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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.

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1 Answer 1

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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 $x\to\infty$.

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  • $\begingroup$ How you get $1$ (or any non-zero number) in the displayed formula? It looks like a circular argument.. but probably I missed something. $\endgroup$ Commented Dec 27, 2022 at 11:38
  • $\begingroup$ 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 $\int_{-\infty}^\infty tVydt = 0$. $\endgroup$
    – JZS
    Commented Dec 27, 2022 at 13:14
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    $\begingroup$ @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) $\endgroup$ Commented Dec 27, 2022 at 18:09
  • $\begingroup$ (cont'd) Obviously, no linear combination of these is square integrable, so $-y''+Vy=0$ doesn't have $L^2$ solutions. $\endgroup$ Commented Dec 27, 2022 at 18:09
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    $\begingroup$ @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. $\endgroup$ Commented Dec 27, 2022 at 22:19

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