# Irreversibility and the Fermi golden rule

When a quantum system is perturbatively coupled to a continuum of states, one uses the Fermi's golden rule to compute the rate of transition form an initial state to a set of states contained in an infinitesimal volume of phase space (the energy density factor). However, I am still puzzled by the fact that the continuum feature of the spectrum implies this non reversible evolution whereas a discrete one would imply Rabi oscillations and finite time reversibility. How does one makes the transition from one to the other?

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One can see the difference by considering the resolvent. Generic matrix elements of $(E-H)^{-1}$ have poles at the discrete spectrum of $H$, but are square integrable (typically smooth) functions of energy in the continuous spectrum, though with large peaks near resonances (poles in the analytic continuation to the nonphysical sheet).

Treating the resonances with small imaginary part as particles is often an excellent approximation, but the imaginary part introduces dissipation. The reason is that $(E-H)^{-1}$ is nonhermitian for nonreal $E$, and remains so in the limit of infinitesimal imaginary part.

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