Is the trajectory of a quantum particle a well defined concept and how does this depend on the interpretation of quantum mechanics?

Question

A common statement about quantum physics is that the "trajectory" of a particle is no longer a well defined concept because of the uncertainty relations for position and momentum.

If one interprets the uncertainty relations as a statement about the simultanous measurement of position and momentum of a given particle, this makes sense.

How does this change if one thinks about ensembles: Ballentine writes in his "The statistical interpretation of quantum mechanics" on page 356 that the following claim is unjustified:

The position and momentum of a particle do not even exist with simultaneously and perfectly well defined (though perhaps unknown) values.

Due to Ballentine is this

conclusion rests on the almost literal identification of the particle with the wave packet...

So does that mean that in the ensemble interpretation you can think of particles with simultaneously well defined position and momentum and thus also of a well defined particle trajectory.

Is this correct and why does it make sense?

Are there other interpretations where you come to different answers to the question if the trajectory of a particle is a well defined concept?

Edit I just realized that one should distinguish between trajectory (classically a map from an interval $I$ to $\mathbb{R}^3$, $\gamma \colon I \to \mathbb{R}^3$ and a path which doesn't involve the time dependence, i.e. path = $\gamma(I)$.

For a path I think it should definitely make sense to talk of an ensemble path for some states.

Answer 1

I think you are correct to center on the Heisenberg uncertainty principle in order to separate the statistical interpretation from the main stream interpretation of the wavefunction applying to an individual particle.

In this question the Heisenberg Uncertainty is discussed for the two interpretations.

In my opinion statistics is a mathematical branch, well defined with given distributions. These are Poisson or Gausian or some others derived mathematically. The standard deviations arising from these distributions are well defined. In a quantum mechanical ensemble the uncertainty does not follow a statistical distribution and this is proven in a recent paper here . The functional form of the distribution depends on the quantum mechanical wavefunction solution for the system under study and not on statistical standard deviations.

Quantum mechanics is about probabilities and in order to measure probabilities one has to use ensembles, ensembles are necessary. It seems he is proposing that they are also sufficient and there is no need for quantum mechanics to rest on the underlying individual systems . The HUP though can be tested on individual particles and systems.

At the moment I have found one publication that claims a violations of the HUP on individual particles, and it seems it has not been repeated.It is behind a pay wall, but I will keep looking for a pdf.

In my opinion if solid measurements of individual particle positions and momenta display a violation of the HUP the statistical interpretation would be validated, where statiscitical means that the distribution comes from the quantum mechanical wavefucntion solution. There are many experiments that validate the HUP on ensembles. Otherwise it is just a claim with no experimental proof.( Solid means the 5 sigma significance of the result usual in particle physics experiments).

So the mainstream interpretation that quantum mechanical wavefunctions apply to individual particles seems still safe.

As my attention was drawn to this question again , I would like to address part of the title:

Is the trajectory of a quantum particle a well defined concept

Photons are elementary particles obeying quantum mechanical rules, and the simple experiment of single photons scattering off double slits shows that same energy photons one at a time have different trajectories, and it is only the ensemble that shows the wave nature of the interaction.

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

For this experiment, for each point hit of a photon a trajectory can be drawn with the uncertainty of the double slit width. A very narrow wavepacket in space.

Answer 2

I think that the uncertainty relations (by which I assume you mean the Heisenberg uncertainty principle) are a red herring here.

Imagine that you know about the position and momentum operators but have never heard of, and have never thought of, the uncertainty principle (that is, the lower bound on the product of the uncertainties). As long as you've noticed that the operators have no eigenstates in common, you'll still be led to reject the classical notion of "trajectory". So that rejection can't depend on the uncertainty principle.