Recognitions:

## Particle in superposition of energy eigenstates and conservation of energy.

 Quote by Demystifier As long as a closed system (e.g., the whole Universe) evolves unitarily, an energy eigen-state must necessarily evolve into an energy-eigenstate. This seems to imply that creation of coherent superpositions of different energies is impossible. But for such a system the state does not change with time, so where does the dependence on time come from? The catch is that in all interpretations of QM, the system as a whole does NOT evolve unitarily. (For example, the wave function collapse is not unitary.) The only exception is many-world interpretation, but even there the physical time needs to be identified with some clock-configuration variable, which again leads to an effective non-unitary "evolution" with respect to the clock time. As a consequence, creation of coherent energy superpositions is possible, in all interpretations.
Incorrect. Hamiltonian of the closed universe includes all the measurements in said universe. That means if you are in an eigen state of Hamiltonian, the measurement won't change that either. It's only when you are considering sub-systems that distinction is relevant.

The only way you can have dynamics in the universe is if universe is not in an eigen state of the Hamiltonian. And why should it be in an eigen state? There are infinitely more states than there are eigen states.

So we don't need to discuss collapse here. The universe is in some super-position of eigen states, and that's enough.

Note that regardless of what kind of superposition you have, [H, H] = 0. So expectation value of energy does not change in time. Yes, energy of the universe is not an eigen energy. But it's always the same, so there are no issues with conservation laws.
 @Demystifier Thanks for that, but I'm still mystified ;-) Consider an isolated H-atom in an excited state (for example 2p). This is an energy eigenstate and stationary. However, we know that the H-atom can emit a photon and go to the ground state. The total energy of the system does not change. So can the system evolve from being excited into a superpositon of excited + (ground state +photon)? If not, is the actual photon emission of an excited atom due to some external disturbance? An interaction with the vacuum state of the photon field?

Recognitions:
 Quote by Sonderval However, we know that the H-atom can emit a photon and go to the ground state. The total energy of the system does not change. So can the system evolve from being excited into a superpositon of excited + (ground state +photon)?
It's not a closed system. In a perfect closed system, an atom in 2p state stays in 2p state. It will never decay. It takes a small perturbation from outside, plus the electromagnetic field to get the decay started. If you place an atom in a perfectly closed box, even if you "nudge" it to decay, it will eventually return into an excited state again. It's only in an open system that decay is irreversible.
 @K^2 Thanks, but I'm still not 100% sure I understand what's going on: Let's assume I "nudge" the system somehow by a quantum process - then it would be in a superposition state. Am I right in assuming that the "nudge-system" itself cannot be in an energy-eigenstate, otherwise it could not change its state and do the nudge?
 Recognitions: Science Advisor referring to the whole universe and its Hamiltonian may be dangerous b/c (as we know from GR and several QG proposals) energy can neither be defined canonically nor via Noether theorem in general; the problem of the construction of a diff. inv. Dirac observable is - afaik - not yet solved; we have H ~ 0 i.e. H does neither correspond to an energy operator, nor to a time-evolution operator

Recognitions:
 Quote by Demystifier Why? If it is not closed then it interacts with something else, but if there is something else to interact with, then the "whole" universe was not really whole in the first place.. In other words, by DEFINITION, the whole universe must be closed.
Usually we define a quantum system by its interaction with a classical surrounding.
What is the classical surrounding of the whole universe?

Recognitions:
@tom.stoer Agreed. Unfortunately, we don't really have a better description yet. We can talk what happens if we can treat universe as a closed system with a definite Hamiltonian, under assumption that such a field theory even exists. Whether it does exist is a separate question.

But yes, this is why all the real work is not being done on true closed systems. They are done on partially isolated sub-systems. Hence the golden rule and pretty much the entire stat mech.
 Quote by Sonderval Let's assume I "nudge" the system somehow by a quantum process - then it would be in a superposition state. Am I right in assuming that the "nudge-system" itself cannot be in an energy-eigenstate, otherwise it could not change its state and do the nudge?
Right. In this case, the "nudge" has to be external to the sub-system.

 Quote by Demystifier As a consequence, creation of coherent energy superpositions is possible, in all interpretations.
Are you talking about exact coherent states, involving an infinite (and not finite) number of photons? If that's the case then we would be able to draw infinite energy from such a state by subtracting single photons each time.. wouldn't we?

Blog Entries: 19
Recognitions:
 Quote by JK423 Are you talking about exact coherent states, involving an infinite (and not finite) number of photons?
No, by "coherent superposition" I meant a pure state, as opposed to a mixed state.

Blog Entries: 19
Recognitions:
 Quote by DrDu Usually we define a quantum system by its interaction with a classical surrounding.
I don't think that it is how we usually define a quantum system, although I admit that some pragmatic physicists do prefer to define quantum systems in that way.
 Quote by DrDu What is the classical surrounding of the whole universe?
Nothing, of course.

Blog Entries: 19
Recognitions:
 Quote by Sonderval Consider an isolated H-atom in an excited state (for example 2p). This is an energy eigenstate and stationary. However, we know that the H-atom can emit a photon and go to the ground state. The total energy of the system does not change.
H-atom in an excited state is an energy eigenstate of the free-atom Hamiltonian. However, it is not an energy eigenstate of the total Hamiltonian including the interaction with quantum electromagnetic field. This is why this system is unstable.

Blog Entries: 19
Recognitions:
 Quote by K^2 Incorrect. Hamiltonian of the closed universe includes all the measurements in said universe. That means if you are in an eigen state of Hamiltonian, the measurement won't change that either. It's only when you are considering sub-systems that distinction is relevant. The only way you can have dynamics in the universe is if universe is not in an eigen state of the Hamiltonian. And why should it be in an eigen state? There are infinitely more states than there are eigen states. So we don't need to discuss collapse here. The universe is in some super-position of eigen states, and that's enough.
So where (or when) do we need to discuss collapse, if not here? And can collapse ever transform an eigenstate of Hamiltonian into a non-eigenstate of Hamiltonian?

Recognitions:
 Quote by Demystifier I don't think that it is how we usually define a quantum system, although I admit that some pragmatic physicists do prefer to define quantum systems in that way.
If you are talking about pure vs mixed states, that is exactly how you are defining your system. The distinction is only relevant statistically, and that implies an external system.
 H-atom in an excited state is an energy eigenstate of the free-atom Hamiltonian. However, it is not an energy eigenstate of the total Hamiltonian including the interaction with quantum electromagnetic field. This is why this system is unstable.
Excited atom + ground state EM vacuum is still an eigen state of such a system.

But stability is a separate issue. 2p is unstable even in basic Hydrogen atom Hamiltonian. 2p + ε 1s already has dipole moment and will radiate. So a small perturbation will result in decay. That's the definition of instability.

Blog Entries: 19
Recognitions: