɛ F (t) c+dɛ Ũ and F typically still have zeroes as some subset of t i 0

18 Example Consider I 4 : x 1 = x 4 t 1, x 2 = x 4 t 2, x 3 = x 4 t 3, x 4 = x 4 I 4 = 1 0 d 3 (1+t t 1 +t 2 +t 3 ) 2ɛ 1 ( s 12 t 1 t 3 s 13 t 2 ) 2+ɛ 0 dx 4 δ(1 x 4(1+t 1 +t 2 +t 3 )) I 4 = 1 0 d 3 t (1+t 1 +t 2 +t 3 ) 2ɛ ( s 12 t 1 t 3 s 13 t 2 ) 2+ɛ x 4

19 Iterated Decomposition Search for a subset of the {t i1,...t ip } such that at least one of Ũ, F 0 as {t i1,...t ip } 0 If no such subset exists then the iteration terminates Else I k = p q=1 I k,q, where I k,q has t i,q > t i,r r Rescale {t i1,...t ip } and factor t i,q out of Ũ, F where possible. Repeat for each new subsector created.

20 Example (1+t 1 +t 2 +t 3 ) 2ɛ Consider I 4 = 1 0 d 3 t : ( s 12 t 1 t 3 s 13 t 2 ) 2+ɛ Numerator is already finite as t 0. Denominator 0 as t 1 and t 2 both 0 Consider I 4,2 (ie t 2 > t 1 ): t 1 = t 2 t 1, t 2 = t 2 I 4,2 = 1 0 dt 1 dt 2 dt 3t 2 1 ɛ (1+t 1 t 2 +t 2 +t 3) 2ɛ ( s 12 t 1 t 3 s 13 ) 2+ɛ I 4,2 = 1 0 t 1 ɛ 2 d 3 t (1+t 1t 2 +t 2 +t 3 ) 2ɛ ( s 12 t 1 t 3 s 13 ) 2+ɛ

21 Subtraction (I) After the iteration terminates and the subsectors are relabelled we have I = #subsectors m=1 I m Each I m is of the form 1 0 ( N 1 j=1 dt j te j +f j ɛ j Ũ and F are O(1) at t 0, so rewrite Ũ(t) a+bɛ g(t, ɛ) = O(1) +... F (t) c+dɛ ) Ũ(t)a+bɛ F (t) c+dɛ

22 Subtraction (II) All the singularities are contained in the N 1 j=1 dt j te j +f j ɛ j If e j > 1 then there is no singularity in t j If e j = 1, subtraction is needed Write g(t, ɛ) g(t j = 0, ɛ) + (g(t, ɛ) g(t j = 0, ɛ)) 1 0 t 1+f ɛ g(0, ɛ)dt = g(0,ɛ) 1 0 dt 1 0 t 1+f ɛ (g(t, ɛ) g(0, ɛ)) = O(1) f ɛ If e j <= 2 then the procedure still works, but with more terms of the Taylor expansion included

23 Example Consider I 4,2 = 1 0 t 1 ɛ 2 d 3 t (1+t 1t 2 +t 2 +t 3 ) 2ɛ ( s 12 t 1 t 3 s 13 ) 2+ɛ (1+t 3 ) 2ɛ I 4,2 = 1 0 dt 1 1dt 3 ( ( s 12 t 1 t 3 s 13 ) 2+ɛ 0 dt 2(t 1 ɛ 2 ) dt 2(t 1 ɛ 2 ( (1+t 1t 2 +t 2 +t 3 ) 2ɛ (1+t 3 ) 2ɛ ))) ( s 12 t 1 t 3 s 13 ) 2+ɛ ( s 12 t 1 t 3 s 13 ) 2+ɛ (1+t 3 ) 2ɛ ) ( s 12 t 1 t 3 s 13 ) 2+ɛ = 1 0 d 3 t(( 1 ɛ + t 1 ɛ 2 ( (1+t 1t 2 +t 2 +t 3 ) 2ɛ ( s 12 t 1 t 3 s 13 ) 2+ɛ (1+t 3 ) 2ɛ ( s 12 t 1 t 3 s 13 ) 2+ɛ ))

24 Numerical Integration I(ɛ) = m I m(ɛ) Perform the Laurent Expansion in ɛ For each order of ɛ the coefficient is a sum of well-behaved integrals over the N 1 dimensional unit hypercube, each of which can be calculated via Monte Carlo integration to yield the full result

25 Example For ease of notation I shall set s 12 = s 13 = 1 (1+t 3 ) 2ɛ ) (1+t 1 t 3 ) 2+ɛ 2 ( (1+t 1t 2 +t 2 +t 3 ) 2ɛ (1+t 3) 2ɛ (1+t 1 t 3 ) 2+ɛ I 4,2 = 1 0 d 3 t(( 1 ɛ + t 1 ɛ I 4,2 = 1 ɛ = log(2) ɛ )) (1+t 1 t 3 ) 2+ɛ 1 0 d 3 1 t + 1 (1+t 1 t 3 ) 2 0 d 3 t 2log(1+t 3) log(1+t 1 t 3 ) + O(ɛ) (1+t 1 t 3 ) 2 + π2 +6log(2) 2 3log(16) 12 + O(ɛ) Full numerical result is 4 4 ɛ 2 ɛ O(ɛ)

26 Outline Linear Divergences Phase Space t t Production 1 What is Sector Decomposition? Why is Sector Decomposition Important? 2 3 Linear Divergences Phase Space t t Production

27 Linear Divergences Phase Space t t Production Complicated loop diagrams yield a lot of variables with t 2+f ɛ Figure: Four-Point Three-Loop Diagram

28 Linear Divergences Phase Space t t Production These divergences rapidly increase computation time for the subtraction, and in many cases the numerical integration becomes unworkable, as eg. 1 log(1+t) 1 1+t t as t 0 but this behaviour is not seen by the numerical integration Taylor Expansions in these variables provide one way around the problem, but this vastly increases both the time and memory required to complete the calculation. For more than 2 of these poles, this method is prohibitively expensive Test new methods to overcome this problem

29 Outline Linear Divergences Phase Space t t Production 1 What is Sector Decomposition? Why is Sector Decomposition Important? 2 3 Linear Divergences Phase Space t t Production

30 Linear Divergences Phase Space t t Production Apply the method to real unresolved radiation Divergences can come from soft/collinear massless particles Figure: γγ q qg

31 Outline Linear Divergences Phase Space t t Production 1 What is Sector Decomposition? Why is Sector Decomposition Important? 2 3 Linear Divergences Phase Space t t Production

32 Linear Divergences Phase Space t t Production We aim to produce the full NNLO cross-section for t t production at the LHC, including two-loop, 1-loop real radiation and double real radiation

33 Further Reading Linear Divergences Phase Space t t Production For an indepth explaination of the method: Sector Decomposition Gudrun Heinrich Arxiv: