Proposed Modification to Afshar's Experiment

[peter0302]

I was considering Afshar's experiment and thought of a possible modification, and I was wondering whether it had been proposed, tried, or thought of. If not I was considering suggesting it to Afshar or posting a paper on it.

The idea is as follows:

Each member of an entangled photon pair is sent either to a lens and detector (signal photon) or a quantum eraser (idler photon). When the idler photon is observed with “which-slit” information intact, the rate of detection of the signal photon is observed in the presence of an Afshar-style interference grid, placed before the lens in the path of the signal photon. It is assumed the grid will have no effect on the detection rate, even when the idler photons are detected with “which-slit” information intact; the only way this can be possible, however, is if the signal photons exhibit interference ([EDIT]wave)-like effects, thereby avoiding the grid via the interference minima, until the instant of detection, at which point their distribution becomes Gaussian. Similarly, for idler photons for which “which-slit” information has been erased, the signal photons are expected to exhibit an interference pattern at all times.

The configuration is essentially that of the Delayed Choice Quantum Eraser, except with the Afshar-style grid placed in front of the lens in the path of the signal photon.

The experiment can result in two possible outcomes.

A. Detections at D0 corresponding to detections at D3 or D4 (which-slit information intact) are significantly fewer in number than detections at D0 corresponding to detections at either D1 or D2 (which-slit information erased). This would only be true if the signal photons corresponding to the detections at D3 or D4 actually traveled a well-defined path between their source and D0, and this path included the possibility of being blocked by the Afshar grid, i.e., they exhibited no self-interference effects.

Since the Copenhagen interpretation states the photons do not have a well-defined path prior to detection, this result would be inconsistent with that aspect of the interpretation, but would preserve the notion of complimentarity, at least insofar as a single entangled photon pair never exhibits both wave and particle-like features simultaneously.

B. Detections at D0 corresponding to detections at D3 or D4 are just as frequent as those corresponding to detections at either D1 or D2. This is the result expected in the absence of the Afshar grid. This would be true only if the signal photons corresponding to detections at either D3 or D4, with which-slit information intact, nonetheless do not have a well-defined path before detection by D0 and, instead, exhibit self-interference until the moment of detection, allowing them to avoid the Afshar grid, yet still ultimately assemble in a Gaussian pattern.

While this preserves the idea that the photons possessed no well-defined path prior to detection, it also would appear to defeat the uncertainty principle, as an unmistakable interference pattern would be (at least) inferable while, simultaneously, which-slit information is obtainable through the joint detection at D0 and D3 or D4. Furthermore, it would be exceedingly difficult to explain this result in any local or realistic model of quantum theory, even in the absence of Bell’s inequality, as the distribution pattern of photons is truly not determined for either member of the entangled pair until both have been detected.

Nevertheless, the “no-signaling” condition would suggest that scenario B must be the correct outcome; the presence of the Afshar grid should have no effect on the results. If it did, it would be possible to communicate with a person observing D0 simply by forcing, or not, the erasure of which-slit information. Forced erasure would result in more detections at D0 than when which-slit information is not erased; no coincidence circuit would be required to assemble the results, as the “message” can be reconstructed merely by observing the statistical frequency of detections at D0.

 

[confusedashell]

I dno much about this but maybe this question has been asked to Afshar himself: http://irims.org/blog/index.php

 

[colorSpace]

I'm not sure whether you are aware that entangled photons display interference only after a process of mathematical analysis. This requires detecting the photons such that the photons of each pair can be matched with each other. A.Zeilinger has described this in an experiment called the "double-double-slit experiment", which studies the interference behavior of entangled photons. (But unrelated to Afshar's experiment and its use of a grid.)

Regarding Afshar's experiment, I would have the question whether the grid doesn't have the same effect as the slits, which is to "uncertainly" bend the path of the photons, meaning they won't fly straight through the grid, and therefore the detectors don't really tell which slit they originally went through.

 

[peter0302]

I'm not sure whether you are aware that entangled photons display interference only after a process of mathematical analysis. This requires detecting the photons such that the photons of each pair can be matched with each other. A.Zeilinger has described this in an experiment called the "double-double-slit experiment", which studies the interference behavior of entangled photons. (But unrelated to Afshar's experiment and its use of a grid.)

I am aware of that, and that is why my experiment still utilizes the coincidence circuit. If you perform the experiment, you obviously won't see any interference pattern at D0 without correlating with the idler photons. But when the grid is present, you should still see the exact same distribution as when the grid is not there. Meaning the photons ignore the grid completely, and signal photons corresponding to "which-path intact" idlers still display the gaussian pattern while photons corresponding to "which-path erased" idlers do not.

Regarding Afshar's experiment, I would have the question whether the grid doesn't have the same effect as the slits, which is to "uncertainly" bend the path of the photons, meaning they won't fly straight through the grid, and therefore the detectors don't really tell which slit they originally went through.

