Hi,
sorry, somehow I completely overlooked this message.
I've recently pushed an example that reproduces this calculation from Peskin to the development version:
<https://github.com/FeynCalc/feyncalc/blob/master/FeynCalc/Examples/EW/EWHiggsToTwoGluonsOneLoop.m>
Furthermore, there is a similar example present in FeynHelpers,
where the final result is compared with the literature (Spira et al.)
<https://github.com/FeynCalc/feynhelpers/blob/master/Examples/EW/EWHiggsToTwoGluonsOneLoop.m>
It is also described in the accompanying preprint:
https://arxiv.org/abs/1611.06793
Sec 5.3
Cheers,
Vladyslav
> Dear developers and users of FeynCalc:
>
> I've calculated the process "Higgs decay into two gluons" at the leading order by FeynCalc. In the final result, I just got a C_0(0,0,m_h^2,m_q^2,m_q^2,m_q^2) which should be -1/(2m_q^2) at the limit of zero Higgs mass (see my code below).
>
> << HighEnergyPhysics`FeynCalc`
>
> onshell = {ScalarProduct[p1, p1] -> 0, ScalarProduct[p2, p2] -> 0,
> ScalarProduct[p1, p2] -> Subscript[m, h]^2/2};
> SetOptions[OneLoop, Dimension -> D];
>
> num1 = -I mq/v (I Subscript[g, s])^2 (I) DiracTrace[
> GAD[mu].(GSD[k] + mq).GAD[
> nu].(GSD[k] + GSD[p2] + mq).(GSD[k] - GSD[p1] + mq)] /.
> DiracTrace -> TR //
> Simplify; num2 = -I mq/v (I Subscript[g, s])^2 (I) DiracTrace[
> GAD[nu].(GSD[k] + mq).GAD[
> mu].(GSD[k] + GSD[p1] + mq).(GSD[k] - GSD[p2] + mq)] /.
> DiracTrace -> TR // Simplify;
> amp1 = num1 FAD[{k, mq}, {k + p2, mq}, {k - p1, mq}]/(2 Pi)^D // FCI;
> amp2 = num2 FAD[{k, mq}, {k + p1, mq}, {k - p2, mq}]/(2 Pi)^D // FCI;
> amp = (OneLoop[k, amp1 + amp2] // PaVeReduce) /. onshell // Simplify
>
> 1/(4 \[Pi]^2 v Subsuperscript[m, h, 2]) I mq^2 Subsuperscript[g, s, 2] (2 p1^mu p2^nu (4 mq^2 Subscript[C, 0](0,0,Subsuperscript[m, h, 2],mq^2,mq^2,mq^2)+\!\(
> \*SubsuperscriptBox[\(m\), \(h\), \(2\)]\ \(\(TraditionalForm\`
> \*SubscriptBox[\("C"\), \("0"\)]\)(TraditionalForm\`0, TraditionalForm\`0, TraditionalForm\`
> \*SubsuperscriptBox[\(m\), \(h\), \(2\)], TraditionalForm\`
> \*SuperscriptBox[\(mq\), \(2\)], TraditionalForm\`
> \*SuperscriptBox[\(mq\), \(2\)], TraditionalForm\`
> \*SuperscriptBox[\(mq\), \(2\)])\)\)+4 Subscript[B, 0](Subsuperscript[m, h, 2],mq^2,mq^2)-4 Subscript[B, 0](0,mq^2,mq^2)+2)+\!\(
> \*SubsuperscriptBox[\(m\), \(h\), \(2\)]\
> \*SuperscriptBox[\(g\), \(mu nu\)]\ \((\((
> \*SubsuperscriptBox[\(m\), \(h\), \(2\)] - 4\
> \*SuperscriptBox[\(mq\), \(2\)])\)\ \(\(TraditionalForm\`
> \*SubscriptBox[\("C"\), \("0"\)]\)(TraditionalForm\`0, TraditionalForm\`0, TraditionalForm\`
> \*SubsuperscriptBox[\(m\), \(h\), \(2\)], TraditionalForm\`
> \*SuperscriptBox[\(mq\), \(2\)], TraditionalForm\`
> \*SuperscriptBox[\(mq\), \(2\)], TraditionalForm\`
> \*SuperscriptBox[\(mq\), \(2\)])\) - 2)\)\)+2 p2^mu p1^nu ((4 mq^2-Subsuperscript[m, h, 2]) Subscript[C, 0](0,0,Subsuperscript[m, h, 2],mq^2,mq^2,mq^2)+2))
>
> msq = 2 (amp (ComplexConjugate[amp] /. {mu -> rho,
> nu -> sigma}) PolarizationSum[mu, rho, p1,
> p2] PolarizationSum[nu, sigma, p2, p1] // Contract) /.
> onshell /. Subscript[g, s] -> Sqrt[4 Pi Subscript[\[Alpha], s]] //
> Simplify
>
> (4 mq^4 Subsuperscript[\[Alpha], s, 2] ((4 mq^2-Subsuperscript[m, h, 2]) Subscript[C, 0](0,0,Subsuperscript[m, h, 2],mq^2,mq^2,mq^2)+2)^2)/(\[Pi]^2 v^2)
>
> \[CapitalGamma]HGG =
> 1/(2 8 Pi) 1/(2 Subscript[m, h]) msq /.
> v -> Sqrt[Subscript[m, W]^2 SW^2/(Pi \[Alpha])]
>
> (\[Alpha] mq^4 Subsuperscript[\[Alpha], s, 2] ((4 mq^2-Subsuperscript[m, h, 2]) Subscript[C, 0](0,0,Subsuperscript[m, h, 2],mq^2,mq^2,mq^2)+2)^2)/(8 \[Pi]^2 SW^2 Subscript[m, h] Subsuperscript[m, W, 2])
>
> On the other hand, we can check the result with the analytical side(for example, Peskin and Schroeder's Final Project 3) and a problem comes. In Peskin, a factor I_f(\tau_q) is defined. By the definition of I_f(\tau_q), it contains an extra factor"3" to make itself become 1 at the limit m_h->0. So that the amplitude squared should contain a factor 1/9.
>
> My problem is that I can't find such a factor 1/9 in the result got by FeynCalc.
>
> Best Regards, Thanks for the help!
> Lingxiao Xu
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