MadGraph5_aMC@NLO

Feyn Rules models and Matrix Elements

Asked by Ekaterina

Hello dear MadGraph experts!
I have 2 questions
1) We created a new model in FeynRules. This model contains a new scalar particle. This particle must decay into two positrons. Charge conjugation was explicitly introduced into the model Lagrangian. However, MadGraph draws only one diagram, although following all the rules, there should be two of them. Could you tell me what could be the problem?
2) Is it possible to somehow display/see the square of the matrix element in a symbolic form (as CompHep/CalcHep does )?

The model we introduced looks like this.

M$ModelName = "DMSM"

FeynmanGauge = True;

M$ClassesDescription = {
  S[5] == {
    ClassName -> X1,
    SelfConjugate -> False,
    QuantumNumbers -> {Q -> 2, LeptonNumber -> -2},
    Mass -> {MX1, 1000},
    Width -> 0,
    PropagatorLabel -> "X1",
    PropagatorType -> D,
    PropagatorArrow -> None,
    ParticleName -> "X1",
    FullName -> "X1"
  }
};

M$Parameters = {
  a1 == {
    ParameterType -> Internal,
ParameterName -> a1,
InteractionOrder -> {QED,1}
  },
  b1 == {
    ParameterType -> Internal,
ParameterName -> b1,
InteractionOrder -> {QED,1}

  }
};
And the Lagrangian :
Lag = X1 anti[CC[e]].(a1 + b1 Ga[5]).e +
  HC[ X1 anti[CC[e]].(a1 + b1 Ga[5]).e ] - ee ebar.Ga[mu].e.A[mu]

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

1) I do not know FeynRules, so could you please provide the UFO model and the process that you are looking at?
     (you can send me the model to my email address: <email address hidden>

2) No we do not use symbolic form for the computation so we can not provide such formula.

Cheers,

Olivier

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Olivier Mattelaer (olivier-mattelaer) said : 		#2

Hi,

I can see the following information in your model:

{
    'id': 2,
    'particles': [-11,-11,-9000005],
    'color': [1 ],
    'lorentz': ['FFS3', 'FFS4'],
    'couplings': {(0, 1): 'GC_1', (0, 0): 'GC_3'},
    'orders': {'QED': 1},
    'loop_particles': None,
    'type': 'base',
    'perturbation_type': None
}

So you can see that your vertex is associated to two lorentz structure and two coupling.
But since they are reported as having the same power of alpha_ew. It is consider as a single vertex for the diagram representation.

MG5_aMC>display lorentz FFS3
in lorentz
structure : Gamma5(2,1)
name : FFS3
spins : [2, 2, 1]
MG5_aMC>display lorentz FFS4
in lorentz
structure : Identity(2,1)
name : FFS4
spins : [2, 2, 1]
MG5_aMC>

details of the couplings:

name : GC_1
value : 2*a1*complex(0,1)
order : {'QED': 1}

name : GC_3
value : 2*b1*complex(0,1)
order : {'QED': 1}

Is this answer your issue?

Cheers,

Olivier

Revision history for this message
Ekaterina (happykate10) said : 		#3

Hello!
Yes, I think this is exactly what I wanted to know.
However, I would like to make sure that despite the fact that the Madgraph presents us with only one diagram, I will get the correct result for the matrix elements taking into account both diagrams.
Thank you!
With the best regards, Ekaterina!

Revision history for this message
Best Olivier Mattelaer (olivier-mattelaer) said : 		#4

Then what I would advise is to set a1 and b1 as external parameter (rather than internal) such that you can compute the amplitude for a1=0, b1=1 and for a1=1, b1=0 (and any other value that you like obviously).

Cheers,

Olivier

Revision history for this message
Ekaterina (happykate10) said : 		#5

Thanks Olivier Mattelaer, that solved my question.