Higgs production through gluon fusion at leading order

Transcription

1 NIKHEF 5-7 Higgs production through gluon fusion at leading order Stan Bentvelsen, Eric Laenen, Patrick Motylinski Abstract The dominant Higgs production mechanism at the LHC involves gluon fusion via an intermediate top-quark loop. The lowest order theoretical cross section is wellknown and used in many LHC studies to determine the experimental discovery sensitivity of the Higgs particle. Although the expression for this cross section can readily be found in the literature, the derivation of this result is usually not given and is not so easy to calculate. In this note we will go step-by-step through the calculation.

2 Introduction In this report we calculate in detail the cross section for the Higgs production process gg H. () at lowest order, assuming the Standard Model. This lowest order process proceeds through a top quark loop between the gluons and the Higgs particle, see Fig.. Although in principle all quarks should be included in the loop, in practice the restriction to just the top quark suffices because the Higgs couples about 35 times more strongly to the top than to the next-heaviest fermion, the bottom quark, leading to a relative suppression of the bottom contribution by a factor 35. The calculation of the matrix element in lowest order should be finite, even though the diagrams feature a loop, which is a common source of infinities. This must be so, because there is no fundamental ggh coupling in the Standard Model which could absorb such infinities. We shall indeed see that the diagrams give a finite answer. This report is organized as follows. We quote first the results of the calculations for the process in the next section, so that it is clear what the final result of the calculation should be. In section 3 we then list the diagrams and Feynman rules for the process, and in 4 we construct the matrix element. In section 5 we reduce the resulting loop integrals to the simplest (scalar) form, and in sections 6 and 8 we calculate these explicitly. In section 7 we put it all together and compute the cross section. For some of the cumbersome calculations we make use of the algebraic manipulation program form []; the relevant code is reproduced in the appendix. Results Before starting the calculation in detail, we first write down the expression for the gluon fusion process in lowest order. It reads [, 3]: ˆσ LO = σ M H δ ( ŝ M H), () σ = G F α s 88 π A(τ), (3) A(τ) = 3 τ [ + ( τ)f(τ)], (4) arcsin / τ τ [ f(τ) = ln + ] τ τ iπ τ <, (5) with M H the Higgs mass, G F Fermi s constant, α s the strong coupling constant defined at some as yet unspecified (see section 8) scale and with τ defined as τ = 4m t M H. (6) For the full cross section for proton-proton collisions the gluon cross section must be folded with the density of gluons in the proton σ (pp H) = τ dx τ/x dx g(x, M H)g(x, M H) ˆσ LO,gg H (ŝ = x x S) (7)

3 where x and x are the momentum fractions of the gluons. The delta-function in the leading order cross section σ LO () ensures that the Higgs particle is on-shell and that the Higgs mass is related to the proton-proton centre-of-mass S as: ŝ = x x S = M H. (8) In the heavy top mass limit, which corresponds to τ, the function ( A(τ) reduces to the constant. This can be seen by expanding the function arcsin / ) (τ) : arcsin / ( ) τ = + τ 3! ( τ) 3 + O ( τ) 5. (9) Inserted in A(τ) in EQ. (4) this yields to order /τ: A(τ) = 3 [ τ + ( τ) + ] ( ) + O = + O 3τ τ ( ). () τ The limit τ corresponds to the heavy Higgs mass limit, and the function A(τ) becomes in this case: A(τ) = 3 [ ln τ ] iπ () 3 Feynman rules and diagrams In lowest order there are two Feynman diagrams contributing to the process gg H, drawn in Fig. 3. The charge flow is indicated by the arrow on the fermion lines, and the direction of the momentum flow is indicated by the accompanying arrow. We use the momentum definitions as given in the Figure. Because gluons are massless, the following ν ν k l k k l + k l l k l + k q = k + k k l k q = k + k µ µ Figure : The two contributing Feynman diagrams with the notation as used in the text. The charge flow is denoted by the arrows on the spinor fields. The momentum flow is indicated by the separate arrows. kinematic relations hold: k = () k = (3) (k + k ) q = ŝ = M H. (4) 3

