# Renormalization conditions for QED self-energies

I'm having trouble understanding renormalization conditions: what I know is that they are the conditions required so that at some point called "renormalization point" (which is usually just the on-shell point) the exact diagram has the same expression as the free diagram.

For example, in the photon self-energy one has that $$G_{\mu\nu}^0=Z_3G_{\mu\nu}=Z_3D_{\mu\nu}(1+...)$$ where $$G_{\mu\nu}$$ is the exact propagator and $$D_{\mu\nu}$$ is the free one, and the superscript $$0$$ is to indicate the bare quantities. Now, performing diagrams calculations one can find that $$G_{\mu\nu}^0=D_{\mu\nu}\frac{1}{1+\Pi^0(q^2)}$$ and if we write $$\Pi^0(q^2)=\Pi^0(0)+\Pi(q^2)$$ separating the divergent and finite parts, then we can write $$G_{\mu\nu}^0\simeq D_{\mu\nu}\frac{1}{1+\Pi^0(0)}\frac{1}{1+\Pi(q^2)}=Z_3D_{\mu\nu}\frac{1}{1+\Pi(q^2)}=Z_3G_{\mu\nu}$$ We found $$Z_3$$, but to make $$G_{\mu\nu}\Big|_{q^2=0}=D_{\mu\nu}$$ one needs to impose the renormalization condition that $$\Pi(q^2)\Big|_{q^2=0}=0$$.

In the electron self-energy instead there are two divergences and one expects to find two renormalization constants and two renormalization conditions, but the process should be about the same. One has $$S^0=Z_2S=Z_2 P^0(1+...)$$ where $$S$$ is the exact propagator and $$P$$ is the free one. Again, performing diagrams calculations one can find that $$S^0=P^0\frac{i}{1-P^0\Sigma^0(p)}=\frac{i}{p\!\!\!/-m^0- \Sigma^0(p)}$$ and if we write the Taylor series $$\Sigma^0(p)=\Sigma^0(m^0)+\frac{d}{dp\!\!\!/}\Sigma^0(p)\Big|_{p^2=(m^0)^2}(p\!\!\!/-m^0)+\Sigma(p)$$ separating the divergent and finite parts, then we can write $$S^0=\frac{i}{p\!\!\!/-m^0-\Sigma^0(m^0)-\frac{d}{dp\!\!/}\Sigma^0(p)\Big|_{p^2=(m^0)^2}(p\!\!\!/-m^0)-\Sigma(p)}=\frac{i}{p\!\!\!/-m-\Sigma(p)-\frac{d}{dp\!\!/}\Sigma^0(p)\Big|_{p^2=(m^0)^2}(p\!\!\!/-m^0)}\simeq\frac{i}{p\!\!\!/-m-\Sigma(p)}\frac{1}{1-\frac{d}{dp\!\!/}\Sigma^0(p)\Big|_{p^2=(m^0)^2}}=\frac{i}{p\!\!\!/-m-\Sigma(p)}Z_2=S Z_2$$ We found $$Z_2$$ and $$\delta_m$$, but to make $$S\Big|_{p^2=m^2}=P$$ one needs to impose the renormalization condition that $$\Sigma(p)\Big|_{p^2=m^2}=0$$.
My question is: where does the other renormalization condition, $$\frac{d}{dp\!\!/}\Sigma(p)\Big|_{p^2=m^2}=0$$ comes from?

• if $f(z)=\frac{1}{z-g(z)}$, then $f$ has a pole at $z=0$ iff $g(0)=0$. The residue of this pole is $\frac{1}{1-g'(0)}$ so if you want the residue to be 1, you need $g'(0)=0$. Related: physics.stackexchange.com/q/211499/84967 Commented Aug 30, 2021 at 17:50
• Why do I want the residue to be 1? Is it a physical requirement (like wanting the pole to be at $z=0$)? Commented Aug 30, 2021 at 20:16
• the residue is arbitrary, it is related to the "wavefunction renormalization". If $\phi$ is an arbitrary field with pole $\langle \phi^2\rangle=\frac{1}{p^2}$, then the rescaled field $\phi'=a\phi$ has pole $\langle\phi'^2\rangle=\frac{a^2}{p^2}$. The normalization of $\psi$ is arbitrary and therefore so is its residue. But this normalization appears in e.g. the LSZ formula so it is simplest to set the residue to 1. If you don't, that's fine, but you need to keep track of it. Commented Aug 31, 2021 at 11:00
• @AccidentalFourierTransform If you post this as a standalone answer, I'd be happy to mark it as accepted. Commented Aug 31, 2021 at 13:20
• Thanks but I like Connor Behan's answer so I don't really think another answer is really necessary. Cheers! Commented Sep 4, 2021 at 20:48

