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In many QFT textbooks, we usually see the calculations of vertex function, vacuum polarization and electron self-energy.

For example, one calculates the vacuum polarization to correct photon propagator $\langle{\Omega}|T\{A_{\mu}A_{\nu}\}|\Omega\rangle$, where $|\Omega\rangle$ is the ground state of an interaction Hamiltonian.

My questions are:

  1. why do people use free photon propagator (and free electron propagator) in QED process instead of corrected one? You calculate those stuff, but you don't use them?

  2. How can we guarantee that all infinites in any QED process can be absorbed by counterterms ($\delta_m, Z_2,...,$ ect.)?

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    $\begingroup$ Which people? Which pages? Which eqs? $\endgroup$
    – Qmechanic
    May 29, 2020 at 15:45
  • $\begingroup$ @Qmechanic For example, <An intro to QFT> by P&S, <QFT&SM> by Matthew. $\endgroup$
    – Sven2009
    May 30, 2020 at 7:35

1 Answer 1

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  1. why do people use free photon propagator (and free electron propagator) in QED process instead of corrected one? You calculate those stuff, but you don't use them?

This depends on what order in perturbation theory you are working to. The corrections to the field strength ($Z$) will be proportional to $\alpha$ so if you are only interested in the leading contribution to any process, this does not need to be taken into account. In general, however, all the corrections need to be included if you are working to higher order.

  1. How can we guarantee that all infinites in any QED process can be absorbed by counterterms ($\delta_m, Z_2,...,$ ect.)?

This is a very good question, and not trivial to see. The fact that this actually works (that a finite number of counterterms can cancel all infinities that arise at any order in perturbation theory) is what it means to be a renormalizable theory. This is something that has to be proved (and for QED or the standard model this has been proved).

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  • $\begingroup$ Thanks for your answer. So in your 1st paragraph, do you mean that we can replace a free photon by a corrected one in any feynman diagram $\Gamma$ (leave other parts unchanged) to get a more accurate amplitude? $\endgroup$
    – Sven2009
    May 30, 2020 at 7:42
  • $\begingroup$ Yes, as long as you include all terms that contribute to that order in perturbation theory $\endgroup$
    – Ihle
    May 30, 2020 at 7:54
  • $\begingroup$ It does not make sense to include just one higher order correction to a process. Doing that does not (in general) make your calculation «more accurate», in many cases you could get completely unphysical results. Perturbation theory is not consistent unless you include all terms at a given order. $\endgroup$
    – Ihle
    May 30, 2020 at 8:00

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