FTL communication via delayed choice measurement

In summary: Alice needs to do is to not have any idler in place when the experiment is done.Yes, it is possible to send a message using entanglement between Alice and Bob even if they are sitting on opposite sides of the planet. Alice needs to not have any idler in place when the experiment is done.
  • #36
Nugatory said:
No, both setups produce the same single-hump pattern. Strilanc's picture is correct

But didn't say it wasn't correct, just said it applies as long as we run Setup A: Kim's set up as it is.

Nugatory said:
If you look carefully at the R01 and R03 curves, you'll see that one has a peak where the other has a trough, so the peaks and troughs don't appear in the sum

Just to make it sure i got your point...are you referring in this quote to the sum of the patterns at d1/2?
 
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  • #37
This is a quote of what's actually described about hits from d1/d2 selected at d0..."One can get an idea of how this works by looking at the graphs of R01, R02, R03, and R04, and observing that the peaks of R01 line up with the troughs of R02"...R01 and R02 are the individual interference patterns we get from hits at d1 and d2, if they overlap, we get the single hump pattern at d0 described in Strilanc's comment under "what Alice sees" column as the erasure graph.

...But, the proposed Setup B assumes no detectors are in place for the idlers path, only d0 for the signal, ...so no detector is measuring position anywhere in the setup and no erasure is done either, ...then, wouldn't it be reasonable to ask: is there anything that could possibly preclude all signal photons from hitting d0 as a wave in a way that a definite interference pattern at d0 emerges?.... just asking
 
  • #38
Just a second thought...but first, thanks to all you guys for taking a slice out of you time and give such an useful insights toward this proposal... it's not mine by the way...saw it in the comments section about a DCQE video and found it quite interesting and enough convincing to be posted in a forum like this to get some feedbacks ...

...first thing noted...the fact that Kim's version needs 2 detectors to measure which way and another pair of detectors to measure which way erasure...forces Alice to go back to Bob in order to make sense out of "what she sees at the screen" for having the results overlapped. But then, the results are overlapped since 2 detectors are needed to measure which way and another pair to measure after erasure.
So this is kind of a loophole.

Let's say Bob just realized that, and thinks, the single-hump pattern Alice always get at the screen is a definite pattern after all if compared with something else...

Then he thinks in a radical way and decides to add a second experiment setup totally different to the one he has used so far and gives Alice a second screen to collect the hits from this second setup.

He thinks, to get a message across to Alice i need to find out the way to arrange this second setup in oder to obtain a pattern that just looks distinctively different from the single-hump pattern she always get from the first setup... so if the photons can travel all along to Alice as a wave then she might get a distinctive interference pattern at her screen and that will be a totally different pattern from the single hump pattern she has received so far.

So he decides to discard all detectors and BS's and replaces all that with a single piece of paper as the sole target for all the idlers photons he gets.

Here is the question...is he going to succeed to have a definite interference pattern at Alice's second screen?
 
  • #39
You might find this article interesting. The point was that it appears impossible to extract information in the manner you are describing, you need information that you can only get at sub/equal-light speeds.

https://arxiv.org/abs/1707.06995
 
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  • #40
Alex Torres said:
So he decides to discard all detectors and BS's and replaces all that with a single piece of paper as the sole target for all the idlers photons he gets.

Here is the question...is he going to succeed to have a definite interference pattern at Alice's second screen?

How many times do we have to say NO IT WON'T MAKE AN INTERFERENCE PATTERN before you'll actually hear the answer?
 
  • #41
SatishR said:
The point was that it appears impossible to extract information in the manner you are describing

Yep...also have seen others rising concerns because on the non communication theorem, but then found out it is conditioned to the use of quantum states as a medium to carry classical bit, there's an obvious reason why you can't use a probability distribution to send any coherent message at all...
 
  • #42
Strilanc said:
How many times do we have to say NO IT WON'T MAKE AN INTERFERENCE PATTERN before you'll actually hear the answer?

Come on dude, this is a discussion forum, ...just saying "No" isn't enough... All answers so far are related to why you can't get an interference pattern given Kim's version as it is...
...but, if the reason to have a double slit in the setup is to produce an interference pattern that can be projected or obtained somewhere in the setup, haven't seen an explanation so far about why it would be impossible to obtain that pattern given there's no measurement anywhere in the modified setup...
 
