# Wavefunction and a central potential $V(r)$ that is singular at origin

I read the following line from Weinberg's Lectures in Quantum Mechanics (pg 34):

As long as $$V(r)$$ is not extremely singular at $$r=0$$, the wave function $$\psi$$ must be a smooth function of the Cartesian components $$x_i$$ near $$x=0$$, in the sense that it can be expressed as a power series in these components.

What does he mean by the potential $$V(r)$$ being 'not extremely singular'? I thought of the electrostatic potential of a charge $$Q$$ placed at $$r=0$$, which is $$V(r)=\frac{kQ}{r}$$. This potential is infinite at $$r=0$$. Is this considered singular or extremely singular?

Also, why is the wave function $$\psi$$ a smooth function when $$V(r)$$ is not extremely singular? I only understand that when $$V(r')=\infty$$, the wave function is $$0$$ at the point $$r=r'$$, like at the barrier of an infinite square well.

Or more generally, what are the requirements for $$\psi$$ to be a smooth function and why?

• For atoms and ions $V(0)$ is infinite but $\psi_{ns}(0) \neq 0$. Jun 26 '20 at 10:31
• $V(0)$ for atoms and ions is negative infinity, which leads to a smooth though not differentiable wavefunction. If it were positive infinity, the wavefunktion would be indeed zero at that point. Jun 26 '20 at 11:06
• Jun 26 '20 at 11:30
• @MartinPeschel But non-differentiable functions cannot be expressed in terms of a taylor expansion/power series right? Why then does the author say that the wavefunction can be expressed as a power series? Jun 27 '20 at 0:58
• that is true. the cartesian form of the ground state wavefunction hydrogen is$Nexp(-\sqrt{x^2+y^2+z^2})$ which has no power series at $(0,0,0)$. Jun 28 '20 at 13:10