Does quantum mechanics imply that particles have no trajectories?

Question

In Classical Mechanics we describe the evolution of a particle giving its trajectory. This is quite natural because it seems a particle must be somewhere and must have some state of motion. In Quantum Mechanics, on the other hand, we describe the evolution of a particle with its wave function $\Psi(x,t)$ which is a function such that $|\Psi(x,t)|^2$ is a probability density function for the position random variable.

In that case, solving the equations of the theory instead of giving the trajectory of the particle gives just statistical information about it. Up to there it is fine, these are just mathematical models. The model from Classical Mechanics has been confirmed with experiments in some situations and the Quantum Mechanics model has been confirmed with experiments in situations Classical Mechanics failed.

What is really troubling me is: does the fact that the Quantum Mechanics model has been so amply confirmed implies a particle has no trajectory? I know some people argue that a particle is really nowhere and that observation is what makes it take a stand. But, to be sincere, I don't swallow that idea. It always seemed to me that it just reflects the fact that we don't really know what is going on.

So, Quantum Mechanics implies that a particle has no trajectory whatsoever or particles do have well defined trajectories but the theory is unable to give any more information about then than just probabilities?

Answer 1

Quantum systems do not have a position. This is intuitively hard to grasp, but it is fundamental to a proper understanding of quantum mechanics. QM has a position operator that you can apply to the wavefunction to return a number, but the number you get back is randomly distributed with a probability density given by $|\Psi |^2$.

I can't emphasise this enough. What we instinctively think of as a position is an emergent property of quantum systems in the classical limit. Quantum systems do not have a position, so asking for (for example) the position of an electron in an atom is a nonsensical question. Given that there is no position, obviously asking for the evolution of that position with time, i.e. the trajectory, is also nonsensical.

You say:

I don't swallow that idea. It always seemed to me that it just reflects the fact that we don't really know what is going on.

and you are far from alone in this as indeed his Albertness himself would have agreed with you. The idea that we don't know what is going on is generically referred to as a hidden variable theory, however we now have experimental evidence that local hidden variable theories cannot exist.

Answer 2

So, Quantum Mechanics implies that a particle has no trajectory whatsoever

It depends what "whatsoever" means and what "particle" means and what "trajectory" means. All these words in physics depend on the framework. For distances larger than nanometers and energies larger than some kilo electron volts or so, the classical framework is what defines these words. A particle has a fixed center of mass that given a momentum describes a trajectory according to the classical mechanics theories.

or particles do have well defined trajectories but the theory is unable to give any more information about them than just probabilities?

For distances larger than nanometers and energies larger than a few keV particles have well defined trajectories.( The sizes depend on the Heisenberg Uncertainty Principle and the very small value of h_bar).

Here is an electron trajectory, the electron is a particle

and the width of its trajectory is smaller than a micron. There is no ambiguity to its particleness , and the trajectory can be computed classically, given the magnetic field which is perpendicular to the plane of the photo.

What brings quantum mechanics in by force is if one accumulates a lot of electron-on-proton scatterings and tries to model mathematically what happens when an electron hits a proton . Classical mechanics fails and the theory of quantum mechanics has been very successful in describing the data at the microscopic level of an electron hitting a proton. The result is that the classical trajectory idea falls down in these small distances. One has instead of a particle meeting a particle, a quantum mechanical entity meeting a quantum mechanical entity and their interaction implies that there is a probability distribution controlling what is happening.

So "whatsoever" is defined as "trajectories exist in the macroscopic dimensions, quantum mechanical probabilities reign at the micro system." The classical trajectory emerges smoothly from the underlying quantum mechanical level.

P.S. There do exist macroscopic manifestations of quantum mechanics , like superconductivity , but that is another story.

Answer 3

Being strict, quantum mechanics doesn't rule out trajectories. In the first place quantum particles can follow classical trajectories under special settings. Anna provided the image of the classical trajectory followed by an electron in a cloud chamber.

Then we have quantum trajectories. Those are radically different from classical trajectories and produce all the 'exotic' behaviors we observer at quantum scales. The next quantum trajectories are plotted for the double slit experiment

The literature on quantum trajectories is very broad, because not only them are used for raising foundational questions about quantum mechanics, but to solve practical problems such as getting rates of chemical reactions, modeling of classical-quantum couplings, and so on. A good introductory book that cites relevant articles is the book by the one of the pioneers on computational quantum trajectory methods

Answer 4

If you have a conserved probability (such as in nonrelativstic quantum mechanics), then you get a conserved current, the probability current.

