In nonrelativistic quantum mechanics, what are the necessary conditions on the potential (or on the hamiltonian in general) for the ground state to be nondegenrate?
If a finite number of nonrelativistic 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! There's probably some sort of fancy entropic argument that you could use to get this result, if that's your thing. 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. Systems containing infinite systems of particles can, and often do, exhibit degeneracy in their ground state. 


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. 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 $lim_{x \to \infty} V(x) = \infty$. 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. 

