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I have read this question:

Changes to chemical properties of elements due to relativistic effects are more pronounced for the heavier elements in the periodic table because in these elements, electrons have speeds worthy of relativistic corrections. These corrections show properties that are more consistent with reality, than with those where a non-relativistic treatment is given. Electrons are not "moving around" a nucleus, but they are instead probability clouds surrounding the nucleus. So "most likely distances of electrons" would be a better term.

Why do "relativistic effects" come into play, when dealing with superheavy atoms?

And this one:

The fact that by measuring the spectra of atoms we know the energy levels, in the Bohr model allows to calculate a velocity for the electron. The Bohr model is superseded by the quantum mechanical solutions which give the probabilistic space-time solutions for the atom, but since it is a good approximation to the QM solution, it can be considered an "average" velocity. There is no way to measure an individual electron's four vector while bound to an atom. One can measure it if it interacts with a particle, as for example "the atom is hit by a photon of fixed energy, with an energy higher than ionization", and an electron comes out and its velocity can be measured. The balance of the energy and momentum four vectors of the interaction "atom+photon" will give the four vector of the electron, and thus its velocity in a secondary way. An accumulation of these measurements would give on average the velocity calculated by the Bohr model.

How do particles that exist only as a cloud of probabilities have actual rates of speed?

Now both of these agree on the fact that electrons are quantum mechanical objects, described by probability densities where they exist around the nuclei (some might say they exist everywhere at the same time with different probabilities), but then the first one says that the relativistic corrections are justified, so that is the correct way.

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Now a free electron can have a classical trajectory, as seen on the bubble chamber image, as the electron spirals in. But why can't the bound electron do the same thing around the nucleus?

As far as I understand, QM is the right way to describe the world of electrons around the nuclei, and they are not classically moving around the nuclei, then they do not have actual classically definable trajectories, but as soon as they are free, they can move along classical trajectories.

Just to clarify, as far as I understand, electrons are not classically orbiting, but they exist around the nuclei in probability clouds, as per QM. What I am asking about, is, if they are able to move along classical trajectories as free electrons, then what happens to these free electrons as they get bound around a nucleus, why are they not able to classically move anymore?

I am not asking why the electron can't spiral into the nucleus. I am asking why it can't move along classical trajectories around the nucleus if it can do that when it is a free electron.

If we can describe the free electron's trajectory with classical methods in the bubble chamber, then why can't we do that with the electron around the nucleus?

Question:

If free electrons have classical trajectories, then why don't bound electrons around the nuclei have it too?

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This is the subject of an underrated classic paper from the early days of quantum mechanics:

Mott's introduction is better than my attempt to paraphrase:

In the theory of radioactive disintegration, as presented by Gamow, the $\alpha$-particle is represented by a spherical wave which slowly leaks out of the nucleus. On the other hand, the $\alpha$-particle, once emerged, has particle-like properties, the most striking being the ray tracks that it forms in a Wilson cloud chamber. It is a little difficult to picture how it is that an outgoing spherical wave can produce a straight track; we think intuitively that it should ionise atoms at random throughout space. We could consider that Gamow’s outgoing spherical wave should give the probability of disintegration, but that, when the particle is outside the nucleus, it should be represented by a wave packet moving in a definite direction, so as to produce a straight track. But it ought not to be necessary to do this. The wave mechanics unaided ought to be able to predict the possible results of any observation that we could make on a system, without invoking, until the moment at which the observation is made, the classical particle-like properties of the electrons or $\alpha$-particles forming that system.

Mott's solution is to consider the alpha particle and the first two atoms which it ionizes as a single quantum-mechanical system with three parts, with the result

We shall then show that the atoms cannot both be ionised unless they lie in a straight line with the radioactive nucleus.

That is to say, your question gets the situation backwards. The issue isn't that "free electrons have classical trajectories," and that these electrons are "not able to move classically anymore" when they are bound. Mott's paper shows that the wave mechanics, which successfully predicts the behavior of bound electrons, also predicts the emergence of straight-line ionization trajectories.

With modern buzzwords, we might say that the "classical trajectory" is an "emergent phenomenon" due to the "entanglement" of the alpha particle with the quantum-mechanical constituents of the detector. But this classic paper predates all of those buzzwords and is better without them. The observation is that the probabilities of successive ionization events are correlated, and that the correlation works out to depend on the geometry of the "track" in a way which satisfies our classical intuition.

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    $\begingroup$ I never knew that paper existed, so thanks for binging it to everyone's attention. It is very very well written. $\endgroup$
    – joseph h
    Jul 4, 2021 at 3:00
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Free electrons do not "have" classical trajectories any more than bound ones do. How definite quantities like position and momentum are is a property of a specific quantum state, not of generic qualities like "free" or "bound".

Bubble chambers are specifically constructed to effectively perform a continuous (more precisely: very rapidly repeated) position measurement on the particles passing through them, so free electrons in bubble chambers look like they have classical trajectories. The particles in bubble chambers also usually are very energetic, so while non-quantitative readings of the uncertainty principle might lead one to believe it must be "high" since the particles are being located rather precisely, the momentum uncertainty is still small relative to the total momentum, see e.g. anna v's answer here.

You can't do this with slow free electrons - they need to be fast enough to create the ionization tracks in the bubble chamber or something similar. You can't do this with bound electrons either, since any interaction strong enough to fix their position like a bubble chamber would likely be an interaction strong enough to change the bound state to something else (i.e. knock the electron off the atom).

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It is not so that free electrons are completely explained through classical trajectories. For some phenomena (e.g. diffraction) you need to resort to a description in terms of quantum mechanical wave functions as well. In the same way, you can explain some aspects of atomic electrons classically. This 'wave-particle duality' is well acknowledged in physics.

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Having abandoned the electron's orbital motion around the nucleus in favour of the probability of residence in certain shaped volumes, the electron is now again attributed a motion within these volumes. This is absolutely unnecessary.

Just take the double-cone volume of the 2px, 2py, 2pz orbitals. In the reality of chemistry, compounds are always sp-hybridised. In the most stable form quadruple, as in methane. Here all 8 electrons are equally involved in the bonds.

Molecules are stable because the atoms share electrons. No electron jumps over to the second part of the lobe. Even more, instead of one 2s and three 2sp orbitals (with 2 electrons each), a distribution of all 8 electrons on the edges of a cube would be the more natural solution for helium and neon. A corresponding spherical harmonic exists. We are only prevented from considering such a solution by thinking in Cartesian coordinates.

Schrödinger's equation can only describe the excited states in the hydrogen atom and in hydrogen-like ions. Using this equation to describe the electron arrangement of other elements does not reflect the reality of chemical compounds.

If free electrons have classical trajectories, then why don't bound electrons around the nuclei have it too?

Free electrons move along orbits. Bound electrons occupy a certain volume in the atom and can be considered immobile. During bonding with another atom, the shape of the volumes changes and the electrons can shift their position a little. This results from empiricism in chemistry.

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