TITLE: Non-degeneracy of ground state in quantum mechanics
QUESTION [12 upvotes]: 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?

REPLY [2 votes]: 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.