  • #43
Alex Torres said:
Come on dude, this is a discussion forum, ...just saying "No" isn't enough... All answers so far are related to why you can't get an interference pattern given Kim's version as it is...
...but, if the reason to have a double slit in the setup is to produce an interference pattern that can be projected or obtained somewhere in the setup, haven't seen an explanation so far about why it would be impossible to obtain that pattern given there's no measurement anywhere in the modified setup...

Your question about "why it would be impossible..." is always going to be tough, as tough as answering why pigs don't fly.

But it might help if you knew that entangled particle pairs - such as used in the Kim et al experiment - do not usually produce interference patterns. See S290, figure 2. And yet the correlated sub-groups within the overall do, as Kim demonstrated.

https://pdfs.semanticscholar.org/3644/6f15507880c629e06391adf9d21aa6d76015.pdf
 
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  • #44
Alex Torres said:
given Kim's version as it is

Please give a specific reference for "Kim's version as it is". I have seen no link to an actual paper in any of your posts.
 
  • #45
Alex Torres said:
if the reason to have a double slit in the setup is to produce an interference pattern that can be projected or obtained somewhere in the setup

No, the reason to have a double slit in the setup is...to see what happens when you have a double slit in the setup. Whether it produces an interference pattern depends on the entire setup. A double slit can't magically make interference where none is possible.

All of the answers you have gotten in this thread amount to repeatedly pointing out to you that none of the changes you have suggested making in the setup, as compared to "Kim's version as it is", change anything about the original setup that affects whether interference is possible. So if you can't get interference in the original setup (which you can't, and you appear to realize that), you can't get interference in any of your modified setups.
 
  • #46
PeterDonis said:
Please give a specific reference for "Kim's version as it is". I have seen no link to an actual paper in any of your posts.

You can scroll up to one of Strilanc's comments above where it clearly explains, with nice graphs, why Alice will never see an interference pattern, given Kim's setup, there's also an acceptable description of the whole experiment in wikipedia...

https://en.m.wikipedia.org/wiki/Delayed_choice_quantum_eraser
 
  • #47
PeterDonis said:
Whether it produces an interference pattern depends on the entire setup. A double

Thought it was demonstrated, the interference pattern at the screen disappears if a measuring device is placed to see which slit the photon went thru.
 
  • #48
Alex Torres said:
the interference pattern at the screen disappears if a measuring device is placed to see which slit the photon went thru

Yes, that's an example of "whether it produces an interference pattern depends on the entire setup".
 
  • #49
PeterDonis said:
Yes, that's an example of "whether it produces an interference pattern depends on the entire setup".
... maybe our only option would be to ask Kim. Did he test the setup for interference pattern before inserting any detector in the idlers path? just to be sure there wasn't something physically destroying the interference pattern before the idlers hit the first pair of detectors d3/4?... otherwise, the presence of those detectors would be totally redundant...
 
  • #50
Alex Torres said:
... maybe our only option would be to ask Kim. Did he test the setup for interference pattern before inserting any detector in the idlers path?
You're forgetting the role of the coincidence counter circuitry in this experiment. No signal photon will be recorded and counted towards the pattern unless its idler is detected by one of the D0-D4 detectors. So if you remove any of these detectors, you're just asking Bob to throw out some of the D0 detections, and any changes in the observed pattern are just the result of not counting all the points in the pattern.
 
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  • #51
Nugatory said:
No signal photon will be recorded and counted towards the pattern unless its idler is detected by one of the D0-D4 detectors

If you leave all detectors in place and decide not to account for the individual patterns from each detector, you still will have an overall pattern at d0, that will be a single-hump pattern

Nugatory said:
So if you remove any of these detectors, you're just asking Bob to throw out some of the D0 detections, and any changes in the observed pattern are just the result of not counting all the points in the pattern.

But, that's exactly the purpose of removing some or all detectors besides d0, to get a distinctive pattern from the single hump pattern we get for leaving all detectors in place.
 
