# Convergence issue with flavor summation over charge-conjugated processes

Dear Whizard authors,

I have a question regarding the process ee->qq e nu_e which does not converge if I use automatic flavor summation for the quarks.

I tried it in two ways:

1. process proc_qqlnu_e = e1, E1 => u:U:c:C, d:D:s:S:b:B , e1:E1, n1:N1

2. process proc_qqlnu_e = e1, E1 => (u:c, D:S:B , e1:E1, n1:N1 ) + (U:C, d:s:b , e1:E1, n1:N1)

In both cases, the same (correct) channels are allowed, so the flavor summation itself is not the issue. But it seems that the phase spaces are different and in case 1 the integration does not converge, while in case 2 it does.

I am wondering if this has a known reason. How can I know in the future in which cases flavor summation can be used in this way and in which cases not?

You can find a minimum working example here: https:/

Cheers,

Ulrike

## Question information

- Language:
- English Edit question

- Status:
- Solved

- For:
- WHIZARD Edit question

- Assignee:
- Wolfgang Kilian Edit question

- Last query:
- 2020-01-10

- Last reply:
- 2020-02-12

Launchpad Janitor (janitor) said : | #1 |

This question was expired because it remained in the 'Open' state without activity for the last 15 days.

Juergen Reuter (j.r.reuter) said : | #2 |

We still need a bit more time on this.

Wolfgang Kilian (whkilian) said : | #3 |

The problem occurs because the flavor sum involves summing over charge-conjugated processes, while the cuts involve specific charges of particles.

This is the original process definition:

process proc_qqlnu_e = e1, E1 => q, qbar, e1:E1, n1:N1

There is a sum over electron and positron in the final state.

This is the relevant part of the cut set:

cuts = ...

and all M < - default_Q_cut [incoming e1, e1]

and all M < - default_Q_cut [incoming E1, E1]

The cuts are selective regarding e+/e-.

Unfortunately, during integration, the flavour assignments within a flavour sum are not well defined -- after all, the process declares them as equivalent. In fact, during integration the lepton will always be taken as e- (if allowed), never e+, therefore the second cut never triggers. As a result, there is a backward singularity which leads to a divergent integral.

The first possible solution is to use charge-agnostic lepton cuts such as

cuts = ...

and all M < - default_Q_cut [incoming lepton, lepton]

The extra backward-scattering cut is probably a negligible effect.

The alternative process setup mentioned in the question (accidentally) allows only either e+ or e- in the final state, never both. Therefore the integration succeeds. If the cuts have to remain charge-selective, I recommend that second solution for the problem.

Juergen Reuter (j.r.reuter) said : | #4 |

Marking as solved.