Agreed; that's what my idea tries to address. Now there's no ambiguity as to which slit the photons went through. When which-slit is available for the idlers, we know exactly which slit the signal photon went through and, yet, simultaneously should be able to deduce the existence of an interference pattern (just for those photons; which is why we need the coincidence circuit) in the fact that no photons are ever blocked by the grid.

 

[colorSpace]

I am aware of that, and that is why my experiment still utilizes the coincidence circuit. If you perform the experiment, you obviously won't see any interference pattern at D0 without correlating with the idler photons. But when the grid is present, you should still see the exact same distribution as when the grid is not there. Meaning the photons ignore the grid completely, and signal photons corresponding to "which-path intact" idlers still display the gaussian pattern while photons corresponding to "which-path erased" idlers do not.

No, when you use an entangled pair, the photons won't fly "around" the grid anymore. The directly visible interference pattern disappears as soon as the photons are entangled in such a way that you can conclude from one photon precise enough what the path of the other is. In order to do that, the light source needs a minimum size, etc, and you loose that behavior as a directly visible behavior, and it can be discovered only later through analysis. This means that then as many photons will hit the grid as corresponds to the thickness of the grid.

Agreed; that's what my idea tries to address. Now there's no ambiguity as to which slit the photons went through. When which-slit is available for the idlers, we know exactly which slit the signal photon went through and, yet, simultaneously should be able to deduce the existence of an interference pattern (just for those photons; which is why we need the coincidence circuit) in the fact that no photons are ever blocked by the grid.

In order to obtain which-slit information, even when using entangled photons, you should have to modify the experimental setup in such a way that the photons don't fly around the grid anymore.

Additional, independently of whether you use entangled photons, the grid should at least to some percentage bend the paths of the photons, and I would expect that this effect is strong enough so that the detectors won't really tell anymore which path they have taken before the grid.

 

[colorSpace]

Another explanation of the original experiment would be that it is likely that along with an interference pattern of positions, there will be an interference pattern of impulses, such that the directions of the photons after the grid will have a regularity independent of the path the photon has taken up to that point.

That is, it could be that the behavior of the photons behind the grid will be defined by the interference pattern of impulses, not by their previous path.

[Edit] And besides, I wonder why pinholes would create an interference pattern of straight lines.

On reading the PDF, there seem to be a lot of rather general assumptions that might not apply to such a specific case. As far as I can tell as a non-physicist, that is.

 

[peter0302]

The directly visible interference pattern disappears as soon as the photons are entangled in such a way that you can conclude from one photon precise enough what the path of the other is

The directly visible interference pattern is irrelevant. When we're dealing exclusively with the subset of photons for which which-slit info was erased, we know from Kim that we'll see an interference pattern on the chart. Can we not therefore infer that, for this subset, the signal photons will have tended to avoid the space corresponding to the interference minima? If not why? In any event, this can eaisly be tested by inserting the grid and comparing the photon count with one slit closed versus both slits open, like Afshar did. If far more than half the photons are blocked when one slit is closed, will that convince you?

No, when you use an entangled pair, the photons won't fly "around" the grid anymore.

Are you saying that the signal photon will behave differentlly at the wire grid than it otherwise would simply by virtue of the _existence_ of an entangled twin, irrespective of whether or how it's detected?

And if you're right, then this is result is still paradoxical but for the opposite reason. In that case, the signal photons _always_ travel in a Gaussian (particle-like) pattern, at which point the corresponding idler photons "choose" either to erase "which-slit" depending on whether the signal photon struck D0 at an interference minimum? That makes no sense either - how do the idler photons know which path of the beamsplitter to take in order to erase or not erase?

Additional, independently of whether you use entangled photons, the grid should at least to some percentage bend the paths of the photons, and I would expect that this effect is strong enough so that the detectors won't really tell anymore which path they have taken before the grid.

That's why you _need_ to use entangled photons - the interferometer, independent of the Afshar grid, tells you which path the photons took. The rate of detection at D0, which is affected by the grid, tells you whether the grid blocked any of the photons. The grid has no effect on the ability to tell which-slit. The grid's presence is solely to tell us whether the signal photon exhibits intference even when which-slit is knowable.

 

[colorSpace]

Are you saying that the signal photon will behave differentlly at the wire grid than it otherwise would simply by virtue of the _existence_ of an entangled twin, irrespective of whether or how it's detected?

My understanding from descriptions of the "double-double-slit" experiment is this:

A light source that produces entangled pairs of photons where the impulses of both photons are precise enough opposite of each other, so that one can conclude which-way-information of one by the other; such a light source will have for example a minimum size and not produce a visible interference pattern in any case.

[Edit added:] A (potentially) visible interference pattern is the requirement for the photons fly around the grid. Without a (potentially) visible interference pattern, many photons will hit the grid.

