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As it pertains to state preparation, but also quite generally, why should the method of waiting until decay to the ground state work at all?

Ballentine writes "It is possible to prepare the lowest energy state of a system simply by waiting for the system to decay to its ground state." What conditions on the system are necessary for this to be true? Does one require $T \to 0$? Are there other conditions for which this method works? Or is this just a "general rule of thumb" as long as we shield our isolated quantum system?

EE18
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  • Yep, the finite temperature of any system with $T>0$ prevents 100% decay – Quantum Mechanic Aug 08 '23 at 14:04
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    the only condition will be that the system is not in a bound state. If it is excited it will decay to the ground state with appropriate time constants for the material . relevant https://www.researchgate.net/figure/Decay-of-the-excited-state-by-a-combination-of-radiative-and-non-radiative-processes_fig1_268397370 , so that the remaining excited states are very small in number – anna v Aug 08 '23 at 14:06
  • @annav I am confused, the ground state itself is often a bound state, no? – EE18 Aug 08 '23 at 14:21
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    I've just checked in the book and Ballentine give justifications for his statement e.g. that the radiated photon heads off to infinity never to return, that the EM field around the system is in its ground state and that the temperature is low. This seems a pretty complete list to me. Can you say what more you think might be required? – John Rennie Aug 08 '23 at 15:02
  • @JohnRennie Ballentine gives specific instances/models, whereas I think I am looking for a more general framework for deciding whether a given system will decay to the ground state. Or is the best we can do is to say “if the system is such that it decays to the ground state…” and then analyze whether that premise holds on a case-by-case basis? – EE18 Aug 08 '23 at 16:13
  • There is always a non-zero probability for a transition to the ground state mainly because there is nothing prohibiting it. So a system will always visit its ground state in a finite time. The question is whether it will stay there. Ballentine's conditions are simply that once in the ground state there is nothing that will excite the system back out of it again. Given the huge range of quantum systems I think the "nothing to excite it back out" requirement will depend on the system. – John Rennie Aug 08 '23 at 16:18
  • Fair enough! I’d be happy to accept an answer to that effect if you’d like. Thanks very much for the help! @JohnRennie – EE18 Aug 08 '23 at 16:26
  • OK, I'll post my comment as an answer. – John Rennie Aug 08 '23 at 16:44
  • You need $k_b T \ll \hbar \omega_{10}$. – DanielSank Aug 08 '23 at 16:53
  • @DanielSank That presupposes that there is a thermal reservoir in contact with our system (that is, that the canonical ensemble is appropriate given the data), right? Probably a pretty fair assumption for most systems? – EE18 Aug 08 '23 at 17:14
  • sorry I meant stable bound state . excited bound states have a lifetime – anna v Aug 08 '23 at 19:17
  • @EE18 yes, correct. The strength of coupling between the reservoir and the quantum system is important as well, as it sets the time scale over which the quantum system relaxes. – DanielSank Aug 08 '23 at 19:17

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In any excited quantum system there is always a non-zero probability for the transition to the ground state. This probability can be small if the transition is prohibited, but if we wait long enough the system will always visit the ground state. The only question is whether the system will stay there.

Ballentine is considering a radiative transition in an isolated system like an atom, and he points out that when the system decays by emission of a photon the photon heads off to infinity making the transition to the ground state irreversible. He also comments that there must be no EM field capable of re-exciting the atom and the temperature must be low enough not to cause thermal excitation.

Given the huge range of quantum systems I think it's hard to make general statements about what restrictions are necessary to prevent the ground state from being excited by its environment.

John Rennie
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  • It is important for that to consider an open system...If the system is in any excited eigenstate of the Hamiltonian, a closed system will remain there forever. You implicitly assumed that, but I think it is worth to spell out. – Tobias Fünke Aug 08 '23 at 17:09
  • When you say “prohibited”, what do you mean? Obviously it’s not truly prohibited since, as you later say, decay eventually occurred? Does prohibited mean prohibited to some order in a perturbation theory argument? – EE18 Aug 08 '23 at 17:16
  • I know the term "prohibited" is used when the spin is not conserved. What is meant, is that the "prohibited transition" is far less likely to take place, since a spin flip has to occur. Look up inter-system-crossing for example. So I think in general "prohibited" means less probability due to breaking of some (non-strict) conservation law. – Martin Aug 09 '23 at 08:19
  • @EE18 This is really a separate question, but we normally consider a transition prohibited when there is a conservation law that (in principle) prevents it. However in the real world there is almost always some mechanism for evading the conservation law so in practice the transition is possible though it may be very slow. For example take the hydrogen line transition. This is forbidden because there is no electric dipole to couple to the electric field of a photon. However it can occur through magnetic dipole coupling but it takes 10⁷ years! – John Rennie Aug 09 '23 at 09:27