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

angular momentumhilbert-spacequantum mechanicsschroedinger equationwavefunction

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?

Best Answer

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.