# Non-degeneracy of ground state in quantum mechanics

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?

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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!

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.

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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 $lim_{|x| \to \infty} V(x) = \infty$.
That's tricky! Allow some handwaving: If the potential goes to $\infty$ that means particles are confined to a bounded region (the probability to find them outside is very low/ can be made arbitrarily low). In a bounded region in classical physics there can be different locations which minimize the potential energy, but in quantum mechanics the ground state describes the probability to find it in any of those -> therefore it is unique in QM. But note that you can get degenerate ground states by dropping the other assumption on V ($V \in \L^2_{loc}$). –  Tim van Beek Jun 4 '10 at 18:03