# p p > z0 z0 with asymmetric decays

Asked by Gilberto Tetlalmatzi on 2021-05-17

Dear MG experts

I am trying to produce the decay

p p to z0 z0, where the first z0 decays leptonically, thus z0 -> l+ l- and the second one according to
z0 -> b b~, so the question is simple: can I produce this entire process by just doing:

generate p p > z z, (z > b b~, z > l- l+),

or this in addition to produce

p p > z z with z > b b~ and z > l+ l-

produces as well the extra channels

p p > z z with z > b b~ and z > b b~
p p > z z with z > l+ l- and z > l+ l-?

Thanks!

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 Revision history for this message Olivier Mattelaer (olivier-mattelaer) said on 2021-05-17: #1

Hi,

my suggestion is to always check the generated feynman diagram.
For decay-chain, you need to do the output first to have the merging of the Feynman diagram.
i.e:
generate p p > z z, (z > b b~, z > l- l+)
output
display diagrams

In the log of the output, you will also have usefull information:

INFO: Processing color information for process: u u~ > z z @1
Decay: z > b b~ WEIGHTED<=2
Decay: z > b b~ WEIGHTED<=2
INFO: Reusing existing color information for process: d d~ > z z @1
Decay: z > b b~ WEIGHTED<=2
Decay: z > b b~ WEIGHTED<=2
INFO: Creating files in directory P1_qq_zz_z_bbx_z_bbx
INFO: Generating Feynman diagrams for Process: u u~ > z z WEIGHTED<=4 @1
Decay: z > b b~ WEIGHTED<=2
Decay: z > b b~ WEIGHTED<=2
INFO: Generating Feynman diagrams for Process: d d~ > z z WEIGHTED<=4 @1
Decay: z > b b~ WEIGHTED<=2
Decay: z > b b~ WEIGHTED<=2

Which is showing that your syntax does only produce Z decay to b b~.
I'm surprised that no warning is raised since your syntax does not make that much sense..
Indeed you ask to decay the Z coming from the Z decay into lepton. Since your Z is decaying only to b b~ the second decay does not happen.

The syntax that you want to use is
generate p p > z z, z > b b~, z > l- l+

Cheers,

Olivier

 Revision history for this message Gilberto Tetlalmatzi (gtx83) said on 2021-05-17: #2

Hi Olver

Thanks, I indeed checked the diagrams before, but the combination of final states was not clear, so I was not sure if it was also producing 4 fermions and 4b's. Your email plus a simple analysis shows that this is indeed the case.

Cheers

Gilberto

________________________________
From: <email address hidden> <email address hidden> on behalf of Olivier Mattelaer <email address hidden>
Sent: 17 May 2021 13:25:46
To: Tetlalmatzi-Xolocotzi, Gilberto, Dr.
Subject: Re: [Question #697096]: p p > z0 z0 with asymmetric decays

Status: Open => Answered

Olivier Mattelaer proposed the following answer:
Hi,

my suggestion is to always check the generated feynman diagram.
For decay-chain, you need to do the output first to have the merging of the Feynman diagram.
i.e:
generate p p > z z, (z > b b~, z > l- l+)
output
display diagrams

In the log of the output, you will also have usefull information:

INFO: Processing color information for process: u u~ > z z @1
Decay: z > b b~ WEIGHTED<=2
Decay: z > b b~ WEIGHTED<=2
INFO: Reusing existing color information for process: d d~ > z z @1
Decay: z > b b~ WEIGHTED<=2
Decay: z > b b~ WEIGHTED<=2
INFO: Creating files in directory P1_qq_zz_z_bbx_z_bbx
INFO: Generating Feynman diagrams for Process: u u~ > z z WEIGHTED<=4 @1
Decay: z > b b~ WEIGHTED<=2
Decay: z > b b~ WEIGHTED<=2
INFO: Generating Feynman diagrams for Process: d d~ > z z WEIGHTED<=4 @1
Decay: z > b b~ WEIGHTED<=2
Decay: z > b b~ WEIGHTED<=2

Which is showing that your syntax does only produce Z decay to b b~.
I'm surprised that no warning is raised since your syntax does not make that much sense..
Indeed you ask to decay the Z coming from the Z decay into lepton. Since your Z is decaying only to b b~ the second decay does not happen.

The syntax that you want to use is
generate p p > z z, z > b b~, z > l- l+

Cheers,

Olivier

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