4 We use the indices i, j to label the color of the quarks in the fundamental SU(3) representation. The color of the gluons in the SU(3) adjoint representation are denoted by the roman indices a, b. The spinor indices are α and β, whereas the indices µ, ν denote the Lorentz group indices for the gluons. We need the Feynman rules to write down the matrix element of the two diagrams. The vertices of the top-quark with the gluon and the top-quark with the Higgs particle are given in the standard model by i β a, µ j α ig s γ µ βα [ta ] ji i β j α iyt δ ij δ αβ with the top Yukawa coupling, y t /, given by the Higgs vacuum expectation value v as y t / = m t /v 75/46. The propagator of the top-quark is given by i α p j β i( p + m) βα δ ij p m + iɛ To construct the matrix element one should follow the lines against the direction of the arrows indicating the charge flow. 4 Matrix element It is straightforward to write down the matrix element M for the two diagrams. We collect the couplings, the imaginary factors i of the propagators, the color factors and the overall minus-sign for the fermion loop in front. The polarization of the external gluons is given by ɛ µ (λ, k) in spin state λ and with momentum k. The momentum in the loop is l, and the loop-integral is regularized by working in d = 4 ε dimensions: ( M = ( ig s ) i y ) t i 3 Tr [t a t b ] ( ) ɛ ν (λ, k )ɛ µ (λ, k ) (5) d d l Tr [( l+ k (π) d + m)γ µ ( l + m)γ ν ( l k + m)+ D D D 3 ( l+ k + m)γ ν ( l + m)γ µ ( l k + m)]. In this expression the contributions from the two diagrams are added. Note that for the crossed diagram the Lorentz-labels µ and ν are interchanged and the loop momentum l reversed sign. The signs of the momenta are chosen such that the propagator terms are identical for the two diagrams. The trace of the spinor indices is taken because the quark lines form a loop. The factor ( ) in the first line is inserted because of the fermion loop. The denominators of the propagators are D = l m (6) D = (l k ) m D 3 = (l + k ) m 4

5 The goal is now to calculate the matrix element M in Eq. (5) by performing the trace and calculate the integral over the loop momentum l. The trace has to be done in d dimensions and can be done with the algebraic manipulation program form []. The relevant form program, together with its output, is listed in the appendix. The result is : Tr [( l+ k + m)γ µ ( l + m)γ ν ( l k + m)+ (7) ( l+ k + m)γ ν ( l + m)γ µ ( l k + m)] = 8m [ k µ kν kν kµ + kµ lν k ν lµ + 4l µ l ν g µν k k g µν l l + g µν m ] This expression still needs to be integrated over the loop integral l, and this will lead to the main complexity of the Higgs production calculation. As it turns out, the terms linear to the loop integral l µ cancel in the final result and only the terms proportional to l µ l ν and g µν l l remain in the calculation. We will be doing them by reducing the integrals to simpler expressions, using the Passarino-Veltman reduction method. We describe this method for this case in the next section. The cross section is then obtained from standard techniques, i.e. square the matrix element, sum over final state spins, average over initial state spins, insert the correct flux factor and insert the one particle phase space. This we will do in section 7. 5 Reduction of loop integrals When inserting Eq. (7) into Eq. (5), the main calculatioal difficulty originates from the tensor integral over the loop momentum d d l l µ l ν C µν =, (8) (π) d D D D 3 with the propagator terms defined in equation (6). Before proceeding to calculate this integral, we shall introduce some common notation: the letter C (as the third letter of the alphabet) is used for this integral because it contains three propagators D, D and D 3. The C µν integral has a tensor Lorentz-structure. Likewise we introduce the vector C µ and the scalar C integrals as follows C µ = C = d d l l µ (9) (π) d D D D 3 d d l. (π) d D D D 3 Similarly we can introduce the B integrals. They contain only two propagators and are defined as: B µ (, ) = d d l l µ (π) d D D B µ (, 3) = B (, ) = d d l (π) d D D d d l l µ (π) d D D 3 B µ (, 3) = B (, 3) = d d l (π) d D D 3 d d l l µ (π) d D D 3 B (, 3) = () d d l (π) d D D 3 We will show that the tensor C µν can be written in terms of the scalar functions C and B. The scalar functions C and B can be then be calculated analytically using the Feynman trick (see later), and thus we obtain the expression for C µν. Note that for this case the dimension d does not appear in the answer. 5

