Compton scattering with off-shell photon

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The discussion centers on calculating the modulus of squared amplitude for the process e-γ*→e-γ, involving a virtual photon. The inquiry specifically addresses how to treat a virtual photon in an external leg within Feynman diagrams. It is emphasized that this scenario typically represents a sub-diagram in a broader Feynman diagram, relevant in vacuum quantum field theory (QFT). The need for context regarding the specific physics for which the off-shell amplitude is being calculated is highlighted as essential for providing a comprehensive answer. Understanding the role of external legs in representing asymptotic free states is crucial for progressing in this calculation.
Marioweee
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Homework Statement
Compute the squared modulus of the summed amplitude over the final and initial polarizations of Compton process with a virtual photon.
Relevant Equations
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How is it treated or what Feymann's rule applies to a virtual photon in an external leg? I would like to calculate the modulus of squared amplitude for the process

e-γ*→e-γ

where the * indicates that the photon is virtual. I've never dealt with virtual particles on a external leg and would like to know how to get started.
Thank you.
 
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I've no idea, what the goal is. This can only be a sub-diagram in a true Feynman diagram, which in "vacuum QFT" always describes S-matrix elements with external legs symbolizing asymptotic free states. So to answer this question, we'd need the context, for which physics such an "off-shell amplitude" should be calculated.
 
(a) The polarisation pattern is elliptical with maximum (1,1) and minimum (-1,-1), and anticlockwise in direction. (b) I know the solution is a quarter-wave plate oriented π/4, and half-wave plate at π/16, but don't understand how to reach there. I've obtained the polarisation vector (cos π/8, isin π/8) so far. I can't find much online guidance or textbook material working through this topic, so I'd appreciate any help I can get. Also, if anyone could let me know where I can get more...

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