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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.

Qmechanic
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Julia
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4 Answers4

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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.

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.

sinlgle phot

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.

anna v
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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.

WillO
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Trajectories in Standard quantum mechanics

Standard quantum mechanics does not involve trajectories for individual particles. For simplicity, let's focus on a one-particle wave function (the argument also applies to the case of several particles). A trajectory of a single would be given by an assignement $$ X: \mathbb T \to \mathbb R^3\,,\enspace t \mapsto X_t\,,$$

where $\mathbb T$ is the time.

Standard quantum mechanics is a probabilistic theory. So the position of the particle $X_t$ is described by a random variable. According to the Born rule, the probability distribution of $X_t$ is given by $$\mathbb P_{\psi_t}( X_t = x) = |\psi_t(x)|^2 dx\,.$$
However, the standard formulation fails to specifiy a joint probability for random variables $\{X_t\}_{t\in \mathbb T}$. If given such a joint probability measure, one could calculate the probability for the particle's trajectory to be $\gamma: \mathbb T \to \mathbb R^3$: $$ \mathbb P( X_\bullet = \gamma_\bullet) = \mathbb P( \forall t: X_t = \gamma_t)\,. $$ Note, that taking the product measure treating the $X_t$ as independent, i.e., $$ \mathbb P( X_\bullet = \gamma_\bullet) = \prod_{t \in \mathbb T} |\psi_t(\gamma_t)|^2 dx\,, \tag{1} $$ does not yield physical trajectories, as the particle would randomly jump between different places. This gets even worse if the many particle case is considered. E.g., the ensemble of particles making up Schrödinger's cat would randomly jump between dead cat configurations and live cat configurations. However, this would not affect the histories we remember as our brain would constantly jump between a brain remembering a cat that died and a brain that remembers a cat that survived.

Trajectories in the phase space formulation of quantum mechanics

In the phase space formulation of quantum mechanics, instead of a wave-function or density matrix the quantum state is described by a quasi probability distribution $W(x,p)$. The time evolution is given by $$ \frac{\partial W}{\partial t} = - \{\{ W, H\}\} = - \{ W, H\} + \mathcal O(\hbar^2)\,, \tag{2} $$ where the Moyal bracket $\{\{\bullet, \bullet\}\}$ is a deformation of the classical Poisson bracket. For $\hbar \to 0$ the equation would reduce to the classical Liouville equation. The difference for non-vanishing $\hbar$ is that $(2)$ is not a flow, because the density of points is not conserved. That is, the evolution of the signed measure $\mu_t(A) = \int_A W_t(x,p) dx dp$ is not given by a flow $\Phi_t : \mathbb R^{2n} \to \mathbb R^{2n}$ such that $$\mu_t(A) = \Phi_t{}_* \mu_0(A) = \mu_0(\Phi_t{}^{-1}(A))\,.$$ this flow would be needed to define the trajectories. In classical mechanics, the trajectories for initial configuration $(x_0, p_0)$ would be defined by $(x_t, p_t) := \Phi_t(x_0, p_0)$, where $\Phi_t$ is the flow corresponding to the classical phase space measure $\rho$.

Trajectories in Bohmian mechanics

Bohmian mechanics is a proposal for a rational completion of quantum mechanics. According to Bohmian mechanics, the wave function always evolves according to the Schrödinger equation $$ \mathrm i \hbar \frac {\partial \psi} {\partial t} = \hat H \psi\, $$ and the particle positions $\vec Q_i(t)$ move according to the law $$ \frac {\mathrm{d} \vec Q_i(t)} {\mathrm{d}t} = \frac {\hbar} {m_i} \operatorname {Im} \frac {\vec \nabla_i \psi} {\psi}\,. $$

As a consequence of the continuity equation, if the initial configurations $\vec Q_i(0)$ are distributed according to Born's rule $$ |\psi_0(x_1, \dots, x_n)|^2$$ the configurations will be distributed according to Born's rule $$ |\psi_t(x_1, \dots, x_n)|^2$$ for any later time.

One way to view this is that Bohmian mechanics provides the joint probability measure that was missing in standard quantum mechanics. The difference between the Bohmian mechanics and the measure defined by $(1)$ is that Bohmian mechanics yields physically reasonable trajectories. In particular, according to Bohmian mechanics, we can trust our memories since Bohmian trajectories do not randomly jump between distinct branches of the wave function. E.g., if my brain is in the state remembering that I observed a cat alive for the last ten minutes, the most likely the Bohmian particles that constitute the cat were in configurations corresponding to a cat alive for the last ten minutes (There are of course caveats due to neuro-science/biology since humans can have false memories).

I hope this gives an overview of the status of trajectories in quantum mechanics.

Lukas Nullmeier
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In Ballentine's view the wave function contains statistical information to predict particle properties. Thus $\langle \psi({\bf r},t) | {\bf p} | \psi({\bf r},t) \rangle$ gives the probability density of finding a particle with momentum $\bf p$ at position $\bf r$ and time $t$. You can only access the information of an ensemble of such particles.

my2cts
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