MadGraph5_aMC@NLO

How does the symmetrization of cross sections work in madgraph?

Asked by Steven Jeremies

Hello,

as an example I looked at the process p p > d t to which only the diagram of the partonic process u b > d t contributes in the sm-no_b_mass model. The difference in the calculated cross section between the processes p p > d t and u b > d t is a factor of 2 so p p > d t contains the symmetrization. How is this symmetrization done, so which process is calculated in terms of the input format?
Is it b u > d t or b u > t d or something like this?

The reason I am asking is, I noticed there is a difference when generating the processes b u > d t and b u > t d, which is that the z-components of the momenta of the particles in the final state are flipped. So in the first case the z component is lets say x and in the second process it then is -x.

I would highly appreciate any kind of input on this topic!

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Olivier Mattelaer
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Olivier Mattelaer (olivier-mattelaer) said : 		#1

Hi,

Did you check this paper: 1106.0522 <https://arxiv.org/abs/1106.0522>
Discussion on optimization is done in the paper, so this can contain the information that you are looking for.

But yes if you ask
b u > t d or
u b > t d
indeed the spectrum will be flipped since we keep the symmetry of the initial state.
If you do p p > t d, then both are present and we actually only integrate one of the two and generate the second by symmetry on the events.

Cheers,

Olivier

> On 16 Sep 2021, at 14:55, Steven Jeremies <email address hidden> wrote:
>
> New question #698753 on MadGraph5_aMC@NLO:
> https://answers.launchpad.net/mg5amcnlo/+question/698753
>
> Hello,
>
> as an example I looked at the process p p > d t to which only the diagram of the partonic process u b > d t contributes in the sm-no_b_mass model. The difference in the calculated cross section between the processes p p > d t and u b > d t is a factor of 2 so p p > d t contains the symmetrization. How is this symmetrization done, so which process is calculated in terms of the input format?
> Is it b u > d t or b u > t d or something like this?
>
> The reason I am asking is, I noticed there is a difference when generating the processes b u > d t and b u > t d, which is that the z-components of the momenta of the particles in the final state are flipped. So in the first case the z component is lets say x and in the second process it then is -x.
>
> I would highly appreciate any kind of input on this topic!
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Revision history for this message
Steven Jeremies (minits) said : 		#2

Hi Olivier,

thanks for the answer. Do you mean by keeping the symmetry of the initial state that in fact you change the roles of t and d and not those of u and b? Because from the inputs it looks like you change the initial momenta.

Revision history for this message
Best Olivier Mattelaer (olivier-mattelaer) said : 		#3

We change the momenta of all particles when applying the symmetry when asking "p p > t d", both the initial and final state.
This is important to keep all the correlation intact.

So to recap,
the ordering of the final state is not important
the ordering of the initial state is important
if you have identical particle in the initial state that are related to the flipping of the initial state, then we generate only one code that performs the phase-space integration/event generation and use the symmetry to flip the correct amount (in some case --like tevatron-- this is not 50% of the events )

Cheers,

Olivier

Revision history for this message
Steven Jeremies (minits) said : 		#4

Thanks Olivier Mattelaer, that solved my question.