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Given the Sturm-Liouville type (time independent Schroedinger) equation

\begin{equation} \frac{d^2 y}{d x^2} - \left(\mu + V(x)\right) y = \lambda \, y,\quad x \in \mathbb{R} \end{equation} where $V(x)$ is symmetric and exponentially decreasing as $|x| \to \infty$, choose the two linearly independent solutions $y_\pm$ such that $y_\pm(x)$ behaves as $e^{\pm\sqrt{\mu+\lambda}\,x}$ as $x \to \infty$.

Bounded solutions to this equation are characterised by standard Sturm-Liouville theory, yielding an ordered sequence of real eigenvalues $\lambda_0 > \lambda_1 > \ldots$ and corresponding eigenfunctions $\phi_i(x)$. The Sturm-Picone Comparison Theorem states that $\phi_n(x)$ has exactly $n$ zeroes; in extension, the oscillation properties of the nonbounded solutions $y_\pm$ are also completely determined for real values of $\lambda$.

Is there anything known for zeroes of $y_\pm(x)$ for complex $\lambda$? In other words, fix $x_* \in \mathbb{R}$ and consider $f(\lambda) := y_+(x_*)$ as a function of $\lambda \in \mathbb{C}$. Can we say anything about the zeroes of $f(\lambda)$?

N.b. Numerical investigations for some concrete $V(x)$ suggest that all zeroes of the corresponding $f(\lambda)$ lie on the real line.

edit: I should add: $\mu \in \mathbb{R}$.

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If $\mu$ is real, and $\sqrt{\mu+\lambda}$ is not pure imaginary, the zeros are real, because they are eigenvalues of one of the problems: $y(-\infty)=0,\; y(x^*)=0$ or $y(+\infty)=0,\; y(x^*)=0$, and these problems are self-adjoint.

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  • $\begingroup$ So, to reformulate your answer: consider the above Sturm-Liouville problem restricted to the halfline $(x_*,\infty)$, with homogeneous boundary conditions. Then, if $y_+(x_*) = 0$ for some $\lambda \in \mathbb{C}$, it follows that $y_+(x)$ solves the above restricted Sturm-Liouville problem, since it obeys the boundary conditions of the restricted problem. However, the restricted problem is itself again of S-L type and self-adjoint, so it follows that $y_+(x)$ solves the restricted problem if and only if $\lambda \in \mathbb{R}$. I should have thought of this myself. Thanks a lot! $\endgroup$ – Frits Veerman Sep 17 '14 at 13:34

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