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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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1 Answers1

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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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