9
$\begingroup$

In order to pose the question, I reproduce an excerpt from this Physics.SE question, about the size of atoms in a Bose-Einstein condensate (BEC):

Heisenberg's uncertainty principle in frames of reference with different velocities.

I heard this from Professor Claude Cohen-Tannoudji himself on a lecture about BECs: as the temperature of the gas decreases, the momentum of the atoms decrease, and thus the uncertainty in the momentum also decreases. Due to the uncertainty principle, the uncertainty in the position has to increase. Atoms grow bigger as they are cooled down.

So, in an atomic BECs the gas temperature is so low, and atoms get so large, that their atomic wavefunctions start to overlap, and begin oscillate in-phase. Thus the condensate behaves as if it was a huge single atom, their wavefunctions oscillating coherently, allowing for applications like atom-lasers (see also here).

Now, my question here is: in which theoretical framework does the size of the wavefunction of an atom depend on temperature - or rather, on its momentum? Do we have any calculations that can reproduce the increase of the size of the atom with lowering the temperature of the gas? I had assumed, for example, that most QM calculations of atoms and molecules were effectively at zero temperature, so I'm confused how this effect is taken into account.

In order to provide futher insight into the question, I quote an excerpt of this article on Nature by James Anglin and Wolfgang Ketterle:

As long as the atoms’ de Broglie wavelength $\lambda_{\textrm{dB}} = \hbar / (2 M k_{\textrm{B}} T)^{1/2}$ is small compared to the spacing between atoms, one can describe their motion with classical trajectories. ($\lambda_{\textrm{dB}}$ is the position uncertainty associated with the thermal momentum distribution, and increases with decreasing temperature $T$ and atomic mass $M$.) Quantum degeneracy begins when $\lambda_{\textrm{dB}}$ and the interatomic distance become comparable. The atomic wave packets overlap, and the gas starts to become a ‘quantum soup’ of indistinguishable particles. (emphasis mine)

$\endgroup$
2
  • $\begingroup$ I edited your post to narrow it down to one question (let me know if I misinterpreted the core of your question). Keeping questions on the site sufficiently narrow and focused makes them more likely to receive an answer that addresses the actual problem. I may actually have an answer that I think will partially address your other points, but if it doesn't (or I wind up not having an answer at all), you can always ask another to question to address these others points. $\endgroup$
    – Tyberius
    Commented Feb 3, 2022 at 3:17
  • $\begingroup$ Hi @Tyberius, your edit is fine, I think it was a bit confusing as it was, now it's more succinct. Eager to see your answer! $\endgroup$
    – Arc
    Commented Feb 3, 2022 at 4:05

1 Answer 1

4
$\begingroup$

In calculations of the electronic energy/wavefunction, temperature typically is not taken in to account (besides cases of orbital occupation smearing, though this is used more as a computational trick than a physical description of the system). So these calculations are effectively zero temperature and often more like no temperature, since temperature is a statistical quantity that isn't well defined when dealing with a single atom or molecule.

So if these simulations are essentially zero temperature, why don't we see this change in atom size you describe? The key is that these are electronic calculations. In principle, the nuclei should also be treated as quantum mechanically and the wavefunction of the atom/molecule should involve both nuclear and electronic degrees of freedom. In practice, these sorts of simulations are much more challenging and thus much less common (though not nonexistent, see usage of nuclear-electronic orbitals, NEOs). In typical electronic structure calculations, nuclei are treated as fixed point charges and under normal chemical conditions this is a very good approximation.

However, as you saw in the Anglin/Ketterle paper, the wave like behavior of nuclei can become apparent at very low temperatures (milli or even micro kelvin), allowing the formation of phases like the Bose-Einstein condensate where the nuclear wavefunctions of neighboring atoms start to overlap.

I'm not nearly experienced enough with this area to recommend software that could do this simulation, but whatever program would have to describe the nuclei in the system quantum mechanically and incorporate effects of temperature using something like the thermostats seen in classical MD simulations. I expect programs that can do this aren't available off the shelf and are specially built to simulate specific systems rather than being general purpose simulation platforms.

$\endgroup$
3
  • $\begingroup$ So, what you are saying is that the size of the nucleus varies with temperature, likely with $T^{-1/2}$ as from the paper, so the electronic calculations we do normally consider pointlike nucleus. But then, finite-size nucleus have an effect which increases with lower temperatures. We do get realistic values calculating with zero-sized nucleus because its not very important at room (and higher temperatures). But at lower temperatures, nucleus size become more and more important and do affect the electronic lengths, and these are only important very near zero kelvin. Did I get it? $\endgroup$
    – Arc
    Commented Feb 4, 2022 at 3:54
  • 1
    $\begingroup$ @Arc that's how I interpret it: the nucleus is basically a point charge until you get to very low temperatures. Something like a Bose-Einstein condensate forming is very much nuclear quantum effect and so if we are approximating the nuclei as point charges, you won't see this effect even if your simulation is at essentially 0 kelvin. $\endgroup$
    – Tyberius
    Commented Feb 4, 2022 at 15:55
  • $\begingroup$ Ok, understood. Just to add things up: the theoretical framework in which the atoms’ de Broglie wavelength goes inversely with the square root of temperature is nuclear physics, say, the standard model? Also, do we agree that whenever the nuclear wavefunctions grow larger in size so do the electronic wavefunctions? The atom has to grow bigger as a whole, correct? $\endgroup$
    – Arc
    Commented Feb 14, 2022 at 23:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .