In QED, the Lagrangian has a term $\bar{\psi}A^\mu\psi$, which gives a correction to the photon propagator, where the loop is made of a pair electron-positron, with the 1st order diagram:

electron loop

In the Standard Model, there are also couplings such as $\bar{u}A^\mu u$ and $\bar{d}A^\mu d$, which would give rise to a similar correction (changing mass and dividing the coupling by 3). but I have never seen this discussion in a QFT textbook. Isn't this correction important? Wouldn't it change how we measure the vacuum polarization? The electron mass is $\sim0.51$MeV while the up quark mass is $\sim2.2$MeV, so it isn't heavy enough to be ignored.

One reason I thought could explain this is that the range of electromagnetic interactions is much larger than that of the strong force, so the quark-antiquark pair would hadronize before they could annihilate, but I'm not completely convinced this is true. Can anybody shed a light on this?

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    $\begingroup$ Can you cite an example of a textbook that calculates the one-loop contribution to the photon propagator in the Standard Model without including quark loops? $\endgroup$ – Chiral Anomaly Nov 19 at 14:08

QED is the theory of electrons and photons only. When extended to the Standard Model quark loops are included.

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    $\begingroup$ Yes, but this is not what most textbooks say. Perhaps I wasn't clear: I ask if the vacuum polarization usually computed in textbooks is accurate with real life measurements of vacuum polarization or there is some term missing due to a quark-antiquark loop $\endgroup$ – Renan Nobuyuki Hirayama Nov 19 at 2:26
  • $\begingroup$ Quark loops should indeed be included, but don't expect accurate answers until you reach the GeV range. Quarks are not free particles, and the pion mass exceeds 100 MeV, so the production threshold is definitely not 4.4 GeV. $\endgroup$ – Bert Barrois Nov 19 at 13:10
  • $\begingroup$ @BertBarrois virtual particles don’t need to be on shell, so production threshold doesn’t apply inside a virtual particle loop. However, heavier particles indeed contribute less to polarization (see my answer). $\endgroup$ – Prof. Legolasov Nov 19 at 20:14
  • $\begingroup$ Don't forget the first lesson of dispersion relationships. The loop does not contribute to the imaginary part of the propagator below the physical production threshold. $\endgroup$ – Bert Barrois Nov 19 at 21:46

The polarized vacuum is effectively described by the Euler-Heisenberg lagrangian. The nonlinear term is proportional to the inverse of the virtual particle mass, so lighter particles like electrons contribute the most.


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