# 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_

k-factor=17.4

(2) sigma_LO= 1.369e-01 +- 5.0e-04 pb, +0.2% -0.7%,

sigma_

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,

Sadek

## Question information

- Language:
- English Edit question

- Status:
- Solved

- Assignee:
- marco zaro Edit question

- Solved by:
- msadek

- Solved:
- 2018-01-08

- Last query:
- 2018-01-08

- Last reply:
- 2018-01-08

marco zaro (marco-zaro) said : | #1 |

Dear Sadek,

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

msadek (msadek) said : | #2 |

Dear Marco,

Thank you very much for your answer.

Cheers,

Sadek