Prodded by several recent articles, I've been trying to follow what it says, and am still quite unsure of the realities.

The analogy usually presented is, if you have a small oil drop on the surface of water, and the water container is subject to a regular pattern of vibration, the water forms standing waves in shapes affected by the edges of the container and any obstructions in the surface of the water. And the oil drop tends to move across the surface of the water following the paths in those waves.

If you look solely at the oil drop, you can't tell which of two equal paths it would follow, but you can predict it will take one of them with equal probability, and predict its motion probabilistically. And if you couldn't see the standing water waves, you could deduce something in that shape exists.

You can even get some analogies for weird quantum behaviour like the an electron passing through two parallel slits and experiencing interference with itself: the water waves form possible channels for the oil drop, and the oil drop goes through one slit or the other, but ends up only at certain places on the far side.

However, the analogy to actual quantum physics is still unclear to me. Not whether it's true, but even what people are suggesting might happen.

Are people suggesting there's some underlying medium like the water? In that case, isn't there some propagation speed? The water waves exist in a steady state once all the obstructions are set up, but they don't respond to changes instantly. If the water trough were miles long, the oil drop would set off following water wave paths that existed at the point it passes through, not the paths corresponding to the obstructions that are going to be in place when the oil drop passes through them.

And yet, as I understand it, no-one expects a propagation delay in quantum experiments. People keep checking it out, but there never is: it always acts like an electron propagates just like it is itself a wave.

I agree, if there WERE some delay, if you changed the slits at this time, and got one result, and changed them at another time, and got another result, that would be massive, massive, evidence of something, possibly of something like pilot wave theory. But AFAIK proponents of pilot wave theory aren't advocating looking for such delays, and don't expect to find any.

Contrariwise, if this is just an analogy, and the quantum equivalent of the water waves (equivalent to the wave function in other interpretations of quantum mechanics) propagates at "infinite" speed, then... that is undetectable, indistinguishable from other interpretations of quantum mechanics. But it raises red-flag philosophical questions about what "infinite speed" means when all the intuition from special (or general) relativity indicates that all physical phenomena are local, and are influenced only by physics of nearby things, and "the same time" is a human illusion like the earth being stationary. Even if you don't expect to detect the pilot wave, can you write down what it should be in a universe where physics is local? Does that in fact provide a way to make QM deterministic and independent of observers, even if you change the reference frame? Because it doesn't sound like it will work.

FWIW, those are very superficial objections, I don't understand what it's saying enough to actually evaluate in depth. But I don't understand why these don't show up on lists of "common objections and rebuttals". Common objections have confident rebuttals in several places, and I've *seen* articles about them, but not understood well enough. Can anyone explain better?

**Digression**

I do agree, the idea that QM equations are an emergent property of something else, ideally a statistical interpretation of a deterministic underlying reality, would be very nice in clearing up a lot of confusion. But AFAIK, the closest candidate to that is Many Worlds, which doesn't appeal to many people who want to get away from QM unpleasantness.

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Chris Purcell(Permanent Link)cartesiandaemon(Permanent Link)ndt1001(Permanent Link)I agree with what Chris says.

I'm not sure what you mean by "actual quantum physics". Maybe you mean that there's a wave function and that when you make a measurement the wave function collapses; That's actually called the Copenhagen interpretation. One of the confusing things about quantum physics is that we can't identify the true nature of thing; We can only make predictions. We have a dozen candidates for the true nature of things which all give the same predictions and we call these candidates interpretations. To say things like "The electron is a wave" is an interpretation in the same way that saying "The electron follows the pilot wave" is an interpretation.

In the pilot wave theory, the pilot wave follows the same equations as the wave function in the Copenhagen interpretation. That's why the two interpretations can give the same predictions. A proponent of pilot wave theory would say that the pilot wave is an underlying medium which controls the movement of the electron (or other particle).

The question about the propagation speed applies to all interpretations. In the Copenhagen interpretation, perturbations to the wave function travel at some speed or other, just like in the pilot wave. If you turn the electron gun on, creating a perturbation, it takes some time for the electrons to hit the screen. Similarly, if you close a slit in a double slit experiment it takes some time for the effect to be seen on the screen. The speed of propagation depends on which equation you're using to model the wave-function/pilot-wave.

