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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?

Deschele Schilder
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TaeNyFan
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  • For atoms and ions $V(0)$ is infinite but $\psi_{ns}(0) \neq 0$. – my2cts 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. – Martin Peschel Jun 26 '20 at 11:06
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    Related: https://physics.stackexchange.com/q/12892/2451 – Qmechanic 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? – TaeNyFan Jun 27 '20 at 00:58
  • 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 $(0,0,0)$. – Martin Peschel Jun 28 '20 at 13:10
  • Finally,what we can say?Is there any mathematical theorem ensures that what Weinberg said is right? – a.p Jul 11 '20 at 19:16

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