How to understand the motion of a particle in Quantum Mechanics?
Solution 1
Having worked with elementary particles all my working life I can assure you that particles have a trajectory.
Another proof is the existence of accelerators which create the beams that we can scatter against targets, as in picture, or against each other and study the results statistically. That is how the standard model of particle physics was built up, studying trajectories.
Trajectories are saved by the HUP, the Heisenberg Uncertainty Principle, which given enough momentum can always be fulfilled as the bubble chamber picture shows. Within its bound lie the bound states of atoms , and there we speak of orbitals, not orbits.
Everything is quantum mechanical, and the HUP is the measure of whether classical concepts and mechanics are applicable or not. So in your example the electron can be generated in a small accelerator and approach with a known trajectory a potential well of an ion up to the point of closenes where the indeterminacy of the HUP destroys the concept of a trajectory and the probabilistic form will apply.
Solution 2
Most people would say that $\Psi(a)^2$ is the probability density that the particle would be found in some small neighborhood of $a$. Most people would not say it is the probability that it is in that location. This is because because that same kind of thinking (that it has a probability of having a property and that a measurement merely reveals the value of that property that already existed) will literally get you in trouble if used in other situations. For instance a spin measurement objectively does not reveal a preexisting value of a component of a vector. And in fact a spin measurement objectively changes the thing you "measure."
But there is a tradition that does do so, and it found a way to make that work for position and it is called the de BroglieBohm (dBB) interpretation (or the pilot wave theory). The price to be paid is that if you say the probability is accurate for an actual position then other so called measurements can't be merely revealing preexisting values.
This is because the measurements can be correlated with position so knowing the position forces the position to determine everything else (and the context and calibration of the exact set up can be a factor too and again this is because those can correlate to position).
If we used different words this wouldn't sound so strange. We could call them spin polarizations instead of spin measurements for instance.
And in the pilot wave theory trajectories are not meaningless. However they aren't what you observe. You observer through interactions. So all you know is the interactions, you don't know the trajectory.
In that sense, how should we understand the motion of a particle in Quantum Mechanics?
There is a notion of probability current (or probability current density). It is, for instance, how we compute the fraction of a beam that reflects or transmits through a barrier. You can imagine streamlines that have that current as their tangents. And for the pilot wave theory a particle has a location and it follows one of those streamlines. And for everyone else it is just one streamline of many.
Almost no one uses pilot waves since the trajectories are much more detailed than is needed just to compute what fraction of your results come out a certain way. But they can still talk about the probability current and that is usually what they mean if they talk about a particular direction of motion. They just don't literally think there is a particle with a location. They are just talking about current.
So you can have locations and if you do then you need to update them. Most people don't imagine locations but still do talk about the current, they just call it a probability current instead of imagining a probability of being located somewhere and then that updating in time.
And you would have a problem if you thought there were x y and z components of a spin vector and that the interactions we call "measuring the z component of spin" merely passively revealed that value. So if thinking they don't exist for position reminds them to not think they exist for components of a spin vector then great. It is great to learn not to make mistakes. And you can still compute all the frequencies you need without imagining particles having actual locations and moving around.
In that setting, how should one understand intuitively and mathematically the idea of motion of a particle in Quantum Mechanics?
You can not worry about it and stick to probability currents, they will track the flow of probability which is all you need. Or you can study pilot wave theory, which doesn't help you make any new predictions and some of those trajectories are weird. For instance the wave function is defined on configuration space so it is the configuration that changes and so they can be highly nonlocal in many senses.
Solution 3
A quantum system is described by a suitable $*$algebra of observables. The quantum states are functionals of the observables, that when applied on observables yield the average value of it in the state. So, given an observable $A$, and a state $\omega$, $$\omega(A)$$ is the evaluation of $A$, i.e. its average value on the state.
Now the state (or equivalently the observables) evolve in time. This means that we have a map $\omega_{(\cdot)}$ that describes how the state changes with time. In other words, $$\omega_t(A)$$ would describe the average value of $A$ at time $t$. If we choose the position operator $x$ as an observable, we get a function of time that we may call effective (averaged) trajectory: $$\bar{x}(t)=\omega_t(x)\; .$$ This averaged trajectory may be interpreted as a notion of motion of a QM particle. Obviously, $\bar{x}(t)=x_0$ it is not a good notion to say that a particle is static (the particle may be moving but be on average on the same place); but a nonconstant function $x(t)$ is a good indication of the motion of a QM particle.
In addition, the classical trajectory is actually $\bar{x}(t)$, in a regime where the quantum effects become negligible.
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Updated on August 01, 2022Comments

