Do Bohmian mechanics trajectories for double slit provide which path information?

  • Thread starter SpectraCat
  • Start date
  • #1
SpectraCat
Science Advisor
1,395
1

Main Question or Discussion Point

In the aftermath of the recent Steinberg paper where they reconstruct average photon trajectories in the double-slit experiment, it has been pointed out several times that the reconstructed average trajectories strongly resemble the single-photon trajectories predicted by Bohmina mechanics. Indeed, the authors themselves mention this point explicitly in the paper, but are careful to say that the reconstructions in their work are interpretation-independent.

Having looked again at the 2001 paper where the Bohmian trajectories for photons in the double slit experiment are calculated (http://arxiv.org/abs/quant-ph/0102071), I found the most striking thing to be that the trajectories appear to never cross the dividing line between the slits. That would seem to mean that the left hand side of the interference pattern arises ONLY from photons passing through the left hand slit, and vice versa for the right hand side of the pattern. But then how can we reconcile this with the usual explanation of the double slit, which says that having "which-path" information about the photons should destroy the interference pattern?

Sorry if this question has been asked before (I guess I am not the first to be confused by this), but it didn't come up in a quick search through the threads, and it seemed relevant in light of the discussion of the latest double-slit experiment with weak-measurements.
 

Answers and Replies

  • #2
Demystifier
Science Advisor
Insights Author
10,603
3,343
the trajectories appear to never cross the dividing line between the slits. That would seem to mean that the left hand side of the interference pattern arises ONLY from photons passing through the left hand slit,
No, it doesn't mean that. You seem to be forgeting that, in Bohmian mechanics, particles are not the only entities that pass through the slits.
 
  • #3
SpectraCat
Science Advisor
1,395
1
No, it doesn't mean that. You seem to be forgeting that, in Bohmian mechanics, particles are not the only entities that pass through the slits.
No, I wasn't forgetting ... I just wasn't aware that the pilot wave had equivalent "reality" to the particles in BM. So it's ok for us to know that if a particle hits a particular spot on the screen, we can map it back to having come through a particular spot in one of the slits, because the pilot wave went through both slits? I thought there was a "random quantum potential" in BM which was responsible for the observed probabilistic results in experiments ... how does that factor in to this experiment? Does it just give the particles a "kick" as they pass through the slits, so you cannot predict the trajectories ahead of time?

Perhaps my question is clearer in the context of the quantum eraser experiment. In the standard interpretation, the particles are "polarization tagged" so that you can tell which one went through each slit, but then the "eraser" removes the "which path" information, restoring the interference pattern. However if you always know "which path" the particles traveled through in BM, how does the "polarization tagging" destroy the interference pattern, and how does the "eraser" restore it?

@Demystifyer .. I completely understand if you don't want to type up full answers to this .. lord knows you've probably done it 20 times already. If you can just give me a link to one of your previous explanations, or to a web resource, that would be fine. I have searched myself without success. Also, I hope it is clear that the tone of this thread is intended to be "I don't understand yet", rather than "I don't think BM is correct".
 
  • #4
Demystifier
Science Advisor
Insights Author
10,603
3,343
So it's ok for us to know that if a particle hits a particular spot on the screen, we can map it back to having come through a particular spot in one of the slits, because the pilot wave went through both slits?
Yes.

I thought there was a "random quantum potential" in BM which was responsible for the observed probabilistic results in experiments ...
No. There is nothing random in BM. Randomness is an illusion, resulting from our ignorance of the initial particle position.

Does it just give the particles a "kick" as they pass through the slits, so you cannot predict the trajectories ahead of time?
No. The slit has a finite width, so particles may have different positions when they pass through the slit. If you don't know what that position is, you can only tell what is a probability for a given position.

However if you always know "which path" the particles traveled through in BM ...
But you don't know that, because you don't know the initial position.

Also, I hope it is clear that the tone of this thread is intended to be "I don't understand yet", rather than "I don't think BM is correct".
Is it anything clearer now?
 
  • #5
SpectraCat
Science Advisor
1,395
1
Yes.


No. There is nothing random in BM. Randomness is an illusion, resulting from our ignorance of the initial particle position.
Fair enough .. I guess should have said "unknowable initial conditions" that give the illusion of randomness ... that is what I was trying to get across.

No. The slit has a finite width, so particles may have different positions when they pass through the slit. If you don't know what that position is, you can only tell what is a probability for a given position.

