MLM matching in QCD process with pre-selection cut in run card

Dear Olivier,

Thanks a lot for the quick response, I got it.

Sorry for my late reply.
I faced similar problem as ccaa when I generate p p > b b~ j j + p p > b b~ j j j.

May I ask, in this case, are there some settings for MLM matching need to be notice? Thanks a lot!!

Best,
Jung

In this case what is the born configuration p p > b b~ (if so you should in principle include the 0 and 1 jet contribution as well)
Otherwise, you might need to specify to pythia8 that it has to not consider the first two jet as part of the matching.

Cheers,

Olivier

Dear Olivier,

Thanks a lot for the reply. May I ask how to specify this in Pythia8?

Thanks!!

Best,
Jung

Hi,

Here is the information that I have:

> ! --------------------------------------------------------------------
> ! Parameters relevant only when performing MLM merging, which can be
> ! turned on by setting ickkw to '1' in the run_card and chosing a
> ! positive value for the parameter xqcut.
> ! For details, see section 'Jet Matching' on the left-hand menu of
> ! http://home.thep.lu.se/~torbjorn/pythia81html/Welcome.html
> ! --------------------------------------------------------------------
> ! If equal to -1.0, MadGraph5_aMC@NLO will set it automatically based
> ! on the parameter 'xqcut' of the run_card.dat
> JetMatching:qCut = -1.0
> ! Use default kt-MLM to match parton level jets to those produced by the
> ! shower. But the other Shower-kt scheme is available too with this option.
> JetMatching:doShowerKt = off
> ! A value of -1 means that it is automatically guessed by MadGraph.
> ! It is however always safer to explicitly set it.
> JetMatching:nJetMax = -1

From this, you can guess that the correct parameter is
> JetMatching:nJetMin = X

Cheers,

Olivier

> On 1 Apr 2019, at 10:14, Chang Jung <email address hidden> wrote:
>
> Question #679219 on MadGraph5_aMC@NLO changed:
> https://answers.launchpad.net/mg5amcnlo/+question/679219
>
> Chang Jung posted a new comment:
> Dear Olivier,
>
> Thanks a lot for the reply. May I ask how to specify this in Pythia8?
>
> Thanks!!
>
> Best,
> Jung
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Dear Olivier,

Thanks a lot for the quick reply!

I got it.

Best,
Jung

Thanks Olivier Mattelaer, that solved my question.

Dear Olivier,

I have another question about c c~ a a process.

In the sense of set nJetMin for b b~ j j case, can I just apply it onto c c~ a a case?
Or, I should follow your suggestion 0)~4) for c c~ a a?

Thanks a lot.

Best,
Jung

In the MLM/QCD context, you can not single out a flavour if that flavour is massless (and therefore is QCD singular).
So you have to forbid any additional jet to come with c-flavor and therefore apply the above procedure.

Cheers,

Olivier

Dear Olivier,

Thanks a lot for the quick response. I got it.

Best,
Jung

Dear Olivier,

I have another question on b b~ j j + b b~ j j j process.

In p p > w, add w+ j, add w+ j j,
as the increase of xqcut, the merged cross section will approach to the cross section of 0j sample right?

In b b~ j j + b b~ j j j case, xqcut will be applied on both 2j and 3j sample right?
But the "0j" sample is actually 2j sample. Although the command in PY8 will allow matching only for > 2j sample.

Will the merged cross section become smaller and smaller as xqcut goes larger?

Thanks.

Best,
Jung

Hi,

> In p p > w, add w+ j, add w+ j j,
> as the increase of xqcut, the merged cross section will approach to the cross section of 0j sample right?

1) The cross-section should be independent of the value of xqcut.
2) The value of xqcut should always be smaller than the merging scale (called sometimes Qcut)
For MLM, xqcut should be at least 10 GeV smaller than Qcut --sometimes more--

So I do not agree with your statement.

> In b b~ j j + b b~ j j j case, xqcut will be applied on both 2j and 3j sample right?
> But the "0j" sample is actually 2j sample. Although the command in PY8 will allow matching only for > 2j sample.

Here you have two strategies (and PYthia needs to know which one you choose)
1) This is a part of the 0j+1j+2j+3j computation.
2) The real "0j" has two jet comming from EW radiation.

The link withe the "0j" cross-section will depend of the above choice.

Cheers,

Olivier

> On 4 Apr 2019, at 09:18, Chang Jung <email address hidden> wrote:
>
> Question #679219 on MadGraph5_aMC@NLO changed:
> https://answers.launchpad.net/mg5amcnlo/+question/679219
>
> Chang Jung posted a new comment:
> Dear Olivier,
>
> I have another question on b b~ j j + b b~ j j j process.
>
> In p p > w, add w+ j, add w+ j j,
> as the increase of xqcut, the merged cross section will approach to the cross section of 0j sample right?
>
> In b b~ j j + b b~ j j j case, xqcut will be applied on both 2j and 3j sample right?
> But the "0j" sample is actually 2j sample. Although the command in PY8 will allow matching only for > 2j sample.
>
> Will the merged cross section become smaller and smaller as xqcut goes
> larger?
>
> Thanks.
>
> Best,
> Jung
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Dear Olivier,

Thanks for the reply.

Sorry I did not write the statement clearly.
I generated with fixing xqcut, and let MG5 automatic choose QCut.

My question is when I change xqcut, the parton level cross section will change, for xqcut is generator level cuts.
In principle, the merged cross section will not affect by xqcut value, but if I put extremely large xqcut, i.e. 500 GeV.
In this case, almost all additional jets contribution cannot pass the xqcut condition.
The w+0j, 1j, 2j generator level cross section will approach w+0j one, so as the merged cross section.

May I ask the definition of merged cross section?
Is it actually the final portion of each 0j, 1j, 2j sample times their generator level cross section?

Thanks.

Best,
Jung

So what you do is taking infinite qcut limit which is the same as keeping the lowest multiplicity.

The merged cross section is the physical cross section after the removal of the double counting

Cheers

Olivier

Ps: I m leaving for Holliday now

Dear Olivier,

Thanks a lot for the reply, I got it.

Best,
Jung