quantum mechanics - Am I missing a trick to solving a 3D potential well problem?

I was playing around with a 3-D potential $V$ such that $V_{(r)} = 0$ for $r0$ otherwise. By using the Schrödinger Equation, I showed that: $$\frac{-\hbar}{2m}\frac{1}{r^2}\frac{d}{dr}\bigl( r^2\frac{d}{dr}\bigr)\psi = E\psi$$

I then used the substitution $\psi_{(r)}=f_{(r)}/r$ and $k=\sqrt{2mE}/\hbar$ to get: $$\frac{1}{r}\frac{d^2f_{(r)}}{dr^2}=-\frac{k^2}{r}f_{(r)} \tag{I}$$

which describes the wavefunction $\psi_{(r)}=f_{(r)}/r$ inside the sphere. Hence, the differential equation has the domain $0\leq r

If I do the same thing to $(I)$, I obtain the equation for simple harmonic motion, but substituting the solution back into $(I)$ as a sanity check gives a division by zero when evalutating for $r=0$. After that, I tried a number of substitutions to make $(I)$ have a more recognisable form - to no avail. Then I had the idea of multiplying my trial solution by some other function of $r$ so that upon substitution into $(I)$, the evaluation of $r=0$ doesn't give an infinity... but I don't know quite how to do that...

Long story short..... my question is: what trick do I need to get a meaningful solution to $(I)$?

Answer

I) The substitution $f=r\psi$ is the standard substitution to get a radial 3D problem to resemble a 1D problem, see e.g. Ref. 1.

II) From the perspective of the normalization of the wavefunction $\psi(r)$, a $1/r$ singularity of $\psi(r)$ at $r=0$ is fine because $|\psi(r)|^2$ is suppressed by a Jacobian factor $r^2$ coming from the measure in 3D spherical coordinates.

III) However in order to keep the kinetic energy $K=\frac{\hbar^2}{2m} \int d^3x~ |\nabla \psi|^2$ finite, a $1/r$ singularity of $\psi(r)$ at $r=0$ is unacceptable, i.e., we must discard the cosine solution and only keep the sine solution. This corresponds to imposing that the wavefunction $\psi(r)$ should be bounded.

References:

1. D. Griffiths, Intro to QM, Section 4.1.3.