# Subtraction in bare perturbation theory on Peskin's book

On chapter 7 of their book on QFT, Peskin and Schroeder derive the vacuum polarization correction to the photon propagator in bare renormalization theory. On page 247, they state that to leading order in $$\alpha$$, $$\Pi(p)=\Pi_2(p)$$. This is obviously simply the definition of $$\Pi_2(p)$$, the one loop correction to the transverse part of the propagator. The problem is that they imediately go on to say

The amplitude for the process will then involve the quantity

$$\frac{-ig_{\mu\nu}}{q^2}\frac{e_0^2}{1-\Pi(q^2)}\underset{\mathcal{O}(\alpha)}{=}\frac{-ig_{\mu\nu}}{q^2}\frac{e^2}{1-[\Pi_2(q^2)-\Pi_2(0)]}$$

(Swapping $$e^2$$ for $$e_0^2$$ does not matter to lowest order.)

which I do not understand: where did the $$\Pi_2(0)$$ come from? In fact, this is clearly the actual renormalization part, and what they are doing is to subtract the infinite part of $$\Pi_2$$ and fix the residue of the propagator at $$1$$ -- but it looks as if I'm missing something since it is not clear from the book at all.

The point is that this is not clear in the text and I cannot make sense of the math without this external input. How does the equality above hold to lowest order and in what sense? What do the authors mean with their presented line of thought?

That part follows from the condition that the residue at the pole should be $$-i$$ (or one, depends if you include the $$i$$ in the numerator of the propagator). In fact it's easy to see that defining $$\Pi(q^2) = \Pi_2(q^2)-\Pi_2(0)$$ one directly gets that, at the pole $$q^2=0 \to \Pi(0)=0$$ and so that the residue at the pole is $$\underset{q^2=0}{\text{Res}}\,D_{\mu\nu}(q^2) = \lim_{q^2\to0}q^2\frac{-i}{q^2}\frac{1}{1-\Pi(q^2)} = -i\frac{1}{1-\Pi(0)} = -i$$
If one didn't subtract $$\Pi_2(0)$$ there wouldn't be any certainty that $$\Pi_2(0)=0$$. And we wouldn't even be able to impose it since $$\Pi_2(q^2)$$ is what it is and comes out directly from the calculation of the vacuum polarization.
• They are taking as given the fact that $\Pi(q^2)=0$ at all orders, and this is due to the physical requirement of the residue at the pole. Then when going to lowest order they won't get only $\Pi_2(q^2)$ but the rescaled quantity. Jul 1, 2020 at 8:37