Two-loop corrections to t t production in the gluon channel. Matteo Capozi Sapienza University of Rome

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1 Two-loop corrections to t t production in the gluon channel Matteo Capozi Sapienza University of Rome November 7, 2016

2 Top quark With a mass of m t = ± 0.13(stat) ± 0.47(syst)GeV, the top quark is the heaviest elementary particle produced so far at colliders. The top quark is produced (at LHC) via two mechanisms: pp t t pp t b, tq ( q ), tw The top quark plays a double role: signal or background for new physics. Top quark does not hadronize, since it decays in about s (one order of magnitude smaller than the hadronization time) almost exclusively in bottom quark and W boson b-tagging to identify top events.

3 Pair production: total cross section pp t t Total cross section computed at NLO by Nason et al. in 1988 (Nucl.Phys. B303 (1988) ). Total cross section computed at NNLO+NNLL by Czakon et al. in 2013 (arxiv: ). At LHC using m t = 173.3GeV s σt t [pb] scales [pb] pdf [pb] 7 TeV TeV TeV Analytical cross action?

4 According to the Factorization theorem we can write the production cross section of the process h 1 + h 2 Q Q + X in the following way: σ h1,h 2 = i,j 1 1 dx 1 dx 2f h1,i(x 1, µ F )f h2,j(x 2, µ F ) ˆσ i,j (ŝ, m t, α s(µ R ), µ F, µ R ) 0 0 s = (p h1 + p h2 ) 2, ŝ = x 1x 2s ˆσ Q, Q M Q Q ij = M Q Q ij,0 + αsmq Q ij,1 + α2 s M Q Q ij,2 2 at LO we have two different virtual channels: q q t t gg t t

5 q q t t corrections at two loops M 2 (s, t, m, ɛ) = 4π2 α 2 s N c [A 0 + ( αs π )A 1 + ( αs π )2 A 2 + O(αs 3 )] A 2 = A 2x0 2 + A 1x1 2 A 2x0 2 = N c C F [Nc 2 A+B + C Nc 2 +N l (N c D l + E l )+N h (N c D h + E h ) N c N c + Nl 2 F l + N l N h F lh + Nh 2 F h 218 two-loop diagrams contributing to 10 different color coefficients D i, E i, F i and A known analytically B and C can be calculated with the same techniques

6 gg t t corrections at two loops M 2 (s, t, m, ɛ) = 4π2 α 2 s N c [A 0 + ( αs π )A 1 + ( αs π )2 A 2 + O(αs 3 )] A 2 = A 2x0 2 + A 1x1 2 A 2x0 2 = (Nc 2 1)[Nc 3 A + N c B + 1 C + 1 N c Nc 3 D + Nc 2 N l E l + N 2 c N h E h + N l F l + N h F h + N l N 2 c G l + N h Nc 2 G h + N c Nl 2 H l + N c Nh 2 H h+ N c N l N h H lh + N2 l N c I l + N2 h N c I h + N ln h N c I lh ] 789 two-loop diagrams contributing to 16 different color coefficients Leading color A, and light-quark E l I l coefficients are known analytically We focus on E h, F h and G h

7 Loop integrals evalutation A generical loop integral has the form: I = d D k 1d D k 2 {p i k j,k 1 k 2 } D 1...D 7 Using the Integration-by-Parts Identities (IBPs) all the dimensionally regularized scalar integrals are expressed as a combination of the Master Integrals (independent scalar integrals). In order to calculate the MIs we use the Differential Equations Method. The MIs satify : xf (x, ɛ) = A(x, ɛ)f (x, ɛ) where f (x, ɛ) is a n-vector containing all MIs and A(x, ɛ) a nxn matrix. Solve the linear DEs is simpler than compute the integrals. In general we look for a solution of the system in Laurent series of ɛ = 4 d ; the coefficients of the series are functions of the kinematic 2 invariants.

