In QFT are unstable particles forbidden from being used as asymptotic states in scattering calculations? I ask because the $S$ matrix has an $I$ term which can propagate the unstable state. Is this term cancelled by the rest of the $S$ matrix?
More specifically, let's say we have a system with 2 massive scalars $\pi$ and $\phi$ with the interaction $\pi^2\phi$, and $m_\phi > 2 m_\pi$. Clearly then the $\phi$ particles are unstable, so can they be used as asymptotic states?