# Wave function boundary condition in scattering problem

The the boundary condition for a wave function in a scattering problem is $$\psi_{\boldsymbol{k}_{1}}(\boldsymbol{r}) \underset{r \rightarrow \infty}{\rightarrow} A\left(\exp \left(\mathrm{i} k_{\mathrm{i}} \cdot \boldsymbol{r}\right)+f(\theta, \phi) \frac{\exp (\mathrm{i} k r)}{r}\right)$$

Why people impose this boundary condition in the wave function? what is the physical intuition to impose this boundary condition? Actually what I am not understanding more, is imposition that the scattered wave should be spherical. Why are we certain that the outgoing wave is spherical?

Scattering problems draw their inspiration from scattering problems in classical physics - for example, a problem of an asteroid passing near the Earth and being deflected by it. Note that even in this classical physical problem, one cannot strictly distinguish the states before, during and after the collision, since the gravitational potential has infinite range. In quantum mechanics this situation is compounded by the fact that the wave solution should exist everywhere in space. The ansatz presented in the question is a solution for a particle with a definite momentum (hence the incident plane wave) being scattered by a centrally symmetric potential (hence the centrally symmetric outgoing solution). In QFT-oriented texts such scattering experssions are motivated by analyzing the time evolution from the distant past, $$t=-\infty$$, to the distant future, $$t=+\infty$$, but introductory texts often do not go into such depths... and sadly do not such problems as scattering by a rectangular barrier or tunnelinga s scattering problems at all.