# Non-isolated ground state of a Schrödinger operator

Question. Does there exist a dimension $$d \in \mathbb{N}$$ and a measurable function $$V: \mathbb{R}^d \to [0,\infty)$$ such that the smallest spectral value $$\lambda$$ of the Schrödinger operator $$-\Delta + V$$ on $$L^2(\mathbb{R}^d)$$ is an eigenvalue, but not an isolated point of the spectrum?

I would expect this to be known, but I could not come up with an example (neither myself nor by browsing some manuscripts about Schrödinger operators).

• You know that there are things like random Schr\"odinger operators which have dense pure point spectrum?-So the answer should be yes. But I guess there are more pedestrian examples. Feb 25, 2019 at 23:10
• Related? Feb 25, 2019 at 23:25
• Set $\phi(x)=(1+x^2)^{-1}$ and $V(x) = (\phi(x))^{-1}\Delta \phi(x) = (1+x^2)^{-2}(6 x^2-2)$ in dimension $d = 1$ (a similar example can be clearly given in any dimension). Then $\phi$ is a $0$-eigenvalue, and $-\Delta+V$ is non-negative definite: the bottom of the essential spectrum is $0$ (because the potential decays at infinity), and, if I am not mistaken, there are no negative eigenvalues. Feb 25, 2019 at 23:27
• @KeithMcClary: Thanks for the link! I found the paper of Simon that is discussed there to be quite helpful.. Feb 26, 2019 at 7:06
• I added a somewhat more detailed answer. I do not think one can refer to Kato's result: if I remember correctly, it is about positive eigenvalues only. Feb 26, 2019 at 14:07

Yes, it is perfectly possible to have an embedded eigenvalue at the bottom of the spectrum. I do not have a reference (although I am quite sure there is one), but here is a simple example in dimension $$d = 1$$. Extension to higher dimensions is immediate.

Let $$\phi(x) = \frac{1}{1 + x^2}$$ and $$V(x) = \frac{\Delta \phi(x)}{\phi(x)} = \frac{6 x^2 - 2}{(1 + x^2)^2} \, .$$ Then:

1. $$\phi \in L^2$$, $$-\Delta \phi + V \phi = 0$$ (by definition of $$V$$), and so $$\phi$$ is an eigenfunction with eigenvalue $$0$$;

2. $$V(x) \to 0$$ as $$|x| \to \pm \infty$$, so that the essential spectrum of $$-\Delta + V$$ is $$[0, \infty)$$;

3. $$-\Delta + V(x)$$ has no negative eigenvalues: if there were any, then the ground state would be orthogonal to $$\phi$$, and so it would necessarily change sign, a contradiction with the Courant–Hilbert nodal domain theorem.

Thus, $$0$$ is the bottom of the spectrum of $$-\Delta + V$$, and it is both an eigenvalue and a point in the essential spectrum, as desired.

• For the nodal theorem don't you need the potential to be non-negative?
– lcv
Mar 26, 2019 at 3:45
• @lcv: I do not think so: all one needs to know is that all nodal parts have equal Rayleigh ratio, and the unique continuation principle. (And one can always replace $V(x)$ by $V(x) + 2$ to have a non-negative potential.) Mar 26, 2019 at 9:16
• I couldn't find the nodal theorem in presence of external potential but I suspect is similar to Perron-Frobenius, in which case you need non-negative diagonal. Of course $V+2$ is positive but then the ground state has energy 2. I still believe your result is correct but..
– lcv
Mar 26, 2019 at 9:38
• Do you have a reference for that result (nodal theorem in presence of external potential)?
– lcv
Mar 26, 2019 at 9:42
• @lcv: Off the top of my head, I do not. A quick google search leads to this paper by Ancona et. al. While it only deals with bounded domains and potentials, it may contain some useful references. Mar 26, 2019 at 10:06