# Why can we "wait for decay to the ground state quite generally"?

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?

• Yep, the finite temperature of any system with $T>0$ prevents 100% decay Commented Aug 8, 2023 at 14:04
• 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 researchgate.net/figure/… , so that the remaining excited states are very small in number Commented Aug 8, 2023 at 14:06
• @annav I am confused, the ground state itself is often a bound state, no?
– EE18
Commented Aug 8, 2023 at 14:21
• 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? Commented Aug 8, 2023 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
Commented Aug 8, 2023 at 16:13