mu -> e gamma operator in FlavorKit

Questions about the interface between SARAH and SPheno as well as the FlavorKit functionality
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WojciechKotlarski
Posts: 16
Joined: 15. Mar 2017, 11:47

mu -> e gamma operator in FlavorKit

Post by WojciechKotlarski » 27. Oct 2017, 02:01

Hi Florian,

We're looking now into the mu -> e gamma process, which goes through operators you call OA2lSL and OA2lSR. As an example, the left one is defined in FlavourKit as

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{OA2lSL,Op[6] Pair[ec[3],k[1]]}
If I compare it with footnote 9 in your paper 1405.1434 I would expect sth like

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{OA2lSL,Op[-6,ec[3],k[1]]}
I guess I'm missing something basic. Could you comment on that?

cheers,
Wojciech

FStaub
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Posts: 822
Joined: 13. Apr 2016, 14:05

Re: mu -> e gamma operator in FlavorKit

Post by FStaub » 27. Oct 2017, 09:31

Hi,

I don't really understand the operator you have written down: ec[3] and k[1] carry Lorentz indices which need to be contracted. Therefore, the Pair[..]. After doing that, this is just a scalar which can be taken out of Op[..]. So, it seems that we actually agree up to an overall sign? This might come from the FermionOrdering.

Cheers,
Florian

WojciechKotlarski
Posts: 16
Joined: 15. Mar 2017, 11:47

Re: mu -> e gamma operator in FlavorKit

Post by WojciechKotlarski » 30. Jan 2018, 01:45

Hi Florian,

Sorry about that (and also about a very late reply). I got confused because I didn't notice that you've used the Gordon identity to express operator with \sigma by the one with a scalar product.

Now, if I look how OA2lSL and OA2lSR are calculated you set all the diagrams which are tagged as wave to 0. Do I miss something, because I think they are not strictly 0. To make it more concrete, I'm now looking into how they are used in the calculation of the muon conversion in the nuclei.

cheers,
Wojciech

FStaub
Site Admin
Posts: 822
Joined: 13. Apr 2016, 14:05

Re: mu -> e gamma operator in FlavorKit

Post by FStaub » 30. Jan 2018, 08:38

Hi,

You need these additional diagrams only to cancel divergences and prove gauge invariance, but in Feynman gauge they have no finite contributions.

There is a review by Jorge Romao where this is shown.

Cheers
Florian

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