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

2.2. Quadruple cuts 38We now must solve for {α, β, δ, γ} using the on-shell constraints l 2 i = 0. Firstlyp 1 · l = 0 implies that β = 0, while putting l on-shell l 2 2 = 0 implies that,0 = s 14 (δρ) (2.16)so that one of δ or ρ is zero. We choose ρ = 0, such thatl µ = αp µ 1 + 1 2 δ〈1|γµ |4]. (2.17)The next two equations are slightly more complicated,0 = −2α(p 1 · P 123 ) − δ〈1|P 23 |4] + s 123 (2.18)0 = −2α(p 1 · P 1234 ) − δ〈1|P 23 |4] + s 1234 (2.19)The difference of these equations determines αWe can now solve for δ,0 = αs 14 − 〈4|P 123 |4] =⇒ α = 〈4|P 123|4]s 14(2.20)0 = − 〈1|P 123|1]〈4|P 123 |4]s 14− δ〈1|P 123 |4] + s 123= 〈1|P 123|4]〈4|P 123 |1]s 14− δ〈1|P 123 |4]δ = 〈4|P 123|1]s 14(2.21)We now have solved for the loop momenta in terms of the external kinematics,l µ = 〈4|P 123|4]p µ 1 + 1 〈4|P 123 |1]〈1|γ µ |4]. (2.22)s 14 2 s 14We can simplify this further,()l µ 1= 〈4|P 123 |4]〈1|γ µ |1] + 〈4|P 123 |1]〈1|γ µ |4]2s 14= 〈4|P 123γ µ |1〉. (2.23)2〈14〉The above can be proven by converting the second term into a trace and commutingp 1 with γ µ . In terms of spinors,|l〉 = |1〉 |l] = |P 123|4〉〈14〉(2.24)