But you don't know that, because you don't know the initial position.
But I am not talking about *before* the experiment .. I am talking about after it. Above you seemed to agree that once you know a particle has hit the screen at a particular spot, you could follow exactly one trajectory back to the place on exactly one of the slits where that particle "originated" (treating the slit as the source). What I still don't understand is why that doesn't constitute "which path" information in the context of the double slit, where having such information makes observing the interference pattern impossible. Is the importance of such "which path" information somehow just an artifact of the CI?

It seems that in the Bohmian picture, we end up with a lot more information than in the CI picture ... we know which slit a particle passed through, and we know that the pilot wave passed through both slits. In the CI picture, we know that the wavefunction passed through both slits, but we cannot say anything about the particle-like properties of that event. What I don't understand is why it is not possible to use the extra information from BM to design an experiment to distinguish between the two interpretations.

Is it anything clearer now?
Not yet .. but perhaps starting to converge ... :wink:
 
  • #6
911
1
No, I wasn't forgetting ... I just wasn't aware that the pilot wave had equivalent "reality" to the particles in BM. So it's ok for us to know that if a particle hits a particular spot on the screen, we can map it back to having come through a particular spot in one of the slits, because the pilot wave went through both slits? I thought there was a "random quantum potential" in BM which was responsible for the observed probabilistic results in experiments ... how does that factor in to this experiment? Does it just give the particles a "kick" as they pass through the slits, so you cannot predict the trajectories ahead of time?

Perhaps my question is clearer in the context of the quantum eraser experiment. In the standard interpretation, the particles are "polarization tagged" so that you can tell which one went through each slit, but then the "eraser" removes the "which path" information, restoring the interference pattern. However if you always know "which path" the particles traveled through in BM, how does the "polarization tagging" destroy the interference pattern, and how does the "eraser" restore it?

@Demystifyer .. I completely understand if you don't want to type up full answers to this .. lord knows you've probably done it 20 times already. If you can just give me a link to one of your previous explanations, or to a web resource, that would be fine. I have searched myself without success. Also, I hope it is clear that the tone of this thread is intended to be "I don't understand yet", rather than "I don't think BM is correct".
Bohmian interpretation is a set of assumptions. One might agree with some and disagree with others and you could have an new interpretation that still works/explains.


For example:

1. particle and wave both happen (particle goes through one of the slits and wave through both)

2. the process is deterministic - there is no connect with 1, it's simply an assumption. one might as well assume that the process is random. The interpretation would still work.

3. there is a pilot wave. even if you assume it's not there, experiments (such as DCQE) are still explainable by sub-samples/coherence.

The wave equations would still (probabilistic) predict the position of the photon/electron, while allowing for the randomness.
 
  • #7
Demystifier
Science Advisor
Insights Author
10,603
3,343
Above you seemed to agree that once you know a particle has hit the screen at a particular spot, you could follow exactly one trajectory back to the place on exactly one of the slits where that particle "originated" (treating the slit as the source).
You could do that if you also knew the wave function of the system. But the interaction with the screen modified the wave function in a complicated way which in practice you cannot know exactly (this effect is better known under the name of decoherence), so you cannot really follow the trajectory back.

Is the importance of such "which path" information somehow just an artifact of the CI?
It is an artifact of a weird interpretation, but not really of the Copenhagen one.


What I don't understand is why it is not possible to use the extra information from BM to design an experiment to distinguish between the two interpretations.
Because any experiment involves decoherence, which for all practical purposes destroys information as I explained above.

More generally, decoherence seems to wash out measurable differences between all interpretations.
 
Last edited:
  • #8
Could this experiment be done using entangled particles? If so, would decoherence still occur?
 
  • #9
Ken G
Gold Member
4,438
332
What gets me about those trajectories is I think they are being completely oversold. It's not at all clear what an "average trajectory" really is, when you put together all those subtle "weak measurements", but it seems to me that the output they publish in their figure could be obtained much more easily, and it would be much clearer what they are actually seeing.

So let me ask this. Can anyone argue that I would not get the exact same figure the following way: I set up their exact apparatus, but I put my detecting wall at various different distances, repeating over and over until I map out not just a 1D detection-rate function, but a full 2D detection-rate field. The field is of course normalized to represent a divergence-free photon flux. Then I simply draw the divergence-free lines of flux of that 2D field. Why am I not getting their exact same figure, using no "weak measurement" at all? Why is their figure nothing but the 2D divergence-free lines of flux of a completely classical wave?

(ETA: I see this is being discussed on a much longer thread, so I present the question there too, it might be more appropriate to answer it there.)
 
