# Why are there two couplings defined for each interaction?

Asked by Ameir Shaa on 2018-02-01

I apologize if this is a trivial question.

In the file, couplings.f, there seems to be two couplings defined for every interaction. For example, take the e- e- z interaction defined by the coupling gzl. In couplings.f, gzl is defined by:

gzl(1) = dcmplx( -ez*(-Half + sin2w) , Zero )
gzl(2) = dcmplx( -ey , Zero )

What is the significance of gzl(1) and gzl(2)?

Many thanks in advance for your answer to this (somewhat trivial) question.

Many thanks,

Ameir

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Olivier Mattelaer
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2018-02-10
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2018-02-10
2018-02-08
 Olivier Mattelaer (olivier-mattelaer) said on 2018-02-01: #1

Hi,

The significance of those number depends of the model that you are using.
Impossible to answer such question without knowing the type of model that you are using.

You should check in the file vertices.py what are the lorentz structure associated to those two couplings
and then you will have the meaning of such coupling.

Cheers,

Olivier

> On 1 Feb 2018, at 17:42, Ameir Shaa <email address hidden> wrote:
>
> New question #663967 on MadGraph5_aMC@NLO:
>
> Hi MadGraph Team,
>
> I apologize if this is a trivial question.
>
> In the file, couplings.f, there seems to be two couplings defined for every interaction. For example, take the e- e- z interaction defined by the coupling gzl. In couplings.f, gzl is defined by:
>
> gzl(1) = dcmplx( -ez*(-Half + sin2w) , Zero )
> gzl(2) = dcmplx( -ey , Zero )
>
> What is the significance of gzl(1) and gzl(2)?
>
> Many thanks in advance for your answer to this (somewhat trivial) question.
>
> Many thanks,
>
> Ameir
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

 Ameir Shaa (ameirshaa) said on 2018-02-02: #2

Hi Olivier,

The MadGraph4 models do not have this file vertices.py. What is the equivalent of this in MG4?

Many thanks,

Ameir

 Olivier Mattelaer (olivier-mattelaer) said on 2018-02-02: #3

Hi,

I actually do not know.
We do not provide support for MG4 model anymore.

If you want to figure it out, You need to check in matrix.f where those couplings are used and in which routine and
then refer to the HELAS manual to know what such routine are.

Cheers,

Olivier

> On 2 Feb 2018, at 03:27, Ameir Shaa <email address hidden> wrote:
>
> Question #663967 on MadGraph5_aMC@NLO changed:
>
> Status: Answered => Open
>
> Ameir Shaa is still having a problem:
> Hi Olivier,
>
>
> The MadGraph4 models do not have this file vertices.py. What is the
> equivalent of this in MG4?
>
> Many thanks,
>
> Ameir
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

 Ameir Shaa (ameirshaa) said on 2018-02-02: #4

Hi Olivier,

I managed to obtain a MG5 version of the model. Here goes. In vertices.py, the e e z coupling is defined like so:

V_107 = Vertex(name = 'V_107',
particles = [ P.e__plus__, P.e__minus__, P.Z ],
color = [ '1' ],
lorentz = [ L.FFV2, L.FFV4 ],
couplings = {(0,0):C.GC_50,(0,1):C.GC_59})

where, in lorentz.py, FFV2 and FFV4 are defined like so:

FFV2 = Lorentz(name = 'FFV2',
spins = [ 2, 2, 3 ],
structure = 'Gamma(3,2,-1)*ProjM(-1,1)')

FFV4 = Lorentz(name = 'FFV4',
spins = [ 2, 2, 3 ],
structure = 'Gamma(3,2,-1)*ProjM(-1,1) + 2*Gamma(3,2,-1)*ProjP(-1,1)')

and in couplings.py, GC_50 and GC_59 are defined like so:

GC_50 = Coupling(name = 'GC_50',
value = '-(cw*ee*complex(0,1))/(2.*sw)',
order = {'QED':1})

GC_59 = Coupling(name = 'GC_59',
value = '(ee*complex(0,1)*sw)/(2.*cw)',
order = {'QED':1})

I still do not know why there are two vertices or what they are for.. I again apologize if this is trivial..

Many thanks,

Ameir Olivier Mattelaer (olivier-mattelaer) said on 2018-02-08: #5

Hi,

So your interaction is written as

GC_50 * Gamma^mu (1-gamma5)/2 + GC_59 Gamma^mu ((1-gamma5)/2 + 2* (1+gamma5)/2)

you can rewrite that as

Gamma^mu (GC_50 (1-gamma5)/2 + GC_59 (3+gamma5)/2)

So if you want to express GC_50 as an expression of the right/left coupling (which is a more typical representation):

Gamma^mu [ (1-gamma5)/2 * (GC_50 + GC_59 ) +2 GC_59 (1+gamma5)/2]

so GC_59 is your left coupling (up to a factor of 2)
and the sum of those two coupling is your right coupling.

Cheers,

Olivier

 Ameir Shaa (ameirshaa) said on 2018-02-10: #6

Many thanks for the clarification Olivier!

 Ameir Shaa (ameirshaa) said on 2018-02-10: #7

Thanks Olivier Mattelaer, that solved my question.

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