# Polarization settings of VBS Same_sign WW process

Asked by xiao jie on 2020-01-23

Dear authors,

I use the newest madgraph5 v2.7.0 to do a quick check for polarization settings of VBS Same-sign WW process.
I generate 300 events for following 4 processes:

1. VBS Same-sign WW inclusive:
generate p p > w+ w+ j j QED=4 QCD=0, w+ > l+ vl @ 1
add process p p > w- w- j j QED=4 QCD=0, w- > l- vl~ @ 2
I got XS1: 0.02667 +- 0.0003334 pb

2. VBS Same-sign WLWL:
generate p p > w+{0} w+{0} j j QED=4 QCD=0, w+ > l+ vl @ 1
add process p p > w-{0} w-{0} j j QED=4 QCD=0, w- > l- vl~ @ 2
I got XS2: 0.001734 +- 3.373e-05 pb

3. VBS Same-sign WTWT:
generate p p > w+{T} w+{T} j j QED=4 QCD=0, w+ > l+ vl @ 1
add process p p > w-{T} w-{T} j j QED=4 QCD=0, w- > l- vl~ @ 2
I got XS3: 0.01467 +- 0.0001696 pb

4. VBS Same-sign WLWT:
generate p p > w+{0} w+{T} j j QED=4 QCD=0, w+ > l+ vl @ 1
add process p p > w-{0} w-{T} j j QED=4 QCD=0, w- > l- vl~ @ 2
I got XS4: 0.004733 +- 6.818e-05 pb

sum of the XS of processes 2-4, XS_Sum=XS2+XS3+XS4=0.02114 +- 0.0001859 pb
Compare XS_Sum with XS1, it's not close.

5. I want to know for VBS Same-sign WLWT, should I consider the like following:
generate p p > w+{0} w+{T} j j QED=4 QCD=0, w+ > l+ vl @ 1
add process p p > w+{T} w+{0} j j QED=4 QCD=0, w+ > l+ vl @ 2
add process p p > w-{0} w-{T} j j QED=4 QCD=0, w- > l- vl~ @ 3
add process p p > w-{T} w-{0} j j QED=4 QCD=0, w- > l- vl~ @ 4
For this process, I got XS5: 0.00917 +- 0.0001623 pb

Then XS_Sum2=XS2+XS3+XS5=0.025574 +- 0.00023716 pb
So XS_Sum2 is close to XS1.

Are my all configurations OK? And what's the correct way to set VBS Same-sign WLWT?

Thanks,
Jie

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2020-01-24
2020-02-05
 Olivier Mattelaer (olivier-mattelaer) said on 2020-01-23: #1

Looks like your XS5 is the correct way to go.
Need to double check if I can make the other syntax the correct one, or if this is not possible for some reason.

Cheers,

Olivier

 xiao jie (talalal) said on 2020-01-24: #2

Dear Olivier,

Thanks,
Jie

 Olivier Mattelaer (olivier-mattelaer) said on 2020-01-24: #3

Hi Jie,

I'm not working for the physics department for the moment (I have only a part time job in the physics department).
So I will not investigate this more up to Monday. In any case this should not be a blocker for you, so I was not even expecting to report anything in this thread.

Cheers,

Olivier

 Olivier Mattelaer (olivier-mattelaer) said on 2020-02-05: #4

Hi Jie,

So to sum up on this, since you seem to be confused here.
in 2.7.0 it seems that we do not always handle correctly the symmetry factor when we have identical particle decaying.

So for the moment the contribution
add process p p > w-{T} w-{0} j j QED=4 QCD=0, w- > l- vl~ @ 4
is not multiply by a factor of 2 like it should (in principle it should not matter if you write w-{T} or w-{0}, it should be the same for MG5aMC)

In 2.7.1, we plan to have two mode one where the ordering matters (such that you can do
add process p p > w-{T} w-{0} j j QED=4 QCD=0, w- > l- vl~, w-> j j @ 4
where the w-{T} will always decay leptonically)
and a second mode which is unordered where
add process p p > w-{T} w-{0} j j QED=4 QCD=0, w- > l- vl~, w-> j j @ 4
will have both w decay in both mode.

In ordered mode, you will need to add w-{T} w-{0} and w-{0} w-{T}
to get the full unpolarised result
while in the unordered mode you will need to include only one.

Thanks to have reported this to us.

Cheers,

Olivier