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

Question

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?

Answer 1

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.

Answer 2

The simplest form of the problem of this kind is the energy eigenstate for free particles wich is a plane wave and diverges if you take the integral of the square from -infinity to +infinitiy. One uses box normalization for practical purposes. The corresponding spectrum in momentum space is delta function. However the thruth is that there are neither delta functions(infinitely narrow functions with finite integral) nor infinitely extended waves in nature but wave packets of finite size however long it may be.