problem with processing batch file

Asked by ahh

Hi CH authers
i used files from HEPMC "minimal extension of Zp model".
i tried to simulate pp->Zp->2 right handed neutrinos and ~n1->neutrino left +Z "SM" which decay to 2 leptons.
the batch file like that :

########################################
# Model Info #
#########################################
Model: B-L (Feyn. gauge)

Gauge: Feynman

#########################################
# Processes #
#########################################
Process : p,p->~n1,~n1
Decay : ~n1->n1,e,E
Decay : ~n1->n1,m,M
Remove : A,H1,H2,W+,W-,b,c,d,e,n1,s,t,u,Z,
Alias : p=u,U,d,D,s,S,c,C,b,B
here is the problem that both Z and Zp have the same decay , so , if i typed exclude Z it will be excluded from both Process and decay .. how can i solve this
Cheers
Ahmed

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Alexander Belyaev (alexander.belyaev) said :
#1

Dear Ahmed,

Remove: *WILL NOT*
exclude particles from decays -- it dows the job only for the production processes

If you like to exclude particles from decays,
you need to use
Remove Decay: statement
(Thanks to Neil!)

So, you do not have a problem with you batch,
it will work as you like

for
 Remove : A,H1,H2,W+,W-,b,c,d,e,n1,s,t,u,Z,Zp

Regards,
         Sasha

Revision history for this message
ahh (ahh) said :
#2

Thanks dr sasha , thanks dr Neil
i did what you said but the output some thing strange ,, almost all the events have finale state 4 electrons while small number have 2electrons and 2 muons !!!!
i ll send asample to your mail
thanks
Ahmed

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Neil Christensen (neil-christensen-qft) said :
#3

Thanks Sasha for your previous answer.

About the second question, the batch is working properly. Since the two ~n1 are indistinguishable, the event_mixer applies all possible decays to each. So, you will get all 4 combinations. You can easily choose the ones you want afterwards.

Best wishes,
Neil

Revision history for this message
ahh (ahh) said :
#4

thanks dr Neil thanks dr sasha
after i get the .lhe file from calchep and pass it to delphes ,, delphes couldnt reconstruct all leptons cuz they are very very close together ...
so , isnt there any way to uncollimate these leptons " from CH i mean"????
Thanks alot for your interst
Ahmed

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Neil Christensen (neil-christensen-qft) said :
#5

Hi Ahmed,

Currently, no. I am sorry, but we do not apply cuts to the decay products at this time. We have started working on this in a development version but it is hard to say when it will be ready. For now, you will have to remove them yourself.

Best wishes,
Neil

Revision history for this message
ahh (ahh) said :
#6

Hi Dr Neil
just excuse me less information , but , how can i remove them my self .. do you mean from .lhe file ???
thanks for your intrest
Ahmed

Revision history for this message
Alexander Belyaev (alexander.belyaev) said :
#7

Hi Ahmed,

first of all, I can reproduce that you have mostly 4 electrons (3/4)
2e2m(1/4) and no 4muons in your short lhe file you have sent me privately.

And this result is absolutely correct. It is not about CalcHEP -- it gives you correct result. This is about physics you should think about.

Delphes has nothing to do wit this.

1. First of all, if you would have looked at
~n1 -> n1, e, E
and
~n1 -> n1, m, M

partial widths, you would have noticed that the first one about 8 times bigger (!!!) then the other.
So one can immediately estimate that the ratio of 4e : 2e2mu : 4mu
events will be 64:16:1
2. The second question you should ask yourself is why partial widths a so different.
The answer is because ~n1 -> n1, e, E process has W-boson as an intermediate particle while ~n1 -> n1, m, M DOES NOT!
This is because you do not have ~n1 -- ~n2 mixing .

So, for the model/process under study you do have what you have got.

If you would like to have symmetry for 4e : 2e2mu : 4mu
you need to include also ~n2 as shown below
into
# Processes #
section.
But the best you will have is
combination
64:16:1
for the first
and
1:16:64
for the second, so their combination would lead to about
65:32:65 ratio for 4e : 2e2mu : 4mu .

So this problem is closed -- it was not about CalcHEP,
it was about you understanding of your results/model.

Regards,
          Sasha

#########################################
# Processes #
#########################################
Process : p,p->~n1,~n1
Process : p,p->~n2,~n2
Decay : ~n1->n1,e,E
Decay : ~n1->n1,m,M
Decay : ~n2->n2,e,E
Decay : ~n2->n2,m,M

Remove : A,H1,H2,W+,W-,Z,b,c,d,e,n1,s,t,u
A#########################################
# Processes #
#########################################
Process : p,p->~n1,~n1
Process : p,p->~n2,~n2
Decay : ~n1->n1,e,E
Decay : ~n1->n1,m,M
Decay : ~n2->n2,e,E
Decay : ~n2->n2,m,M

Remove : A,H1,H2,W+,W-,Z,b,c,d,e,n1,s,t,u
Alias : p=u,U,d,D,s,S,c,C,b,B
Alias : p=u,U,d,D,s,S,c,C,b,B

#############################

Revision history for this message
Alexander Belyaev (alexander.belyaev) said :
#8

Hi Ahmed,

could let us, please, if your problem is solved?

Sasha

Revision history for this message
ahh (ahh) said :
#9

Hi dr sasha
Of course solved
but , there is one thing : dr neil said the problem is ~n1 are the same ,so , its realible to take all possible decays ,while ,BR( 3/4 : 1/4) ~n1->W,l : ~n1->Z,nu so its realible to find three events of electrons from W and one event for Z . i think if i need symmetry i have to process it
Decay : ~n1->W,l
Decay : W->l,nu
But , my second question to Dr neil , if there is , any way to put cuts on the out goinng particles l , nu which came from decay ???????
he said i have to remove these events my self , is that mean i have to scape this events from my analysis ???
and really thanks alot for your help,
Cheers
Ahmed

Revision history for this message
Neil Christensen (neil-christensen-qft) said :
#10

Yes, you have to remove the events you do not want yourself (from the LHE file).