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

Demystifier
Gold Member
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.

Demystifier
Gold Member
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.
What is the classical surrounding of the whole universe?
Nothing, of course.

Demystifier
Gold Member
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.

Demystifier
Gold Member
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?

K^2
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.

Demystifier