Hadronic production of a Higgs boson in association with two jets at ...

5.2. One-loop results 144+ 1s 123[−m 4 φ 〈1 ]3〉2 〈2 3〉〈1 2〉 〈2 |p φ | 4] 〈3 |p φ | 4] + 〈4 |p φ| 2] 2 〈4 |p φ | 1]〈4 |p φ | 3] [1 2][1 3]× F 2mh4F (s 14, s 123 , s 23 , m 2 φ[)]+ 1 m 4 φ 〈1 3〉3s 123 〈1 2〉 〈3 |p φ | 4] 〈1 |p φ | 4] − 〈4 |p φ | 2] 3〈4 |p φ | 3][1 2][2 3]× F 2mh4F (s 24, s 123 , s 13 , m 2 φ[) ]+ 1 m 4 φ 〈1 3〉2−s 123 〈2 |p φ | 4] 〈1 |p φ | 4] + 〈4 |p φ| 2] 2)}F 2mh4F (s 34 , s 123 , s 12 , m 2[1 3][2 3]φ){ }+ 3 ↔ 4 , (5.36)where the function containing poles and associated logarithms is conveniently writtenas,V 5 (s 12 , s 34 , s 13 , s 24 ) = − 1 [( ) µ2 ǫ ( ) µ2 ǫ ( ) µ2 ǫ+ − −ǫ 2 −s 12 −s 34 −s 13( µ2−s 24) ǫ ].(5.37)We note that the apparent double pole in ǫ in Eq. (5.37) is cancelled upon expandingabout ǫ = 0.This result for the φ amplitude is particularly simple, containing neither bubblecontributions nor rational terms. This is also true for the helicity amplitudeA 4;3 (φ, 1¯q , 2 q , 3 − g , 4+ g ) 2 , which can easily be checked using the previously calculatedresults in ref. [109]. It is therefore more efficient to program the full result for A 4;3 ,rather than to program the individual primitive amplitudes using Eq. (5.5).Furthermore, for the case of two negative gluon helicities calculated here one cancheck using Eq. (5.5) that the corresponding φ † amplitude is zero. Therefore wehave,A 4;3 (H, 1 −¯q , 2 + q , 3 − g , 4 − g ) = iA 4;3 (A, 1 −¯q , 2 + q , 3 − g , 4 − g ) = A 4;3 (φ, 1 −¯q , 2 + q , 3 − g , 4 − g ) . (5.38)2 The amplitude A 4;3 (φ, 1¯q , 2 q , 3 + g , 4 − g ) is not independent and is obtained by swapping labels 3and 4.