**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).



Non-isolated ground state of a Schrödinger operator
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.




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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.


