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Consider the transition rate of the evolution of the state $|i\rangle$ to a state $|f\rangle$ - the Fermi golden rule: $$ d \omega_{if} = 2\pi |\mathcal{M}_{fi}|^{2} \delta(E_{f} - E_{i})d\nu, \quad \mathcal{M}_{fi} = \mathcal{V}_{fi}+\sum_{n\neq i}\frac{\mathcal{V}_{fn}\mathcal{V}_{ni}}{E_{i}-E_{n}}+\dots $$ Here $V$ is the perturbation operator, with $V_{fi}(t) = e^{i\omega_{fi}t}\mathcal{V}_{fi}$, $|i,f\rangle$ are initial and final states being the eigenstates of the free Hamiltonian $H_{0}$ and $|n\rangle$ is the intermediate state.

The energy is conserved in the transition. However, it is not conserved for the intermediate states $|n\rangle$. The total Hamiltonian $H = H_{0}+V$ itself is a generator of time translations corresponding to the conservation of the energy. What is a formal reason for this? Has this anything to do with the approach of the perturbation theory (within which we divide the total Hamiltonian into two parts)?

Name YYY
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    I don't think this question has been asked clearly enough yet. Do you mean time-dependent perturbation theory? Could you write down the expression you are asking about? – Andrew Steane Dec 11 '19 at 22:34
  • @AndrewSteane : thanks, I have updated the question. – Name YYY Dec 11 '19 at 22:38
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    You’re still not consistent. If you’re talking about time-dependent perturbation, it literally means that energy is not conserved. – David Saykin Dec 11 '19 at 23:02
  • @DavidSaykin : how about the Fermi golden rule explicitly stating that the energy is conserved in presence of time-dependent perturbations? – Name YYY Dec 11 '19 at 23:05
  • @NameYYY Fermi's Golden rule states that $E_f = E_i + \omega$, where $\omega$ is the frequency of perturbation. If you set $\omega=0$ it means that perturbation is time independent. – David Saykin Dec 11 '19 at 23:07
  • @DavidSaykin : Tending $\omega$ to zero, we still have transitions and the intermediate states in which the energy is not conserved. – Name YYY Dec 11 '19 at 23:10
  • @NameYYY yes, if you want to calculate transition probability in the presence of time-independent perturbation, you can still use formulas from time-dependent perturbation theory and set $\omega=0$, I agree with you. Answer in first order in perturbation in the limit of large evolution time would result in famous Fermi's Golden rule with transition amplitude $M_{fi}=V_{fi}$ and factor $\delta(E_f-E-i)$ which can be interpreted as "only transitions between states with the same bare energies are non-vanishingly small (as $t\to\infty$)". Now you want second--order correction...to be continued.. – David Saykin Dec 11 '19 at 23:30
  • @NameYYY ...you somehow guess that transition amplitude would have the form above, but you guess wrongly that that time--dependent factor in front of the second term would be the same. It is not. – David Saykin Dec 11 '19 at 23:32
  • "However, it is not conserved for the intermediate states $|n\rangle$" $-$ the "intermediate states" are mathematical fictions. They are internal parts of the formalism and they are not assigned any physical reality by the formalism itself. Why would they energy need to be conserved "for" them? What does that statement even (really) mean? – Emilio Pisanty Dec 11 '19 at 23:54
  • @DavidSaykin : the next summand will have exactly the form I mentioned in the limit $t\to \infty$ and $\omega \to 0$, up to the time-dependent factor which will then give the golden rule. Anyway, I need to rephrase the question in order to make it mode clear. – Name YYY Dec 12 '19 at 08:17
  • @NameYYY I disagree with your edit. Second term would also acquire t-dependent factor which also a function of $E_n$. But even if your expression above is correct as is I still don’t understand what concerns you. As it is written right now $\delta$-function requires $E_f=E_i$ and $E_n$ could be arbitrary. – David Saykin Dec 12 '19 at 08:27
  • @EmilioPisanty : using this theory, we can consider, for example, particles scattering. Then the intermediate states can be considered as a particles (as they have correct dispersion relations). The behavior of the intermediate states often determines physical behavior of the scattering. For example, for the proton-electron scattering the intermediate state signals about the bounded state - the hydrogen atom (see a question https://physics.stackexchange.com/questions/519100/feynman-rules-and-the-hydrogen-atom). – Name YYY Dec 12 '19 at 08:27
  • @EmilioPisanty : The next important point is that such perturbation theory distinguish temporary and spatial coordinates, so we can state that the intermediate state exists during some amount of time, although being unmeasured. – Name YYY Dec 12 '19 at 08:27
  • @DavidSaykin : I have added the definition of $V_{fi}$ to the question. Using it, you can reproduce the formula in the limit $t\to \infty$. About the second question, see my answer to the comment of Emilio Pisantly. – Name YYY Dec 12 '19 at 08:38
  • @EmilioPisanty : The next important point is that such perturbation theory distinguish temporary and spatial coordinates, so we can state that the intermediate state exists during some amount of time, although being unmeasured. Finally, even if they are pure formal tool, we still need to explain why the energy is not conserved. – Name YYY Dec 12 '19 at 08:40
  • @NameYYY now that you made V time-dependent I disagree with your formula even more. If you want, I can write down correct one as an answer. But I feel like it’s not what you’re confused about. – David Saykin Dec 12 '19 at 08:40
  • @NameYYY maybe your confusion is that you believe that energy conservation that follows from the fact that Hamiltonian is t-independent and energy conservation from Golden rule are the same? They are not. First one is exact and it deals with exact energies (not $E_i$ and $E_f$). The second conservation of energies is approximate and deals with bare energies. – David Saykin Dec 12 '19 at 08:44
  • @DavidSaykin: the formula I've written is indeed correct in the limit $\omega\to\infty$. You may derive it or may not, but it really does not matter. Next, the fact that the energy is conserved "approximately" has nothing to do with the non-conservation of the energy for particular states where the non-conservation is not approximate. – Name YYY Dec 12 '19 at 09:23
  • @NameYYY I’m sorry but where did condition $\omega\to\infty$ come from? I was thinking that $\omega=0$. I still disagree with your expression even in that limit, don’t you think you miss some integrals? – David Saykin Dec 12 '19 at 09:28
  • @DavidZaykin: excuse me, this is a misprint. i am completely sure that the expression is correct, I have derived it when I was the 3rd year student. – Name YYY Dec 12 '19 at 09:35

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