Weyl–Titchmarsh theory

Remarks

The quotient does not depend on the choice of $\psi$. Also $\psi(z, 0) \ne 0$, since otherwise the non-real number $z$ would be an eigenvalue of the self-adjoint operator with the Dirichlet condition $y(0) = 0$. The function $m$ is analytic on $\C \setminus \R$, satisfies $m(\overline{z}) = \overline{m(z)}$, and maps the upper half-plane into itself. So $m$ is a Herglotz function, and the measure in its Herglotz representation is the spectral measure of the operator with the condition $y(0) = 0$. Example: for $q = 0$, the solution $\psi(z, x) = e^{i\sqrt{z}\, x}$, with the branch of $\sqrt{z}$ that has positive imaginary part, is square integrable on $[0, \infty)$. So $m(z) = i \sqrt{z}$; for $z = i$ this is $e^{3\pi i/4}$, which lies in the upper half-plane. Non-example: for $z \in \C \setminus \R$ the solution $e^{-i\sqrt{z}\, x}$ grows exponentially, so it cannot be used for $\psi$. Some books use the other sign convention, or the boundary condition $y'(0) = h y(0)$ at $0$, which changes $m$ by a Möbius transformation.