In almost all situations it will do for you what you want a velocity to do. Just don't try to get it to do more than you want by expecting it to be too classical. For instance the expectation value of momentum could be zero in an energy eigenstate, even if the probability current is zero everywhere.

Furthermore, the probability current is not unique. If you add the curl of an arbitrary vector field to your probability current it will accomplish just as much. So you don't want to read too much into it.

But it might be what you need if you are looking for something.

Answer 5

Quantum physics has momentum. It's just not something fundamental. Saying that quantum physics doesn't have momentum because it's just the waveform evolving is like saying that an airplane doesn't have wings because it's really just a bunch of atoms.

Answer 6

Question: Does quantum mechanics imply that particles have no trajectories?

Answer:

The location of a particle is probabilistic. Generally, the probability of finding the particle in a neighborhood around one point in space is different than the probability of finding the particle in a neighborhood around a distinct point in space. Further, generally the probability of finding the particle in a neighborhood around one point in space may vary from time to time. Perhaps more information regarding the trajectory of a particle might be gleaned from the concept of a probability current [1].

Yet, a particle may have a trajectory in Hilbert space. For example, each energy eigenvector in a Hilbert space is represented by a point in the Hilbert space. Similarly, the location of a soccer ball can be represented by a point in physical space. Just as soccer ball's position in physical space might evolve along a trajectory from an in initial point at time $t_i$ (i.e., $r(t_i)$) to a final point at time $t_f$ (i.e., $r(t_f)$), a particle's "position" in Hilbert space might evolve along a trajectory from an in initial point at time $t_i$ (i.e., $\Psi(\mathbb{r},t_i)$) to a final point at time $t_f$ (i.e., $\Psi(\mathbb{r},t_f)$).

Thus, I claim as follows:

In the first instance I include time as a variable. Then, I write, for example, that $\Psi : \mathbb{R}^3\times\mathbb{R} \to \mathcal{H}$ is the rule that assigns to any $\left(\mathbf{r},t\right)$ in the domain the vector $$ \Psi(\mathbf{r},t) $$ in the Hilbert space. Finally, I can denote a piecewise smooth curve $\mathcal {C} \subset \mathcal{H}$ where $\mathbf {\Psi} \colon \mathbb{R}\times[t_i,t_f]\to \mathcal{C}$ is a bijective parametrization of the curve $\mathcal {C}$ such that $\Psi(\mathbf{r},t_i)$ and $\Psi(\mathbf{r},t_f)$ give the endpoints of $\mathcal {C}$ and $t_i < t_f$.

In the second instance I include time as a parameter. Then, I write, for example, that $$\Psi_t : \mathbb{R}^3 \to \mathcal{H} \qquad \textrm{for}~t_i\leq t\leq t_f$$ is the rule that assigns to any $\mathbf{r} $ in the domain the vector $$ \Psi_t(\mathbf{r}) $$ in the Hilbert space. I can reinterpret the family of states of the variable $\mathbf{r}$ indexed by the other variable $t$ as $$ \Psi_\mathbf{r}(t). $$ Then, I can denote a piecewise smooth curve $\mathcal {C} \subset \mathcal{H}$ where $\mathbf {\Psi}_\mathbf{r} \colon \mathbb{R} \to \mathcal{C}$ is a bijective parametrization of the curve $\mathcal {C}$ such that $\Psi_{\mathbf{r}}(t_i)$ and $\Psi_{\mathbf{r}}(t_f)$ give the endpoints of $\mathcal {C}$ and $t_i < t_f$. Moving back to the former interpretation, $\Psi_{t_i}(\mathbf{r})$ and $\Psi_{t_f}(\mathbf{r})$ give the endpoints of $\mathcal {C}$.

[1] https://en.wikipedia.org/wiki/Probability_current

Answer 7

“In quantum mechanics, due to Heisenberg's uncertainty principle, the notion of a particle trajectory does no longer make sense. QT is probabilistic not deterministic.. ...no definite trajectory.