Polarization at EPA

Asked by serdar on 2021-02-15

Hi
I know that the polarization of the electron and positron in an e-e+ collision is written using the % symbol (For example e%,E%-> ...). But how will the polarization of the electron be written in the equivalent photon approximation as follows? Will the % symbol be used again? But where will it be used?

Process: A,A->W+,W-
pdf1: Equiv. Photon
pdf2: Proton Photon
Photon particle : e^-

Thank you, Serdar

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Alexander Belyaev (alexander.belyaev) said :
#1

Yes, exactly
for polarised photons, you need to enter

A%,A%->W+,W-
process for example

The helicity range is [-1,1]
-1 is for 100% left polarised photon
1 is for 100% right polarised photon

Best,
Alexander

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Alexander Belyaev (alexander.belyaev) said :
#2

P.S.
Correction from my side:
I have replied how to use photon polarisation in general,
but it is not actually clear how the polarisation of the initial electron is related to the polarisation of the EPA photon emitted. This should be explored.

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serdar (serdarxx) said :
#4

Hi
Does putting % symbol on particle-1 like the one below makes the coming electron in EPA polarized? I think the answer to this question is not clear. I will continue to search. Thanks.

Process: A%,A->W+,W-
pdf1: Equiv. Photon
pdf2: Proton Photon
Photon particle : e^-

Revision history for this message
Alexander Belyaev (alexander.belyaev) said :
#5

No,

  the coming electron in EPA is not affected by
setting polarization
for
  A%,A%->W+,W-
process
Regards
Alexander

On 19/02/2021 14:15, serdar wrote:
> Question #695540 on CalcHEP changed:
> https://answers.launchpad.net/calchep/+question/695540
>
> serdar posted a new comment:
> Hi
> Does putting % symbol on particle-1 like the one below makes the coming electron in EPA polarized? I think the answer to this question is not clear. I will continue to search. Thanks.
>
> Process: A%,A->W+,W-
> pdf1: Equiv. Photon
> pdf2: Proton Photon
> Photon particle : e^-
>

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