If you're using the Schrödinger equation then some perturbations travel at infinite speed, which makes that equation non-relativistic. If you're using the Dirac equation then perturbations cannot travel faster than the speed of light and the equation is compatible with special relativity.

I'm pretty sure you can formulate pilot wave theory using the Dirac equation rather than the Schrödinger equation, making QM deterministic and independent of reference frame. I've not seen it done, but I don't see any major problems. However, it would become a non-local theory when you start to model more than one particle in order to handle entanglement. There would be non-local effects behind the scenes, but you wouldn't be able to send information faster than the speed of light. In that respect it's much like the Copenhagen interpretation.

The pilot wave theory does have one feature which I find uncomfortable: Although the pilot wave affects the momentum and position of the particle, the particle has no effect on the pilot wave. What that's saying is that the actual position of the particles in the system can have no effect on any interference effects you might see. In that sense, the actual position of the particles seem like optional extras. I really don't like that feature, but others think it's acceptable.

Cheers,

Neil.

cartesiandaemon(Permanent Link)I'm not sure what you mean by "actual quantum physics"I'm not sure where the confusion is? I explain the oil drop on vibrating water analogy. That's often used as an analogy for pilot wave theory, but AFAIK is only a very loose analogy? Then I switch to talking about quantum physics. I guess I could leave out "actual" and just say "quantum physics", would that be better?

A proponent of pilot wave theory would say that the pilot wave is an underlying medium which controls the movement of the electron (or other particle).But is that just an analogy? Or do they mean medium, as in, it has some physical existence even if we can't detect it (but that might suggest we could one day)? Or some people think one and some people think the other?

If you turn the electron gun on, creating a perturbation, it takes some time for the electrons to hit the screen.I may need to work this through a little. I wasn't really familiar with the Dirac equation before this. But I assumed that if you had a wave equation like the schrodinger equation you *could* have a version of it with a propagation speed, whether or not we were able to write down a mathematical form for it or not.

But that seems to be the propagation of the wave function (which non-pilot-wave interpretations mostly equate with the actual object). Which commonly travels at the speed of light (like we expect most things to).

But imagine an incredibly large version of a simple experiment like the double slit experiment. AFAIK the copenhagen or many worlds interpretation has the wave propagate locally from the origin, through each slit, and to the eventual detector. If one slit is closed or the wave passing through it is affected in some way, that half of the wave doesn't arrive or doesn't exhibit interference cancellation with the other slit. But what happens at each slit is independent.

But pilot-wave theory seems to suggest there's a particle with a specific position which is travelling at the speed of light, following "ripples" in the pilot wave. But those ripples then need to travel backwards in time, in order to ensure there's a path for the particle towards whichever slit is open when it finally reaches one, not which is open at the time its emitted.

That's what seems extremely dodgy to me. I guess that's not what pilot wave tries to say? But why not, what's the difference?

Even if that is true, then I agree that if you can't necessarily transmit information like that, it's not necessarily ruled out, but it sounds very wrong as a suggestion for how the universe is likely to be set up.

ndt1001(Permanent Link)Thanks for your thoughts.

I think it's best not to get bogged down about whether it's "actual quantum physics" or just "quantum physics". It was a minor point and I generally understood your post.

But is that just an analogy? Or do they mean medium, as in, it has some physical existence even if we can't detect it?I see what you're asking now. I think the wave travelling in a medium would mostly be seen as an analogy, much like I would say that electromagnetic waves are analogous to water waves. Like you suggest though, some people may see things differently. One way in which the analogy breaks down is that the propagation speed of a pilot wave (or wave-function wave) can vary between waves and between different parts of the same wave. More on that below...

I wasn't really familiar with the Dirac equation before this.I'm glad I've introduced you to something new. I must admit I'm a bit rusty on this. Maybe I'll have a go at solving the two equations for a fairly localised moving particle. That would be instructive and help to identify anything I've misremembered.

