Consider the QED Lagrangian


where the 0 subscript denotes bare fields. The bare fields are related with the renormalized fields via



with these redefinitions the Lagrangian takes the form


it is customary to define




Moreover, the $Z$ renormalization constants are defined to be


it is often said that this leaves the QED Lagrangian in the following form (see page 2 of these notes http://isites.harvard.edu/fs/docs/icb.topic1146665.files/III-5-RenormalizedPerturbationTheory.pdf).


Nonetheless, the carefull reader will notice that one term coming from $mZ_2Z_m\bar{\psi}\psi$ is completely ignored, namely



  • $\begingroup$ Your $\delta_m$ is defined wrong. Peskin/Schröder (and most others) define it as $\delta_m = Z_2 m_0 - m$, in your dimensionless treatment, it would be $\delta_m = Z_2 Z_m - 1$, not $\delta_m = Z_m - 1$ as you write. This also makes the $\delta_2$ in the mass term disappear, leaving you with $m\delta_m\bar\psi\psi$. $\endgroup$ – ACuriousMind Oct 28 '15 at 14:15
  • $\begingroup$ @ACuriousMind ok, You are right about Peskin but check equations (8), (9) and (12) of these notes isites.harvard.edu/fs/docs/icb.topic1146665.files/… . He does use the same formulas I have written above, yet the $\delta_m\delta_2$ part is ignored. Do you have any idea why? $\endgroup$ – Yossarian Oct 28 '15 at 16:12
  • 1
    $\begingroup$ I think there is an expansion (in powers of $e^2$) implicit there that means that $\delta_2\delta_m$ is second-order and hence neglected. $\endgroup$ – ACuriousMind Oct 28 '15 at 16:22

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