If there is a degenerate quantum state there is no preferred basis. If a field is applied, say a Zeeman effect, the degeneracy is removed and there are states with different energies. We might be quick to say the lowest energy state establishes the ladder of states. However, if we have a QED field in a perfect cavity there will then be a superposition of different energy states and photons. In effect there is no preferred basis. We have a preferred basis with deccoherence, but that only pushes the barrier back. In effect we are still left with the problem of how classical states are robust against quantum noise or decoherence so there is some reference used to define a preferred basis.
LC On Wednesday, October 9, 2019 at 1:28:38 AM UTC-5, Brent wrote: > > > > On 10/8/2019 9:20 PM, Alan Grayson wrote: > > I've argued this before, but it's worth stating again. It's a > > misintepretation of superposition to claim that a system described by > > it, is in all the component states simultaneously. As is easily seen > > in ordinary vector space, an arbitrary vector has an uncountable > > number of different representations. Thus, to claim it is in some > > specific set of component states simultaneously, makes no sense. Thus > > evaporates a key "mystery" of quantum theory, inclusive of S's cat and > > Everett's many worlds. AG > > No. It changes the problem to the question of why there are preferred > bases. > > Brent > > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/86364ad2-d096-4db1-894f-b265667f20df%40googlegroups.com.
