# discrete spectrum of Schrödinger operator

Let $u \in C^{\infty}_c(\mathbb{R})$ and consider the Schrödinger operator $$L = - \frac{\partial^2}{\partial x^2} + u(x).$$ I am looking for bound state solutions to $L \psi = - \kappa^2 \psi$ as explained below. Since $u(x)$ has compact support, for large $|x|$ the eigenvalue equation looks like $$- \frac{\partial^2 \psi}{\partial x^2} + \kappa^2 \psi = 0,$$ the solutions to which are linear combinations of $\mathrm{e}^{\pm \kappa x}$. I am interested in the specific solution which is asymptotic to $\mathrm{e}^{-\kappa x}$ as $x \to +\infty$. Now, at the other end of the real line, such a solution should be asymptotic to $$\alpha(\kappa) \mathrm{e}^{\kappa x} + \beta(\kappa) \mathrm{e}^{-\kappa x}, \qquad x \to -\infty,$$ and I wish to require that $\beta(\kappa) = 0$. I have a feeling that there should only be finitely many values of $\kappa$ for which this is satisfied, but I don't really have a clue how to prove it. Krein-Rutmann Theorem-like arguments I think may give you that there are only countably many such eigenvalues, but that's a general theorem and I don't see how it should say anything about the particular solutions I am considering.

• There are only finitely many negative eigenvalues. This question is particularly discussed in connection with the inverse scattering method as applied to the Korteweg-de Vries equation. – ifw Jan 29 '17 at 21:05
• Yes, I am trying to understand this in connection with the inverse scattering transform. Could you point to a reference that explains this issue? – onamoonlessnight Jan 29 '17 at 21:40
• This is e.g. explained in Chapter 2 of Solitons, Nonlinear Evolution Equations and Inverse Scattering by Ablowitz and Clarkson. See there in particular Lemma 2.2.2 and its proof. – ifw Jan 29 '17 at 22:18

The number of bound states is indeed finite. This can be proved as follows. First of all, for a bound state your eigenvalue $-k^2$ must be real. This is because your
operator with real $u$ and zero boundary conditions at $\pm\infty$ is self-adjoint. Next, if real $\kappa^2$ is very large by absolute value, then you cannot have a bound state: if $\kappa^2$ is very negative, $\psi$ oscillates at $\infty$, if $\kappa^2$ is very positive, $\phi$ cannot be zero on both infinities by the Sturm comparison theorem. It remains to notice that your function $\beta(\kappa)$ is analytic, so it cannot have infinitely many zeros on a finite interval.
• $\beta(\kappa)=W(e^{-\kappa x}, \psi)/(2\kappa)$ is actually not holomorphic at $\kappa =0$, which is exactly the point that matters here, but of course there are alternative arguments that show that there are only finitely many eigenvalues (oscillation theory works fine). – Christian Remling Jan 30 '17 at 22:17
• Just a pole at $\kappa=0$ does not alter my argument. – Alexandre Eremenko Jan 31 '17 at 7:55