8 Decoupled case At each order of the Laurent expansions around ɛ = 0: the differential equations for a set of integrals f can be written as: xf i = c 0(x)f i,0 + inhomogneous terms The problem is solved once we solve the homogeneous DE c c 0(x) = c 21 c 22 0 c 31 c 32 c 33 One has to solve only first order DEs. We can define a basis of MIs (the canonical basis ) such that the system assumes a simple form (Henn arxiv: ): xf (x, ɛ) = c(x, ɛ)f (x, ɛ) xf (x, ɛ) = ɛc(x)f (x, ɛ) We solve in power series around ɛ = 0 f (x, ɛ) = i=0 f i(x)ɛ i f i (x) x = ɛa(x)f i 1 (x) The solution is very simple: f i (x) = x A(x )f i 1 (x )dx Decoupled MIs are present both in q q and gg channels

9 Coupled case At each order of the Laurent expansions around ɛ = 0: xf i = d 0(x)f i,0 + inhomogneous terms d 0(x) = d 11 d 12 d 13 0 d 22 d 23 0 d 32 d 33 It is not possible to choose a basis to triangulize the matrix and decoupling the equations we have to solve a 2 X 2 system of ODE xh1(x) = d11(x)h1x + d12h2(x) { xh 2(x) = d 21(x)h 1x + d 22h 2(x) x 2 h 1(x) = p(x) xh 1(x) + q(x)h 1(x) Coupled MIs are present only in gg channel

10 gg t t We focus on the contributions to gg t t coming from Feynman diagrams with at least 1 closed massive fermionic loop. 256 diagrams with fermionic loops 117 diagrams with at least one top loop evaluate their Feynman integrals to determine E h, F h and G h

11 Decoupled subtopologies Coupled subtopologies We study the coupled subtopologies

12 Sunrise Two coupled MIs Homogenous differential equation solved in d=2 by Weinzierl et al. (arxiv: ): Sun 0(d = 2, t) = K(m(t)) 4 (t 9)(t 1) 3 Complete solution in d=2 computed by Euler s variation of costant Transport of the solution in d=4 by Taraasov dimensional relations

13 Sun 1(t) = Sun 1, 2(t) (d 4) 2 Solution in four dimension + Sun 1, 1(t) (d 4) + Sun 1,0(t) + Sun 1,1(t)(d 4) + O((d 4) 2 ) S 1, 2 = 3 8 Sun 1, 1(t) = 18 t 32 Sun 1,0(t) = C 0,Ec (t)e(m c (t)) 4 + C 0,Kc(t)K(m c (t)) 4 + E(m(t))C 0,E (t) 4 + K(m(t))C 0,K (t) 4 + C 0,0(t) S 1,1(t) = C 1,Ec (t)e(k c (t) 2 ) 4 + C 1,Kc(t)K(k c (t) 2 ) 4 + K(k(t)2 )C 1,K (t) 4 + C 1,e (t)e(k(t) 2 ) 4 + C 1,0(t) C 0,i dt a t5 +b t C 1,i dt u t 5 +v t d(t ) K(m(t)), K(m(t )) dt a t5 +b t K(m(t)), t2 6t+ (t 9)(t 1) 3 3 = (t 9)(t 1) 3, m(t) = (t 9)(t 1) 3 2, m c (t) = 1 m(t) 2 Sun 2(t) = d Sun 1(t)+tSun 1 (t)+3sun 1(t) 3

14 Vertex V (t) = V (t) 1 t + Sun(t) 1 t At each order of the Laurent expansion: V 2(t) = C i t0 Sun 2, 2 (x) dx 1 t 1 t V 1(t) = C i t0 Sun 2, 1 (x) dx 1 t 1 t Polinomial V 0(t) = C i t0 Sun 2,0 (x) dx 1 t 1 t V 1(t) = C i t0 Sun 2,1 (x) dx 1 t 1 t Integration over elliptic kernel Non-elliptic homogenous solution, elliptic inhomogeneous solution

