WHIZARD

How to gen "real" single-W?

Asked by Leonhard Reichenbach

Hello Whizards,

I am currently trying to generate the process (only one charge/quark flavor for simplicity)

e1, E1 => e1, N1, u, D

I am mostly interested in the case where the resulting e1, N1 is _not_ coming from a W- and u, D coming from a W+.

I tried the following (with whizard version 3.1.2)

```
?vis_diags = true
process enuqq_all = e1, E1 => e1, N1, u, D
process enuqq_ww = e1, E1 => e1, N1, u, D { $restrictions = "3+4~W- && 5+6~W+" }
process enuqq_ww_or_sw = e1, E1 => e1, N1, u, D {$restrictions = "5+6~W+" }
process enuqq_sw = e1, E1 => e1, N1, u, D { $restrictions = "3+4~!W- && 5+6~W+" }
process enuqq_sw2 = e1, E1 => e1, N1, u, D { $restrictions = "!W- && 5+6~W+" }
compile
```

but both enuqq_sw and enuqq_sw2 are said to be forbidden and no diagrams are generated, while the other processes result in the diagrams that one would expect.

Am I doing something wrong? Is there another way to do this?
(I am also not sure if it makes sense at all physics wise, I am just an experimentalist...)

Question information

Language:
English Edit question
Status:
Solved
For:
WHIZARD Edit question
Assignee:
Juergen Reuter Edit question
Solved by:
Juergen Reuter
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Juergen Reuter (j.r.reuter) said : 		#1

Hi Leonhard,
the so-called single-W topology is mostly a jargon where one of the Ws (here the hadronic one) is fully on-shell, and the second one can be far off-shell. Indeed, the particle combination e1, N1, u, D can come only from W-, W+ in the SM (this is different for loop matrix elements, but even then you will find virtual Ws). You will find that the production is very much enhanced for mostly collinear splitting e- -> nue W-. however for moderate collider energies (365 GeV and below) the enhancement is not so strong because of the large W mass. (Only if the W mass becomes small compared to sqrt(s)/2. So I would generate "single-W" by the process e+e- -> u,dbar,e-,nuebar but then do not impose any kinematic constraints on the lepton. A typical selection for LEP you can find e.g. in
https://inspirehep.net/literature/441927
or
https://arxiv.org/abs/hep-ex/0209015,
which is something like e+ e- -> jj e- nuebar
for
"The single W signal is defined from e+e− → e+νef¯f′ Monte Carlo events that satisfy the following
phase-space requirements [4, 5]:
| cos θe+ | > 0.997
 min(Ef , E¯f′ ) > 15 GeV"
Does this help?
Cheers,
    JRR (Juergen)

Revision history for this message
Leonhard Reichenbach (zehvogel) said : 		#2

Hi Jürgen,

thanks a lot for your answer!

This is more or less what I expected.

However, I am still a bit confused by this statement:

> Indeed, the particle combination e1, N1, u, D can come only from W-, W+ in the SM

What are the other diagrams that do not have two Ws then, e.g. the other 9 in enuqq_ww_or_sw that are not the two s-channel WW or the one t-channel WW diagram? And where is the difference to e2, N2, u, D?

Cheers,
Leonhard

Revision history for this message
Best Juergen Reuter (j.r.reuter) said : 		#3

Ok, maybe I was a bit fast and did not look into the diagrams that you generated with your example above yourself. There are actually diagrams where you have a single-W situation (shame on me), e.g. e+e- -> qqbar, where the quark radiates a single W that then decays leptonically. Howver, that would not be considered single-W (in my understanding), but electroweak radiation (at least at high energies). In the same way you can have e+e- -> neutrino pair or e+e- -> muon pair where a neutrino or a muon radiates a hadronically decaying W. W pairs constitute the standard diagrams (neutrino t-channel exchange and s-channel gamma/Z to WW). Then you have diagrams that could be considered WZ or Wgamma fusion to u,dbar. You could generate a "single-W" by forbidding all W- propagators but such diagram selections are always dangerous and will very likely break gauge invariance and hence lead to wrong numerical results. That's I would generated the full 4-fermion process and enrich "single-W" phase space by kinematic selection.

Revision history for this message
Leonhard Reichenbach (zehvogel) said : 		#4

Thanks Juergen Reuter, that solved my question.