# Spectrum near zero of $-\partial^2_x + V : L^2(\mathbb{R}) \to L^2(\mathbb{R})$, where $V = O(|x|^{-2 - \delta})$

Let $$H = -\partial^2_x + V(x) : L^2(\mathbb{R}) \to L^2(\mathbb{R})$$ be a one dimensional Schrödinger operator, where the potential $$V$$ is real-valued, belongs to $$L^\infty(\mathbb{R})$$, and, as $$|x| \to \infty$$, decays as $$|x|^{-2 - \delta}$$ for some $$\delta > 0$$. Since $$V \in L^\infty$$, by the Kato-Rellich theorem, the operator $$H$$ is self-adjoint with respect to the domain the Sobolev space $$H^2(\mathbb{R})$$.

I would like to show that $$0$$ cannot be an eigenvalue of $$H$$, that the spectrum of $$H$$ on $$(-\infty, 0)$$ consists only of eigenvalues, and finally that there exists $$\mu > 0$$ sufficiently small so that $$(-\mu, 0) \subseteq \rho(H)$$.

These properties of $$H$$ are asserted but not proved in the article by Jensen and Nenciu, Rev. Math. Phys. 13(6) (2001). It seems that the above properties are quite classical, because I have consulted many of the prominent modern references (the books of Reed--Simon and Yafaev, articles by Jensen, Kato, Gesztesy, etc.), but have not found any concrete discussion about them. Hint, solutions, or pointers to relevant references are greatly appreciated. I would like to establish each of these properties without using too much abstract machinery, if possible.

Letting $$H_0 = -\partial^2_x$$, I think I have worked out that, due to the decay of $$V$$, $$V(H_0 - \lambda^2)^{-1}: L^2(\mathbb{R}) \to L^2(\mathbb{R})$$ is compact in $$\text{Im} \, \lambda> 0$$. We should also have $$(H - \lambda^2)^{-1} = (H_0 - \lambda^2)^{-1}(I + V(H_0 - \lambda^2)^{-1})$$ exists for $$\text{Im} \, \lambda \gg 1$$ and (by the analytic Fredholm theorem) is meromorphic in $$\text{Im} \, \lambda> 0$$. But this is as far as I have got toward proving any of the above properties.

As a general rule of thumb, it's usually most convenient in one-dimensional problems to work with solutions of the ODE $$-y''+Vy=Ey$$ rather than operator theoretic methods.

Here, everything follows from the behavior of the $$E=0$$ solutions. To analyze the spectrum below $$E=0$$, we can replace the actual potential by a smaller potential (by min-max). The borderline case for finite spectrum below $$E=0$$ is $$V_0=-(1/4)|x|^{-2}$$. This we can see by solving $$-y''-cx^{-2}y=0$$ explicitly (for $$x\ge 1$$, say), which is possible since this is an Euler equation. For $$c<1/4$$ the solutions have only finitely many zeros, and then the claim follows from oscillation theory.

Finally, $$E=0$$ cannot be an eigenvalue because a square integrable solution $$y$$ would also solve $$y(x)=1+\int_x^{\infty} (t-x)V(t)y(t)\, dt ,$$ but then $$y\to 1$$ as $$x\to\infty$$.

• How you get $1$ (or any non-zero number) in the displayed formula? It looks like a circular argument.. but probably I missed something. Commented Dec 27, 2022 at 11:38
• Thank you for your answer. I am also struggling to see how $1$ or any nonzero constant appears in the display. Here is my approach. If $y'' = -Vy$ is in $H^2$, then $y'$ and $\int_x^\infty Vydt$ have the same distributional derivative, thus $y' = c + \int_x^\infty Vydt$ for some $c$. Sending $x \to \pm \infty$ gives $c = 0 = -\int_{-\infty}^{\infty} Vydt$. Performing a similar "antidifferentiation" calculation gives $y = - \int_x^\infty (t -x)Vydt$ and I found $\int_{-\infty}^\infty tVydt = 0$.
– JZS
Commented Dec 27, 2022 at 13:14
• @GiorgioMetafune: You are both right, I was a bit sloppy. I didn't really think about this but just adapted the usual approach to discuss Jost solutions to $E=0$ (the $1$ on the RHS is what one needs to obtain the solution with $y=1$ for large $x$ for compactly supported $V$). But I think essentially the argument is fine: View the equation I wrote as an integral equation for $y$. This has a solution (by iteration) $y$, and $y\to 1$. Moreover, $y$ also solves $-y''+Vy=0$. A similar argument with the modified equation $y=x+ \ldots$ gives a second solution $y\simeq x$ (cont'd) Commented Dec 27, 2022 at 18:09
• (cont'd) Obviously, no linear combination of these is square integrable, so $-y''+Vy=0$ doesn't have $L^2$ solutions. Commented Dec 27, 2022 at 18:09
• @JZS: Yes, right again, nothing is very obvious here. The statement (in a more general version) is Lemma 3.1.4 in Marchenko, and it has a lengthy proof. The integral equation to consider is $y=x+\int_0^x \ldots$, this avoids the convergence issues you pointed out. Commented Dec 27, 2022 at 22:19