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"Unstable" or "metastable" just don't feel like the right descriptions of this material. Two other symmetries which limit decays are the conservation of "baryon number" and "lepton number." A baryon is a proton or a neutron, or an excitation which decays eventually into a proton or a neutron; a lepton is a particle, like the electron or its ...
the one-particle state $|\Psi\rangle$ in section 24.3 of Ref. 1 corresponds to an asymptotic state in the Hilbert space, not an unstable (possible composite) particle. the optical theorem relates the total decay rate $\Gamma$ to the imaginary part of the self-energy $\widetilde{\Sigma}$, cf. eqs. (24.13)+(24.48).
The average lifetime of an unstable particle $\tau$ is related to the decay constant $\tau = \dfrac 1 \lambda$. As is discussed in Particle lifetimes from the uncertainty principle , "the uncertainty principle in the form $\Delta E \Delta t > \hbar/2$ suggests that for particles with extremely short lifetimes, there will be a significant ...
0. In Griffith's Introduction to Quantum Mechanics, he said that " For unstable particle, that spontaneously disintegrates with a lifetime τ, the total probability of finding the particle somewhere should not be constant." So, my question is, why disintegration of a particle leads to non constant value of probability, because according to me ...
1. Suppose one wants to describe an unstable particle that spontaneously disintegrates with a life time say "tau". In that case the total probability of finding the particle is not constant. But should decrease at some rate, say exponential rate. Now my question is that the textbook asks me calculate the rate of change of probability by ...
7. I am searching for not too old literature on the quantum description of unstable particles. I am referring to something beyond the ad-hoc S-matrix description based on the optical theorem common to textbooks such as those given by Peskin and Schröder or Weinberg etc. The book "Open Quantum Systems and Feynman Integrals" by Exner seems to go ...
Edit: Any mass of an unstable particle is complex and defined as the pole of a propagator. The mass of a particle like Higgs is determined quite indirectly, as it takes lots of scattering experiments to reliably determine the relevant cross sections.
On Peskin & Schroeder's QFT, section 7.3, the book discusses the unstable particle. Before (7.57), the book gives the formula of a scalar particle propagator For unstable particle, the book defines the particle's mass by condition $$ m^2-m_0^2-\operatorname{Re} M^2\left(m^2\right)=0 \tag{7.58}$$ Then the book says "the pole in the propagator is ...
The particle is in Unstable Equilibrium when a slight displacement of the particle from equilibrium position makes it go farther away from the position (which means it cannot come back to the mean position) or the forces acting on the particle are in the direction of displacement."
Just to complete Hritik Narayan's answer: the lifetime of an unstable particle is by definition the average time before it decays in its rest-frame. So whatever the frame used, the lifetime remains the same. Now, as already mentioned, what you measure experimentally does depend on the frame and thus on the velocity of the particle.