If a Hamiltonian is unbounded from below, say the hydrogen atom where the Hamiltonian is -∞ at r=0, is there a way to tell if the ground state is bounded (e.g. hydrogen is -13.6 eV and not -∞ eV)?(adsbygoogle = window.adsbygoogle || []).push({});

It seems if the potential is 1/r^2 or less, then the energy will be finite as:

[tex]\int d^3 r (1/r^2) P(r) = \int r^2 dr (1/r^2) P(r)=1[/tex]

where P(r) is the probability density.

Or is this too naive and you have to work out P(r) which can help the integral converge or make it diverge?

There are two things at play it seems, the potential and kinetic energies, and Heisenberg's uncertainty principle which keeps the hydrogen atom from falling into the nucleus.

Is there a way to minimize H(p,r)=p^2/2m+V(r) with respect to r and p, using the constraint ΔrΔp<h/2, and seeing if the solution is finite?

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# Unbounded Hamiltonian leading to finite ground state

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