Gold 10 months
In Classical Mechanics when we talk about the motion of a particle it is the same as talking about the idea of trajectory. The fact is that in Classical Mechanics, a particle has a definite position given by a point $a\in \mathbb{R}^3$ while in Quantum Mechanics the best we can get is a probability amplitude $\psi : \mathbb{R}^3\to \mathbb{C}$ of the particle presence around a particular point $a\in \mathbb{R}^3$, i.e., such that $\psi(a)^2$ is the probability density that the particle is located on some small neighborhood of $a$.
Because of that in Quantum Mechancis the idea of trajectory is meaningless. But nonetheless a particle move around, otherwise it would stay always where it is. In that sense, how should we understand the motion of a particle in Quantum Mechanics?
For instance, I've seem some books talking about "a particle propagating from left to right along the $Ox$ axis" or "a particle comes from the left" when talking about potential barriers. This is of course connected to the idea of motion of the particle, but since we don't have trajectories, I don't know how to understand those statements.
In that setting, how should one understand intuitively and mathematically the idea of motion of a particle in Quantum Mechanics?

pianyon over 5 yearsare these trajectories determined by expectation values of the position operator as a function of time? how should one think of these trajectories as arising from states and operators? you mentioned the HUP, but is it possible to elaborate on the how exactly the HUP gives you classical trajectories?

anna v over 5 years@qmarv These trajectories are generated by consecutive interactions of the particle, in this case a pi+, with the atoms in the bubble chamber, which then become little bubbles following the foot print of the pi+ according to its momentum as it was generated at the interaction point, pi+proton scattering, the basic quantum mechanical scattering has happened. The millions of interactions as the pi+ turns are within the heisenberg uncertainty classical points. statistical accumulations of such events give the quantum mechanical crossection for pi+protn scattering.

Tom B. over 5 years@annav, when you say "The millions of interactions as the pi+ turns are within the heisenberg uncertainty classical points", are you saying that measurements of position and momentum, from the trajectories in the bubble chamber, are not more precise than HUP allows?

Conifold over 5 yearsIsn't this misleading? Cloud chamber track is no more a trajectory of a particle than the exhaust trail is the trajectory of a plane, and indeed much less considering the relative sizes of the "trajectory" and the supposed object.

anna v over 5 years@Conifold Classical point measurements always are larger than the Heisenberg uncertainty,. In the above picture one can see this, accuracy is in microns for space, and the Mev/c ionisation momentum is of order of keV/c. h is a very small number. The exhaust of the airplane does measure the trajectory of the plane, within the accuracy of the width of the exhaust . It is the only "proof" of a mathematical trajectory after the plane has left the scene. The same with the pi+, except the rules of widths are quantum mechanical.

Conifold about 5 yearsWe do not call a spurt of water the trajectory of a water molecule. But it would have been relatively harmless classically since we can trace the latter trajectory more precisely and see that it is confined to the spurt's tube. But Heisenberg uncertainty is no error bound. The sloppy talk of tracks as trajectories only encourages people to push classical analogies where they do not belong, which comes to haunt them later when they try to form a coherent picture of quantum mechanics. Isn't it better to say that quantum particles have no trajectories because they are no particles?

anna v about 5 years@Conifold No it is not better. The picture above shows trajectories. The theory that fits them is quantum mechanical and describes "quantum mechanical particle entities" . When you have scanned thousands of pictures with pions and kaons you call them trajectories of a particle, in the same way that you call the footprints of a dog in the dry mud of the path, the trajectory of the dog. You can plot the way the dog went from here to there..

anna v about 5 yearsDealing with quantum mechanical entities it is a disservice to students to deny the existence of the tracks in the above picture. You can explain it with quantum mechanics and QFT wave packets once they have the basic experimental data in the head and have studied QFT.

Conifold about 5 yearsOf course you can, that is my point. Denying the existence of the tracks is denying the obvious, denying that they relate to underlying trajectories the "same way" as dog to its tracks or plane to its exhaust trail is just prudent. Cloud chambers and colliders are only one context where quantum effects manifest, and pictures suggestive for them are context limited, just as Maxwell's cogwheels for describing electromagnetic field. Picturesque asif metaphors are certainly useful, but they do not make pions and kaons into particles with classical trajectories, the governing laws are different.

anna v about 5 years@Conifold To replace the "particle" with a blanket "there are no particles" imo is wrong. There are QFT modeled wave packets in four dimensional space, identified one to one with the macroscopic tracks; we call those wave packets " particles " and describe their properties. We call virtual particles "particles" because they carry the quantum numbers of the free particles. One needs labels and the label "particle" for the macroscopic manifestations of the electron wave packet imo is appropriate.

Conifold about 5 yearsNo need to change established terminology, quantumparticle is a fine idiom. But I have seen it become a learning obstacle unless proper care is taken early on. Textbooks are forced to gloss lack of it with "waveparticle duality", which does not do the job and often confuses matters further. Like you I am speaking from experience: the particle heuristic does not work for everybody everywhere, and "neither waves nor particles with limited analogies to them in special cases" should at least be mentioned as a "more precise" option.