Difference between W-boson inclusive decay and individual decay in process

Hi,

by default the “l+” tag does not contains the tau. So

> generate p p > w+ w- j j QED=4 QCD=0 NP=1, w+ > l+ vl, w- > j j
is equivalent to 1+2

define l+ = l+ ta+
> generate p p > w+ w- j j QED=4 QCD=0 NP=1, w+ > l+ vl, w- > j j

is equivalent to 1+2+3

Using the “l+” mulitparticle allows us to realize that the e+ and mu+ matrix element are the same and therefore to be faster.

Cheers,

olivier

> On Sep 25, 2016, at 17:52, Ramkrishna <email address hidden> wrote:
>
> New question #402265 on MadGraph5_aMC@NLO:
> https://answers.launchpad.net/mg5amcnlo/+question/402265
>
> Hi,
>
> I would like to ask if I generate the process:
>
> generate p p > w+ w- j j QED=4 QCD=0 NP=1, w+ > l+ vl, w- > j j
>
> where the W+ can decay to any of the three leptons.
>
> Now, If I generate this process into three steps,i.e.
>
> 1) generate p p > w+ w- j j QED=4 QCD=0 NP=1, w+ > e+ ve, w- > j j
>
> 2) generate p p > w+ w- j j QED=4 QCD=0 NP=1, w+ > mu+ vm, w- > j j
>
> 3) generate p p > w+ w- j j QED=4 QCD=0 NP=1, w+ > ta+ vt, w- > j j
>
>
> And add them together. Then, Is it same as the inclusive generated sample or we might loose some contribution from cross-section in this process? If we loose some of contribution then what type of those contribution and is it significant contribution or not?
>
> with regards,
> Ram
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Dear Olivier,

I defined l as

define l+ = e+ mu+ ta+
define l- = e- mu- ta-
define vl = ve vm vt
define vl~ = ve~ vm~ vt~

So, it will contain all three of them. So, As you said, If I use "l+" then it will be just faster no any change will be there from Physics point of view. Am I correct?

Thanks Olivier Mattelaer, that solved my question.