 

[colorSpace]

I think we have a misunderstanding about "visible interference pattern". When I say for entangled photons there is no visible 'visible interference pattern', I mean that they won't fly around the grid. That already would be a 'visible interference pattern' for me.

 

[peter0302]

Well, Kim's experiment managed to show interference minima 1 mm apart on the detector, and that was with a lens that created a far field condition by narrowing the beam - the minima at the lens LS would be even further apart. So I still fail to see how you know these photons would not "fly around" (I'd prefer to say "tend to avoid") the grid?

And, again, I recognize that the grid will ALWAYS block some photons. Kim's minima are not null points like (I believe?) Afshar's are. But I submit that, at least when which-slit is unavailable, the grid will block even more photons when one slit is closed, meaning the signal photons tend to avoid the grid when the slits are open, even though they don't always avoid it completely. This in itself would demonstrate a wave-like path before detection when which-slit info is unavailable.

But then I'm going even further and suggesting they'll tend to avoid the grid the same way even when which-slit info IS available, which would be evidence of a pilot-wave, Bohmian type picture.

 

[colorSpace]

Well, Kim's experiment managed to show interference minima 1 mm apart on the detector, and that was with a lens that created a far field condition by narrowing the beam - the minima at the lens LS would be even further apart. So I still fail to see how you know these photons would not "fly around" (I'd prefer to say "tend to avoid") the grid?

I don't know Kim's experiment, but entangled isn't the same as entangled. According to A.Zeilinger's descriptions, or what I understood of them, in order to have the photons impuls-entangled such that it the impuls is always exactly opposite, the source of light needs to have a minimum size. And when it has such a minimum size, there will be multiple overlapping interference patterns. This is what later-on needs to be sorted out by analysis, yet at first there is always the overlap of patterns, and so no single "black" areas will be in the patterns where you could place the grid. This will be the case even when none of the photons are used to obtain path information.

But then I'm going even further and suggesting they'll tend to avoid the grid the same way even when which-slit info IS available, which would be evidence of a pilot-wave, Bohmian type picture.

That's speculation on your part, even more so than in Afshar's experiment, thinking that with a grid placed, the detectors would still give correct path information. I think that assumption can't be made, since the photons are free to bend around the grid. In this unique case where the grid is placed in an interference area, it may just be that there will be an interference of impulses such that the photons (still being in superposition) may bend just as if they came from the respectively "other" pinhole. That may be just what the wave function turns out to be in such a unique case.

 

[peter0302]

Read Kim. The only reason I can find for the pattern to be always Gaussian without the coincidence circuit is the 180-degree phase shift between D1 and D2. Thus, the sum of D1 and D2 detections is always Gaussian. According to the paper this is due to photons striking D2 having been reflected one additional time by the beamsplitter than those striking D1.

Whether this is unavoidable I don't know. Couldn't they have introduced a mirror before D1 to put the detections back into phase, resulting in a _visible_ interference pattern at D0 any time the idler photons chose the erase path)?

At any rate, the pattern emerges as soon as they filter out joint detections at D0 and D1 or D2, respectively. No tricks required. So I don't think multiple-overlapping interference patterns is an issue that can't be solved in my experiment the same way Kim solves it - with a coincidence circuit. Either way, a certain detectable subset of photons _will_ avoid certain regions of D0; we can use Afshar's grid to see if they avoid those regions in mid-flight as well. We can likewise use Afshar's grid to see if the "which-slit-known" photons do the same thing.

 

[colorSpace]

Read Kim.

I'd need a pointer to some webpage about Kim.

At any rate, the pattern emerges as soon as they filter out joint detections at D0 and D1 or D2, respectively. No tricks required. So I don't think multiple-overlapping interference patterns is an issue that can't be solved in my experiment the same way Kim solves it - with a coincidence circuit. Either way, a certain detectable subset of photons _will_ avoid certain regions of D0; we can use Afshar's grid to see if they avoid those regions in mid-flight as well.

So what I was saying is that you probably have a lot of photons hitting the grid, but apparently if you are using the coincidence circuit such that these are filtered out (mathematically, so to speak) anyway, then you might be ok.

But that might not be easy if you place the grid mid-flight, as you probably need to. How do you make sure that those photons hitting the grid are exactly in the same subset as the ones you are filtering out?

We can likewise use Afshar's grid to see if the "which-slit-known" photons do the same thing.

You shouldn't be able to successfully filter out an interference pattern for photons in a situation where you know the path. I doubt that quantum physics has an "error" here, even if it has errors. Rather, I would assume that you perhaps don't really know the path, due to things like possible bending at the grid in a fashion that is itself determined by the interference (or that you don't really have an interference pattern).

 

[colorSpace]

(It seems that Kim's experiment is using interference patterns anyway, but that the grid is placed in an additional location, rather than where these usually occur.)

