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Consider the three Feynman diagrams below (images my own):

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From what I have read (from various different sources) we have two options to consider the contribution from the Feynman diagrams:

  1. Consider the contribution from each with a constant vertex factor $g_{em}\approx 1/137$ and $g_w\approx 1/29$ and add them separately.

  2. Consider only the first diagram with a 'vertex function' (or equivalently a 'running coupling constant') $g_{em}(q^2)$ and $g_{w}(q^2)$ and do not include the others when summing over all possible Feynman diagrams.

Assuming this interpretation is correct (please correct me if I am wrong) I am under the impression that for the strong force we have no option but to include diagrams of the second and third form (and analogous). For the weak and EM force however, it seems to me that we can drop the first and second Feynman diagrams and simply consider the first with a constant coupling constant as given in point 1 above.

Assuming everything I have said so far is correct (which I doubt) my question is therefore when is this allowed? i.e.

When can we consider only the first Feynman diagram above, ignoring the second and third and use the constant coupling factors $g_{em}\approx 1/137$ and $g_w\approx 1/29$?

innisfree
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Quantum spaghettification
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  • Your dichotomy 1 vs 2 isn't right. How about calculate as many diagrams as possible AND use running coupling at judicious scale for best result? – innisfree Feb 15 '17 at 10:10
  • @innisfree are you saying include the contribution from all three diagrams, and still use a running coupling constant? From what I have read (e.g. Griffiths pg 265 2nd ed) the contribution from the second and third diagram is taken care of by the running coupling constant. Therefore including all three diagrams and using a running coupling would seem to me like double counting. – Quantum spaghettification Feb 15 '17 at 10:20
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    Yes. I see the confusion. With running coupling at judicious scale, you minimize particular terms in the higher-order diagrams (log terms, but other terms, finite terms, are still present). So you get a more accurate result at fixed order. However, to get an even better result, go to higher order still. – innisfree Feb 15 '17 at 10:26
  • @innisfree Oh, ok. So it is kind of like that Hartree-Fock method in atomic physics. Using a running coupling constant means that my second and third diagram contribute much less (although still a finite amount) and therefore dropping them is much more appropriate. In theory though I could still include all diagrams and use a constant coupling constant. – Quantum spaghettification Feb 15 '17 at 10:41
  • Possible duplicate: http://physics.stackexchange.com/q/201303/50583 – ACuriousMind Feb 15 '17 at 13:04
  • @innisfree sorry just a point of clarification. Do the log terms vanish or are they simply minimized (but still present)? – Quantum spaghettification Feb 15 '17 at 13:53
  • @ACuriousMind thanks for your comment, although the question you linked to is highly related (particularly to the discussion the the comments) and is therefore of great use. I think it falls short of been a duplicate since it does not (explicitly at least or for what I can see) address the question in bold within my answer. – Quantum spaghettification Feb 15 '17 at 13:57

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