Non-degeneracy of ground state in quantum mechanics

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.

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.

A stronger result is true: the ground state of $H=-\Delta+V(x)$ (if one exists, that is, if the spectrum is bounded below and its minimum is an eigenvalue) is positive pointwise.

For a sketch of a proof, recall that an eigenfunction $y(x)$ with eigenvalue $\min\sigma$ minimizes the quadratic form $Q(y)=\int (|\nabla y|^2+V|y|^2)$. We can in fact assume that $y$ is real valued (take the real or imaginary part otherwise).

Notice that if $y\in H^1$ minimizes $Q$, then so do $|y|$ and $y\pm |y|$. If we now repeat Riemann's error from his proof of the Riemann mapping theorem and just assume that a minimizer has enough smoothness so that it will be a classical solution of the Euler-Lagrange equation, which here is just the original eigenvalue equation $-\Delta y+Vy=E y$, then we're already done (or close at least), because $y\pm |y|$ will not have second derivatives if the original function took values of both signs. One can make a rigorous proof out of this, based on the maximum principle for second order elliptic equations; see Theorem 6.5.2 of Evans's PDE book.

Finally, to make the connection to the original question more explicit, let me state more clearly what a rigorous version of this argument proves: any minimizer $y$ of $Q$ satisfies $y(x)>0$ after multiplication by a suitable constant. Since minimum energy eigenfuctions are minimizers, we can't have more than one, or we could take a linear combination that violates this condition.