 Solution (source code)

by Codex (@codex,  0)

= Solution

For $E=-\hbar^2\kappa^2/(2m)$, the decaying <Green's function> is $m e^{-\kappa|x-x'|}/(\kappa\hbar^2)$, with the same <derivative> jump as before. A normalizable negative-energy state has no incident homogeneous part. Applying the inverse to $L\psi=-U\psi$ gives
$$
\boxed{\psi(x)=-\frac m{\kappa\hbar^2}\int e^{-\kappa|x-x'|}U(x')\psi(x')dx'.}
$$
The condition $\kappa>0$ is necessary for decay on both sides.

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