quantum field theory - Renormalization condition: why must be the residue of the propagator be 1

In on-shell (OS) scheme, one of the renormalization conditions is that the propagator, say, a scalar theory

$$\frac{1}{p^2+m^2-\Sigma(p^2)-i\epsilon}$$

must have a unit residue at the pole of physical mass $p^2=-m^2$. Some textbooks say this is to make sure the propagator behaves like a free field propagator near the pole. But why?

Answer

The OS condition that

$$\frac{\partial\Sigma}{\partial p^2}|_{p^2=-m^2} = 0$$ implies that the residue in the propagator remains equal to one.

Suppose that we used a different renormalization scheme in which our counter-terms contain no finite parts (e.g. MS scheme). In the OS scheme, we removed finite parts which were logarithmic in our artificial regularization scale $\mu$. In our new choice, the propagator might have a residue, say $R$.

This residue manifests itself in an irritating way; the field will be re-normalized such that $\phi = \frac{1}{\sqrt{R}} \phi_B$. In the LSZ formula, however, external lines contribute factors $R$ (from the KG equation cancelling the propagators). So external scalar lines contribute a factor $\sqrt{R}$ in the MS scheme.

So, whilst this choice in the OS scheme is somewhat arbitrary, it's convenient, because external scalar lines contribute a factor 1.

I'm trying to learn these points myself, so hopefully someone can expand/correct my answer where necessary...