Last edited:
  • #10
Demystifier
Science Advisor
Insights Author
10,603
3,343
  • #11
Demystifier
Science Advisor
Insights Author
10,603
3,343
So let me ask this. Can anyone argue that I would not get the exact same figure the following way: I set up their exact apparatus, but I put my detecting wall at various different distances, repeating over and over until I map out not just a 1D detection-rate function, but a full 2D detection-rate field. The field is of course normalized to represent a divergence-free photon flux. Then I simply draw the divergence-free lines of flux of that 2D field. Why am I not getting their exact same figure, using no "weak measurement" at all? Why is their figure nothing but the 2D divergence-free lines of flux of a completely classical wave?
If I understood you correctly, you measure only the particle positions, not the velocities or momenta. By repeating it many times, you obtain the probability density scalar rho(x) everywhere. What I don't see is how you can use rho(x) to determine the (divergence-free) photon flux?
 
  • #12
Ken G
Gold Member
4,438
332
The photons are essentially all the same wavelength, or could be for all the difference it would make. So there's no difficulty converting between rho(x) and an energy flux. We already do this all the time, every time we interpret a classical energy flux as a quantized photon flux. What I'm saying is equivalent to a completely classical experiment, where we simply measure the wave energy flux everywhere, and convert it into lines of flux, something we know we can do if the energy flux is divergence free (i.e., if the classical wave suffers little dissipation). The only reason I suggested it the way I did is because if one does a completely classical experiment, one is not sending the photons through one at a time, and someone might wonder if that makes a difference somehow. So the way I suggested it is (I claim) going to give the exact same figure as the completely classical way, but the process is carried out one quantum at a time.
 
  • #13
DevilsAvocado
Gold Member
751
91
Could this experiment be done using entangled particles? If so, would decoherence still occur?
Demystifier has already answered, but here is more info; it has been done in the http://en.wikipedia.org/wiki/Quantum_eraser_experiment" [Broken]:
550px-2SlitApparatus_w3_pol.svg.png

But nature is a "tricky b**ch" :smile:, you can’t do any real measurements without breaking the wave-particle duality... see:

http://www.sciencemag.org/content/307/5711/875.abstract" [Broken]
 
Last edited by a moderator:
  • #14
SpectraCat
Science Advisor
1,395
1
You could do that if you also knew the wave function of the system. But the interaction with the screen modified the wave function in a complicated way which in practice you cannot know exactly (this effect is better known under the name of decoherence), so you cannot really follow the trajectory back.
Ok, then what is the significance of the trajectories computed in figures like this:

Doppelspalt.jpg


I guess those are "perfect" trajectories .. i.e. predictions for what would happen in the absence of decoherence? Otherwise, I don't see why I wouldn't be able to connect a point on the detection screen with a point on a particular slit. Also, the fact that the trajectories don't cross the dividing line between the slits seems like a much more qualitative feature of the prediction ... does decoherence destroy that as well, so we can no longer be sure that particles contributing to the left-hand side of the interference pattern arise from the left-hand slit, and vice versa?

Has anyone simulated how those predictions are affected by the decoherence experienced in a "real" experiment?
 
  • #15
Ken G
Gold Member
4,438
332
I don't think decoherence is really the issue here, because no matter how much docoherence you have, you still never know the trajectory that a photon will follow, you only can ever reconstruct a trajectory it followed after the fact. So the concept of "trajectory" in QM is always a purely after-the-fact notion, which means it is never something you can actually test because testing requires prediction. Someone else could always ask "how do you know it followed that trajectory" and the answer would be "it had to given my assumptions, but they're just assumptions." That's inherently true, it doesn't matter about the amount of coherence, it's just the inherent limitation of the "trajectory" concept that makes it not work as well as it does classically. Classically, trajectories are predictable.
 
  • #16
I don't think decoherence is really the issue here, because no matter how much docoherence you have, you still never know the trajectory that a photon will follow, you only can ever reconstruct a trajectory it followed after the fact. So the concept of "trajectory" in QM is always a purely after-the-fact notion, which means it is never something you can actually test because testing requires prediction. Someone else could always ask "how do you know it followed that trajectory" and the answer would be "it had to given my assumptions, but they're just assumptions." That's inherently true, it doesn't matter about the amount of coherence, it's just the inherent limitation of the "trajectory" concept that makes it not work as well as it does classically. Classically, trajectories are predictable.
got an experiment for u
"consider the same interference experiment,the interference pattern produced is used to illuminate the next source in series,which then produces a new interference pattern,this gets cascaded say 10 times and the final interference pattern is obtained on the wall.
now ,the observer, observing the final interference pattern is not known of the fact that there's another observer measuring the particle position in the first stage of interference,and viceversa

"will ur answer b the same(no interference pattern) or something else?
and if u think that the experiment is not possible,state the reason along
 
  • #17
Demystifier
Science Advisor
Insights Author
10,603
3,343
The photons are essentially all the same wavelength, or could be for all the difference it would make. So there's no difficulty converting between rho(x) and an energy flux.
I still have no idea what you are talking about. Please write an EQUATION that gives the relation between rho(x) and flux.
 
