# Why is the solution of the radial Schrödinger equation valid at $r=0$?

The Schrödinger equation for a particle in a central potential is $$\left[\frac{p_r^2}{2m}+\frac{\ell(\ell+1)}{2mr^2}+V(r)\right]\psi(r,\theta,\varphi)=E\psi(r,\theta,\varphi).$$ This gives solutions of the form: $$\psi_\ell^m(r,\theta,\varphi)=\frac{y_\ell(r)}rY_{\ell m}(\theta,\varphi)$$ Where $$Y_{\ell m}$$ are the spherical harmonics and $$y_\ell(r)$$ is the solution to the equation: $$-\frac{\hbar^2}{2m}\frac{d^2y_\ell}{dr^2}+\ell(\ell+1)\frac{\hbar^2}{2mr^2}y_\ell(r)+V(r)y_\ell(r)=Ey_\ell(r)$$ The book that I am using (Messiah) states that the solutions are valid at the origin by disregarding solutions of the type $$br^{-\ell}$$ for constants $$b$$ thereby assuring that $$y_\ell(0)=0$$. My question is, how does this ensure that $$\psi_\ell^m(r,\theta,\varphi)$$ is a valid solution of the Schrodinger equation at the origin? Is it because $$y_\ell$$ goes to $$0$$ faster than $$r^{-1}$$?

If $$\psi$$ is to be a normalizable wavefunction, then the function $$y_l(r)/r$$ needs to be square-integrable. If $$y_l(r) \propto r^{\alpha}$$ near $$r = 0$$, then the radial portion of the integral for $$\psi^2(r,\theta,\phi)$$ in the region $$a \leq r \leq b$$ will be $$\int_a^b \frac{y_l(r)^2}{r^2} r^2 \, dr \approx \int_a^b r^{2 \alpha} \, dr = \frac{1}{2 \alpha} \left[ b^{2\alpha + 1} - a^{2 \alpha + 1}\right].$$ If $$\alpha < -1/2$$, then this can be seen to be divergent in the limit $$a \to 0$$, which would mean that the integral of $$\psi^2$$ over all of space (including the origin) would diverge and $$\psi$$ would not be a valid wavefunction. So we reject any solutions with this behavior as $$r \to 0$$.
• That clears a lot up. So basically, the reason $\psi$ is valid is because the singularity in $r^{-1}$ gets cancelled out by $y_l$. Have I got that right? Commented Dec 22, 2020 at 18:19