FeynRules-Branching_Ratio

Asked by daniel camargo

Hi, my apologies for ask here about the feynrules package tools, but maybe you can help me. I was implemented a model with new fermions and new scalars in FeynRules, and I want to find the branching ratio of for example the H scalar decaying in A B with the command "BranchingRatio[{H, A, B}, decays]" such as lies in the manual of FeynRules. The package show me a lot formulas, but all of these are equal of the PartialWidth[{H, A, B}, decays] and TotWidth[H, decays]. Therefore I am confused with the validity of these expressions for the branching ratio. Can you said me the manner of obtain the Branching ratio with FeynRules as well as the Partial Width?. Thanks!!!

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

Did you look at the dedicated paper?
 arXiv:1402.1178

They are more information that in the FR manual.
In top of that I have open a ticket on the FR website with a copy-paste of your question:
https://feynrules.irmp.ucl.ac.be/ticket/104

Cheers,

Olivier

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Benjamin Fuks (fuks) said :
#2

[cc of the answer of the FR website]

Hi Olivier, Daniel,

You need to closely follow what is in the manual:
 1) rules = FeynmanRules[yourlagrangian];
 2) decay = ComputeWidth[rules]
 3) BranchingRatio[yourprocess] with yourprocess being of the form {A,B,C}.
Then, you get a number. If this is not the case, there is maybe an issue with the model file or some parameters may not be provided with a numerical value.

Cheers,

Benjamin

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daniel camargo (dacamargov) said :
#3

Hi, yes I was trying to do it, but the outcome is the following :

>BranchingRatio[{HG, Y1, HC[Y1]}]

Decay::NumMass: Mass is not numerical. Cannot decide whether decay channel is open.

(2.76845*10^26 NumericalValue[Null])/(1.87337*10^23 -
   5.57152*10^18 NumericalValue[
\!\(\*SubsuperscriptBox[\("p"\), \(Ext[2]\), \(PRIVATE`eext[
       2]\)]\)] NumericalValue[
\!\(\*SubsuperscriptBox[\("p"\), \(Ext[2]\),
SubscriptBox[\("\[Mu]"\), \(2\)]]\)] - 5.57152*10^18 NumericalValue[
\!\(\*SubsuperscriptBox[\("p"\), \(Ext[3]\), \(PRIVATE`eext[
       3]\)]\)] NumericalValue[
\!\(\*SubsuperscriptBox[\("p"\), \(Ext[3]\),
SubscriptBox[\("\[Mu]"\), \(3\)]]\)] + 1.67511*10^14 NumericalValue[
\!\(\*SubsuperscriptBox[\("p"\), \(Ext[2]\), \(PRIVATE`eext[
       2]\)]\)] NumericalValue[
\!\(\*SubsuperscriptBox[\("p"\), \(Ext[2]\),
SubscriptBox[\("\[Mu]"\), \(2\)]]\)] NumericalValue[
\!\(\*SubsuperscriptBox[\("p"\), \(Ext[3]\), \(PRIVATE`eext[
       3]\)]\)] NumericalValue[
\!\(\*SubsuperscriptBox[\("p"\), \(Ext[3]\),
SubscriptBox[\("\[Mu]"\), \(3\)]]\)])^1.

What do you think? Thanks.

Daniel

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Benjamin Fuks (fuks) said :
#4

Hi Daniel,

It seems that not all the numerical value of the masses of your particles are provided. Please make sure that all the masses are numerically fixed. This will solve the problem.

Cheers,

Benjamin

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daniel camargo (dacamargov) said :
#5

Hi Benjamin,

I reviewed all the parameters values and coupling constants in the implementation of the model and up to now I am sure that all of them have numerical values. Now, I ran again the FeynRules package and the result was for the BranchingRatio[{HG, X1, HC[X1]}] the following:

Decay::NumMass: Mass is not numerical. Cannot decide whether decay channel is open.
Out[20]=
                 0.000263142 NumericalValue[Null]

Similarly to the PartialWidth[{HG, X1, HC[X1]}] that is,
Decay::NumMass: Mass is not numerical. Cannot decide whether decay channel is open.

Thanks for any hint...

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Benjamin Fuks (fuks) said :
#6

Hello,

Could you please provide some material to reproduce your issue (mathematica notebook, model file, etc...)

Cheers,

Benjamin

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daniel camargo (dacamargov) said :
#7

Hi,

of course! what do you need? and which mail I can send it?

Cheers,

daniel

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Benjamin Fuks (fuks) said :
#8

Hi

I need a mathematica notebook which shows the problem + the FR model file.

Best,

Benjamin

PS: you can use my CERN e-mail address (fuks_at_cern.ch)

Revision history for this message
Best Benjamin Fuks (fuks) said :
#9

Hi David,

You should use X1bar instead of HC[X1] for the antiparticle. In this case it works like a charm.

Cheers,

Benjamin

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Benjamin Fuks (fuks) said :
#10

I mean, Daniel, not David... Sorry.

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daniel camargo (dacamargov) said :
#11

Thanks a lot!

Revision history for this message
daniel camargo (dacamargov) said :
#12

Thanks Benjamin Fuks, that solved my question.