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In the derivation of Fermi’s Golden rule for the application of a sudden constant perturbation, we get the following formula for the rate:

$$ P_{f \leftarrow i}(t) = |\langle f|V|i\rangle|^2 \frac{4\sin^2\left(\frac{\omega_{fi}t}{2}\right)}{(E_f - E_i)^2}$$

It is then standard to show that as $t \rightarrow \infty$, the sinc function turns to a delta and energy is conserved. However I have questions about the seeming ‘non-energy conservation’ at finite times.

I understand that since there has been a sudden time dependent change in the Hamiltonian so energy has no reason to be conserved but certain aspects of this seem puzzling.

  1. The sinc function has multiple ‘smaller peaks’ around the main peak. What makes these other energies ‘favoured’ for the system to jump to compared to the other energies?

  2. Where does the initial energy supplied by the perturbation go after infinite time. Am I right in thinking that it is returned back from the system to the perturbation in the infinite time limit so that energy can be conserved after all?

  3. Is anything different in the situation where the perturbation has an infinitely slow turn on proportional to $e^{\epsilon t}$. Does this slow (not sudden) turn on suddenly make things adiabatic and remove some of this energy conservation confusion?

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    $\begingroup$ Linked. $\endgroup$ May 3 at 23:58
  • $\begingroup$ I don’t think that question answers too much of my question but thanks $\endgroup$
    – Alex Gower
    May 4 at 0:03
  • $\begingroup$ Isn't $t$ the length of time for which the perturbation is applied? How can $t \to \infty$ and the perturbation be sudden as well? $\endgroup$ May 4 at 3:28
  • $\begingroup$ Because the perturbation still switches on instantly and abruptly at t=0 $\endgroup$
    – Alex Gower
    May 4 at 9:16
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Hidden part of the Fermi Golden rule iceberg
I think part of the problem may go away, if we put energy conservation in more rigorous terms, appropriate for doing calculations. Conservation laws follow from the symmetries of space-time, and manifest themselves in the mathematical form of the physical equations. In classical mechanics they appear as the first integrals of the equations, whereas in quantum mechanics as the commutation between the conserved quantities and the Hamiltonian. In this sense the energy of the system as a whole always conserved, unless the Hamiltonian is explicitly time-dependent.

Fermi golden rule is a useful tool for calculations and is easily derived using basic quantum mechanics, which gives impression of simplicity. There is however a lot what is hidden under the carpet in these derivations (see, e.g., this answer), usually in the form of assumptions rather vaguely stated in derivations. Let me make a few specific points:

  • When dealing with Fermi Golden Rule, the Hamiltonian is explicitly time-dependent, so the energy conservation does not apply.
  • The system of interest is only part of the whole - the other part is the driving field, which is considered classical, and which can add and remove the energy from the system.
  • There is an implicit dephasing mechanism, which localizes the system in one of the states - this mechanism usually appears either in terms of the finite density of the final states (introduced ad-hoc) or as taking the limit $t\rightarrow+\infty$
  • The dephasing is strong enough to prevent transitions $f\rightarrow i$, i.e., the Rabi socillations.
  • Another way to clarify the dephasing: we consider only $P_{i\rightarrow f}(t)=\frac{d}{dt}|c_f(t)|^2$, i.e., only one element of the density matrix, neglecting its non-diagonal terms.

Full description
Most of these issues disappear, if we consider interaction of the system with a quantized field:

  • the energy conservation is then the conservation of the whole energy of the system + photon field
  • the density of the final states and the irreversibility of the transition appear as the result of taking the thermodynamic limit - the infinite number of photon modes (otherwise we would observe collapse and revival).
  • during any final time we are dealing with the superposition of the system states and the photon field, so energy of the system alone is not an integral of motion, and we are not in an eigenstate of the system Hamiltonin.

Intermediate level description of transitions
An intermediate level of description is achieved via Bloch equations, which explicitly include the non-diagonal density matrix elements.

Answers to OP
Let me formulate how this applies more specifically to the questions formulated in the OP:

  1. $P_{i\rightarrow f}$ is a mathematical object not governed by the energy conservation (it is only one element of the full density matrix). The interpretation in terms of energy conservation appears only after taking the limit $t\rightarrow \infty$.
  2. If we consider a periodic force driving a classical oscillator, then, unless the force is in resonance, during some parts of the cycle it accelerates the oscillator, and during other parts of the cycle counteracts the oscillations. Similarly the field driving a two-level system can both add and take away the energy (as in the already mentioned Rabi oscillations)
  3. Adiabatic switching does facilitate the derivations, in my opinion. It does not always correspond to the same phsyical situation, but it gives the same result in the limit $t\rightarrow\infty$ - another reminder that the Fermi Golden rule is meaningful only in this limit; it is not intended for describing the transient process after switching on the perturbation.
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  • $\begingroup$ Thank you! I don't quite understand what you meant about the rephrasing bit, could you elaborate? $\endgroup$
    – Alex Gower
    May 7 at 9:10
  • $\begingroup$ And is your main point that the specific shape of the peaks (e.g. Sinc here) on the graph are purely artifacts of the exact time dependence of the perturbation (e.g. Sinusoidal or constant but only from t=0) and ultimately don't really matter because they will always be exactly compensated for with the energy distribution of the driving field? $\endgroup$
    – Alex Gower
    May 7 at 9:23
  • $\begingroup$ If one honestly solves the Hamiltonian as given in time-dependent perturbation theory, one obtains the never-ending oscillations between the two levels (Rabi oscillations). It requires some more work to get these oscillations explicitly and even than it is approximate. FGR assumes that these oscillations are very quickly damped and the system finds itself localized in the excited state - it requires some kind fo thermodynamic process, which is not in the Hamiltonian. It is only under such an assumption that the rate $P_{i\rightarrow f}$ is meaningful. $\endgroup$ May 7 at 9:35
  • $\begingroup$ Hmm I'm just trying to exactly see the difference because I've derived Rabi flopping for a two level system just using the TDPT formalism (although there was no truncation there so I guess it wasn't a perturbation theory). Is it the case that if we included an infinite number of orders here we would still get the Rabi flopping? So it's not that the formalism excludes it but that taking the result only to first order (for Fermis Golden Rule) does? $\endgroup$
    – Alex Gower
    May 7 at 9:43
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    $\begingroup$ S-matrix originates in wave scattering problems, which are essentially static in time. The adiabatic switching is used to assume that the initial and the final states are solutions of the Hamiltonian in distant past and the distant future. FGR studies dynamic situation - the rate of transitions. But sudden switching is also a mathematical trick, since one always takes the long-time limit. In your question you are interested in the transient situation, which is really not the point of the FGR. $\endgroup$ May 7 at 10:30

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