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Consider the QED Lagrangian

$$\mathcal{L}=\bar{\psi}_0(i\gamma^{\mu}\partial_{\mu}-e_0\gamma^{\mu}A_{0\mu}-m_0)\psi_0-\frac{1}{4}(\partial_{\mu}A_{0\nu}-\partial_{\nu}A_{0\mu})^2$$

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

$$\psi_0=\sqrt{Z_2}\psi\qquad{}A_{0\mu}=\sqrt{Z_3}A_{\mu}$$

$$m_0=Z_mm\qquad{}e_0=Z_ee$$

with these redefinitions the Lagrangian takes the form

$$\mathcal{L}=Z_2\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-eZ_eZ_2\sqrt{Z_3}\bar{\psi}\gamma^{\mu}A_{\mu}\psi-mZ_2Z_m\bar{\psi}\psi-\frac{1}{4}Z_3(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2$$

it is customary to define

$$Z_1\equiv{}Z_eZ_2\sqrt{Z_3}$$

leaving

$$\mathcal{L}=Z_2\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-eZ_1\bar{\psi}\gamma^{\mu}A_{\mu}\psi-mZ_2Z_m\bar{\psi}\psi-\frac{1}{4}Z_3(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2.$$

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

$$Z_1=1+\delta_1\qquad{}Z_2=1+\delta_2\qquad{}Z_3=1+\delta_3\qquad{}Z_m=1+\delta_m$$

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).

$$\mathcal{L}=\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi+\delta_2\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-e\bar{\psi}\gamma^{\mu}A_{\mu}\psi-e\delta_1\bar{\psi}\gamma^{\mu}A_{\mu}\psi-m\bar{\psi}\psi-m(\delta_m+\delta_2)\bar{\psi}\psi-\frac{1}{4}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2-\frac{1}{4}\delta_3(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2.$$

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

$$m\delta_2\delta_m\bar{\psi}\psi.$$

Why?

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  • $\begingroup$ 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
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    $\begingroup$ 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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