  • #18
Demystifier
Science Advisor
Insights Author
10,603
3,343
Ok, then what is the significance of the trajectories computed in figures like this:

Doppelspalt.jpg


I guess those are "perfect" trajectories .. i.e. predictions for what would happen in the absence of decoherence?
Yes. (By the way, a weak measurement does not involve decoherence, which is exactly why weak measurements are so interesting.)

Also, the fact that the trajectories don't cross the dividing line between the slits seems like a much more qualitative feature of the prediction ... does decoherence destroy that as well, so we can no longer be sure that particles contributing to the left-hand side of the interference pattern arise from the left-hand slit, and vice versa?
Yes.

Has anyone simulated how those predictions are affected by the decoherence experienced in a "real" experiment?
Of course. The effect of decoherence can, for most practical purposes, be modeled by the old good collapse postulate. So if you are familiar with textbook QM, you don't really need to see the result.
 
  • #19
SpectraCat
Science Advisor
1,395
1
Yes. (By the way, a weak measurement does not involve decoherence, which is exactly why weak measurements are so interesting.)


Yes.


Of course. The effect of decoherence can, for most practical purposes, be modeled by the old good collapse postulate. So if you are familiar with textbook QM, you don't really need to see the result.
Well .. I think I *do* need to see the result in this case, although I am quite familiar with textbook QM, because I still don't understand the answer to this question:

"What (if any) is the physical significance of trajectories like the ones in the image I posted?"

If such trajectories are only meaningful in the absence of decoherence, and *complete* experiments necessarily involve decoherence (I understand about weak measurements, but you need a strong measurement to finish the experiment), then who cares about those predictions? Furthermore, doesn't that mean that the trajectories reconstructed by Steinberg et al. can't really be related to the Bohmian trajectories, since they were generated from the combination of weak and strong measurements? In that paper, they really do reconstruct the trajectories by correlating average momenta with individual points on the pattern, and tracing them back to the slits. The interference pattern in this experiment is just as "fully decohered" as any other pattern, since it is a strong measurement, so I guess I must still be missing some significant point somewhere.
 
  • #20
Demystifier
Science Advisor
Insights Author
10,603
3,343
Furthermore, doesn't that mean that the trajectories reconstructed by Steinberg et al. can't really be related to the Bohmian trajectories, since they were generated from the combination of weak and strong measurements?
I think it is the crucial question. My answer consists of 2 points:
1) Mathematically, weak trajectories and Bohmian trajectories are identical.
2) Ontologically, weak trajectories and Bohmian trajectories are totally unrelated.
 
  • #21
Ken G
Gold Member
4,438
332
I still have no idea what you are talking about. Please write an EQUATION that gives the relation between rho(x) and flux.
Sorry, I thought it would be clear that flow vectors could be found from knowing the flux field. Let's say you had a compressible gas flowing through a 2D pipe. We then have the concept of a "flow vector," whose direction is along the streamline and whose magnitude is proportional to the mass flux along that direction. By conservation of mass, the flow vectors are divergence-free. Now imagine drawing parallel slices across the pipe, and plot along those slices a 1D "flux function", which gives the local mass flux density across the slice at each point. The question you are asking is tantamount to solving for the flow vectors given the flux functions, because once I have the flow vectors, I can draw "flow trajectories" along them. Note that these flow trajectories, also called streamlines, could equally be called "average trajectories" of the flow particles, once we recognize the flow is made of quanta.

The problem of finding the divergence-free flux vectors from the flux functions is called the "Neumann boundary problem", see for example http://en.wikipedia.org/wiki/Neumann_boundary_condition
and note that if we define y by saying that the flow vector field is the gradient of y (possible for a divergence-free flow), we have exactly the Laplace problem used as an example in that Wiki.

I claim that we have no evidence that the "average trajectories" are anything different from what I would certainly expect them to be: the streamlines of the flux vectors determined from the flux functions found by moving the detecting wall around and doing the experiment over and over.

