10
$\begingroup$

As a particle travels to a screen, it is traveling through 3-dimensional space. In the oil droplet experiment, there are only two dimensions of any importance—the droplet merely moves along the surface of a wavey fluid. It seems like a 3-dimensional wavey superfluid would result in far more complex dynamics. So how does the fluid surface in the oil droplet experiment correspond to the actual space a particle travels through?

Links, as requested:

$\endgroup$
2

5 Answers 5

9
$\begingroup$

In the Faraday pilot-wave fluid droplet dynamics, the fluid wave is meant as an analogy for the wavefunction. More specifically, the experiments are constructed as physical implementations analogous to the de Broglie-Bohm theory, where a particle with discrete coordinates is 'guided' by a pilot wave which follows the Schrödinger equation.

To be a bit more explicit, the de Broglie-Bohm theory works with a standard wavefunction $\psi(\mathbf q,t)$ on a single- or multi-particle configuration space $\{\mathbf q\}$, which obeys the Schrödinger equation $$ i\hbar\frac{\partial}{\partial t}\psi(\mathbf q,t)=-\frac{\hbar^2}{2m}\sum_i\nabla_i^2\psi(\mathbf q,t)+V(\mathbf q)\psi(\mathbf q,t). $$ This wavefunction then guides an actual 'particle' with coordinates $\mathbf q(t)$ on the configuration space (so it may represent the coordinates of multiple particles) by matching its momentum to the local momentum of the wavefunction, understood as $$ m_k\frac{\mathrm d\mathbf q_k}{\mathrm dt}=\hbar \operatorname{Im}\left(\frac{\nabla_k\psi(\mathbf q,t)}{\psi(\mathbf q,t)}\right) $$ for the $k$th particle. The particle is then distributed on the configuration space according to $|\psi(\mathbf q,t_0)|^2$ at an initial time $t_0$, and it retains that distribution; upon measurement it is the configuration-space particle that gets detected.

The water droplets behave similarly but not exactly in this way. The fluid surface acts in a wave-like way, and it influences the particle-like droplet, exchanging energy and momentum with it. The analogy is, however, nowhere near exact, and it is described in considerable (but still readable) detail in

Faraday pilot-wave dynamics: modelling and computation. P.A. Milewski et al. J. Fluid Mech. 778 361 (2015).

In short, the droplet moves balistically between bounces, and during the bounces its horizontal component $\mathbf X(t)$ obeys the Newton equation $$ m\frac{\mathrm d^2\mathbf X}{\mathrm dt^2}+\operatorname{drag}(t)\frac{\mathrm d\mathbf X}{\mathrm dt}=-F(t) \nabla \overline\eta $$ where $\overline\eta=\overline\eta(x,y)$ is the fluid surface height you'd have if the droplet wasn't bouncing.

This means, to begin with, that there's two big differences between the Faraday-wave droplets and the de Broglie-Bohm theory. For one, the wave's influence on the mechanical system is of a rather different character. To tack on to this, the droplet can influence the Faraday wave, which is unthinkable in Bohmian mechanics.


So, to summarize: the liquid wave is an (imperfect) physical analogue for the Bohmian wavefunction. What is the fluid surface itself an analogue for? Nothing. That's an over-reading of the analogy. It is just an analogy, and not all elements of the analogy need to mean something on the other side. The status of the analogy is well summarized by J.W.M. Bush in the introduction of

The new wave of pilot-wave theory. J.W.M. Bush. Physics Today 68 no. 8, 47 (2015). U. Arizona eprint.

The bouncing-droplet experiments are indeed important. They show that particle-like system can indeed display wave-like behaviour, like two-slit interference or mode quantization, from more fundamental interactions which we might have missed at first. In this sense, they are an encouragement to keep looking for a Bohmian-like explanation for quantum mechanics.

On the other hand, bouncing-droplet experiments are fundamentally limited. They have a hard time simulating the three-dimensional motion of a single-particle, and they mostly cannot simulate the quantum mechanics of multiple particles (even two particles on one dimension) since the wavefunction is a wave on the (#particles)$\times$(#dimensions)-dimensional configuration space. In other cases, such as Mandel dips, the quantum wave interference occurs over even more abstract spaces.

The bouncing-droplet experiments, like other quantum analogues such as elastic rods, are therefore not proof of anything, and they have no new implications on the foundations of quantum theory. They are an interesting testing ground, and it would be good to see them take on interesting foundations issues, like Bell inequalities and contextuality, where they could produce interesting new questions and takes which could then be mirrored on the quantum side. As they are now, though, they're simply an interesting curiosity, I'm afraid.

$\endgroup$
0
0
$\begingroup$

As a particle travels to a screen, it is traveling through 3-dimensional space.

No, what happens is the configuration space of the system changes. This is essential for explaining interactions that destroy the interference. If your right slit deflected downwards and the your left slit deflected upwards then the two waves wouldn't overlap and hence when the wave goes through both slits there would be no interference. Just a patch to the right and down and another patch to the left and up and no fringes. No brighter strips, no fainter strips, just two distinct disjoint patches. Similarly, when you get which way information when you go though one slit you do get the information by changing a different particle and hence you deflect that other particle and thus the whole beam as a function of configuration space is deflected to then not overlap, hence you get no interference.

Quantum mechanics seems way more mysterious when people don't tell you the wave assigns complex numbers to configurations, not to locations in physical 3d space.

So how does the fluid surface in the oil droplet experiment correspond to the actual space a particle travels through?

It's not a surface. You have a wave. It assigns a complex number (and a spin vector) to every configuration. The (change in) phase tells you how the configuration is changing if the configuration is at that given configuration. The magnitude of the wave at a given configuration is what corresponds to the surface of the fluid. It produces an additional force on the particles in the configuration. The configuration would naturally have a classical force based on how the parts interact when they are that far away from each other. And the (change in) magnitude tells you an additional quantum force.

So the wave evolves as it always does. So when the wave has nonzero parts corresponding to parts going through each slit, you might get interference. The location of each particle (something you don't know) jointly gives the configuration of the entire system. The (spatial variation of the) phase of the wave then tells you the velocity of each particle. And the (spatial variation of the) magnitude of the wave then tells you an additional quantum potential that also changes the velocity as much as the classical potential corresponding to that configuration does.

$\endgroup$
0
0
$\begingroup$

Water surface plays the role of energy and impulse transfer,while in vacuum the particle is exposed by forses of the unknown energy transfering fields that fluctuate by the magnitude and diversion,thus making particle motion chaotic and intermittent. The main difference is that the transfer of energy from vacuum to particle and within the vacuum are not devided from each other as in the walking droplets bath.

$\endgroup$
0
$\begingroup$

Surface is property of 3D world, so question is flawed from beginning. In walking droplet experiment, water is the medium, which transfers water surface waves. In quantum mechanics, corresponding medium is electromagnetic field, which transfer electromagnetic waves. EM waves are created by short random moves of charged particles caused by vacuum energy, see https://en.wikipedia.org/wiki/Casimir_effect#Possible_causes:

In a simplified view, a "field" in physics may be envisioned as if space were filled with interconnected vibrating balls and springs, and the strength of the field can be visualized as the displacement of a ball from its rest position. Vibrations in this field propagate and are governed by the appropriate wave equation for the particular field in question. The second quantization of quantum field theory requires that each such ball-spring combination be quantized, that is, that the strength of the field be quantized at each point in space.

PS. I'm amateur.

$\endgroup$
-4
$\begingroup$

In the oil droplet experiments that suggest de Broglie’s pilot wave theory might be accurate, what does the fluid surface correspond to?

The three dimensional space that particle is moving through. But might I add that you shouldn't think of the particle as the oil droplet. Instead think of the particle as something more like a hurricane. The eye of the storm corresponds to the oil droplet. And that's where there is no wind.

As a particle travels to a screen, it is traveling through 3-dimensional space. In the oil droplet experiment, there are only two dimensions of any importance—the droplet merely moves along the surface of a wavey fluid.

Have a play around with Falaco solitons. IMHO it's a better analogy than the oil droplet. It's still not ideal mind.

It seems like a 3-dimensional wavey superfluid would result in far more complex dynamics.

True enough. The dynamics of 3-dimensional wavey water are complicated enough:

enter image description here GNUFDL image by Kraaiennest, see Wikipedia

So how does the fluid surface in the oil droplet experiment correspond to the actual space a particle travels through?

IMHO the fluid surface corresponds to the space, and the oil droplet corresponds to the middle of the particle. Don't think of an electron as some little ball that has a field, instead think the electron's field is what it is. And this field doesn't have an edge. But it is spherically symmetric, so it does have a centre.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.