# How does the collision between two atoms work?

Considering the quantum mechanical model for an atom, what exactly happens when two atoms (say, two Ca2+ ions in a Brownian motion) collide with each other? As I know, this collision is not like a regular elastic or inelastic collision between two macroscopic objects. Is it mainly due to the coulombic repulsion between the electrons of the two atoms? And, how is the trajectory of the two atoms after collision determined, and what factors contribute to it? Are these trajectories and the angle by which the atoms get deviated deterministic, or fuzzy just as the atoms themselves?

• Hint: there's an entire subfield of quantum mechanics called scattering theory. – Emilio Pisanty Apr 21 at 10:35

## 1 Answer

One can make a simplistic quantum model of the two atoms by treating them as point particles with appropriately fuzzy, but as high-information as possible, positions and momenta directed toward each other. You do not need relativistic quantum field theory in this case - though you might need at least a crude version if you want to also include photon emission, namely a coupled EM field, but not a full QFT for the electrons and such, just the EM, because if this is meant to simulate a chemical process, this is suitably low-energy that we aren't creating or destroying any known massive particles.

And you are right in your hunch: The collision will be, as you say, fuzzy. The available information in position falls steadily with time as the probability distributions broaden during approach and then even more in collision - if one starts out broader than the other it will spill over the other on each side. Moreover, the positions after collision will be entangled, or correlated: you cannot simply and totally faithfully write down two separate wave functions

$$\psi_\mathbf{r}(\mathbf{r}_\mbox{atom 1})$$

and

$$\psi_\mathbf{r}(\mathbf{r}_\mbox{atom 2})$$

to describe their fuzzy positions independently after collision. Instead, you need a six-dimensional wave function

$$\psi_{\mathbf{r}_1, \mathbf{r}_2}(\mathbf{r}_\mbox{atom 1}, \mathbf{r}_\mbox{atom 2})$$

for both and so can't also entirely faithfully picture the result. Nonetheless, if you finally then query the system at a suitably late time as to the outgoing directions of the atoms (thus increasing the information), your result will be quite random with the usual randomness of quantum queries.

There is a little animation that attempts to visualize something like this here:

https://www.youtube.com/watch?v=exy2twNRhzQ

There it is not two free atoms but one particle (could be another atom) hitting a bound state (likely meant to represent an atom with orbiting electron modeled) that is fixed in place as a simplification. Look at that little thing wnoozle - mwrrp's energy is fuzzy at this point now thanks to the presence of motion information, which comes with a cost in energy information. We would need to include an EM field as mentioned above if we want to see that decay to its ground state, which it would in "real life", leaving a photon with fuzzy existence and also likewise entangled, since if you see or otherwise register that photon you then also gain information that the atom is now in a low state, otherwise if you don't, you gain information that it might still be in a high state (unless it is actually a miss, but we can imagine having an all-encircling photon detector if we want).

Quantum pool, as you can tell, will be significantly harder than regular pool, and truly worthy of the casino, for it is fundamentally impossible to predict if you will find your balls in the holes or to aim them so as to guarantee it!

• Thanks. Just to make it finalized, suppose two Hydrogen atoms are moving toward each other on the x-axis, and they collide at t=0. As I understand you, there are multiple possibilities for the angle by which they deviate from the x-axis, right? And is this angle determined by the positions of the electrons within the electron cloud around the nuclei of the two atoms (if you suppose electrons are point charges)? – Ali Lavasani Apr 21 at 16:59
• @Ali Lavasani : Yes. Regarding how the electrons go, I'd think that pretty infeasible to simulate completely ab initio, like here, due to the exponential nature of quantum systems with increasing particle number. As said by Emilio Pisanty under your question though, there is scattering theory, which is basically designed to work around this and I believe deals with the questions you mention. However, I haven't looked into it, so would not know how exactly that changes the probabilities for the deviation. – The_Sympathizer Apr 22 at 6:17