most recent 30 from http://mathoverflow.net 2013-06-19T04:35:07Z http://mathoverflow.net/feeds/question/27016 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27016/non-degeneracy-of-ground-state-in-quantum-mechanics Onkar 2010-06-04T05:41:04Z 2010-06-23T03:11:25Z

<p>In non-relativistic quantum mechanics, what are the necessary conditions on the potential (or on the hamiltonian in general) for the ground state to be non-degenrate? </p>

http://mathoverflow.net/questions/27016/non-degeneracy-of-ground-state-in-quantum-mechanics/27030#27030 Jamie Vicary 2010-06-04T09:01:22Z 2010-06-04T09:01:22Z

<p>If a finite number of non-relativistic particles are moving in an infinite potential well, then the combined system has a nondegenerate ground state, regardless of the symmetry of the hamiltonian. I remember this from a long time ago, and I always thought it was impressive. I also remember I was always annoyed that I didn't know how to prove it, or know a reference where I can look it up. If you find one, let me know!</p> <p>There's probably some sort of fancy entropic argument that you could use to get this result, if that's your thing.</p> <p>If the potential was bounded above, I can't see immediately why this should create degeneracy on the ground state --- so it's plausible that the theorem holds in this case as well.</p> <p>Systems containing infinite systems of particles can, and often do, exhibit degeneracy in their ground state.</p>

http://mathoverflow.net/questions/27016/non-degeneracy-of-ground-state-in-quantum-mechanics/27045#27045 Tim van Beek 2010-06-04T13:09:36Z 2010-06-04T13:19:13Z

<p>I think you can find an answer to your question in the book of Simon/Reed, "Methods of Mathematical Physics", vol.4 "Analysis of Operators". They have a chapter devoted to the question of the existence of nondegenerate ground states, chapter XIII.12.</p> <p>One relevant theorem would be XIII.47, which says that the Schrödinger operator has a nondegenerate strictly positive ground state if the potential V is in $L^2_{loc}(\mathbb{R}^n)$ and $lim_{|x| \to \infty} V(x) = \infty$.</p> <p>I don't think that there is a simple necessary condition on the potential, but only several sets of sufficient conditions, but could be wrong about that.</p>