  • #52
Nugatory said:
So if you remove any of these detectors, you're just asking Bob to throw out some of the D0 detections, and any changes in the observed pattern are just the result of not counting all the points in the pattern
Let me correct the answer to this second quote, should read: But, that's exactly the purpose of removing some or all detectors besides d0, to get an overall distinctive pattern from the overall single hump pattern we get for leaving all detectors in place and not counting towards the individual patterns
 
  • #53
Alex Torres said:
But, that's exactly the purpose of removing some or all detectors besides d0, to get a distinctive pattern from the single hump pattern we get for leaving all detectors in place.
You don't get a different pattern by removing detectors. You just lose some information.
 
  • #54
mfb said:
You don't get a different pattern by removing detectors. You just lose some information.

The proposal has a basic assumption: ...if all detectors, besides d0, are removed and all the idlers hit a single target where no measurement takes place, the modified setup will turn out to be a Standard DS experiment with no detector in place at the slits and d0 mimicking the screen behind the slits, should get a definite interference pattern.
 
  • #55
Alex Torres said:
if all detectors, besides d0, are removed and all the idlers hit a single target where no measurement takes place, the modified setup will turn out to be a Standard DS experiment with no detector in place at the slits and d0 mimicking the screen behind the slits, should get a definite interference pattern.

Of course you will not get an interference pattern that way. All of the DCQE experiments and variants are based on the fact that you use incoherent light for illuminating the double slit, so you do not get an interference pattern behind the double slit under any circumstances unless you perform a posteriori filtering.
 
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  • #56
Cthugha said:
Of course you will not get an interference pattern that way. All of the DCQE experiments and variants are based on the fact that you use incoherent light for illuminating the double slit, so you do not get an interference pattern behind the double slit under any circumstances unless you perform a posteriori filtering.

...got it! You just nailed it!...but there's still something i don't get, what could possibly be in Kim's mind when he designed the setup, given he should have known in advance the double slit in his setup had no chance of rendering an interference pattern anywhere in the setup, that he might still get an interference pattern at d1 or d2?
 
  • #57
Said "d1 or d2" , if for example, you only cherry pick the hits from d1 at d0 and plot them into a graph, you get a definite interference pattern for all the hits you accounted for at d1 only...the same should stand if you cherry pick from d2 only
 
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  • #58
Alex Torres said:
...got it! You just nailed it!...but there's still something i don't get, what could possibly be in Kim's mind when he designed the setup, given he should have known in advance the double slit in his setup had no chance of rendering an interference pattern anywhere in the setup, that he might still get an interference pattern at d1 or d2?
That's what a calculation from the laws of quantum predict will happen - the calculation is in his paper. The point of the experiment was to confirm that this prediction is accurate, and to demonstrate that the relative ordering of the signal and idler detections doesn't matter (in Kim's experiment the idlers were detected about eight nanoseconds after the signal photons).
Alex Torres said:
Said "d1 or d2" , if for example, you only cherry pick the hits from d1 at d0 and plot them into a graph, you get a definite interference pattern for all the hits you accounted for at d1 only...the same should stand if you cherry pick from d2 only
Yes, and Kim's paper has a picture of the d0-d2 coincidence pattern right next to the d0-d1 one.
 
  • #59
Cthugha said:
All of the DCQE experiments and variants are based on the fact that you use incoherent light

...would you please give us a link to a source backing up that statement?...just made a little research...and found out they used a laser, a coherent source of light...
 
  • #60
Alex Torres said:
if all detectors, besides d0, are removed and all the idlers hit a single target where no measurement takes place, the modified setup will turn out to be a Standard DS experiment with no detector in place at the slits and d0 mimicking the screen behind the slits

No, it won't. Look at the diagram in the Wikipedia article you linked to. If you take out all the detectors except d0, is what is left just a standard double slit experiment, with d0 in the same position as the detector screen in that setup? I don't think so.
 
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  • #61
Cthugha said:
All of the DCQE experiments and variants are based on the fact that you use incoherent light for illuminating the double slit

I'm not sure the light is incoherent. In the Kim experiment described in the Wikipedia article @Alex Torres linked to (and also in the Kim paper referenced in that article), the light is said to be produced by an argon laser. That would be coherent light, not incoherent light.
 
