Monday, 20 April 2015

quantum field theory - Why an unstable particle is not the exact eigenstate of the exact hamiltonian?


According to Srednicki's QFT textbook, page151 "According to our development of LSZ formula in section5, each incoming and outgoing particle should correspond to a single-particle state that is an exact eigenstate of the exact hamiltonian. This is clearly not the case for a particle that can decay."


Why an unstable particle is not the exact eigenstate of the exact hamiltonian? Because its eigenvalue of the full hamiltonian is the $m_\phi$. Where is my fault?




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