# k-factor, scale uncertainties and pp->z>tt~ at NLO

Hi,

I am interesting in the NLO QCD correction to the production of top quark pairs through the Z boson exchange (pp---> z---> tt~). In MadGraph, one can generate this reaction by
(1) generate p p > t t~ QCD<=0 / a w- w+ [QCD]
or by
(2) generate p p > z> t t~ [QCD]

These commands allow to generate the same Born, the same one-loop and the same qq~->tt~g real emission Feynman diagrams. The only difference is in the qg->tt~q and q~g>tt~q~ real emission diagrams, where some diagrams are discarded if we use the method (2).

The cross sections, the uncertainties and the k-factor obtained by (1) and (2) are:
(1) sigma_LO=1.373e-01 +- 5.0e-04 pb,+0.2% -0.7%,
sigma_NLO=2.392e+00 +- 1.1e-02 pb, +17.6% -14.2%
k-factor=17.4
(2) sigma_LO= 1.369e-01 +- 5.0e-04 pb, +0.2% -0.7%,
sigma_NLO=1.890e-01 +- 6.3e-04 pb,+3.0% -2.2%
k-factor=1.37
In the first one, we get huge k-factor and the scale uncertainties at NLO are not reduced.

Which method is the correct one? and why the k-factor is very large in (1)?

Cheers,

## Question information

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English Edit question
Status:
Solved
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Assignee:
marco zaro Edit question
Solved by:
Solved:
2018-01-08
Last query:
2018-01-08
2018-01-08
 marco zaro (marco-zaro) said on 2018-01-08: #1

thanks for submitting this question.

Before event starting, let me warn you that what you are trying to do Is not well defined. QCD corrections to pp > z > t t~ are at order alpha^2 alpha_s, which is the same order of EW corrections to the (interference) at order alpha_s alpha. Then, since the z is off-shell, you may have gauge-invariance issues, specially in real emissions (see below).

Then, the correct way to veto particles is
generate p p > t t~ / a w- w+ QCD<=0 [QCD]
i.e. to veto them before specifying the orders
This way the difference between the two approaches is only the inclusion or not of diagrams with a t-channel z boson in the real emission (they are included in (1) but not in (2)). Since in (2) the Z is highly off-shell, these diagrams may be responsible for the large K-factor you get in (1).

Now, to go back to your problem, in view of what I wrote at the beginning of my answer, I do not know what solution is the best, nor if any solution is correct.

Best wishes,

Marco