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When solving the Schrodinger equation in 2D polar coordinates, one has to deal with various Bessel functions. In the most simple example, the infinite circular potential well, the solutions to the radial differential equation are the Bessel functions of first $[J_m(kr)]$ and second $[Y_m(kr)]$ kind. One usually discards the $Y_m(kr)$ functions on account of their asymptotic behavior at $r = 0$, $$Y_m(kr) \sim (kr)^{-m}$$ and so they are not square integrable functions. However, in the case of zero angular momentum, $m=0$, the Neumann function of zeroth order, $$Y_0(kr) \sim \ln (kr),$$ although infinite at the origin, is square integrable! So why do we have to discard it as well? What are the boundary conditions that have to be satisfied by a radial wave function at the origin?

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In order to keep the kinetic energy $\langle K\rangle =\frac{\hbar^2}{2m} \int d^2x~ |\nabla \psi|^2$ finite, a logarithmic $\ln(r)$ singularity of the radial wave function $\psi(r)$ at $r=0$ is unacceptable. This conclusion holds even if we take the potential energy $\langle V \rangle$ into account:

  1. A non-negative potential $V\geq 0 $ will only make the total energy $E= \langle K \rangle + \langle V \rangle$ bigger.

  2. A potential $V$ with a power law singularity $V(r) \sim r^p$, $p>-2$, at $r=0$ would only have a finite contribution.

  3. A negative potential $V<0$ with a power law singularity $V(r) \sim r^p$, $p\leq -2$, at $r=0$ would lead to a spectrum for the Hamiltonian $H=K+V$ that is unbounded from below.

The general principles for imposing boundary conditions of the wavefunction $\psi$ at $r=0$ in various dimensions are also e.g. outlined in this related Phys.SE post and links therein.

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    $\begingroup$ OK, I see, thank you. But, is this a general principle? I understand that with no potential energy, the kinetic energy has to be finite. However, if there is some central potential, then the kinetic energy doesn't have to be finite, right? $\endgroup$ – user17116 Mar 1 '17 at 10:34
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    $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Mar 1 '17 at 13:01
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Generally, the solution must solve the Schrödinger's equation at each point of the configuration space. This in particular means that you should take the singular solution as a distribution. Neumann functions don't solve Schrödinger's equation in polar coordinates in every point: they would only solve it if it had an additional (source) term proportional to Dirac delta with its singularity at the origin. See my answer to a related question for extended discussion.

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  • $\begingroup$ Excellent answer! Can you recommend some further literature on the subject? $\endgroup$ – user17116 Mar 1 '17 at 15:18
  • $\begingroup$ Sorry, I don't know of any literature discussing this issue. I only saw guides to solving TISE which just looked for regular solutions without elaborating why. $\endgroup$ – Ruslan Mar 1 '17 at 16:53

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