# How do atoms repel each other?

1. Jul 14, 2014

### jaydnul

Take the first orbital given to us by the time independent SWE, which is a sphere around the nucleus. Conventional teaching is that the electron is smeared across the whole surface area of that sphere. If the electron doesn't appear until after we measure it, how then do atoms repel each other? Or is the electric charge of the electron uniformly distributed across the surface of the spherical orbit? If the latter is true, doesn't that prove that quantum objects truly are matter waves that spread out in space?

Please don't take this as crack-pottery. I realize I am wrong about something, but can't put my finger on it.

Thanks!

2. Jul 14, 2014

### phinds

I wish I could give you an answer, but can't ... I just wanted to say I find your attitude healthy and refreshing. We get so many posters who think they have found a flaw in modern physics that it's nice to see someone like me who knows they don't know it all [Don't tell Drakkith about this, though. My story to him is that I DO know it all ]

3. Jul 14, 2014

### WannabeNewton

I'm not aware of this as the conventional teaching. It's certainly not correct.

It's not that the electron doesn't "appear until after we measure it". Unfortunately these kinds of things tend to be interpretation-dependent and we don't want to get into that can of worms here but it's definitely safe to think of the situation as that of the electron simply not possessing a position until a position measurement is made, if you wish to invoke wave-function collapse. But not possessing a position until measurement is definitely not an issue when it comes to electric repulsion. It's not even relevant.

In the hydrogen atom problem if we take the proton to be in a momentum eigenstate, which is true to a good approximation, then we can go to the rest frame of the proton and center our spherical coordinates on the origin of this rest frame; note the notion of a rest frame of the proton requires it to be in a momentum eigenstate.

The stationary states of the Schrodinger equation $i\hbar \partial_t |\Psi \rangle = \hat{H}|\psi \rangle$ are obtained after separation of variables into spherical harmonics (eigenfunctions of $\hat{L} = i\hbar \partial_{\varphi}$) and radial functions from a radial equation $-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + (\frac{\hbar^2 l (l + 1)}{2m r^2}-\frac{e^2}{4\pi\epsilon_0 r})u = Eu$. In words, the Hamiltonian $\hat{H}$ and consequently the radial equation for $u$ (which is related to the stationary states) already includes the Coulomb attraction $-\frac{e^2}{4\pi\epsilon_0 r}$ between the electron and proton so the Coulomb attraction is already implicitly built into the stationary states $u$.

The same logic applies to a system of interacting atoms. The many-body Hamiltonian for the interacting atoms will already include the Coulomb repulsions between them (electron-electron interaction terms) and as a result the stationary states will already implicitly include the Coulomb repulsions.

No.

4. Jul 14, 2014

### jaydnul

I learned from the mistake of doing exactly what you're preaching against. Most of the time it was followed by ZZ's legendary vigilantism against crack-pottery.

5. Jul 14, 2014

### jaydnul

Perfect. Thanks WannabeNewton

6. Jul 15, 2014

### jaydnul

Wouldn't the exact position of the electrons effect the magnitude of the Coulomb repulsions between the two atoms? For instance, with two hydrogen atoms, both electrons could be in front of their protons (the closest distance between them at the moment of "collision"), resulting in a stronger repulsion, or both electrons could be on the far sides of their atoms, resulting in a smaller magnitude of repulsion.

Another way to put this is if I have two hydrogen atoms and the distance between the nuclei is 1 μm, then the repulsive force between the protons would be:
$$F=\frac{q^2}{4πε_{0}(10^{-12})}$$
But the $r$ between the two electrons could be much less than 1 μm or much more than 1 μm depending on where they are in their clouds. (There's also the attractive force between the respective protons-electrons but I don't need that to raise the question).

Last edited: Jul 15, 2014
7. Jul 16, 2014

### WannabeNewton

The Coulomb interaction is given in terms of the position operator so it already takes into account the fact that an electron has no definite position in a stationary state of the Hamiltonian. There is no "exact position" of the electron. You measure its position and you can calculate the magnitude of the Coulomb interaction at the position you measured it at but until you make a measurement it doesn't make sense to talk about the position of the electron. Yes the Coulomb interaction will vary depending on the measured position simply because it is a function of the position operator but so what? What's the issue there?

8. Jul 17, 2014

### jaydnul

K, I worked out the many body Schrodinger for two hydrogen atoms by hand and now I get it. (bad joke)

No, but I do think I understand. For the hydrogen atom, there is a probability that the coulomb attraction between the nucleus and the atom could be a range of different values, and we only find a definite value when we measure.

Similarly, with two hydrogen atoms, there is a probability that the coulomb repulsion between the two could be a range of different values, but we only get a definite value when we measure it.

The part I'm still a little fuzzy on is this. Atoms are "colliding" all of the time, so doesn't that mean that the system is forced to choose a definite value for the coulomb force?

Tell me if this is right. Atoms are always repelling with a definite value that is allowed by the SWE, but that range of different possible magnitudes is too small to notice macroscopically? In other words, if I set up an experiment for two hydrogen atoms to collide, they would repel at different magnitudes each time I performed the experiment, but the variation in those values would not be a noticeable change in our macroscopic viewpoint.

9. Jul 17, 2014

### WannabeNewton

In QM, elastic and inelastic collisions are done through the S-matrix of scattering cross sections. In scattering cross section calculations we can still only calculate the probability of an incident beam of atoms in some initial free particle momentum eigenstate scattering into some solid angle $d\Omega$ into some density of final free particle eigenstates when interacting through the Coulomb force with some target atom. There is nothing definite here period, except for the fact that the in and out states of the scattered particles are free particle momentum eigenstates, but the scattering itself has no element of definiteness to it. There is only probability and this probability is not due to our ignorance, it is inherent in nature.

10. Jul 17, 2014

### jaydnul

But can't we measure (as opposed to calculate) the definite outcome after it happens? Isn't the HUP about what we can predict, not about what we can measure?

Yes, but the probability is just the odds of a certain definite outcome, right?

Edit: Sorry if I seem thick. Just can't seem to tackle this problem with my feeble brain.

Last edited: Jul 17, 2014