6 In order to show that the tensor C µν can be written in terms of the scalar functions C and B we first write the Lorentz-structure of the tensor C µν explicitly in most general terms, using the available vectors k, k and the metric g µν, noting that C µν is symmetric under interchange of the indices µ and ν: C µν = k,µ k,ν C + k,µ k,ν C + {k, k } µν C 3 + g µν C 4 () with the symmetry operator {k, k } µν k,µ k,ν + k,ν k,µ. Contracting the tensor C µν with the momenta k gives C µν k ν s = k,µ C s + k,µ C 3 + k,µ C 4 () C µν k ν s = k,µ C s + k,µ C 3 + k,µ C 4 It is customary to define the projection operators P µ k and P µ k. When operated on the vectors k µ i, they are required to behave as A representation of these operators is P µ k i k j,µ = δ ij. (3) P µ k = s kµ (4) P µ k = s kµ Using these projection operators we can map the individual coefficients for the expansion of C µν as defined in equation (). It is convenient to define the sets of scalar functions R i as follows R 3 P µ k C µν k ν = s C 3 + C 4 (5) R 4 P µ k C µν k ν = s C R 5 P µ k C µν k ν = s C R 6 P µ k C µν k ν = s C 3 + C 4 after which we will express these functions R i in terms of reduced integrals. We now start to rewrite the expression of C µν () in terms of the B integrals (which have two denominators only). We do this by contracting C µν with the vecors k and k. First note that k l = (D D k ) (6) k l = (D 3 D k ) where we leave the term k and k in for later use. It means that when contracted with k µ i, i =,, one of the propagators of C µν cancels in the denominator and turns into a B function (note that here we can put the gluon on mass-shell, i.e. k = ): C µν k ν = (B µ(, 3) B µ (, 3)) (7) C µν k ν = (B µ(, ) B µ (, 3)) 6

7 Here we have the expression for the integral C µν k ν with three denominators reduced to integrals B µ with two denominators. We can perform the same expansion of the function B µ (i, j) in terms of its Lorentzstructure as we did for C µν in equation (). We will first do this for the B µ (, ) and B µ (, 3) functions, as they are the easiest. The only terms available for them are B µ (, ) = k,µ B (, ) (8) B µ (, 3) = k,µ B (, 3). The function B (i, j) can be written in terms of the scalar integrals of equation (), by applying the contraction of equation (4) again, such that k µ B µ (, ) = k B (, ) (9) = [ ( d d l ) ] d d l k (π) d D D (π) d D D the first integral in this expression is zero, which can be seen by applying a suitable shift in momentum l. We are left, after dividing by k on both sides, with B (, ) = B (, ) and similarly (3) B (, 3) = B (, 3) Now we turn to the expansion of B µ (, 3), which is a little more involved. writing our the integral gives d d l l µ B µ (, 3) = (π) d [(l k ) m ] [(l + k ) m ] Explicitly and we apply a shift to the momentum l, l = l + k, and integrate over l, such that we get two terms B µ (, 3) = B µ (, 3) + k,µb (, 3) (3) The first term has now the equivalent structure of B µ (, ) and B µ (, 3) with momenta k = k + k and becomes: ( B µ(, 3) = (k,µ + k,µ ) ) B (, 3) (33) such that the answer for B µ (, 3) becomes: (3) B µ (, 3) = (k,µ k,µ )B (, 3) (34) At this stage we are ready and have expressed the vector integrals B µ (i, j) in terms of scalar integrals B (i, j). We can use this result in equation (7), in order to write the contraction of C µν k ν in terms of scalar integrals B (i, j), as follows: C µν k ν = 4 k,µb (, 3) + 4 (k,µ k,ν )B (, 3) (35) C µν k ν = 4 k,µb (, ) 4 (k,µ k,ν )B (, 3). Note that in this case we keep track of the term k, which we have set to zero earlier in equation (6). We need this small off-shellness to avoid dividing by zero in the determination of the coefficient B, where k cancels in the result. For the C functions this problem does not appear. 7

