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What does it mean for a particle to decay in low energy quantum physics? To be more precise, let us consider either a fermionic Fock space $\mathscr{F}$ generated by a single-particle Hilbert space $\mathscr{H}$. Let $|0\rangle$ denote the Fermi sea (ground state of the Fermi gas Hamiltonian $H_0=\sum \epsilon_k c_k^*c_k$). Now when I think of decay, I presume that we are talking about excited states, i.e., states with particles and holes added to the Fermi sea, which evolve as $e^{iH t} \psi \propto e^{-t/\tau} (\cdots) + (1-e^{-t/\tau})|0\rangle$ where $H=H_0+H_1$ and $H_1$ is some perturbative interaction term.

The question is that why is $e^{iH t}\psi \to |0\rangle$ true? If my understanding of "decay" is incorrect, please correct me.

Andrew Yuan
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    I'm not familiar with the details in the specific context you're describing. But, in general, you can model decay using a non-Hermitian Hamiltonian operator. Then the energy spectrum is complex. The imaginary part of the energy eigenvalues are the inverse decay times. This kind of model arises by starting from a full system, and then averaging over some of the degrees of freedom. For instance, we can describe the process where a muon decays into an electron and two neutrinos using a Hermitian Hamiltonian with all particles, or we can use a non-Hermitian one with only the muon and electron. – Andrew Mar 18 '21 at 03:20
  • I see. That also makes sense. I guess in that case could you elaborate or point to some references in which such a non-Hermitian operator arises from hermitian operators (possibly with examples in condensed matters) – Andrew Yuan Mar 18 '21 at 03:53

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Term decay is rather rare in condensed matter physics, so I will assume that what you really mean is the finite lifetime.

Condensed matter studies complex many-paryicle systems. For example, description of a crystal in terms of electron Bloch waves is valid only under assumption of a a rigid lattice and absence of Coulomb interaction. Once the interactions are taken into account, we end up with complex many-body states, which are usually interpreted in terms ofquasiparticles (see my answer here, for example) which may or may not be similar to fundamental particles studied in the relativistic QFT. In many cases they are similar - such as Landau quasiparticles in metals or "conduction electrons" and holes in semiconductors. However, not being the true eigenstates of the many-body Hamiltonian, these quasiparticles are unstable, i.e., with time they decompose (decay) into other excitations. This is just another way of saying that they have finite lifetime

Roger V.
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