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? 
 A: 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...
A: The pole corresponds to an on-shell particle going from one point to another. Then, the residue effectively tells you how many of those particles are being transmitted. Since in your physical/renormalized theory, the propagator should correspond to $1$ quantum of the renormalized field being transmitted, you set the residue at the pole to $1$.
A: *

*It should be stressed that the fact that the propagator has a pole at the physical mass $k^2=-m_{\rm ph}^2$ is the very definition of the physical mass, i.e. it does not depend on the renormalization scheme, cf. e.g. Ref. 1. The on-shell (OS) renormalization scheme equates the renormalized mass $m\equiv m_r$ and the physical mass $m_{\rm ph}$. 

*The unit residue condition (in the $k^2$ variable) is strictly speaking just a natural/convenient convention in the OS renormalization scheme. It is in general incompatible with other renormalization schemes, such as, e.g. the $\overline{MS}$ renormalization scheme.
References: 


*

*M. Srednicki, QFT, 2007; eq. (27.7). A prepublication draft PDF file is available here.

