# 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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Great, similar questions have been bothering me too for quite some time. Arnold, I do not understand your post. The question is about the Fermi golden rule, but I did not see any mention of imaginary spectrum in its derivations. Can you please explain what did you mean? Or post a link to a paper, if possible.

I agree with Vladimir, there is really no dissipation unless the Hamiltonian contains some other interaction with the environment. In the light of this, I think the Fermi rule cannot be satisfactorily derived only from the Schroedinger's equation of the system in driving external field. Some other interactions are needed, and their effect may be described well by the Fermi rule, but I do not know whether there is some paper on this - I would like to read one.

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Arnold Neumaier won't see your question unless you post it as a comment on his post. (Unfortunately I don't think he's an active user of this site anymore either.) – Nathaniel Apr 13 '13 at 12:14

In order to obtain irreversibility in case of discrete spectrum, you have to add another channel of transition responsible for irreversibility (absorption of excitations). It is often done with adding level widths (imaginary parts of energy levels). If you "imply" that there are no widths, then your time evolution contains a superposition of oscillations, even in case of a (quasi) continuum final states.

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