# Centrifugal force in the Hydrogen atom for $L=0$

I have found the following interesting article: http://arxiv.org/abs/0706.0924

The authors examine the radial momentum operator in detail, in particular its time evolution due to the forces acting on the electron in the Hydrogen atom. Their key result is equations (31), (32) and (33). Here they derive explicit expressions for the centripetal force (resulting from the Coulomb interaction between electron and nucleus) and the centrifugal force (which appears to be the result of conservation of angular momentum).

The authors then demonstrate that, upon averaging, the effects of the centripetal force and the centrifugal force become identical. Hence on average there is no force, and therefore the radial momentum remains constant -- precisely as expected for a time-independent state.

Unfortunately the authors do not comment on the special case of the Hydrogen ground state (where $N=1$ and $L=0$), in which case there is no angular momentum. So how can there be a centrifugal force associated with it?

The formulas by the authors are in terms of a numerator and a denominator, both of which become zero for $L=0$. The quotient of numerator and denominator is tacitly assumed to yield the same (constant) value as in the more general case $L > 0$.

My questions: Is there indeed a centrifugal force acting on the electron in the Hydrogen atom, which on average balances the centripetal Coulomb force? If so, how can we understand this force in the case of $L=0$, given that the only obvious source (= angular momentum) equals zero in this particular case?

• Could you provide abbreviated versions of the proofs you're talking about? It's much easier for readers here to look at your text than to click-through and find the correct equation & read the accompanying text. – BMS Nov 10 '14 at 20:23
• – rob Nov 10 '14 at 23:53
• – Qmechanic Nov 29 '14 at 13:53

The "rotating particle" picture does not work for $L=0$, unfortunately. Both the electron and the nucleus are waves who create a standing spherically symmetric wave of their relative motion.

For an intuitive description, let's go to a semi-classical picture of the electron movement in the atom.

I order to have an angular momentum, the electron should rotate around an axis of symmetry. Then, if in the plane perpendicular to the axis of symmetry the electron rotates, say, clockwise, the angular momentum will point along that symmetry axis. But, what to do, the electron in the atom is a non-standard creature. The level n = 1 has SPHERICAL SYMMETRY, i.e. ANY axis in the space is a symmetry axis for this level. The electron rotates on a circle around whatever axis you pick. So, in which direction can point the angular momentum?

However, with the centrifugal force the situation is better. On whichever circle rotates the electron, the centrifugal force points outwards, compensating the nuclear attraction.

I hope that this helps,

Sofia

• In a semi-classical picture or in the famous Bohr model the electron rotates around the nucleus in an elliptical or circular orbit. However, these models have been superseded by the contemporary QM view. In the L=0 case, the electron has a finite probability of being very close to the nucleus. Is this perhaps because there is no centrifugal force? If we look at the Hamiltonian for the Hydrogen atom, then in the L=0 case there is no term present that contributes to a centrifugal force..... – M. Wind Nov 11 '14 at 1:36
• If in the case L=0 there weren't a centrifugal force to compensate the centripetal force of electrical attraction, the electron wouldn't only have a probability to be close to the nucleus, it would FALL completely on the nucleus. So, the centrifugal force is there. The n=1 shell is stabile, does not emit, s.t. the electron is a standing wave, not an outgoing wave (neither an ingoing one) but a superposition of both. So, it is in equilibrium. – Sofia Nov 11 '14 at 15:34
• The Schroedinger equation does not tell us what happens if the electron hits the nucleus. However, the electron may well pass straight through the proton, appearing unharmed on the other side. This would in fact be a perfectly symmetric process: converting potential energy to kinetic energy and then back to potential energy. The equilibrium would then result from a balance between potential and kinetic energy. – M. Wind Nov 11 '14 at 22:45
• Please see: As far as I know, the solution for spherically symmetrical potential is of the form (1/r)\F(r), and F(r) is of SINUSOIDAL form, e.g. sin(kr) or a superposition of sinusoids, in order to avoid the wave-function going to infinity at r = 0. – Sofia Nov 12 '14 at 0:52
• (continuation): the probability that the electron be in the nucleus is VERY small. You still have to explain while the biggest probability for the electron is to be OUT of the nucleus. – Sofia Nov 12 '14 at 0:54