Ultimately, $$\frac{d}{dp\!\!/} \Sigma(p) \big |_{p^2 = m^2} = 0$$ is a choice of scheme. What QFT allows us to do is solve for renormalized amplitudes as finite functions of the renormalized couplings. But how the latter relate to the bare couplings is up to us. Ideally, the way to use this would go something like:

1. Perform an experiment at some energy $$\Lambda_1$$ to measure a cross section.
2. Use the action to compute a loop level cross section for the same process where the bare parameters $$\{g_0 \}$$ and their corresponding counterterms $$\{ \delta_g \}$$ are left arbitrary.
3. Set the two equal each other to solve for $$\delta_g$$. It will have an infinite part cancelling the divergence and a finite part to match the observed value.
4. Repeat this until all counterterms are fixed. A renormalizable theory is one for which this only has to be done a finite number of times.
5. From now on, the predictions your theory makes for all of the other (infinitely many) cross sections are unambiguous. Anyone else who wants to test the theory can do so at $$\Lambda_2$$ and all they have to do to compare is look up what solution for the counterterms you got and what your $$\Lambda_1$$ was.

The secret is that even at tree level you were already doing this! We started with a QED action having some a priori unknown parameter $$m_0$$ and some counterterm for it $$\delta_m$$. Then experimentalists told us that the electron propagator had a pole at $$p^2 = (511 \mathrm{keV})^2$$ and this renormalization condition told us that $$m_0$$ and $$\delta_m$$ were not independent... they had to belong to a one dimensional locus. As a tree level coincidence, it was possible along this locus for $$\delta_m$$ to be finite so we just made the simplest choice which was $$\delta_m = 0$$ and $$m_0 = 511 \mathrm{keV}$$.

Loops lead to divergences, but it is still possible to tune $$\delta_m$$ so that the renormalized mass is $$m = 511 \mathrm{keV}$$. There are other schemes that yield different values for $$\delta_m$$ which might be better for studying strong coupling or critical exponents. For these schemes, $$m$$ would not be where the pole of the propagator is... it would just be some arbitrary value that you need to write down along with some $$\Lambda$$ for future experimentalists to be able to compare to your theory. But the scheme you are using called the on-shell scheme is usually the best one to start with because it leads to renormalized masses that still have a simple physical interpretation.

• Thank you for your post, it sheds a light on the ideas behind renormalization... but I don't see how it answer my question: using the on-shell scheme, why should $\Sigma'(p)$ be zero? Commented Aug 30, 2021 at 19:37
• Because that leads to the greatest similarity with tree level conventions. You can renormalize $m$ however you want. But if your choice makes $m^2$ a pole of the propagator with residue 1 (just like it is at tree level), that is by definition the on-shell scheme. Commented Aug 30, 2021 at 20:18
• Hi. I have a question. Why do experimentalists tell us that the electron propagator has a pole at $p^2=(511 keV)^2$? I mean, yeah, experimentalists can tell us the rest mass of the electron, but why does this mass have to be the pole of the propagator (with residue one)?? Commented Jun 6, 2023 at 11:58
• The residue of one is a choice of course. But the pole of a correlation function is the value of mass that controls which Klein-Gordon equation the particle satisfies far away. There would be a deep inconsistency with other methods if electron scattering amplitudes were best fit by correlation functions with a pole somewhere other than $511 keV$. For the most short lived particles, observing a resonance which comes from the propagator's pole is the only known way to measure mass. Commented Jun 6, 2023 at 12:42
• Thanks @ConnorBehan. Now, it makes sense. Commented Jun 7, 2023 at 8:26