# Effective potential of radial equation of Hydrogen

In QM, the study of the hydrogen atom, why does it follow that for the radial equation given as $$-\frac{\hbar^2}{2m}\frac{d^2 u}{d r^2} + \bigg( -\frac{e^2}{4 \pi \epsilon_0} \frac{1}{r} + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2} \bigg)u = Eu,$$ the term $\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}$ is an effective potential that tends to keep the wave function with $l \neq 0$ away from the origin?

• In short, because the centrifugal/effective potential goes to plus infinity for $r\to 0$. Feb 24, 2017 at 10:06
• @Qmechanic What centrifugal term are you referring to? Could you elaborate a bit, more detail is better since I am not well schooled in classical mechanics and only learning the basics of QM.
– user100411
Feb 24, 2017 at 10:09

The term

$$V_\mathrm{centripetal}=\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}$$

is an effective potential because it appears in the radial equation in the same way (coefficient of $u$) as the Coulomb potential $$V_\mathrm{Coulomb}=-\frac{e^2}{4\pi\epsilon_0r}$$

The Coulomb potential is attractive for opposite charges which you can see from the minus sign in the equation. Taking only the Coulomb potential into account energy gets smaller the smaller $r$ is.

The effective potential on the other hand is always positive making it repulsive (energy gets smaller for large $r$). Specifically if you get close to the origin, $r\to 0$ the centripetal potential $V_\mathrm{centripetal}\to +\infty$. In physics, systems tend to minimize their energy, so a position where the potential energy is very large is not attractive for the system and you can say that the centripetal potential is "keeping the wave function away from the origin".

In a classical mechanics analogy, a potential like this would result in a force $F=-\frac{dV}{dr}>0$, i.e. a force which is positive, pushing the "particle" away from the origin.

Since the centripetal potential is $\propto r^{-2}$ and the Coulomb potential $\propto r^{-1}$, at small $r\to 0$ the centripetal potential will dominate the total potential $V_\mathrm{eff}=V_\mathrm{Coulomb}+V_\mathrm{centripetal}$. At large $r\to\infty$ the Coulomb potential will dominate. In-between there is an optimum $r$ at which the energy gets minimized. In a picture: • This is a really good answer thanks. Just to confirm one thing, I have not come across potential energies being described as attractive or repulsive. Is the idea that, as you stated since a system naturally tends towards a minimized energy, if this minimized state is reached with decreasing $r$ with regard to some reference point (in this case the origin) then the potential is characterized as attractive? Or is the definition of attractive and repulsive potentials with regard to the related conservative force $F = \frac{d V}{d r}$?
– user100411
Feb 24, 2017 at 11:07
• @JohnDoe: Usually you would say the potential is attractive/repuslive (not the "potential energy is attr/rep). There is a minus sign missing in your force, $F=-\frac{dV}{dr}$. Otherwise yes and yes. If the potential energy is decreased with decreasing $r$ this is called attractive potential. If the potential energy is increased with decreasing $r$ this is called repulsive potential. But this is the same in the analogy with the conservative force (only that in QM you don't speak about forces usually). Feb 24, 2017 at 11:39
• @JohnDoe: Of course it makes only sense to speak of attractive/repulsive in cases where the physical situation corresponds to something (here the electron) being attracted/repulsed from something (here the proton). Feb 24, 2017 at 11:43
• @JohnDoe: Later you will encounter other potentials like "barrier potentials" or "potential wells". Look them up and see whether their shape corresponds to what you associate with a barrier/well. Feb 24, 2017 at 11:46
• Suggestion to the answer (v2): Replace the word centripetal potential with the word centrifugal potential in various places. The potential pushes the electron radially outwards not inwards. Feb 24, 2017 at 12:53