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Consider a group of charged point(at least considered as such in this non-relativistic limit) particles such electrons,protons , nucleii alone in an empty infinite universe and NOT considering any internal structure nor spin nor magnetic interaction though can consider spin as far as symmetry, antisymmetry requirements such belonging to a certain Young diagram. And also assume atleast one pair of particles has opposite charge eg not all particles have the same sign of charge. Consider only coulomb potential and normal non-relativistic momentum. Has anyone seen a proof that for a group of charges of which it may be considered no bound states exist, which may be somewhat controversial, that the bound states are only continuous and there is no finite increment no matter how small but NOt zero between the whole system eigenstates. By bound states here I mean all states which have a NEGATIVE energy eigenvalue of which there definitely MUST be obviously. Here i am considering the zero point of energy as that of particles an infinite(in the limit) distance apart eg of infinitesimal wave function density at rest. For example such a zero point energy could also(and in fact more clearly obvious) be the case of a group of particles all of the same charge sign. I think you know what I mean.

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  • $\begingroup$ to be more clear the charged particles can be any hypothetical point charges of any finite positive mass and any finite charge- that is not necessarily restricted to electrons ,protons, nucleii $\endgroup$
    – user158293
    Sep 16, 2020 at 0:44
  • $\begingroup$ " for a group of charges of which it may be considered no bound states exist, which may be somewhat controversial, that the bound states are only continuous and there is no finite increment no matter how small but NOt zero between the whole system eigenstates" italics to point out two opposite statements in your question. Also "all of the same charge sign" make no sense for bound solutions in a schrodinger equation, by construction. In the schrodinger solutions the high n states tend to be continuous close to the free limit. $\endgroup$
    – anna v
    Sep 16, 2020 at 3:25
  • $\begingroup$ the "all of the same charge sign" was an example of my definition of what could be zero energy as one could more easily envision I specifically stated bound states with respect to at least one pair of particles of opposite charge for bound state. In fact i proclaim at least non-relativistic when all are of the same charge sign then this is in fact the only case when no bound states exist. I say in all other cases bound states must exist - i am defining bound state to mean less than zero energy eigenvalue with the above definition of the zero energy level. $\endgroup$
    – user158293
    Sep 16, 2020 at 5:18

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