What this also means is, we can get the same results entirely classically. We simply turn up the intensity of the light so high that we can do local flux measurements, and map out the flux functions that way.

Or, we can even look at the polarizations, classically, and form the same ratios that tell us an "average direction" for the local "average trajectories."

Or, we can solve the Bohm pilot-wave equations, which have to give the same results if they are to mean anything.

All of these roads lead to the same place: the concept of an "average trajectory" is a statistically aggregated concept, and hence is purely classical. If this is not true, I would find it quite surprising, and certainly no one has given any argument that it is not true.

Thus, I conclude what this experiment is actually saying is, "we can recover the boring classical result using a much more complicated weak measurement approach", which may tell us something about weak measurement but nothing surprising about the behavior of this system. What the experiment is not telling us is anything about the "trajectory" concept as it applies to single quanta, other than that when you aggregate the weak measurements, you do in fact get the classical answer.
 
  • #22
Demystifier
Science Advisor
Insights Author
10,603
3,343
Sorry, I thought it would be clear that flow vectors could be found from knowing the flux field.
But how can you know the flux field? (An equation might help.)

We then have the concept of a "flow vector," whose direction is along the streamline
But what is the "direction of streamline" for given rho(x)?
 
  • #23
Ken G
Gold Member
4,438
332
But how can you know the flux field? (An equation might help.)
It's the LaPlace equation, applied to a scalar function whose gradient is the divergence-free flow vectors. The entire problem consists in solving for that scalar function, given flow functions across the various locations of the detector walls. This is basic LaPlace, done to death in many fields of physics, it is merely a different boundary condition from what is normally seen. Let's say you were a sound engineer and you had measurements of how loud a particular speaker sounded at various places in an auditorium. Do you think you'd have any trouble calculating the sound energy flux pattern in that auditorium? Would it be a big stretch to call your results the "average trajectories" of the phonons?
 
  • #24
DevilsAvocado
Gold Member
751
91
... does decoherence destroy that as well, so we can no longer be sure that particles contributing to the left-hand side of the interference pattern arise from the left-hand slit, and vice versa?
Yes.
¿Qué?

Are there different dBB 'schools'...

According to http://www.bohmian-mechanics.net/", it’s the other way around??
http://www.mathematik.uni-muenchen.de/~bohmmech/BohmHome/gallery/double/slide4.html

The double slit experiment, Picture 4
[URL]http://www.mathematik.uni-muenchen.de/~bohmmech/BohmHome/gallery/double/double4.gif[/URL]
These are the paths, one of which the actual particle follows, according to Bohmian mechanics. Each path passes through one of the slits (left hand side); and hits the screen (right hand side) at a random position that appears |psi|2 distributed, displaying the famous diffraction fringes. Although one cannot know in advance through which slit the particle will pass, one can decide afterwards: it passes through the upper slit if and only if it hits the screen on the upper half, that is, *above the symmetry axis*.
 
Last edited by a moderator:
  • #25
Demystifier
Science Advisor
Insights Author
10,603
3,343
It's the LaPlace equation, applied to a scalar function whose gradient is the divergence-free flow vectors. The entire problem consists in solving for that scalar function, given flow functions across the various locations of the detector walls. This is basic LaPlace, done to death in many fields of physics, it is merely a different boundary condition from what is normally seen. Let's say you were a sound engineer and you had measurements of how loud a particular speaker sounded at various places in an auditorium. Do you think you'd have any trouble calculating the sound energy flux pattern in that auditorium? Would it be a big stretch to call your results the "average trajectories" of the phonons?
OK, now I understand your idea. However, there are at least two problems with it:

1) The solution is not unique. Namely, if you have found a solution corresponding to the velocity field v(x), then any other velocity field v'(x) of the form
v'(x)=v(x)+u(x)/rho(x)
where u(x) is an arbitrary divergence-free vector field, is also a possible solution. Indeed, such alternative solutions correspond to alternative trajectory interpretations of QM. Weak measurement, by contrast, leads to a single velocity field equal to the Bohmian one.

2) There are complications when two or more particles are involved. For simplicity, consider 2 particles. A naive extension of your idea would require two functions rho1(x1) and rho(x2). However, instead of this you must deal with a single function in a double-dimensional space rho(x1,x2). The resulting waves can hardly be thought of as "classical".
 
Last edited:

Related Threads for: Do Bohmian mechanics trajectories for double slit provide which path information?

  • Last Post
Replies
1
Views
632
Replies
5
Views
3K
Replies
1
Views
2K
Replies
8
Views
1K
Replies
10
Views
2K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
1
Views
3K
Top