Does MadGraph include decay branching ratios?

Asked by Maksym Ovchynnikov on 2021-01-13

I have a model of some scalar particle a + a mediator V coupled to the baryon current J^mu_B = sum_q bar q gamma^mu q implemented in MadGraph via FeynRules. The Lagrangian is

L = 1/3 *g *J^mu_B V_mu - g_B V^mu (a* d_mu a + a d_mu a*)

Next, I compute the production of a particle in MadGraph using

p p > V, V > a a~

fixing the V and a mass, and the coupling constant g. My question is the following. In the output, there is the corresponding cross-section.

My question is the following. How does MadGraph include the branching ratio of the decay V > a a~ in this cross-section? Does it involve the numerical value of the decay width of V written in parameters.py? P.S. I am using a very small decay width Gamma_V.

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English Edit question
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Answered
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MadGraph5_aMC@NLO Edit question
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Last query:
2021-01-13
Last reply:
2021-01-15

Hi,

The symtax that you use (the one with the comma, that we dubbed the decay syntax)
does not use the BR computation. It computes the full amplitude square (a 2>2 matrix-element in your case) and the propagator of the V is integrated over the invariant mass up to some cut-off preventing the too off-shell region.
This propagator does include the total width of the particles (as includes in the param_card).

If your width is very_small (less than 1e-6 or 1e-8 times the mass --do not remember the threshold--), you will see a warning that such small width can creates numerical issue. In such case to avoid the numerical issue we do use an unphysical width (set to 1e-6 or 1e-8 times the mass) in the denominator and the cross-section is re-scaled following the Narrow-Width approximation such that the cross-section returned is the physical one.

> Does it involve the numerical value of the decay width of V written in parameters.py?
The computation only care about the total width.
If you set the total width to Auto, then the information of decays.py will be used to compute the total width.
if you set the total width to a numerical value we will then use that one.

Cheers,

Olivier

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