# Physical coupling as a measurable quantity

I have a question about Preskill's quantum field theory note. He want to argue that the renormalized coupling is something measurable.

First let me introduce some backgroud. Suppose the Lagrangian can be written as, \begin{align} \mathcal{L}=&\frac{1}{2}\partial_\mu\phi_{ B}\partial^\mu \phi_{ B} -\frac{1}{2}m_0^2\,\phi_{ B}^2-\frac{1}{3!}\lambda_0\,\phi_{ B}^3 \nonumber\\[5pt] =&\frac{1}{2}\partial_\mu\phi_{ R}\partial^\mu \phi_{R} -\frac{1}{2}m^2\phi_{ R}^2-\frac{1}{3!}\lambda_{ R}\phi_{ R}^3 +\mathcal{L}_{\rm ct} \end{align} where the counterterms are \begin{align} \mathcal{L}_{\rm ct}=\frac{1}{2}(Z-1)\big( \partial_\mu\phi_{R}\partial^\mu \phi_{ R}-m^2\phi_{R}^2\big) -\frac{1}{2}Z\delta m^2\phi_{R}^2-\frac{1}{3!}\delta\lambda\phi_{R}^3 \end{align} The renirmalized couplimh is defined as \begin{align} \Gamma(m^2,\,m^2,\,m^2)=\lambda_{R} \end{align}

In order to argue $$\lambda_R$$ is measurable, he consider the 2-2 scattering process,

This amplitude with external legs on shell has the form,

so $$-i\lambda_{R}^2$$ is the exact residue of each pole. The poles occur for physical values of the momenta, so their residues can be determined only by a continuation of the measured amplitude away from physical momenta. I do not quite understand this last argument. The amplitude $$i\mathcal{M}$$ is observable has pole structure with residue $$-\lambda_R^2$$. Why do we need analytical continution away from physical momentum? Away from physical momentum, $$i\mathcal{M}$$ is no longer obserable, and the 3-pt 1PI diagram is no longer $$\lambda_R$$. For me, it is also strange we can measure some divergent quantity (pole structure). How can we read pole structure from machine? Sorry I do not no particle experiments.