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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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4 Answers 4

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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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  • $\begingroup$ I am very interested in your answer. Could you provide details (or examples or references) why a (typical) matrix element $\langle \psi | (E-H)^{-1} | \phi \rangle$ is typically smooth but have large peak as a function of $E$ near the resonance? (And how is it related to poles in the analytic continuation to the non-physical sheet?) $\endgroup$
    – Laplacian
    Mar 17, 2023 at 4:45
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    $\begingroup$ @eigenvalue: see, e.g., en.wikipedia.org/wiki/Resonance The Green's function is here called transfer function. $\endgroup$ Mar 17, 2023 at 11:10
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    $\begingroup$ A detailed treatment is in the book by Economou (2006), Green's functions in quantum physics. $\endgroup$ Mar 17, 2023 at 11:20
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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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    $\begingroup$ 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.) $\endgroup$
    – N. Virgo
    Apr 13, 2013 at 12:14
  • $\begingroup$ You don't need a driving external field for the Fermi golden rule to hold, you just need coupling from an initial state to a continuum. See my other answer. $\endgroup$ Jul 28, 2023 at 13:37
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Sorry for resurrecting this old thread, but I don't feel the accepted answer addresses the question exactly, because the resolvent interpretation already assumes a preferred direction of time (the "physical sheet", which comes from the one-sided Fourier transform of the propagator). This has also been bugging me quite a bit and I think it's a very deep question (it's a concrete and relatively simple version of the more fuzzy question "how can dissipation arise from quantum mechanics"), so let me give several possible viewpoints on this apparent paradox. Hope that helps any confused reader who finds this question on google...

First, note that the Fermi golden rule itself does not break reversibility: it says what happens to a particularly prepared state (the initial state), but applies to positive as well as negative times. If you're able to prepare a state in the right way, you'd be able to see the reverse process, ie a transition to the initial state. But of course to do that you have to prepare the continuum amplitudes in the right way. In many (all?) applications, the continuum states come from an unbounded physical domain: "information" (eg the set of amplitudes on the initial states) is not "lost", it just becomes weirdly encoded in the relative amplitudes of the waves that propagate at infinity.

Second, regarding Rabi oscillations, note that oscillations at different frequencies can cancel each other by destructive interference, as long as their amplitudes are coherent. You can plausibly understand and possibly even derive (I haven't tried) the Fermi golden rule from the Rabi oscillations by considering a superposition of infinitely many Rabi oscillations, just because $\int f(\omega) e^{i \omega t}$ can be localized if $f$ is continuous. If you have a system without continuous spectrum, you will get recurrence, ie approximate periodicity (see eg https://en.wikipedia.org/wiki/Poincar%C3%A9_recurrence_theorem), but the recurrence time will grow as the spectrum densifies and approximates continuous spectrum.

Third, mathematically what the Fermi golden rule and such things are doing is focusing the attention to a subsystem, ie computing in effect an approximation to $P e^{-iHt} P$ where $P$ is the projector on the "interesting" states (eg the initial state). This effective dynamics (on the subsystem of interest) may not share the properties of the full dynamics (eg it's not unitary, so it can model things like dissipation). This is particularly clear in the context of the Fermi golden rule, because then you can approximate this "non-autonomous" dynamics $P e^{-iHt} P$ (the state of the subsystem at times $t$ does not depend only on the state of the subsystem at times $t'$, $t' < t$) quite neatly in terms of a simple non-Hermitian effective Hamiltonian ("self-energy").

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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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