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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.