Particle vs Wave Interpretations of QM

[Roberto Pavani]

I apologize if my original description was unclear.

The detector screen stays flat and at the same distance from the slits. I simply replace a small area around an interference maximum with a thick piece of track-recording plastic whose front face is part of the detector plane. What would the ensemble of recorded tracks inside the plastic look like after many events?

 

[PeterDonis]

The detector screen stays flat and at the same distance from the slits. I simply replace a small area around an interference maximum with a thick piece of track-recording plastic whose front face is part of the detector plane

This doesn't help. What is "track-recording plastic"? What is it supposed to do that the detector screen itself doesn't already do? And how does it do it?

 

[Roberto Pavani]

I think I may still be explaining this poorly.

I am not proposing a new detector. I am simply using renormalize's earlier statement that a plastic film can record the trajectory of a charged particle passing through it.

My question is purely conceptual. Suppose that, instead of the usual detector material, a small region around an interference maximum is replaced with a sufficiently thick piece of the "track-recording plastic" mentioned by renormalize. The screen remains flat and at exactly the same position as before.

After many detected protons, what would the resulting 3D pattern of recorded tracks look like inside that piece of plastic?

 

[PeterDonis]

I am simply using renormalize's earlier statement that a plastic film can record the trajectory of a charged particle passing through it.

Do you mean what @renormalize said in post #44? I'll assume you do in what follows.

a small region around an interference maximum is replaced with a sufficiently thick piece of the "track-recording plastic" mentioned by renormalize.

So you're thinking that the plastic will record, basically, which direction the protons going through it are coming from? (Just take the track recorded in the plastic and look at the direction it points back to.)

Then, as I said, the plastic is recording which-path information, so it will remove the interference (and the tracks it records will not show any interference). However, since the plastic only covers a portion of the detector screen, it will not affect the interference pattern in the rest of the detector screen, where the plastic is not present. So you'll end up with mixed results: an interference pattern on the part of the detector screen that the plastic doesn't cover, and tracks in the plastic that show no interference (just a mixture of tracks coming from one slit and tracks coming from the other).

 

[Roberto Pavani]

Why should the tracks extrapolate back to one slit or the other?

At the central maximum (the one aligned with the source), I would naively expect the incoming probability current to be associated with the coherent superposition of both slits, rather than with two distinguishable populations corresponding to the individual slits.

In fact, my (probably wrong) intuition would be that the tracks might look as if the protons had arrived more or less straight from the source, with the two slits acting transparently as part of a single coherent process rather than as two distinguishable origins.

But I am not an expert, which is precisely why I was asking the question rather than suggesting an answer of my own, which could easily be wrong or misleading.

 

[gentzen]

Then, as I said, the plastic is recording which-path information, so it will remove the interference (and the tracks it records will not show any interference).

No, retrodiction is neither bound by uncertainty relations (think about the EPR paper), nor does it necessarily destroy interference.
A track of the proton in the plastic can reveal information about its momentum. But because the of the uncertainty relation between position and momentum accuracy, this information won't be able to reveal enough which-path information to destroy the interference.

 

[PeterDonis]

Why should the tracks extrapolate back to one slit or the other?

Because measuring a track means measuring a track--not just a point of impact, but a trajectory. So, heuristically, by measuring the track you are forcing the particle to commit to a particular trajectory, which must come from one or the other slit, since those are the only places a trajectory of the particle can come from, and the measurement forces it to come from one or the other.

my (probably wrong) intuition would be that the tracks might look as if the protons had arrived more or less straight from the source

No, this is impossible since it would require the protons to pass through the non-slit part of the screen in between. If it were not impossible for the protons to do that, you would never get an interference pattern at all, because the presence of the interference pattern depends on it being impossible for the particles to reach the detector screen by any other path than the paths that go through the slits.

This is a good illustration of why you need to use math to do physics, not intuition.