But that seems to be the propagation of the wave function (which non-pilot-wave interpretations mostly equate with the actual object). Which commonly travels at the speed of light (like we expect most things to).I wouldn't expect perturbations in the wave function to propagate at the speed of light. I would expect them to travel at the same speed as the particle (well actually the range of possible speeds of the particle since the speed isn't usually well defined). That's a major difference between water waves and wave-function waves; water waves always travel at (roughly) the same speed, but wave-function waves can travel at any speed (but limited to be lower than the speed of light in the case of the Dirac equation). Having said that, high energy electrons will travel at close to the speed of light and the wave-function of light propagates at the speed of light.

The wave-function equations needs to approximate Newton's laws when they are applied outside the realm of the very small. The wave-function of the centre of mass of a tennis ball needs to have a peak which is much less than 1mm wide since I could measure the position with that accuracy and momentum which is restricted to a range of less than 50g.mm/s since the mass is about 50g and the speed can certainly be measured to within 1mm/s. That wave needs to propagate without spreading out significantly at a speed of momentum/mass. As I understand it, both the Schrödinger equation and the Dirac equation can be used to model a tennis ball like that.

That's what seems extremely dodgy to me. I guess that's not what pilot wave tries to say? But why not, what's the difference?I agree it's dodgy. I think what actually happens is that the particle gets carried along by the pilot wave and sometimes hits a closed slit. There's nothing to say that it must get to the other side. When you close one of the slits, the number of particles hitting the screen will go down, by a factor of two if there are two identical slits. Most of the particles in a double slit experiment will hit the material which surrounds the slits.

However, this raises another uncomfortable aspect of the pilot wave theory. You now have a pilot wave which continued through one of the slits, but the particle got left behind because it tried to go through the slit which was closed. I suppose that's really just a reflection of the fact that the pilot wave pays no attention to what actually happened to the particle.

Cheers,

Neil.

cartesiandaemon(Permanent Link)Maybe I'll have a go at solving the two equations for a fairly localised moving particle. That would be instructive and help to identify anything I've misremembered.I'm interested if you do; despite a maths background, I've not approached the actual maths here very much.

I wouldn't expect perturbations in the wave function to propagate at the speed of light. I would expect them to travel at the same speed as the particleSorry, yes, I meant, "speed of light, or less", I was tripped up because I so often use light as an example. I meant to contrast "speed of light or less" with "faster than light/infinite speed/backwards in time", not with slower velocities.

The wave-function of the centre of mass of a tennis ball needs to have a peak which is much less than 1mm wide since I could measure the position with that accuracyI have no worthwhile experience with the numbers, but that sounds roughly how I understood it.

I think what actually happens is that the particle gets carried along by the pilot wave and sometimes hits a closed slit. There's nothing to say that it must get to the other side.Doh. I thought through this more. You're quite right, that all works fine (I think).

If I understand enough, it sounds like pilot-wave theory is equivalent to a hidden variable theory, with the hidden bit being "where the particle really is". That means most of the simple experiments work as expected, but the Bell Inequality experiments rule out the idea that the true-particle-position travels locally (or if it travels locally with respect to the pilot wave, then the pilot wave must travel back in time). Is that right?

ndt1001(Permanent Link)I'm interested if you do; despite a maths background, I've not approached the actual maths here very much.So you won't be scared by partial differential equations, complex numbers and exponentials?

Sorry, yes, I meant, "speed of light, or less",I see. We're on the same page then.

If I understand enough, it sounds like pilot-wave theory is equivalent to a hidden variable theory, with the hidden bit being "where the particle really is".Exactly.

the Bell Inequality experiments rule out the idea that the true-particle-position travels locally (or if it travels locally with respect to the pilot wave, then the pilot wave must travel back in time). Is that right?Yes to the former; The true particle position is non-local if you have more than one particle. You would actually track the true state of the system rather than tracking particles individually and the rate of change of the state variables of one particles would depend on the states of all the other particles.

Cheers,

Neil.

cartesiandaemon(Permanent Link)So you won't be scared by partial differential equations, complex numbers and exponentials?That's fine.

