1-dimensional_coulombic_potential
### 1-Dimensional_Coulombic_Potential

### Frame 19

### Frame 20

In slide 19 we can see the graph of the Coulombic potential energy between an electron and a proton. Since Coulomb's law is an inverse square law, at the point where the distance between the electron and proton is 0, the potential energy is infinitely negative. On the graph, this is shown as a clear vertical asymptote.

This results in the Psi function having a very sharp curvature at zero on the graph. In fact, the curvature is essentially infinite, but the Erwin meets Goldilocks program gives only an approximation. At this point where the curve is so sharp, the kinetic energy is very large. (A very high positive curvature of psi) / (relatively small, negative value of psi) is a large negative number. This, when multiplied by -C/m, gives a large positive value for the KE of the electron).

The graph of the probability density is symmetrical, as would be expected from a symmetrical PE graph, and the nodes are very close together, but still at the extremes of the Psi function graph Coulumbic potential is approaching a limit as on the right of the Morse potential. In one dimension it is difficult to handle the numerical simulation near the origin where curvature must be so sharp. curiously this system is simpler in three dimension than in one dimension and psi has a simple analytical form, as we shall soon see.

In frame 20, we see that as the total energy in increased, the Psi function (and probability density graph) spreads way out. There is still a very sharp curvature at d=0. The highest probability density area is still at the extremes, as would be expected from the ball on a spring analogy discussed in class. As in the Morse potential graphs, the waves spread out more than they would in a Hooke's Law graph. This is because as the total energy and the number of nodes increases, the potential energy well widens more and thus the wavelengths become longer and the successive energy will be lower than would be expected for Hooke's Law where E=k(n-0.5). Here, instead, E = -k/(n^2), where n is the number of nodes in the psi function.

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