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I'm having a hard time understanding how the cancellation of the infinities works in QED at 1 loop level.

I think the aim of this procedure is to obtain a Lagrangian such that all the diagrams that contain 1 loop are finite. To do that we consider the 3 divergent diagrams:

  1. Vacuum polarization diagram (a)

  2. Electron self-energy diagram (b)

  3. Vertex graph (c)

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I believe those are the graphs that we need to consider because they constitute the first correction respectively to the photon propagator, electron propagator and QED vertex which are the building blocks of any diagram. So if those are finite every diagram containing 1 loop should be finite. We can calculate these diagrams with dimensional regularization. For example in the case of the vacuum polarization diagram we have (ignoring the external propagators):

$\Pi_{\mu \nu}=(k^{2}\eta_{\mu \nu}-k_{\mu}k_{\nu})\Pi(k^2)$ with

$\Pi(k^2)=-\frac{(e\mu^{\frac{-\epsilon}{2}})^2}{6\pi}\mu^{\epsilon}[\frac{1}{\epsilon}+\frac{1}{2}\gamma_{E}-\frac{1}{2}ln(4\pi)]+\Pi_{f}(k^{2})$

where $\Pi_{f}(k^{2})$ is constant.

Now comes the part which I don't understand: We introduce a counter term in the Lagrangian: $\Delta L_{1}=-\frac{e^{2}}{24\pi}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2[\frac{1}{\epsilon}+\frac{1}{2}\gamma_{E}-\frac{1}{2}ln(4\pi)]$.

This counter term gives rise to a new two-photon vertex contribution of the form:$-\frac{e^{2}}{6\pi}(k^{2}\eta_{\mu \nu}-k^{\mu}k^{\nu})[\frac{1}{\epsilon}+\frac{1}{2}\gamma_{E}-\frac{1}{2}ln(4\pi)]$. Diagrammatically this is represented with a photon propagator (wiggly line) with a cross in the middle.

It is then said that this new vertex will contribute to the vacuum polarization with a term: $\Delta\Pi_{\mu \nu}=\frac{e^{2}}{6\pi}(k^{2}\eta_{\mu \nu}-k_{\mu}k_{\nu})[\frac{1}{\epsilon}+\frac{1}{2}\gamma_{E}-\frac{1}{2}ln(4\pi)]$.

My question is: Why does this counterterm give a contribution to the vacuum polarization function?

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It contains the same kinetic term $F_{\mu\nu}F^{\mu\nu}$ as the original QED Lagrangian, so it contributes to the two point function just as that term does.

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  • $\begingroup$ I agree with that, but then what is been made finite is the two point function (i.e the full photon propagator at 1 loop level). But, after this renormalization procedure, if I want to compute another diagram that contains a fermion loop between two photon propagator I don't see why it should be finite. Don't we want all diagrams at 1 loop level to be finite? $\endgroup$
    – Mathew
    Jun 3 at 12:37
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In difference to the usual terms in the Lagrangian, the counter terms are perturbative terms, i.e. they are only taken into account if they are in same order of perturbation as the actual perturbative correction that is going to be calculated.

So if for instance in the perturbation series one reaches the correction where vacuum polarization is needed apart from the Feynman-diagram with the 1-loop vacuum polarization one also considers the Feynman-diagram which belongs to the counterterm, i.e. the one with the cross amidst the photon line.

Each counterterm of fixed order gets a new Feynman-diagram associated (in this example the one with a cross midst the photon line) which is taken into account only if the same order of perturbation is computed as the order of the counterterm. This procedure guarantees that the divergences in the loop diagrams are cancelled.

It is important that they are of the same order. When going to even higher order than just 1-loop, there will be also counterterms necessary for the (higher order) corrections of the same concept -- i.e. vacuum polarization, self-energy and vertex correction -- but again these counterterms of higher order which will only used if the corresponding order in the perturbation series computation is reached.

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  • $\begingroup$ Does it mean that, after the renormalization, when I see the vacuum polarization diagram I should (implicitly) consider it as the sum of the one "unrenormalized" I had at the beginning plus the counter diagram? Also what do you mean with order? The number of loops or the power of e? $\endgroup$
    – Mathew
    Jun 3 at 12:57
  • $\begingroup$ @Mathew: Yes, then the counterterm and the divergent part of the vacuum polarization will mutally cancel leaving only the finite part uncancelled. And when I speak of orders I mean powers of $e$. $\endgroup$ Jun 3 at 13:27

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