Decaying only one of two identical final state particles

Asked by Patrick Schaefers

Hello,

I am working on the SM process w+ w- > z z > z e+ e- with vector-boson fusion in particular and encountered two problems.

First, if I use the above notation, i.e.
    generate w+ w- > z z > z e+ e-
it will also generate two diagrams with a Z boson in the s-channel, which I do not want. I tried the syntax
    generate w+ w- > z z > z e+ e- $z
to resolve this issue, but nothing changed.
Is there a way to remove only the Z boson s-channel diagrams?

Second, I tried to use a proper decay, i.e.
    generate w+ w- > z z, (z > e+ e-)
but this will always decay both Z bosons, as it is also stated in "help generate" ("Decay chain syntax: > Note that identical particles will all be decayed.").

Is there any way to decay only one Z boson and keep the other?
I tried things like
    define z1 = z
    define z2 = z
    generate w+ w- > z1 z2, (z1 > e+ e-)
but this did not change anything, too.
What is working is
    generate w+ w- > z z, (z > e+ e-), (z > z)
but it prints out a warning and sets the weight of z > z to 0, so I think I should better not use this.

Thank you very much for your help in advance!

Cheers,
Patrick

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Olivier Mattelaer
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Best Olivier Mattelaer (olivier-mattelaer) said :
#1

Hi,

This is the kind of process which is ambiguous by definition.
The problem is related with the S-matrix formalism which assumes to have plain-wave for the external particles.
Having one Z in plain wavefunction and not the other breaks the symmetry between the two Z and the symmetry factor
is ill-defined.

On top of that,
1) > z z > does not have any special meaning in madgraph. “> z w+ >” means to have at least one “z” and at least one “W+” in s-channel.
and “> z z > “ means to have at least one “Z” and at least one “Z” in s-channel. For both of them it can be the same “Z” and therefore the syntax is equivalent to
“> z >”.

2) “$” is a on shell cut, you still have the diagrams but the particle are forbidden to be on shell. (following the on shell definition of the run_card)

3)
> What is working is
> generate w+ w- > z z, (z > e+ e-), (z > z)
> but it prints out a warning and sets the weight of z > z to 0, so I think I should better not use this.

This is technically working but as above, I would not guaranteed that symmetry factor are correct for such weird syntax.

Note that for this syntax, both Z are assumed to be on shell (following the on shell definition of the run_card)

4) the various syntax that you tried probed different regime of production (some with on shell Z, some with off-shell Z).
Please check carefully the regime that you want.

Cheers,

Olivier

> On Apr 26, 2016, at 12:42, Patrick Schaefers <email address hidden> wrote:
>
> New question #292421 on MadGraph5_aMC@NLO:
> https://answers.launchpad.net/mg5amcnlo/+question/292421
>
> Hello,
>
> I am working on the SM process w+ w- > z z > z e+ e- with vector-boson fusion in particular and encountered two problems.
>
> First, if I use the above notation, i.e.
> generate w+ w- > z z > z e+ e-
> it will also generate two diagrams with a Z boson in the s-channel, which I do not want. I tried the syntax
> generate w+ w- > z z > z e+ e- $z
> to resolve this issue, but nothing changed.
> Is there a way to remove only the Z boson s-channel diagrams?
>
>
> Second, I tried to use a proper decay, i.e.
> generate w+ w- > z z, (z > e+ e-)
> but this will always decay both Z bosons, as it is also stated in "help generate" ("Decay chain syntax: > Note that identical particles will all be decayed.").
>
> Is there any way to decay only one Z boson and keep the other?
> I tried things like
> define z1 = z
> define z2 = z
> generate w+ w- > z1 z2, (z1 > e+ e-)
> but this did not change anything, too.
> What is working is
> generate w+ w- > z z, (z > e+ e-), (z > z)
> but it prints out a warning and sets the weight of z > z to 0, so I think I should better not use this.
>
> Thank you very much for your help in advance!
>
> Cheers,
> Patrick
>
>
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Revision history for this message
Patrick Schaefers (patrick-schaefers) said :
#2

Thanks Olivier Mattelaer, that solved my question.