Particle in superposition of energy eigenstates and conservation of energy.

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.

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.

Sonderval

@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?

K^2

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.

Sonderval

@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?

tom.stoer

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

DrDu

Usually we define a quantum system by its interaction with a classical surrounding.
What is the classical surrounding of the whole universe?

K^2

@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.
Right. In this case, the "nudge" has to be external to the sub-system.

JK423

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?

Demystifier

No, by "coherent superposition" I meant a pure state, as opposed to a mixed state.

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.
Nothing, of course.

Demystifier

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.

Demystifier

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?

K^2

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.
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.

Demystifier

DrDu was talking about a CLASSICAL surrounding, while what you say above refers to a QUANTUM surrounding.

It would be so if there was no interaction term in the Hamiltonian describing atom and EM field. But the interaction term is there, so what you say above is not correct.

Time_

The notion of a closed universe is quite interesting. Demystifier to expand on what you are saying, if energy eigenstates, so the different superpositions of a particle, must evolve into a coherent state, where does all this excess energy from these states go? Our universe,in accordance with things such as Pauli Exclusion can only have on outcome of a solution, and the physical existence of a particle in two different states seems impossible. Could it be that many worlds theory holds valid in that all these different energy eigenstates come together to form an infinite number of possibly universes, each branching from another, as a wave function collapses and the other superpositions become irrelevant to our world?