 

[PeterDonis]

retrodiction

Is not what is going on in the scenario @Roberto Pavani is describing. Measuring the track is basically equivalent to measuring the momentum of the particle at the detector screen, which means measuring the direction the particle is coming from. And we already know from other experiments (for example, experiments that place telescopes pointed at each slit, when doing the double slit experiment with light) that making such a direction measurement at the detector provides which path information and destroys interference.

because the of the uncertainty relation between position and momentum accuracy

You're contradicting yourself, since you said retrodiction is not subject to uncertainty relations.

In the viewpoint I'm taking, where, as above, retrodiction is not involved, position-momentum uncertainty doesn't play a role, because we're not trying to measure the position of anything at the slits. All we're doing is assuming that the momentum measurement using the tracks is accurate enough to be able to distinguish which slit the track came from. That doesn't require a position measurement of anything at the slits. It just requires fine enough angular resolution of the tracks.

 

[Roberto Pavani]

I would like to clarify something.

In my original question, although the setup may not have been described clearly, I was not proposing any particular shape for the tracks in the plastic. I was simply asking what one would expect to see at the end of the experiment by examining the plastic under a microscope.

Later I mentioned an intuition of my own, while explicitly noting that it could be wrong. If that introduced confusion, I apologize.

What I am still struggling with is this: if recording tracks in the plastic provides enough which-path information to destroy the constructive interference that would otherwise bring the protons into that region, then I do not understand what tracks would actually be recorded there.

Conversely, if no protons reach the plastic, then no tracks are formed, and I again do not understand what the expected observation would be.

My intention was not to suggest an answer, but to understand what the experiment would show. That is why I asked the question in the first place.

So I would prefer to withdraw my tentative intuition, since it may have been more misleading than helpful, and return to the original question: what would the plastic actually record?

 

[PeterDonis]

if recording tracks in the plastic provides enough which-path information to destroy the constructive interference that would otherwise bring the protons into that region, then I do not understand what tracks would actually be recorded there.

Tracks that come from one slit or the other.

Perhaps it might help to think of things in terms of the path integral formulation. In the normal double slit experiment, the probability amplitude for the particle hitting any particular point on the detector screen is the sum of the amplitudes for paths coming through each slit (and no others, since it's impossible for the particle to get to the detector screen any other way). But this assumes that we measure the point of impact at the detector screen, and nothing else.

In your alternate formulation, in the area where the thick plastic is present that records tracks, we are no longer measuring the point of impact at the detector screen. We are instead measuring the track the particle takes through the plastic, which will be some straight line--because there are no magnets or anything else present that would cause the particle to take a curved path, and because, just as in a bubble chamber, measuring the track forces the particle to choose just one of the many possible paths it could have taken (i.e., only one of the many possible paths that would otherwise appear in the path integral). Since the only way for the particle to get to the plastic is to go through the slits, the straight line track will point towards one of the slits. It can't point to both because a straight line can only point in one direction.

In path integral terms, the amplitude for any particular track to appear in the plastic is not the sum of amplitudes for paths coming from both slits; the sum now only includes paths coming from one slit (the slit the track points to). That is what removes the interference.

 

[PeterDonis]

the constructive interference that would otherwise bring the protons into that region

You seem to be assuming that it is impossible for particles to reach that region without inteference being present. That's not correct. The probability of a particle reaching that region without interference is just the sum of the probabilities of it coming from each of the slits, and those probabilities are low, but not zero.

 

[gentzen]

That doesn't require a position measurement of anything at the slits.

We all agree that the plastic is not at the slits, but at the screen. So in this scenario, any "supposed" measurement of the position at the slits is a retrodiction.

It just requires fine enough angular resolution of the tracks.

Good, if you really measure the momentum accurate enough such that the angular resolution allows you to determine which slit the proton must have gone through, then the position accuracy of the track was so coarse that you indeed lost the interference.

But not because of some mysterious retrodictive effect, but because your position resolution was too coarse to resolve the interference. (And because a proton has a very small wavelength, the interference pattern will often indeed be finer than what the plastic can resolve.)

 

[Roberto Pavani]

Thanks for the answer, although it is still not clear to me what pattern would be observed in the plastic.

Please note that the analysis of the tracks takes place only after the experiment has ended and no more protons are being sent.

I am asking about the pattern found in the plastic after the fact. During the experiment itself, no information is being extracted from the recorded tracks.

I think this changes the scenario somewhat, because the plastic is not being used as an active which-path detector during data collection. Any inference about trajectories or slit origins would only be attempted afterwards by examining the accumulated tracks.

 

[Matterwave]

Sorry to interject, but it might help to draw out the exact experiment in mind. What is placed where, when do you do what, and what kind of detector each thing is. This helps to get everyone on the same page. It's why experimental papers tend to have lots of diagrams! :)

 

[jeffn1]

Because the branches are decohered, and because in the overall wave function you are entangled with the cat--and both you and the cat are entangled with everything else in the environment--in each decohered branch, you experience whatever is consistent with the state of the cat and everything else within that branch.

I am trying to understand where, from just a plausibility perspective, I would deviate from MWI. So, I think this is it, but I want to make sure.

So, Peter, even under a less extreme view of MWI, whenever a person looks at a detector showing the location of decohered photon (what other theories might refer to as the "collapse"), there is another "me" (out of nearly infinite "me's"?) viewing the photon in one of the other locations included within the (Schrodinger) wave function?

Were these new "me's" created when the photon hit the wall?

Does it have to be a conscious person, or would this apply whenever (for example) a photon hits a wall (and decoheres)? Does it matter if I glanced at the wall when the light hit, but did not pay much attention to it? What if I was out of the room where the light hit the wall?

[I am sorry if I am so off base and these are not coherent (no pun intended!) questions].

 

[PeterDonis]

any "supposed" measurement of the position at the slits is a retrodiction.

There is no "supposed" measurement of position at the slits. There is only a measurement of momentum in the plastic. The momentum measurement, with sufficient angular resolution, tells you which slit the particle came through, but that's not a position measurement at the slits, not even a "supposed" one. It's just a restriction on the possible paths that can go into the path integral.

if you really measure the momentum accurate enough such that the angular resolution allows you to determine which slit the proton must have gone through, then the position accuracy of the track was so coarse that you indeed lost the interference.

I don't see how this follows at all. The presence of interference at the detector screen does not depend on the accuracy of any position measurement. It depends on having contributions to the probability amplitude coming from both slits. The track measurement, assuming sufficient angular resolution, eliminates that regardless of how accurately the track position is captured. (And note that "sufficient" angular resolution here is quite coarse, since the slits have a finite separation that is large compared to the wavelength of the momentum states involved.)

I would suggest that you write down the actual math that you think supports your claim here.

 

[PeterDonis]

the plastic is not being used as an active which-path detector during data collection.

Doesn't matter. Recording the tracks is enough in itself regardless of whether or how that information contributes to anything else during the running of the experiment.

 

[PeterDonis]

Were these new me's created when the photon hit the wall?

I'm going to emphasize this because I've said it already but it does not appear to have sunk in:

Unitary evolution cannot create or destroy anything.

There are no "new me's". There is one wave function. It has branches because of entanglement interactions followed by decoherence (which is the general answer to all your questions about when branching happens), but it's still just one wave function.

 

[renormalize]

Sorry to interject, but it might help to draw out the exact experiment in mind. What is placed where, when do you do what, and what kind of detector each thing is.

Here is a diagram from https://physics.stackexchange.com/q...een-done-using-a-track-chamber-or-even-contem that illustrates the experimental configuration being discussed, as I understand it:

The momentum measurement, with sufficient angular resolution, tells you which slit the particle came through, but that's not a position measurement at the slits, not even a "supposed" one.

Does this necessarily follow though? Couldn't all the tracks point back directly to the source rather than to one or the other slit, with the density of those tracks exhibiting the familiar 2-slit interference pattern? I'd have to see an actual calculation for this configuration (or better yet an experiment!) to decide.
(I do believe, however, that if the detection chamber is extended all the way back to the slit-plane, then the tracks will all originate at specific slits and that the interference pattern will disappear.)

 

[PeterDonis]

Couldn't all the tracks point back directly to the source

How? In path integral terms, what contribution to the amplitude would such paths make? My answer is, zero, since the experiment is set up to make it impossible for particles to reach the detector screen (or the plastic recording tracks) except by going through one slit or the other. So the only paths with a nonzero contribution to the path integral will be paths that go through one of the slits. And those paths can only produce tracks that point at one slit or the other.

 

[jeffn1]

I am sorry I interfered with the more detailed discussion you are having. At some point (after your technical discussion is concluded) I would love a better understanding of this point (the more conceptual and less mathematical, the better)

The you that is in the "cat is alive" branch sees the cat as alive, and the you that is in the "cat is dead" branch sees the cat as dead.

and this:

Bear in mind that, if you find it hard to believe that the MWI actually says this because it seems so outlandish, you're not alone. I find it outlandish too. So do many others. But it is what the MWI actually says, outlandish and all. People who believe the MWI is true do actually believe what I described. If you really push about how outlandish it seems to you, many of them will probably waffle. But if you actually do believe the MWI, those outlandish things are what you're committed to, whether you like it or not.

[UPDATE: I think I have a common sense understanding now. So basically every time there is a branching things split off from there. So, there is no "big bang" when this occurs. Rather the gazillions of times every second (in every room )there is decoherence the branch splits off and becomes separated and inaccessible to the branch it split off from. So, there would be a new you (and everything and everyone else) with all your existing experiences and histories up to this point. From there things would likely deviate in the split off branches ......I'll pass.]

 

[PeterDonis]

At some point (after your technical discussion is concluded) I would love a better understanding of this point (the more conceptual and less mathematical, the better)

Please start a separate thread if you want to discuss a separate question.

 

[renormalize]

So the only paths with a nonzero contribution to the path integral will be paths that go through one of the slits. And those paths can only produce tracks that point at one slit or the other.

Without a detailed calculation that's just personal speculation. Yes, the classical paths entering the quantum path integral all point to the slits but that doesn't mean that the Mott tracks calculated from evaluating that integral must do so. In fact they don't, as is easily seen from the analogous electromagnetic example of Young's double slit experiment. Here is the result for the Poynting vectors from a 2D FEM simulation of that experiment (red lines added by me):

Clearly, the directions of the of Poynting trajectories at the "screen" (the top of the image) all point to the center of slit-plane and not to the individual slits (or to the source, as I mistakenly suggested). And the standard interference pattern is evident as well. So I see no reason to think that replacing the 2D screen by a 3D trajectory-tracking chamber in any quantum double-slit experiment should negate the expected interference pattern. Do you have references that say otherwise?

 

[PeterDonis]

In fact they don't, as is easily seen from the analogous electromagnetic example of Young's double slit experiment

Is this a peer-reviewed paper? It doesn't look like one. Not to mention that the red lines are added by you.

the Poynting vectors

Is there a peer-reviewed paper that explains how the classical Poynting vector is even relevant to calculating the quantum amplitude for a track in a particular direction appearing within the thick plastic in the scenario under discussion?

Do you have references that say otherwise?

So far I haven't seen any peer-reviewed references that support the claim you're making.

 

[PeterDonis]

the Mott tracks calculated from evaluating that integral

I don't think your Mott track analogy works here, at least not the way you are trying to use it.

In the case of the standard double slit, where we have a detector screen that shows individual dots for each particle impact, the path integral is to calculate the amplitude for a dot appearing at a particular point on the screen. The paths contributing to this amplitude come from both slits since there are possible paths from the source to that point on the screen that go through both slits. Hence, there will be interference.

In the case of the plastic "track recorder" that @Roberto Pavani wants to use in his version of the experiment, the path integral is to calculate the amplitude for a track going through the plastic along a particular line, to within a particular angular resolution. The paths that contribute to this amplitude will be paths that come from whatever is within that particular angular resolution of the line in question. Which will be, either one of the two slits, or nothing (since there is no way for a particle to get to the plastic other than by going through one of the slits). So the only tracks that will have a nonzero amplitude at all are those for which one of the two slits is within the angular resolution, and for any given track, only one of the two slits will be within that angular resolution. So the amplitude for every track that has a nonzero amplitude at all will only have a contribution from one slit. Hence, no interference.

That's a basic description of the math involved as I understand it.

 

[gentzen]

(And note that "sufficient" angular resolution here is quite coarse, since the slits have a finite separation that is large compared to the wavelength of the momentum states involved.)

My "actual math" rather tells me that "the slits have a finite separation that is small compared to the wavelength of the momentum states involved".

I would suggest that you write down the actual math that you think supports your claim here.

Good. Let me parameterize the relevant geometric situation as follows
λ: wavelength of proton
D: distance between the slits plane and the plastic detector screen
w: finite separation between the slits
d: distance between the "intensity" maxima of the interference pattern on the plastic detector screen

Let us suppose that D ≫ w, and that we are in a paraxial situation (this second assumption simplifies the math, but is otherwise not necessary). Then we have d ~ λ and d ~ D. We also have d ~ 1/w (here we seem to disagree), so that d ~ D λ / w. More explicitly
d = α D λ / w with α not far away from 1. (For a grating instead of a double slit, α would be 1.)

For the math of the momentum uncertainty, it is sufficient to assume one measurement near the surface of the detector screen with finite uncertainly Δp in momentum and Δx in position. (The momentum (direction) after that first measurement can in theory be determined as accurately as desired, by a sufficiently far away second measurement with high position accuracy.)

So the angle uncertainty is Δp / (h/λ), which gives a position uncertainty D Δp / (h/λ) near the slits. For being able to know which slit the proton went through, we should have
w/2 > D Δp / (h/λ)
which is equivalent to
Δp < w/2 h/(Dλ)
For being able to resolve the interference, we should have
Δx < d/2 = α/2 D λ / w

So we get
Δx Δp < α/2 D λ / w * w/2 h/(Dλ) = α/4 h

We see that the uncertainty relation doesn't allow us to resolve the interference and know which slit the proton went through at the same time.

 

[PeterDonis]

Let us suppose that D ≫ p

What's p?

 

[gentzen]

What's p?

Should have been D ≫ w.
I fixed it.

 

[PeterDonis]

My "actual math" rather tells me that "the slits have a finite separation that is small compared to the wavelength of the momentum states involved"

Mine doesn't. Let's suppose we apply a voltage of 1000 V to accelerate the protons; that gives them a speed of about 400,000 meters per second. The de Broglie wavelength associated with a proton with that speed is $h / m v$, or about 1 picometer. Even if we take an extremely small slit separation of 1 micrometer (much too small to even be seen with the naked eye), that's still a million times larger than the proton wavelength.

 

[gentzen]

Mine doesn't. Let's suppose we apply a voltage of 1000 V to accelerate the protons; that gives them a speed of about 400,000 meters per second. The de Broglie wavelength associated with a proton with that speed is $h / m v$, or about 1 picometer. Even if we take an extremely small slit separation of 1 micrometer (much too small to even be seen with the naked eye), that's still a million times larger than the proton wavelength.

But then the interference pattern becomes much too fine to be resolved:

(And because a proton has a very small wavelength, the interference pattern will often indeed be finer than what the plastic can resolve.)