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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).

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    $\begingroup$ 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. $\endgroup$
    – Sascha
    Commented Feb 25, 2019 at 23:10
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    $\begingroup$ Related? $\endgroup$ Commented Feb 25, 2019 at 23:25
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    $\begingroup$ 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. $\endgroup$ Commented Feb 25, 2019 at 23:27
  • $\begingroup$ @KeithMcClary: Thanks for the link! I found the paper of Simon that is discussed there to be quite helpful.. $\endgroup$ Commented Feb 26, 2019 at 7:06
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    $\begingroup$ 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. $\endgroup$ Commented Feb 26, 2019 at 14:07

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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.

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  • $\begingroup$ For the nodal theorem don't you need the potential to be non-negative? $\endgroup$
    – lcv
    Commented Mar 26, 2019 at 3:45
  • $\begingroup$ @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.) $\endgroup$ Commented Mar 26, 2019 at 9:16
  • $\begingroup$ 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.. $\endgroup$
    – lcv
    Commented Mar 26, 2019 at 9:38
  • $\begingroup$ Do you have a reference for that result (nodal theorem in presence of external potential)? $\endgroup$
    – lcv
    Commented Mar 26, 2019 at 9:42
  • $\begingroup$ @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. $\endgroup$ Commented Mar 26, 2019 at 10:06

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