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A quantum harmonic oscillator, or a quantum model of the hydrogen atom, have discrete energy spectrum. However, both models involve a potential, which means that the system is not isolated. One could think that in principle, the external environment responsible for the potential could be integrated into the model, so as to obtain an isolated system again. In the case of the hydrogen atom, this would be the wave function of nucleus+electron and the photon they could exchange. My questions are the following:

1/ is this idea correct? Is it possible to have a quantum model of nucleus+electron as an isolated system, without external potential, or more generally, to eliminate potentials by considering larger systems?

2/ if this is the case, would the energy spectrum still be discrete in such a larger system? Intuitively I would say no: an isolated system must have continuous energy spectrum but I'm not sure this intuition is correct.

3/ if this intuition is correct, isn't there a puzzle in the fact that the energy spectrum becomes discrete when the nucleus is represented as a potential rather than as a quantum system? Shouldn't both representations give the same result?

4/ if the intuition of (2) is not correct, where does discreteness "comes from" in an isolated complex system? Does it come from spliting the system in parts and having an interaction hamiltonian with discrete spectrum between these parts for example?

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  • $\begingroup$ Related: physics.stackexchange.com/q/39208/2451 and links therein. $\endgroup$ – Qmechanic Dec 23 '17 at 11:55
  • $\begingroup$ Thanks. I had read it before posting my question but it didn't really answer it, since boundary conditions are apparently involved for retrieving discreteness according to the answers, so the puzzle remains: why couldn't these boundary conditions be incorporated into the model, and then would we still have discreteness? $\endgroup$ – Quentin Ruyant Dec 23 '17 at 12:05
  • $\begingroup$ either classically, the potential between two charges, or quantum mechanically the potential is between the two charges, it is not external. $\endgroup$ – anna v Dec 23 '17 at 12:16
  • $\begingroup$ @annaV in Newtonian gravitation, in a model of free fall for example, we can introduce a potential associated with the Earth, but this is because the Earth is not part of the model. We could in principle come up with a larger model containing the Earth, and we would have gravitational forces between two objects instead of a linear potential applying to the object in free fall. Isn't it the same in quantum mechanics? $\endgroup$ – Quentin Ruyant Dec 23 '17 at 12:52
  • $\begingroup$ You are wrong. The gravitational potential of the earth and apple system is a 1/r potential, except that the mass of the earth is enormous with respect to the mass of the apple and one can use the approximation that leads to f=mg. instead of f~m1m2/r^2 . en.wikipedia.org/wiki/… (The potential is the same with a 1/r form.) en.wikipedia.org/wiki/… $\endgroup$ – anna v Dec 23 '17 at 14:50

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