In one dimension, motion in a Coulomb-like attracting potential $U=-\frac{\alpha}{|x|}$ is quite ill-defined. I'll now try to explain this for classical particles, then will say some words about quantum mode.

Let's consider two cases:

- 3D motion of a particle
*nearly* free falling into the attracting center, but with nonzero angular momentum $L$.

In this case we know from Kepler laws that the solution is an elliptic trajectory. Its area will go to $0$ as $L\to0$, leaving major axis length finite. Thus, in the limit of $L\to0$ we have a free falling particle with trajectory being a straight line, and, surprisingly, the particle will effectively repulse from the attracting center after reaching it. But this motion is nothing than 1D motion in Coulomb-like potential, and the particle doesn't ever go past attracting center, having asymmetric trajectory.

- 1D motion in a more general potential $U=\frac{-\alpha}{\sqrt{x^2+\varepsilon^2}},$ which will become usual Coulomb-like potential when $\varepsilon=0$.

Let $\varepsilon\ne0$ and starting speed of particle $v_0=0$. At $x\approx0$ the potential behaves like a harmonic oscillator, so the particle will go through it. Eventually the particle will slow down at the point opposite to starting point, and continue moving in the opposite direction. Now make $\varepsilon$ smaller. The particle will still move in a trajectory, symmetric with respect to origin. As you take the limit $\varepsilon\to0$, you'll get your particle moving in a symmetric trajectory.

You can see that both cases show one-dimensional motion in Coulomb-like potential, *but* they arrive at contradicting results. So, one should draw the conclusion that the solution of the problem depends on from which side you take your limit: either nearly free-falling particle with $L\to0$, or nearly Coulombic potential with $\varepsilon\to0$.

When you take quantum equation, i.e. Schrödinger equation, you will in fact arrive at similar results with both approaches because you no longer have definite trajectories, and your approach in the first case is putting all the 3D system into a quantum wire with attracting Coulombic center in the middle and shrinking its walls.

The result in both cases in quantum mode will be that your ground state energy is unbounded from below as you approach 1D Coulombic regime, and the wavefunction will converge to something like a Dirac delta function, though the limit might not be strictly it. Of course, this result is quite nonphysical: you need infinite energy to excite the particle.

As for the solution given in the answer you link to, I'd say it's at least incomplete: it doesn't include boundary conditions on joining of $u^+$ and $u^-$ functions, and even if it does give you the solution, it doesn't include ground state.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Ruslan