What is the reason for assuming the scattered far-field wave function to be in the form of $f(\theta,\phi) \frac{e^{ik r}}{r}$?

Question

In many quantum mechanics textbooks and solid-state texts, the authors usually use the standard form of the scattered wavefunction $\psi_i(r) + f(\theta,\phi) \frac{e^{ik r}}{r}$ where $\psi_i$ is the incoming wavefunction and $f(\theta,\phi)$ some function of the detection angles. My question is how do I justify the choice of $\frac{e^{ikr}}{r}$ as the r-dependence part of the scattered wave?

The only justification that I have encountered so far is that in the far-field limit, the scattered wave should be an outward propagating wave. What puzzles me is, why wouldn't I write down something like $\frac{e^{ikr}}{r^2}$ or some superpositions of these functions ($\frac{e^{ikr}}{r}$, $\frac{e^{ikr}}{r^2}$, $\frac{e^{ikr}}{r^3}$, etc.).

Answer 1

You're almost there. You're asking the right questions, and you almost certainly have the facts you need to deduce the answer for yourself.

Just from geometry the outward wave intensity should fall as $1/r^2$ in 3D, right? And you form the intensity as $\psi^*\psi$, which means you want the $1/r$ term from among those that you suggest.

There is no reason the general solution can't include the stronger terms you list, but the far-field can be defined as the regime where those terms have become negligible.

Answer 2

While dmckee's answer is a very good one, I would add that the wave function $\psi(r) = \frac{1}{r} \, e^{i k r}$ is also an eigenfunction of the Laplacian in spherical coordinates (I'm removing the angular parts for simplicity): $$\tag{1} \nabla^2 \psi = \frac{1}{r^2} \, \frac{\partial}{\partial r} \Big( r^2 \, \frac{\partial \psi}{\partial r} \Big) = -\, k^2 \, \psi.$$ The equation $\nabla^2 \phi + k^2 \phi = 0$ is a wave equation. We could add a time function $e^{- i \omega t}$ to $\psi(r)$, so the function $\psi(t, r)$ satisfies $\square \, \psi = 0$, where $\square$ is the D'Alembertian.