MadGraph5_aMC@NLO

Output total virtual contribution to cross section?

Asked by Andrew Larkoski

Can MadGraph evaluate the total virtual contribution to the cross section for a process? Specifically, I generate

> generate p p > Z g [ virt= QCD ]

and would just like the total cross section contribution; i.e., the coefficients of the epsilon poles and the epsilon^0 value, when the complete final state phase space is integrated over (with some minimum cut on the Z boson pT, say). Launching this command just seems to output individual virtual contributions to fixed points on phase space.

My question is motivated by studying final-state parton fragmentation. I have a factorization theorem and correspondingly have renormalization factors, and I would like to evaluate the hard virtual contribution to the factorization theorem, and then remove divergences with the known renormalization factors. Is it possible to just evaluate the bare hard function in MadGraph; i.e., the total virtual contribution to a process, expanded in powers of epsilon?

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Olivier Mattelaer (olivier-mattelaer) said : 		#1

Hi,

Out of the box, this will not be possible.
My first idea would be to use a reweighting approach
where you first generate events according to the tree level amplitude
and then you perform a reweighting of the amplitude thanks to the (four) value given by the above output.

The built-in reweighting interface does not have a syntax/interface to get the 1/eps and 1/eps^2 term, so you will either have to hack that interface (or most likely) implement such reweighting yourself.

If you want to use the built-in interface, the first thing that you will need to do
is to change the fortran code to offer new python API to get the 1/eps and 1/eps^2
and then define a plugin reweighting interface to use those value instead of the standard one.

Cheers,

Olivier

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Andrew Larkoski (larkoa) said : 		#2

Hi Olivier,

Thanks for the reply and information! I'll see if I can hack something or use the real NLO contribution to approach the problem.

Best,
Andrew