For the radial equation

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Central potentials and canonical three-dimensional systems · Central potentials and canonical three-dimensional systems

The Structure of Quantum Mechanics, Ch Central potentials as a rotation problem

Problem

For the l=0l=0 radial equation −ℏ22μu′′(r)+V(r)u(r)=Eu(r)-\dfrac{\hbar^2}{2\mu}u''(r) + V(r)u(r) = Eu(r), with V(r)V(r) finite everywhere (including at r=0r=0), someone argues:"The equation is second order, so near r=0r=0 it has two independent solutions: one with u(0)=0u(0)=0, and one with u(0)≠0u(0)\ne 0 (equivalently R(r)=u(r)/r→const/rR(r)=u(r)/r \to \text{const}/r as r→0r\to 0). Normalisability is the only thing that can rule a solution out, and ∫0∞∣R(r)∣2r2 dr=∫0∞∣u(r)∣2 dr\int_0^\infty |R(r)|^2 r^2\,dr = \int_0^\infty |u(r)|^2\,dr is perfectly finite for the u(0)≠0u(0)\ne 0 solution too, because the factor r2r^2 in the measure exactly cancels the 1/r21/r^2 in ∣R∣2|R|^2 near the origin, leaving no divergence there. So both solutions are equally acceptable on normalisability grounds, and setting u(0)=0u(0)=0 is just a convention." ((a)) Is the claim that ∫0∞∣u∣2dr\int_0^\infty|u|^2 dr stays finite for the u(0)≠0u(0)\ne 0 solution correct? ((b)) Find the actual error in the argument, and state the correct reason u(0)=0u(0)=0 is required.

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