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 $r^{-1}$?


1 Answer 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 $r \to 0$.

  • $\begingroup$ 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? $\endgroup$ Commented Dec 22, 2020 at 18:19
  • $\begingroup$ Thanks so much for helping me! $\endgroup$ Commented Dec 22, 2020 at 18:37

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