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

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.).

• Feb 12 at 15:50

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.
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.