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I have some trouble understanding the lowest order feyman diagrams for compton scattering.

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Does two feyman diagrams mean that compton scattering can happen through two processes? What is the specific process in each case?

TaeNyFan
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  • Those are not particularly illustrative of the difference between the $s$-channel and the $t$-channel processes. For better laid-out Feynman diagrams, see e.g. http://www.personal.soton.ac.uk/ab1u06/teaching/qft/qft1/christmas_problems/2014/xmas_problem_solution.pdf – probably_someone Dec 07 '18 at 14:07
  • It goes without saying that these electrons are free electrons, not bound ones in atoms? – Peter Bernhard Oct 30 '22 at 18:12

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Here is a clearer sum of the two lowest order diagrams:

compt

It shows the two geometric ways energy and momentum can be exchanged between the two incoming particles to produce the two outgoing, to first order in a series expansion.

Does two feyman diagrams mean that compton scattering can happen through two processes? What is the specific process in each case?

Crossections are calculated in quantum field theory as sums of a convergent series expansion of the original scattering amplitude. The series has diminishing constants in front of each order, as in all series expansions. These are the first order diagrams corresponding to two integrals over the available momentum and energy phase space. There is a one to one correspondence of all the elements in the feynman diagram to terms within the integral. Have a look here.

That there are two lowest order diagrams does not mean that they can be separated, they contribute to the calculations. The true value needs the sum of all orders, but for most experimental checks, the first order is sufficient. Here is how a calculation for these two diagrams goes.

anna v
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    The series is not convergent, it is famously asymptotic. Each term is accompanied by a smaller and smaller factor $\alpha^n$, but the increasing number of diagrams $\sim n!$ means the series diverges no matter how small $\alpha$ is - the radius of convergence is zero. See https://mcgreevy.physics.ucsd.edu/s13/final-papers/2013S-215C-Bond-Justin.pdf for instance –  Dec 07 '18 at 17:00
  • @JulianIngham I thought that renormalization took care of that, but it is too much for the level ot this question. – anna v Dec 07 '18 at 17:10
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    No it does not; renormalisation redefines the coupling $\alpha$ at some scale $\mu$ to ensure perturbative corrections are as small as possible (or alternatively to absorb dependence on UV physics into local counterterms), but the multiplicative factor $n!$ is just due to combinatorics. Another clear way to see that is that the standard Dyson argument for the series diverging isn't invalidated by redefining $\alpha$ by fixing its value at some scale. –  Dec 07 '18 at 17:15
  • @annav See here and links therein. – isometry Dec 08 '18 at 09:50