ungrouping subprocesses

Asked by Angela

In model sm, I've defined a multiparticle m composed of quark and antiquark. A gluon interacts with it.
It creates two sub-processes with 3 diagrams each.
I would like the final cross section computation to include the terms involving the cross-multiplication of the diagrams in the two sub-processes. How do I go about it?
  If I do a "set group_subprocesses False", it does not work. It continues to show two subprocesses of 3 diagrams each.

Question information

Language:
English Edit question
Status:
Answered
For:
MadGraph5_aMC@NLO Edit question
Assignee:
No assignee Edit question
Last query:
Last reply:
Revision history for this message
Olivier Mattelaer (olivier-mattelaer) said :
#1

Can you give more details of what you want.
What is the process under-consideration?

and which cross-term you want that are missing.

Cheers,

Olivier

> On 11 May 2018, at 16:08, Angela <email address hidden> wrote:
>
> New question #668861 on MadGraph5_aMC@NLO:
> https://answers.launchpad.net/mg5amcnlo/+question/668861
>
> In model sm, I've defined a multiparticle m composed of quark and antiquark. A gluon interacts with it.
> It creates two sub-processes with 3 diagrams each.
> I would like the final cross section computation to include the terms involving the cross-multiplication of the diagrams in the two sub-processes. How do I go about it?
> If I do a "set group_subprocesses False", it does not work. It continues to show two subprocesses of 3 diagrams each.
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Revision history for this message
Angela (angela0192s) said :
#2

As an example, let me take a simple process where a gluon interacts with a meson. An incoming meson absorbs a gluon and then emits a guon. i.e.
define m = c c~
generate g m > m g.
This gives me two sub-processes 1) gluon interaction with c, which has 3 diagrams and 2) gluon interaction with c~, which again has 3 diagrams. This means that the amplitude square = M^2 = sum of squares of 1st sub process + sum of squares of 2nd sub process. This means M^2 is a sum of 3! + 3! = 12 terms.
What I actually need is M^2 = (3 diagrams of 1st subprocess + 3 diagrams of 2nd subprocess)^2. This would have a total of 6! = 720 terms.
The cross terms here are actually the cross product of the diagrams of the 1st sub-process with the diagrams of the 2nd sub-process.

Is this possible?
If I can somehow ungroup the 3 + 3 diagrams into a set of 6 diagrams, and then square, I believe I would get the result I require.

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

Hi Angela,

Since c and c~ are distinguishable, the cross-term between those two amplitude are zero.

Now if you have a practical case where you need such cross-term, I'm worried that
S-matrix formalism does not apply to your scenario and therefore you can not use our tool.

Cheers,

Olivier

> On 12 May 2018, at 09:23, Angela <email address hidden> wrote:
>
> Question #668861 on MadGraph5_aMC@NLO changed:
> https://answers.launchpad.net/mg5amcnlo/+question/668861
>
> Angela posted a new comment:
> As an example, let me take a simple process where a gluon interacts with a meson. An incoming meson absorbs a gluon and then emits a guon. i.e.
> define m = c c~
> generate g m > m g.
> This gives me two sub-processes 1) gluon interaction with c, which has 3 diagrams and 2) gluon interaction with c~, which again has 3 diagrams. This means that the amplitude square = M^2 = sum of squares of 1st sub process + sum of squares of 2nd sub process. This means M^2 is a sum of 3! + 3! = 12 terms.
> What I actually need is M^2 = (3 diagrams of 1st subprocess + 3 diagrams of 2nd subprocess)^2. This would have a total of 6! = 720 terms.
> The cross terms here are actually the cross product of the diagrams of the 1st sub-process with the diagrams of the 2nd sub-process.
>
> Is this possible?
> If I can somehow ungroup the 3 + 3 diagrams into a set of 6 diagrams, and then square, I believe I would get the result I require.
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Revision history for this message
Angela (angela0192s) said :
#4

For a bound state meson, only the meson is detected by the detector; not the individual c and c~. There is no way to distinguish the gluon from interacting with c and c~. Hence the amplitudes for the two would add, i.e. the cross terms would not be zero.

If nothing can be done, then this thread can be closed. If there is no workaround, then I would not be able to use this tool.

Revision history for this message
Olivier Mattelaer (olivier-mattelaer) said :
#5

Hi,

The hypothesis here is that all particles are asymptotically free.
So this requires that all quark/gluon are free particles (i.e. that Q^2 is much larger than the QCD scale ~ 2 Gev^2.) Then we interface to parton-shower to simulate QCD radiation to lower scale and then the hadronization occurs.

From what I read from your question you seem more interested in NRQCD. That we do not support in MG5aMC. I would suggest to check for HELACONIA.

Cheers,

Olivier

> On 12 May 2018, at 21:57, Angela <email address hidden> wrote:
>
> Question #668861 on MadGraph5_aMC@NLO changed:
> https://answers.launchpad.net/mg5amcnlo/+question/668861
>
> Angela posted a new comment:
> For a bound state meson, only the meson is detected by the detector; not
> the individual c and c~. There is no way to distinguish the gluon from
> interacting with c and c~. Hence the amplitudes for the two would add,
> i.e. the cross terms would not be zero.
>
> If nothing can be done, then this thread can be closed. If there is no
> workaround, then I would not be able to use this tool.
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Can you help with this problem?

Provide an answer of your own, or ask Angela for more information if necessary.

To post a message you must log in.