Thermal vibration looks impossible, but does a single atom in vacuum vibrate due to the electrons' or subatomic particles' actions? If yes, how much is that vibration and is there a way to stop it?

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    $\begingroup$ Atomic nuclii do vibrate. "Nuclear energy states", "nuclear energy levels", and "nuclear resonance" would be useful search terms. I'm not sure what the classical physics interpretation would be, or whether or not the vibration of the nucleus corresponds to a vibration of the electron shells. $\endgroup$
    – g s
    Sep 25 at 4:25

Vibration is a mechanical phenomenon whereby oscillations occur about an equilibrium point. The word comes from Latin vibrationem ("shaking, brandishing"). The oscillations may be periodic, such as the motion of a pendulum—or random, such as the movement of a tire on a gravel road.

Vibration is a classical mechanics phenomenon. Atoms may be considered classical mechanics points when in bulk and a model with mechanical vibrations used, but a single atom can only be described quantum mechanically about its center of mass.

The quantum mechanical wavefunction has electrons in orbitals and nuclei within the nucleus in energy levels depending on the quantum mechanical model used. All the periodic functions describing the atom have to do with the probability of measurement, which means many atoms must be measured in order to see the phase space of the possible positions, and any periodic function has to do with the frequencies displayed in the probability distributions, not space distributions for one atom.

Within the envelope of the Heisenberg uncertainty, one cannot know exactly the position of the single atom, because then the uncertainty in its momentum would be very large, but that does not mean it vibrates, it just means that the probability of its location and momentum are correlated.

  • $\begingroup$ So, I get it like the center of mass of a single atom in vacuum does not change its position due to random vibrations, but -while measuring- there is some uncertainty (due to HUP) which resembles vibration. Right? $\endgroup$
    – Xfce4
    Oct 5 at 3:09
  • $\begingroup$ right, except the vibrations are in probability space, it needs a lot of measurements under the same boundary conditions to see them. $\endgroup$
    – anna v
    Oct 5 at 3:15

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