8 We now use the projection operators (3) to obtain the functions R i, as defined in equations (5), in terms of the scalar integrals B : R 3 = 4 B (, 3) (36) R 4 = 4 B (, 3) + 4 B (, ) R 5 = 4 B (, 3) + 4 B (, 3) R 6 = 4 B (, 3). We are now ready to determine the Lorentz factors of the expansion of C µν as in equation (), using the expression of R i as given in (5): C = s R 5 (37) C = s R 4 C 4 = ( ) B (, 3) + m C R 3 R 6 d C 3 = s (R 6 C 4 ). The expression for C 4 and C 3 are a little less trivial to obtain, since they cannot be obtained from equation (5) by direct inversion. Instead, consider the projection operator P µν defined as: P µν = ( g µν P µ k d k ν P µ ) k k ν. (38) When acted on C µν it indeed selects the coefficient C 4 P µν C µν = d [sc 3 dc 4 s C 3 C 4 s ] C 3 C 4 = C 4, (39) and when acted on the integral expression of C µν gives the result as given above in equation (37), after noting that d g µν l l m d d l C µν = + (π) d D D D 3 (π) d m = B (, 3) + m C. D D D 3 (4) We are now ready to calculate the scalar integrals B and C. We will first calculate B, then compute the cross section in terms of the scalar integral C and keep the calculation of this integral for desert. 6 Calculation of B So far, the result of the loop integral of the matrix element of equation (5), after the Passarino-Veltman reduction, contains the scalar integrals B and C in d = 4 ε dimensions 3. The dependence on the dimension d originates from the expression of C 4, as given in equation (37). ɛ µ. 3 Do not confuse the regulator ε of dimensional regularisation with the polarization vector of the gluons, 8

9 We expand the dependence on the dimension d in terms of ε as /(d ) (/)( + ε + ε +...). The complete integral of the trace, obtained from form, becomes: d d l M = Tr [( l+ k (π) d + m)γ µ ( l k + m)] = D D D 3 (ɛ ɛ ) [ ] 8m(ε + ε )B (, 3) 4mMHC + 6m 3 ( + ε + ε )C [ ] 6m +(k ɛ )(k ɛ ) ( ε ε )B MH (, 3) + 8mC + 3m3 ( ε ε )C MH This result can be written as M = (ɛ ɛ )a + (k ɛ )(k ɛ )b with (4) a = 8m(ε + ε )B (, 3) 4mM HC + 6m 3 ( + ε + ε )C b = a M H Note that the scalar function B only appears in this expression multiplied with the regulator ε or higher orders. As the regulator ε will be set to zero at the end of the calculation, we are only interested in eventual poles in ε of B. Any constant terms or terms proportional to ε in B will vanish. For the scalar integral C we need to keep track of the constant term as well, as it will not vanish for ε. Note that we do not expect poles in B proportional to /ε or higher inverse order, as this would make the cross section unrenormalisable. The same is true for pole terms /ε or higher inverse order for the scalar intergral C. In order to keep the correct dimensions we introduce µ the dimensional regularization (mass) scale µ, and write for B B = µ ε d d l (π) d (l m )((l + k + k ) m ) We shall do this integral using Feynman s trick, i.e. by rewriting a fraction as an integral as follows: AB = dx [xa + ( x)b]. (43) This technique can be extended to higher powers of the fraction: (4) A α A αn n = Γ(α + + α n ) Γ(α ) Γ(α n ) dx dx n δ( x x n ) x α x αn n [x A x n A n ] (α + +α n). (44) In addition we need the integral over d dimensions d d l (π) d [ l M + iɛ ] s = ( ) s Γ ) ( s d Γ(s) i(4π) ε 6π [ M iɛ ] d s (45) Now we have all the tools to calculate the scalar function B. We will use the Feynman trick to cast the scalar function B in the form suitable for equation (44). Then the relevant pole /ε is identified, which we can use in equation (4). 9

10 First apply the Feynman trick to obtain d B = µ ε d l dx (π) d [( x)(l m ) + x(l + k + k ) m ]. (46) Denote k = k + k and rewrite the expression as d B = µ ε d l dx (π) d [l + xl k + xs m ]. (47) We shift the variable l by defining l = l xk such that d B = µ ε d l dx (π) d [l + xs x s m ] (48) (we have dropped the prime from l) and we have cast the integral into a form in which it can be calculated using expression (45), with M = xs x s m. The result for B becomes ε Γ(ε) i(4π) ε B = µ dx [ x s xs + m ] ε (49) Γ() 6π In this expression we need to identify the pole /ε. It occurs at only one place, which is in Γ(ε), as it can be written as: Γ(ε) = ε Γ( + ε) ; Γ( + ε) e εγ E ; γ E.577 (5) For the integral over x we are therefore only interested in the constant term (as any other terms will vanish in the final result of the matrix element). Note that a ε = ε ln a + O(ε ). (5) Hence the whole integral over x can be approximated as. Using this we obtain ε i(4π)ε B = µ = i 6π 6π e εγ E ( ε ε γ E + ln 4π ) µ ε. (5) We are ready to substitute this result 4 in the expression of the matrix element, equation (4), and take the limit ε. The result for the matrix element can be written in equation (4) with the coefficients: a = 8m 6π b = a M H with τ defined in equation (6). ( M H ( τ)6π C ) 4 The combination of terms /ε γ E + ln 4π occurs often in the dimensional renormalisation. In many situations in higher order gauge theories the real and virtual contributions both contain these poles, that have to be subtracted to reach the final finite result. It is common use to not only subtract the pole /ε (minimal subtraction, or MS scheme), but to subtract the γ E and ln 4π terms as well (the MS scheme). (53)

11 7 Calculation of the cross section Our next step is to determine the cross section in terms of the scalar integral C, without calculating this integral explicitly yet. Any cross section can be written in general terms as ˆσ = dp S ŝ spins,color M (54) For our process, the one-particle phase space is just d 4 q dp S = (π) (π)δ +(q M 4 H )(π)4 δ 4 (k + k q) = πδ(ŝ MH ) (55) Now note that averaging over the spins in the initial state gives a factor ( ), averaging over color gives a factor (. 8) Also we use that (g s ) = 6π αs yt and that ( = m t = ) v GF m t. The color factor trace equals Tr[t at b ] = δ ab and hence the color contribution is 4 δ abδ ab =. The cross section becomes ˆσ = α sπ 3 G F 6M H m t MH spins M MH δ(ŝ M H ) (56) where we redefined the matrix element M with all constants, coupling constants etc, left out. M = a(ɛ ɛ ) a(ɛ MH k )(ɛ k ) (57) with ɛ i = ɛ µ (λ i, k i ). We have to sum over the spin states λ, λ and one normally uses the completeness relation for this. Note however that in this case we only sum over the polarization states, the gluon being massless on-shell. One way of implementing this is by introducing the auxillary vector n µ in the completeness relation, with n = : It follows that so that λ i ɛ µ (λ i, k i )ɛ ν (λ i, k i ) = g µν + n µk i,ν + n ν k i,µ n k i P i,µν (58) P µν i k i,ν = P µν i n ν = λ,λ M = a P µν P,µν + b (P µν k,µ k,ν )(P ρσ k,ρ k,σ ) + m HR(a b) = a ( + 4 4) = a (59) where we used the notation for b as in equation (53). The factor a is defined in equation (53) ˆσ = α s GF τ 6π M H ( τ)6π C MH δ(ŝ M H ) (6) We now have the complete expression for the cross section as given at the start in equation (), in terms of the integral C. The calculation of this integral is our remaining task.

12 8 Calculation of C The explicit form of C is d d l C = (π) d (l m )(l k l m )(l + k l m ) We will use the Feynman trick twice to write the integral in the form (use equation (44) for the second line): d d l C = dx (π) d [l xl k m ] (6) [l + k l m ] d d l = dx dy Γ(3) (π) d Γ()Γ() y [ l x( y)k l + yk l m ] 3 again shift the integration parameter l to l = l + K, with K = x( y)k yk (such that K = x( = y)ys) to obtain: d d l Γ(3) C = dx dy (π) d Γ()Γ() y [ l (m + K ) ] 3 (63) Here C is cast in form of equation (45), and we can perform the integral. Since the power in the denominator is 3, we will find the term Γ( + ε) in the answer, which contains no pole in ε. The answer is therefore finite in the limit ε and we can take this limit here straightaway. The result becomes: C = ( ) 6π i y dx dy (64) (m xsy( y)) This integral is tedious but can be done with conventional techniques. First integrate over x: C = i dy 6π ( y)( s) ln ( m sy( y)x ) = i 6π = i 6π s dy ln ( ξy( y)) ( y)( s) (6) dy ln ( ξy( y)) (65) y with ξ = s. In the third line we changed integration variable y y. Note that the m logarithm in the argument becomes imaginary for ξ > 4. The calculation of the integral in (65) now follows along the lines of Chapter 7 of the book by Smith and De Wit. Integration by parts gives I = ξ = = dy ln( ξy( y)) y ( y) ln y dy y( y)ξ ( y) ln y dy (y y + )(y y ) dy( y) (ln y) β ( y y + y y ) (66)