Yes to the former; The true particle position is non-local if you have more than one particle.Right. That's what it sounded like. But that makes it hard for me to see what's attractive about the theory.

ndt1001(Permanent Link)Right. That's what it sounded like. But that makes it hard for me to see what's attractive about the theory.It's attractive because it's deterministic; The randomness comes from not knowing the initial true-state variables. But I'm with you since I don't find that reason to be sufficiently attractive.

Coincidentally, Looking Glass Universe has just done a video on pilot wave theory. She points out that pilot wave theory solves the measurement problem (how the wave-function collapses and why) by not having any wave-function collapse.

Cheers,

Neil.

ndt1001(Permanent Link)Here's the big picture for solving these equations. First, I'm a wimp so I'm only going to tackle a particle moving in free space. So the time dependant Schrödinger equation looks like this:

iℏ∂Ψ(r,t)/∂t = -(ℏ

^{2}/2m)∇^{2}Ψ(r,t)As usual, i

^{2}=-1. The symbol ℏ represents the reduced plank's constant. For the purposes of this discussion, it's a small constant. We're trying to solve for the wave-function, Ψ.This is a linear partial differential equation, so if we find two or more solutions then a linear combination of them will also be a solution. It's often a good idea to look for fixed energy solutions as the equation is simpler when doing that. These are solutions which obey the following equation, where E is the energy:

Ψ(r,t) = Ψ(r,0) e

^{Et/iℏ}Substituting this into the original equation allows us to evaluate the partial derivative with respect to time leaving only the spatial derivatives. This new equation is the time-independent Schrödinger equation (tiSe).

In free space I happen to know that the solutions of the tiSe look like:

Ψ(r,0)=e

^{p.r/-iℏ}Here, p is the momentum vector of the particle. Handily, these look like Fourier components. So the strategy is to take the initial wave-function, split it into Fourier components, work out the value of E for each component, multiply by e

^{Et/iℏ}and recombine the components to produce the final solution.More details later.

Cheers,

Neil.

Edited at 2017-08-19 08:05 pm (UTC)ndt1001## Deriving the time-independent Schrödinger equation

(Permanent Link)iℏ∂ψ(r,t)/∂t = -(ℏ

^{2}/2m)∇^{2}ψ(r,t) + V(r,t)ψ(r,t)This includes an extra term involving V(r,t) which I didn't include before. This V(r,t) is the potential energy of the particle at position r and time t.

I should have said that the wave-function, ψ(r,t) is a function which takes a complex value over all of time and space. The r variable represents position and the t variable represents time. ∇ is the Laplace operator. If anything else isn't clear then please ask.

Whenever V(r,t) is independent of t, we can convert the tdSe to the time-independent Schrödinger equation (tiSe) by assuming ψ takes the form:

ψ(r,t) = ψ(r,0) e

^{Et/iℏ}.Although this looks like I'm pulling a rabbit out of a hat, this particular rabbit often hangs out around Schrödinger equations. As a side note, integrating functions and solving differential equations often seems to me like pulling rabbits out of hats. Yes, there are tools which work in subsets of cases, but nothing which works generically.

The assumed form of V is:

V(r,t) = V(r,0).

Back to the tdSe. I'm going to simplify the left-hand side by substituting the assumed form of ψ

iℏ∂ψ(r,t)/∂t = iℏ∂(ψ(r,0) e

^{Et/iℏ})/∂t= iℏψ(r,0) ∂(e

^{Et/iℏ})/∂t= iℏψ(r,0) (E/iℏ) e

^{Et/iℏ}= E ψ(r,0) e

^{Et/iℏ}.Then, I'm going to simplify the right-hand side in the same way.

-(ℏ

^{2}/2m)∇^{2}ψ(r,t) + V(r,t)ψ(r,t)= -(ℏ

^{2}/2m)∇^{2}(ψ(r,0) e^{Et/iℏ}) + V(r,0)ψ(r,0) e^{Et/iℏ}= -(ℏ

^{2}/2m)(∇^{2}ψ(r,0)) e^{Et/iℏ}+ V(r,0)ψ(r,0) e^{Et/iℏ}.Equating the simplified left-hand side with the simplified right-hand side, we have:

E ψ(r,0) e

^{Et/iℏ}= -(ℏ^{2}/2m)(∇^{2}ψ(r,0)) e^{Et/iℏ}+ V(r,0)ψ(r,0) e^{Et/iℏ}.We can divide through by the common factor e

^{Et/iℏ}.E ψ(r,0) = -(ℏ

^{2}/2m)∇^{2}ψ(r,0) + V(r,0)ψ(r,0).This is the time-independent Schrödinger equation.

