 Solution

For a decaying bound state take $\alpha >0$. Substituting $F=f{e}^{-\alpha r}$ gives $F^{\prime \prime }=e^{-\alpha r}(f^{\prime \prime }-2\alpha f^{\prime }+{\alpha }^{2}f)$. The term involving the energy cancels ${\alpha }^{2}f$, leaving the displayed equation withAt the lowest power of a series from the Frobenius method $f={\sum }_{k\ge 0}{a}_k{r}^{k+c}$, the indicial equation is $c(c-1)-\ell (\ell +1)=0$. Its roots are $\ell +1$ and $-\ell$. The latter makes $F\sim r^{-\ell }$ and the radial probability density integral $\int |F{|}^{2}dr$ diverge for positive integral $\ell$. Hence $\boxed{c=\ell +1}$.

Equating the remaining coefficients givesUnless the numerator vanishes, the regular confluent-hypergeometric series contains a large-radius growing branch with the exponential factor $e^{2\alpha r}$. Multiplication by the external $e^{-\alpha r}$ then leaves exponential growth, so it cannot be a normalizable bound state. Equivalently, the recurrence at large $k$ has $a_{k+1}/a_k\sim 2\alpha /k$, indicating that growing exponential scale; regularity at the origin has already fixed the combination of solutions, leaving termination as the way to remove it.

If the polynomial ends at degree index $N\ge 0$, its last recurrence numerator gives $A=2\alpha (N+\ell +1)$. With $n=N+\ell +1>\ell$,For the separate angle-independent exponential, normalization includes the full solid angle:Thus one can choose $\boxed{N=\sqrt{{\alpha }^3/\pi }}$, up to an overall phase. Angle independence corresponds to the $\ell =0$ extension, distinct from the positive-$\ell$ family used in the indicial discussion.