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

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2.4. Sp<strong>in</strong>or <strong>in</strong>tegr<strong>at</strong>ion 512.4.3 A double cut exampleAs a more concrete example <strong>of</strong> the above method, we describe the calcul<strong>at</strong>ion <strong>of</strong> abubble coefficient <strong>in</strong> the s 12 channel for the one-loop amplitude A (1)4 (φ, 1− , 2 + , 3 − , 4 + ).The tree products have the follow<strong>in</strong>g form,∑i=±A (0)L (li 1, φ, 3 − , 4 + , l −i2 )A (0)R (li 2, 1 − , 2 + , l −i1 ) =−〈l 1 3〉 4 〈l 2 1〉 4 − 〈l 1 1〉 4 〈l 2 3〉 4〈l 1 3〉〈34〉〈4l 2 〉〈l 2 1〉〈12〉〈2l 1 〉〈l 1 l 2 〉 2.Us<strong>in</strong>g the Schouten identity we can re-write the numer<strong>at</strong>or <strong>in</strong> terms <strong>of</strong> three functions(which will be def<strong>in</strong>ed <strong>in</strong> chapter 3), for now, we concentr<strong>at</strong>e on one <strong>of</strong> these threefunctions,(2.74)F s12 = − 〈13〉2 〈l 1 1〉〈l 2 3〉〈34〉〈l 2 4〉〈12〉〈l 1 2〉 . (2.75)Next we use momentum conserv<strong>at</strong>ion to remove l 1 <strong>in</strong> favour <strong>of</strong> l 2 ,F s12 = − 〈13〉2 [l 2 |l 1 |1〉〈l 2 3〉〈34〉〈l 2 4〉〈12〉[l 2 |l 1 |2〉= − 〈13〉2 [l 2 |P 12 |1〉〈l 2 3〉〈34〉〈l 2 4〉〈12〉[l 2 |P 12 |2〉We rescale l 2 = tl and perform the trivial t <strong>in</strong>tegr<strong>at</strong>ion,F s12 =∆F s12 ==〈13〉 2 [l 2 2]〈l 2 3〉〈34〉〈l 2 4〉〈12〉[l 2 1]∫t 〈13〉 2 [l2]〈l3〉〈l|P 12 |l] 〈34〉〈l4〉〈12〉[l1]∫s 12 〈13〉 2 [l2]〈l3〉〈l|P 12 |l] 2 〈34〉〈l4〉〈12〉[l1](2.76)(2.77)Here we use ∆F s12 to represent the double cut <strong>in</strong>tegral (for simplicity and to highlightthe oper<strong>at</strong>ions on the <strong>in</strong>tegrand we have suppressed the details <strong>of</strong> the measure). Nextwe wish to rewrite l <strong>in</strong> terms <strong>of</strong> p and η, recall th<strong>at</strong> p + η = P, here P = P 12 so anobvious choice for p and η is p = p 1 and η = p 2 , so we write th<strong>at</strong> |l] = |1] + z|2] and<strong>in</strong>tegr<strong>at</strong>e <strong>in</strong> z dropp<strong>in</strong>g log terms.∫〈13〉 2 〈3l〉∆F s12 =〈34〉〈4l〉〈12〉〈2l〉(〈2l〉 − z〈1l〉)F<strong>in</strong>ally we replace |l〉 and take residues <strong>in</strong> z,∫〈13〉 2 (〈13〉 + z〈23〉)∆F s12 =〈34〉(〈14〉 − z〈24〉)〈12〉(1 − zz)(2.78)

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