 Solution

Let $\lambda =m{U}_0{a}^2/{\hslash }^2\ll 1$ and $ka\ll 1$. Inside a region of size $a$, the Schrödinger equation estimates the curvature as $a^2{\psi }^{\prime \prime }=O(\lambda +(ka{)}^2)\psi$. Matching to the incoming and reflected waves constrains the slope as well. To leading order the interior wave has nearly constant value $C$, while its small accumulated derivative change encodes the potential's integrated strength.

This is not a Born assumption that $C$ is close to one. At very low $k$ the wave can be strongly reflected even by a weak potential, and $C$ can be small. The matching solution in the next part gives $C\sim k/(k+im\gamma /{\hslash }^2)$. It implies a fractional variation across the range of order $ka+m|\gamma |a/{\hslash }^2$, which is small under the two specified conditions. ThusThe approximation requires a finite short-range potential with a controlled width and integral; it does not describe arbitrary long-range tails.