4
$\begingroup$

My focus is on condensed matter physics, so I've never really explored this question although it always seemed curious to me. My "immediate reaction" intuition would dictate that cold metal atoms would come together to form a metal, but obviously this isn't the case AMO physicists would have a lot less to work with.

If I understand correctly what I've read, typically in a high vacuum atoms are separated from their source (a piece of metal) by laser ablation, vaporization or some other method that produces "hot" atoms, and then the atoms are trapped and cooled through a combination of laser cooling and evaporative cooling. That explains how one can prepare gaseous metal atoms, but it doensn't yet explain why they stay that way.

Why don't the atoms condense again when they come into contact? If they have no net charge and hence don't attract one another, are their collisions just too infrequent for it to matter over experimental timescales, even when many atoms are in the same trap?

Another explanation I could think of is like activation energy. Is there a repulsive barrier that the neutral atoms would need to overcome to reunite, but which they cannot overcome because they've already been cooled?

$\endgroup$
2
  • 2
    $\begingroup$ Well, one thing is that any cluster formed (even of two atoms) would no longer respond appropriately to the cooling lasers so they would 'evaporate' away from the region of interest. $\endgroup$
    – Jon Custer
    Commented Jan 7, 2019 at 21:41
  • 1
    $\begingroup$ Alkali atoms tend to form dimers in the gas phase. $\endgroup$
    – my2cts
    Commented Jan 7, 2019 at 22:34

1 Answer 1

5
$\begingroup$

Good question - there is a whole subfield of AMO devoted to studying inelastic collisions and engineering chemical reactions in trapped diffuse gases.

You are correct that it is often energetically favorable for gas-phase atoms (metallic and otherwise) to form bound states; however, in order to remain bound, they would have to get rid of their binding energy during the very short period ($\sim\mathrm{ps}$) while they are colliding. In a diffuse gas in high vacuum, there are only a couple of mechanisms to accomplish this, and they are typically quite unlikely.

One way is for the colliding species to emit one or more photons; however, the spontaneous emission lifetime of the bound state is typically much longer than the collision time, so this does not happen often.

Another way is for a third atom to join the collision and carry away some of the binding energy in the form of kinetic energy. This is often the dominant loss mechanism from traps. However, the probability of three atoms colliding at once goes as the cube of the number density, which by design is kept very low in gas-phase atomic traps.

I believe you are also correct about the role of low temperatures and reaction barriers. When the colliding atoms have nonzero angular momentum relative to each other ($\vec{r}\times\vec{p}\neq 0$, where $\vec{r}$ is the interatomic separation and $\vec{p}$ is the atomic momentum), they experience a "centrifugal barrier" that prevents them from reaching the reaction zone if they have insufficient kinetic energy. See e.g. Fig. 2 of this Physics Viewpoint and this review of scattering theory, especially the section titled "The Hard Sphere." This means that at ultracold temperatures, the only energetically allowed entrance channel to a chemical reaction is the 2-particle quantum state with zero angular momentum. This is why the ultracold temperature regime is sometimes called the "s-wave scattering regime" or the "single partial wave" regime (e.g. this paper and this book).

$\endgroup$
1
  • 2
    $\begingroup$ Nice answer. I would add that there was an effort to make hydrogen Bose-Einstein condensate (and indeed succeeded at MIT, several years after the Rb and Na BEC) , and I believe this was inspired by the fact that hydrogen would be stable against formation of solid at low temperature. $\endgroup$
    – wcc
    Commented Jan 25, 2019 at 3:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.