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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.

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    $\begingroup$ Since quantum mechanical particles don't behave like classical ones, every interpretation has a slightly different twist so that none of them give 'the trajectory of a particle'. In Copenhagen you don't have the 'trajectory'. In path integral, you don't get 'the trajectory', instead you get a ton of them. In Bohm, you get a particle trajectory, but you also have this pilot wave thing which doesn't have a trajectory. $\endgroup$ – knzhou Mar 17 '16 at 20:21
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    $\begingroup$ "the almost literal identification of the particle with the wave packet..." But in the formalism, are they not literally the same? I will have to read Ballentine's paper more closely, but this strikes me as somehow thinking that it's really CM "under the hood", instead of the simpler alternative that the world looks like it behaves according to QM because it behaves according to QM. $\endgroup$ – Robin Ekman Mar 18 '16 at 0:25
  • $\begingroup$ Have a look at the discussion here physics.stackexchange.com/q/15539 $\endgroup$ – anna v Mar 18 '16 at 7:17
  • $\begingroup$ Related physics.stackexchange.com/q/186170 $\endgroup$ – juanrga Oct 18 '16 at 17:35
  • $\begingroup$ @RobinEkman, they are not. Think of psi function for system proton + electron. There is one function, but there are two particles. $\endgroup$ – Ján Lalinský Aug 6 '18 at 20:19
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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.

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    $\begingroup$ I think you are confusing the statistical interpretation with a hidden variable model. Ballentines article states that the statistical interpretation is neutral towards thr existance of a hidden variable model. $\endgroup$ – lalala Oct 9 '17 at 17:25

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