13 where in the second but last line we used the principal value integral ( ) x y iɛ = P x y ± iπδ(x y) (67) with the poles of the denominator given by y ± = ± 4/ξ ± β ; β = We now have to determine the integral for the various cases. 4m s (68) 8. Imaginary part for ξ > 4 Lets start with the imaginary part of I, which only appears for ξ > 4, when the argument of the logarithm becomes negative, ln( x) = ln x + iπ. The result becomes iii = iπ β dyδ(y y + )( y) ln y + iπ β = iπ β [( y ) ln y + ( y + ) ln y + ] = iπ β [β ln y β ln y + ] dyδ(y y )( y) ln y = iπ ln y + y = iπ ln + τ τ (69) where τ = 4m /s. 8. Real part For the real part of the integral I, with ξ > 4, we need the trick to first determine the derivative to ξ: di dξ = y dy (7) y( y)ξ 3

14 We change the integration variable to u, defined as y = (+u) and hence y = ( u). The expression for the derivative becomes 5 di dξ = = βξ P = βξ ln u du 4 ξ + ξu ( du ( ) + β β ( u + β u β )) (7) We now need to integrate over ξ. We change again variables as dξ di dξ dβ dξ di dβ dξ (73) where dξ/dβ = βξ /. For ξ > 4 the integral becomes: ( ) I = dβ βξ + β βξ ln β ( ) ( ) + β d + β = dβ ln β dβ ln β = ( ) + β ln + Cnst. (74) β Whereas for ξ < 4 the integral becomes ( ) ( )] (75) dξ arcsin ξ/4 = dx( 4) [arcsin arcsin x = ξ/4 ξ(4 ξ) x The integration constant is chosen such that the integral is continuous at ξ = 4, i.e. Cnst = π. All results can be summarized for the scalar integral C as C = i 6π s [ ln ( ) + β iπ] β = τ, τ < β arcsin ( τ ) τ > This result can be substituted in Eq. (56), and we re done. 5 The principal value integral is defined as integral with infinitesimal distance to the pole [ P du β δ = lim du ] u β δ u β + du β+δ u β [ ( ) ( )] δ β = lim ln + ln δ β δ ( ) β = ln + β (7) 4

15 Appendix In this appendix the form algebraic manipulations, used in several places in the text, are presented. The input program to arrive at equation (7) is as follows: FORM by J.Vermaseren,version 3.(Sep 9 3) Run at: Wed Sep 4 3::44 3 * * Calculation of gg->h via top quark loop * Result using the algebraic manipulation program FORM * * Declarations Symbol m, d; Vector k k l; Index mu=d nu=d; * Local M = ( ( g_(,l)+g_(,k)+m)*g_(,mu)*( g_(,l)+m)*g_(,nu)*( g_(,l)-g_(,k)+m) + (-g_(,l)+g_(,k)+m)*g_(,nu)*(-g_(,l)+m)*g_(,mu)*(-g_(,l)-g_(,k)+m) ); tracen,;.sort Time =. sec Generated terms = 4 M Terms in output = 8 Bytes used = 4 bracket m; print;.end Time =. sec Generated terms = 8 M Terms in output = 8 Bytes used = 7 M = + m * ( 8*k(mu)*k(nu) - 8*k(nu)*k(mu) - 6*k(nu)*l(mu) + 6*k(mu) *l(nu) + 3*l(mu)*l(nu) - 8*d_(mu,nu)*k.k - 8*d_(mu,nu)*l.l ) + m^3 * ( 8*d_(mu,nu) ); References [] J. A. M. Vermaseren, (), math-ph/5. 5

16 [] H. M. Georgi, S. L. Glashow, M. E. Machacek, and D. V. Nanopoulos, Phys. Rev. Lett. 4, 69 (978). [3] M. Spira, A. Djouadi, D. Graudenz, and P. M. Zerwas, Nucl. Phys. B453, 7 (995), hep-ph/