 

[colorSpace]

Does this page use the same terms (D0-D3) and setup as you do?

http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser

 

[colorSpace]

OK, I'm getting a better picture now, sorry it took me a while to get up to speed on this one. :) It seems that this text doesn't explain all too well how the filtering works. I think you don't correctly understand the filtering for the case where the path information is "intact".

Each member of an entangled photon pair is sent either to a lens and detector (signal photon) or a quantum eraser (idler photon). When the idler photon is observed with “which-slit” information intact, the rate of detection of the signal photon is observed in the presence of an Afshar-style interference grid, placed before the lens in the path of the signal photon. It is assumed the grid will have no effect on the detection rate, even when the idler photons are detected with “which-slit” information intact; the only way this can be possible, however, is if the signal photons exhibit interference ([EDIT]wave)-like effects, thereby avoiding the grid via the interference minima, until the instant of detection, at which point their distribution becomes Gaussian. Similarly, for idler photons for which “which-slit” information has been erased, the signal photons are expected to exhibit an interference pattern at all times.

1. The first case, where the path info is intact: In this case the filtering by D3 and D4 doesn't select a subset of photons which would have an interference pattern. So you would get a reduced detection rate since some photons in this subset hit the grid (compared to the same without the grid).

2. When the path information has been erased, filtering with D1 and D2, you should see an interference pattern, and since unlike in the Afshar case, there is no other path determining apparatus at D0 anyway, there is no change, and no surprise, of any kind, due to placing the grid.

This would be my understanding of the situation.

 

[colorSpace]

In order to get a more interesting experiment, you might place the lens and split detectors from the Afshar experiment at D0.

Still, in the first case (path info intact) you would have the same reduction in detection. In the second case, you'd probably get the same result as in the Afshar experiment, but perhaps now you have additional possibilities of counting...

 

[peter0302]

Sorry for not replying sooner. You can get their paper here:

http://arxiv.org/abs/quant-ph/9903047

I think the Wikipedia article is accurate but probably lacks some of the finer details found in the paper, particularly the reason for the interference pattern only emerging when one erase-path or the other (but never both) is examined.

 

[peter0302]

1. The first case, where the path info is intact: In this case the filtering by D3 and D4 doesn't select a subset of photons which would have an interference pattern. So you would get a reduced detection rate since some photons in this subset hit the grid (compared to the same without the grid).

2. When the path information has been erased, filtering with D1 and D2, you should see an interference pattern, and since unlike in the Afshar case, there is no other path determining apparatus at D0 anyway, there is no change, and no surprise, of any kind, due to placing the grid.

Definitely yes on number 2, although this, itself, should be of interest, since it means that the interference pattern does have physical meaning aside from the probability of detection at D0.

As for number 1, that's the $64,000 question. But consider this: if, in the presence of the grid, you get a reduced detection rate for D3+D4, but not for D1+D2, doesn't this allow the transmission of signals? That's why I assume the detection rates will be the same, meaning the interference pattern will be seen even for the D3+D4 subset.

 

[colorSpace]

As for number 1, that's the $64,000 question. But consider this: if, in the presence of the grid, you get a reduced detection rate for D3+D4, but not for D1+D2, doesn't this allow the transmission of signals? That's why I assume the detection rates will be the same, meaning the interference pattern will be seen even for the D3+D4 subset.

Unfortunately, I think it is again the same principle: You can recognize a reduced detection at D0 only after bringing in the information from D3 and D4 which selects the subset that shows a reduction. Photons are hitting the grid in either case, you'll just be looking at different subsets in hindsight. The entanglement is nevertheless there in (instantly) producing the information that will later-on allow you to do that, after you have collected the information classically. But please keep trying... it just ain't that easy. :)

 

[peter0302]

But you know ahead of time what the detection rate for D1 + D2 should be without the grid and what the detection rate for D3+D4 should be without the grid. If you did something at the other end (like flipped a switch forcing the photons always to go to D1+D2 or D3+D4 a la Cramer), you could then receive a message in the reduced detection rate at D0.

So no signal means D1+D2 must equal D3+D4 or else someone at D0 could differentiate between them just by looking at the overall rate. So the grid _must_ have the same effect on detections regardless whether which-slit is intact.

 

[colorSpace]

Whether the summary detection rate is reduced depends only on whether the grid is placed or not, which is a decision you make at D0.

Otherwise, you are looking at subsets, and you need that data from D1-D4, in order to select the subset of photons, first.

(Those photons pairs where one is hitting the grid will not be filtered out by the coincidence counters at all.).

 

[peter0302]

I'm starting to think that DCQE might not be the perfect set-up for this test after all because, by the very nature of the eraser itself (i.e. the 90-degree phase shift between D1 and D2) it's impossible to get a consistent interference pattern with which to test against D3 or D4.