15 Crossed vertex Two coupled differential equations: C 1 (s) = 2 s C 1(s) + 4 s C 2(s) + Ω 1 C 2 (s) = 1 s(s+16) C 1(s) 2(s+8) s(s+16) C 2(s) + Ω 2 Ω 1, Ω 2 So we obtain an homogenous second order differential equation for the scalar integral: C 1,0 (s) = s(5s+64)c 1,0 (s)+4(s+9)c 1,0 (s) s 2 (s+16) One can prove that the the homogenous solution for the scalar is: C 1,0 (s) = K(m(t)) s 3/2 a K(mc (t)) 1 + s 3/2 a 2 s m(t) = i 4, mc (t) = 1 m(t) 2 Particular solution: C 1,p (s) s ds 2K(m(t)) 0 W (s) πs 3/2 Ω i (s ) +... Integration over the product between GPLs and C 2,0 (s) 1 4 elliptic integrals. ( ) sc 1,0 (s) + 2C 1,0(s)

16 Box Tree MIs, two MIs homogenous equations coupled. We have two homogenous second order differential equation in s and t: ( ( ) ) ) 2 Box 0,1 2 3s 4 t 5t 2 42t 27 2s(t 1) 2 24t 3 137t t (t 1) 4 t s 2 = [ (s 4)s 2 (s(t 9) 4t + 4) ( (s 2)t + t ) + (s(5t + 3) + 4(t 1)t) ( ) ) 2s 3 13t 4 162t t t 27 + s 2 12t 5 247t t 3 762t t + 57 (s 4)s 2 (s(t 9) 4t + 4) ( (s 2)t + t ) (s(5t + 3) + 4(t 1)t) ]Box 1,0 (s, t)+ s[ ( ) ) 6s 4 t 5t 2 42t s(t 1) 2 9t 3 76t t (t 1) 4 t (s 4)s 2 (s(t 9) 4t + 4) ( (s 2)t + t ) (s(5t + 3) + 4(t 1)t) + ( ) ) s 3 53t t t 2 552t s 2 6t 5 151t t 3 606t 2 34t + 93 (s 4)s 2 (s(t 9) 4t + 4) ( (s 2)t + t ) ] Box 0,1 (s(5t + 3) + 4(t 1)t) s

17 ( ) ( ) 2 Box 0,1 2s 4 t 3 + 5t 2 33t s(t 1) 2 7t 3 27t 2 183t (t 1) 4 (t + 3) t 2 = [ (t 1) 2 (s(t 9) 4t + 4) ( (s 2)t + t ) + (s(5t + 3) + 4(t 1)t) ( ) ( ) s 3 4t 4 + 5t t 2 567t s 2 2t t t 3 838t t 609 (t 1) 2 (s(t 9) 4t + 4) ( (s 2)t + t ) ]Box 1,0 (s, t)+ (s(5t + 3) + 4(t 1)t) ( ) ( ) s 4 t 2 + 6t 27 8s(t 1) 2 t 2 2t (t 1) 3 (t + 3) s[ (t 1) 2 (s(t 9) 4t + 4) ( (s 2)t + t ) (s(5t + 3) + 4(t 1)t) ( ) ( ) s 3 2t 3 t 2 116t s 2 t 4 16t 3 62t t (t 1) 2 (s(t 9) 4t + 4) ( (s 2)t + t ) (s(5t + 3) + 4(t 1)t) ] Box 0,1 t We are looking to find an ellptic homogenous solution How can one solve the homogenous equations? Two possible ways: By computation of the maximal cut (A. Primo and L. Tancredi arxiv: ) By parametrization of the MI variables (Bonciani et al. arxiv: )

18 Conclusion and outlook We want to determine the total analytical cross section for the pair production Evalutation of color coefficients E h, F h, G h Coupled subtopologies Solutions in terms of elliptic integral