# Is the spherical outgoing wave solution to the Schrodinger equation not a member of $L^2$?

I was reading a discussion about the Mott problem, where the authors discuss the outgoing spherical wave solutions to the Helmholtz equations $$\nabla^2 f = - k^2 f$$. This equation can also be identified with the time-independent Schrodinger equation for a particle subjected to spherically symmetric potential. The solution is given as $$f = \frac{e^{i {\bf k.R}}}{R}$$ where $${\bf R} = (x,y,z)$$.

The authors further discuss that this solution is not in $$L^2$$, and the probability interpretation fails for $$|f|^2$$, which you can find in the attached figure.

Can someone explain why the outgoing spherical wave solution to the Helmholtz equation is not in $$L^2$$ or the probability interpretation fails, as the authors discuss?

• @VaibhavK, Could you elaborate a bit? What exactly is not clear? Commented May 17 at 10:24
• Oops, my bad, I thought the image was a part of your question and thought you wanted to add some more after the "we write.." sentence. Commented May 17 at 10:30

Integrating in spherical coordinates, we have $$\|f\|_2 = \int |f|^2d^3\mathbf{R} = \int_0^\infty\left(\frac{1}{R^2}\right)\left(R^2dR\,d^2\Omega\right) = 4\pi\int_0^\infty dR = \infty.$$ So $$f$$ is not in $$L_2$$. Similarly, since $$|f|^2$$ cannot be scaled so that it integrates to $$1$$, there's no way to interpret it as a proper probability distribution.
• Thanks. It absolutely makes sense why $f$ is not in $L^2$. In my mind, I was always thinking that this integral will converge. Still, we can define some sort of relative probability using f, right? Gamov has used this wavefunction to describe the tunneling of $\alpha$ particles. Without some sort of probability description, using $f$ as a wavefunction would not make sense. I will look into this. Commented May 17 at 15:06