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The Coulomb's force is given by $$ F = {k q^2 \over r^2} $$ When $ r \rightarrow 0 $, $ F \rightarrow \infty $ Does this mean two electrons never touch each other?

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Define touch. Electrons are standing waves and are known to show interference patterns. I don't know what you mean by them 'touching'. –  Gerard Aug 4 '13 at 12:05
    
@user1305192 can there not be free electron in space at some given postion? –  hasExams Aug 5 '13 at 15:57
    
Yes, there can be an electron at a given position. But then you can't measure its momentum. So, for all intents and purposes, you don't know where it is. –  Gerard Aug 8 '13 at 15:04
    
Pauli Exclusion . . . –  Dimensio1n0 Aug 9 '13 at 11:55
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Pauli Exclusion won't save them from touching if they have different spins –  Ruslan May 22 at 6:17
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Yes, two electrons or two charged particles are not allowed to overlap – the interaction energy goes like $1/r$ so it would diverge in the strict limit and the world doesn't have the infinite energy to do that. Two electrons are forbidden to overlap for another reason, the Pauli exclusion principle.

On the other hand, two particles of opposite signs attract by a force that goes to infinity in the $r\to 0$ limit, too. Nevertheless, even in this case, they won't end up overlapping due to the uncertainty principle of quantum mechanics.

For example, take the hydrogen atom (or any atom). The electron can't sit exactly at the nucleus even though it would save an infinite amount of energy. The reason is that a very low $r$ implies a very low $\Delta x$ of the same order and, therefore, a high $\Delta p \geq \hbar / \Delta x$. A high $\Delta p$ means a high average $\Delta p^2$, and therefore high kinetic energy, and this one actually wins if the proton-electron distance is too short. If the electrostatic force increased faster than $1/r^2$ at short distances, one could actually beat the kinetic energy and oppositely charged particles would choose to sit on top of each other.

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@ Lubos Motl what dou you mean by electrostatic force increased faster than 1/r^2 .please clarify on that. –  AaKASH Mar 31 '13 at 12:42
    
The answer is wrong. Correct one is the opposite - see my answer. –  Ruslan May 22 at 7:40
    
@AaKASH - I mean if the potential energy were $K/r^n$ for $n\gt 2$ or as any function that is greater than that for all $r\lt \varepsilon$. For example $K/r^3$. This is so huge near $r=0$ that even the huge kinetic energy you get from the inevitable $\Delta p$ is smaller and it's energetically favored for the electron to sit strictly at $r=0$. Of course, it's just a mathematical limit - you won't find those things exactly in Nature. –  Luboš Motl May 30 at 4:52
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Although this is quite an old question, I have to disagree with answer by Luboš.

First, Pauli exclusion principle says that no two fermions can share the same state. But, if the electrons have different spins (i.e. are in so called spin-singlet state), then they can be in the same positional state.

Next, indeed, in classical case, two charged particles with equal signs cannot touch, because their potential energy $\sim1/r$ will go to infinity at the collision point. But, as electrons are quantum particles, they obey uncertainty principle, which allows them, in particular, to tunnel to classically restricted locations. To more precisely describe their motion, one has to use Schrödinger equation.

Suppose two electrons with different spins are in a parabolic 3D potential well, while interacting by Coulomb force between each other. Schrödinger equation for them can be simplified by separation of variables*. Then the part for inter-electron wave function would look like (neglecting dimensional constants):

$$-\frac{\text{d}^2\phi}{\text{d}r^2}-\frac2r\frac{\text{d}\phi}{\text{d}r}+\frac{l(l+1)}{r^2}\phi+\left(Ar^2+\frac2r\right)\phi=E\phi.$$

The boundary conditions for $\phi(r)$ are boundedness at $r=0$ and $r\to\infty$.

Solving** this equation with $l=0$ (otherwise electrons won't touch because of centrifugal force, which grows faster than Coulomb force), we get the wavefunction for ground state:

enter image description here

As you can see, the wavefunction doesn't go to zero at $r=0$. Neither will it vanish for excited $S$-states (i.e. excited states with $l=0$). Instead, there's a cusp with a local minimum of probability density at the point of collision. Were the potential to go to infinity at a higher rate, like $r^{-2}$, the wavefunction would indeed vanish at that point. This is what happens for $P, D$ and other states with $l>0$.

Note that the reason for this is very similar to reason why electron in the atom doesn't have infinite binding energy in atom in $S$ states: the wavefunction there has a local maximum cusp, but it still is bounded. For potentials $\sim -r^{-2}$ the electron would fall onto the nucleus and have infinite binding energy.


* See the (paywalled) article for details on separation of variables.
** I solved it via Frobenius method, limiting to 1000 series terms. I'm not sure if there's a closed-form solution for this BVP.

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An interesting answer, @Ruslan, but I don't know why it implies that they will touch. The probability of strictly $r=0$ is still zero, right? Would you say that the electron touches the nucleus at the 1s state? –  Luboš Motl May 30 at 4:54
    
@LubošMotl Of course, probability at $(r,\theta,\phi)=(0,?,?)$ is zero, but so it is for any other $(r,\theta,\phi)$ (because of integration over zero volume). The difference with centrifugal potential is that in the Coulomb case relative probability $P(0,?,?)/P(r,\theta,\phi)$ is non-zero for all $r$, $\theta$ and $\phi$. So, yes, I'd say the electron does touch the nucleus in 1s state. Using point-likeness of particles as a reason for them to not touch would be somewhat vacuous (and then your answer taking potential into account would be redundant). –  Ruslan May 30 at 8:34
    
It is not redundant! For attractive potentials stronger than $1/r^2$, like $-1/r^3$, the electron would touch the nucleus even in my, strong sense, as the wave function would be proportional to the delta-function! So the probability for the electron to be strictly at the origin would be positive, finite. I would agree with you that the probability density for the $1s$ state is parametrically higher than for $l\neq 0$ states but it isn't enough to call these things "touching" (in the approximation of a pointlike nucleus; for a finite realistic nucleus, the touching occurs for any $l$). –  Luboš Motl Jun 3 at 11:26
    
Indeed. Although, I would't call it a touch, instead it'd be a (permanent and irreversible) fusion :) Also it seems wavefunction won't be proportional to delta function per se, although it would be something qualitatively very similar to it. –  Ruslan Jun 3 at 11:41
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