MadGraph5_aMC@NLO

Easy way to set branching ratio equal to 1?

Asked by Jeff Asaf Dror

Suppose I want to consider an amplitude with an on-shell intermediate particle. If I understand correclty Madgraph uses the width of the particle to calculate the branching ratio for the process. Now suppose that I just want to set the branching ratio equal to 1. I could do this in two steps. - calculate the width in a seperate launch and then insert this into the full amplitude calculation as a parameter.

But this is silly since I just want Madgraph to not divide by whatever value it calculates for the width of the intermediate particle. Is there an easy way to specify this? I tried setting the width parameter to zero or negative values, but this greatly increased the time to do the calculation and didn't yield any results.

Question information

Language:
English Edit question
Status:
Solved
For:
MadGraph5_aMC@NLO Edit question
Assignee:
No assignee Edit question
Solved by:
Rikkert Frederix
Solved:
Last query:
Last reply:
Revision history for this message
Best Rikkert Frederix (frederix) said : 		#1

Dear Jeff,

MadGraph does not consider separate production and decay, but rather the full matrix elements. In there, the width is a parameter of the propagator; there is no explicit multiplication by the Branching Ratio or any other division. Indeed, you method with using a separate run and inserting the width is the way to do it.

Best regards,
Rikkert

Revision history for this message
Jeff Asaf Dror (asafjeffdror) said : 		#2

Thanks Rikkert Frederix, that solved my question.

Revision history for this message
Jeff Asaf Dror (asafjeffdror) said : 		#3

Hi Rikkert,

I have recently realized that there is an option to set the width of particles to "Auto". While this doesn't exactly set the branching ratio to 1, does this effectively do this if I only have 1 decay mode?

Thanks,
Jeff

Revision history for this message
Rikkert Frederix (frederix) said : 		#4

Hi Jeff,

No, it computes the decay width on the fly using all possible decays available in the Model.

Best,
Rikkert