I've introduced a new constant, E. This is actually the energy of the particle, but I'm not sure how to show that. For now, I'm going to plow on and find solutions to the equation.

Cheers,

Neil.

ndt1001## Solving the time-independent Schrödinger equation

(Permanent Link)E ψ(

r,0) = -(ℏ^{2}/2m)∇^{2}ψ(r,0) + V(r,0)ψ(r,0).I'm going to solve this for a particle in free-space which, having no forces acting on it can neither gain nor lose potential energy, so the function V(

r,0) is constant. The value of the constant doesn't matter since it only produces a constant offset in E. The resulting wave-function will be different, but not in a measurable way.For simplicity, I'm going to set V(

r,0) to zero yielding:E ψ(

r,0) = -(ℏ^{2}/2m)∇^{2}ψ(r,0).This is definitely a tricky step when V(

r,0) is not constant, but in this case it's not too bad. What it's saying is that taking the second derivative of ψ is the same as multiplying by a constant. One way to achieve this is to say that if you take the first derivative, the result is the same as multiplying by a constant.That's a bit imprecise as the ∇

^{2}operator produces a scalar, but∇on its own produces a vector. So maybe our new constant is a vector. Let's try it:∇ψ(r,0) =vψ(r,0).Both sides of this equation are vectors and I'm going to separate the components. I'll also split the

rparameter to ψ into its components:∂ψ(x,y,z,0)/∂x = v

_{x}ψ(x,y,z,0),∂ψ(x,y,z,0)/∂y = v

_{y}ψ(x,y,z,0),∂ψ(x,y,z,0)/∂z = v

_{z}ψ(x,y,z,0).Each of those equations is solved by some kind of exponential since the exponential function remains the same when you take the derivative. The solutions are:

ψ(x,y,z,0) = ψ(0,y,z,0) e

^{vxx},ψ(x,y,z,0) = ψ(x,0,z,0) e

^{vyy},ψ(x,y,z,0) = ψ(x,y,0,0) e

^{vzz}.Putting these together, we have:

ψ(x,y,z,0) = ψ(0,0,0,0) e

^{f(x,y,z)}where

f(x,y,z) = v

_{x}x + v_{y}y + v_{z}z =v.r.So,

ψ(r,0) = ψ(

0, 0) e^{v.r}.We can now substitute this into the right-hand side of the simplified tiSe:

-(ℏ

^{2}/2m)∇^{2}ψ(r,0)= -(ℏ

^{2}/2m)∇^{2}ψ(0,0) e^{v.r}= -(ℏ

^{2}/2m)v^{2}ψ(0,0) e^{v.r}= -(ℏ

^{2}/2m)v^{2}ψ(r,0).Now, we've got a constant multiple of ψ(

r,0) on both the left-hand side and the right-hand side, we can equate the two constants (since ψ must be non-zero somewhere).E = -(ℏ

^{2}/2m)v^{2}This is a bit troubling because the energy is supposed to be positive. The resolution is that

vis an imaginary vector. If we make this assignmentv= ik, we get:E = (ℏk)

^{2}/2m.This equation looks somewhat like one of the classical equations for kinetic energy:

E = p

^{2}/2m,where

pis the momentum. In fact, -ℏ∇is the momentum operator in quantum mechanics and ℏkis the momentum value of this solution.Finally, we can plug everything back together again and get:

ψ(

r,t) = ψ(0,0) e^{(Et-p.r)/iℏ}.The analysis above might not have produced all of the solutions to the equation, but with a bit of knowledge about Fourier analysis it should be possible to show that these solutions are the only ones as it is possible to form any function at t=0 with a linear combination of these functions. The resulting values at t≠0 will be forced by the tdSe.

I hope this is clear and perhaps a little enlightening.

Cheers,

Neil.