 Solution

Apply the Frobenius method at the origin. The lowest power in $f^{\prime \prime }-\ell (\ell +1)f/r^2$ is $r^{c-2}$, with coefficientThe terms $-2\alpha f^{\prime }$ and $2mKZf/({\hslash }^{2}r)$ start at power $r^{c-1}$ and therefore cannot cancel a nonzero $r^{c-2}$ coefficient. Because $a_0\ne 0$, the indicial equation isIts roots are $c=\ell +1$ and $c=-\ell$. Since $\psi =r^{-1}f{e}^{-\alpha r}$, their leading behaviors are respectively $\psi \sim r^{\ell }$ and $\psi \sim r^{-\ell -1}$. The physical choice isThis is the Coulomb radial origin condition. For $\ell \ge 1$, the other branch is not even locally normalizable: its radial probability integral behaves as ${\int }_0^{\epsilon }{r}^2{r}^{-2\ell -2}dr={\int }_0^{\epsilon }{r}^{-2\ell }dr$, which diverges.

For $\ell =0$, the singular behavior $1/r$ is locally square-integrable with the measure $r^{2}dr$, so local wavefunction normalization alone would not reject it. The ordinary three-dimensional Coulomb potential problem additionally requires the regular boundary condition $u(0)=0$. A $1/r$ branch violates that condition and has divergent kinetic energy integral ${\int }_0^{\epsilon }{r}^2|{\partial }_r(1/r){|}^{2}dr={\int }_0^{\epsilon }{r}^{-2}dr$. Thus it is excluded from the physical energy domain as well. The regular choice gives $u\sim r^{\ell +1}$ in every case.