# Tag Info

Starting from $$\tag1 \mathcal{P}\sim\frac{ \mathrm{i} }{ p ^2 - m _0 ^2 + M ^2 ( p ^2 )}$$ and as Jeff points out, by the Optical theorem*, $M^2$ (for a particle that decays) can have a nonzero imaginary part. Hence one defines the physical mass $m$ of the particle, not through m ^2 - m _0 ^2 - M ^2 ( m ^2 ) ...
You are certainly correct that there are other terms in the sum. However, the derivative term is zero by the renormalization conditions for the scalar field and the other terms are assumed to be small (when $p^2$ is far from $m^2$ the diagrams as small anyways). For completeness here the derivation: By performing an infinite sum over $1PI$ diagrams we ...