Renormalization conditions for QED self-energies

Question

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?

Answer 1

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.