# 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 Schr?dinger 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 Schr?dinger 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. – Sascha Feb 25 at 23:10
• Related? – Keith McClary Feb 25 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. – Mateusz Kwa?nicki Feb 25 at 23:27
• @KeithMcClary: Thanks for the link! I found the paper of Simon that is discussed there to be quite helpful.. – Jochen Glueck Feb 26 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. – Mateusz Kwa?nicki Feb 26 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 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.) – Mateusz Kwa?nicki Mar 26 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 at 9:38
• Do you have a reference for that result (nodal theorem in presence of external potential)? – lcv Mar 26 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. – Mateusz Kwa?nicki Mar 26 at 10:06