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  • #62
PeterDonis said:
No, it won't. Look at the diagram in the Wikipedia article you linked to. If you take out all the detectors except d0, is what is left just a standard double slit experiment, with d0 in the same position as the detector screen in that setup? I don't think so

Agree, but then... (given we already solved the issue about the incoherent source of light) we have to go back and ask, what could possibly be in Kim's mind for inserting a double slit in the setup, if he wasn't expecting to get an interference pattern somewhere in the setup?
 
  • #63
Alex Torres said:
...would you please give us a link to a source backing up that statement?...just made a little research...and found out they used a laser, a coherent source of light...
The pump laser is a source of coherent photons, but these are not the photons we're talking about here. The experiment is using the photons produced by parametric downconversion in the BBO crystal after the double slit and these are not coherent.
 
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  • #64
Nugatory said:
The pump laser is a source of coherent photons, but these are not the photons we're talking about here. The experiment is using the photons produced by parametric downconversion in the BBO crystal after the double slit and these are not coherent

Yep.. ... that could solve the whole issue, but if that's the case it would rise concerns about another issue...how did he get an interference pattern by cherry picking only from d1 at d0, or cherry picking only from d2 at d0?
 
  • #65
Alex Torres said:
...would you please give us a link to a source backing up that statement?...just made a little research...and found out they used a laser, a coherent source of light...

Just to clarify: the kind of coherence we talk about here is spatial coherence (that is what a double slit measures) and not any higher order coherence that is typically associated with lasers. This is basically a measure of how point-like the light source is. Accordingly, the degree of coherence depends on the distance between the slit and the light source and the size of the active area of the light source. One can generally show that spatial coherence and entanglement in the spatial/angular degree of freedom are mutually exclusive, see Phys. Rev. A 63, 063803 (2001): https://journals.aps.org/pra/abstract/10.1103/PhysRevA.63.063803 .

An explicit measurement of the visibility of the interference patterns of light emitted from BBO crystals at different distances to the slit has been done in Zeilinger's group. Unfortunately the thesis containing the results is only available in German, but maybe some people here are able to understand it: https://www.univie.ac.at/qfp/publications/thesis/bddiss.pdf
The calculations of the necessary distances are around page 45.
 
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  • #66
Given that ..."During an experiment, detector D0 is scanned along its x-axis, its motions controlled by a step motor."... End quote

Can detectors d1-4 have the same mechanical characteristics?
 
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  • #67
Alex Torres said:
Given that ..."During an experiment, detector D0 is scanned along its x-axis, its motions controlled by a step motor."... End quote

Can detectors d1-4 have the same mechanical characteristics?
Yes, but why? There's only one position where they can detect anything.
 
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  • #68
Nugatory said:
Yes, but why? There's only one position where they can detect anything.

The issue of the BBO generating entangled pairs without capacity to reflect interference at the screen was solved in the Walborn's DCQE version...a more simplified version...
https://files.acrobat.com/a/preview/0b5148b2-be98-4fdb-9f7c-da1f58490d73

The SPDC occurs first, the signals are sent directly to Alice at d0 without going thru the slits. The idlers are sent to Bob who has a standard DS setup that reflects a definite interference pattern at the screen, thus solving the issue of having overlapped results when measuring interference patterns or clumps.

Here's the the assumption...
Alice and Bob have screen type detectors that moves across the x-axis with step motors controlled by individualy coordinated atomic clocks, the rate at which the laser pumps is also controlled by the clocks ...

The question:...will their results correlate don't matter how far appart Alice and Bob are from each other??
 
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  • #70
Alex Torres said:
The SPDC occurs first, the signals are sent directly to Alice at d0 without going thru the slits. The idlers are sent to Bob who has a standard DS setup that reflects a definite interference pattern at the screen, thus solving the issue of having overlapped results when measuring interference patterns or clumps.

No, it does not. You only get the interference pattern directly in the non-delayed version, where the light arriving at the double slit is linearly polarized. Walborn states that at the beginning of section III. In the polarization entangled BBO version of the experiment he uses afterwards, the incoming light is not polarized linearly anymore and there is no direct interference pattern left. This is easier to see in Walborn's popular physics article from American Scientist, where he explains the experiment in a more layman-friendly way:

http://www.mat.ufmg.br/~tcunha/2003-07WalbornF.pdf
 

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