FTL communication via delayed choice measurement

[Alex Torres]

<Moderator's note: 2 threads merged as it is an identical topic.>

Given Kim's version of DCQE, let's say Bob in on Earth looking at d0, Alice is on Mars holding the prism that deflects the idler coming from (a) to d3 or (b) to d4. After the experiment run ends Bob should get clump at d0 without a feedback from Alice. In the actual experiment Bob receives each idler at d0, 8 nanoseconds before Alice

 

[Alex Torres]

Given Kim's version of DCQE is it possible FTL communication by having Bob on Earth looking at d0, and Alice on Mars holding the prism that deflects the idler (a) to d4 and the idler(b) to d3 so by the time the experiment is done Bob would have a definite clump pattern at d0 without the need to receive any feedback from Alice. (To avoid noise from Bob to Alice we can use the same channel used by Victor, Alice and Bob in the Delayed choice entanglement swapping)

 

[Strilanc]

No, it's not. Bob sees the same thing on d0 regardless of Alice's choice. You need to split the measurements of d0 into two groups in order to see the interference or not, but in order to do the splitting you need Alice's measurement results.

 

[Alex Torres]

Yes, know that will happen if you leave all 4 detectors at Alice. In Kim's papers you see a single band for all hits mixed in a single graph at d0, but if you see d3/d4 only there's a bell shaped pattern that even if you overlap for not cherry picking at them, due to symmetry you also get a bell shaped pattern at d0, assuming a setup with d3/d4 only.

 

[Alex Torres]

Let me make just one correction. By Alice having only d3/d4 in place until the experiment is done, Bob don't even has to plot anything into a graph to see a clump at d0. Next day another run of the experiment starts at the same hour, but this time Alice has nothing in place so her twin idler gets lost in the space, after the experiment is done Bob should get fringes, since no measurenent was done.

 

[Nugatory]

By Alice having only d3/d4 in place until the experiment is done, Bob don't even has to plot anything into a graph to see a clump at d0. Next day another run of the experiment starts at the same hour, but this time Alice has nothing in place so her twin idler gets lost in the space, after the experiment is done Bob should get fringes, since no measurenent was done.

That's not how it works. The signal photons whose corresponding idlers will interact with D3 or D4 if they are there contribute to the "clump" subset of the total detections at D0; the signal photons whose corresponding idlers would interact with D1 or D2 if they are there contribute to the "pattern" subset of the total detections at D0. Either way, the total pattern at D0 is the same.

(There are also practical problems separating any signal at D0 from background noise if any of D1-D4 are not providing input to the coincidence counter).

 

[Alex Torres]

But... what pattern might you get at d0, by not detecting the idler at all and letting it just hit the wall?

 

[Nugatory]

But... what pattern might you get at d0, by not detecting the idler at all and letting it just hit the wall?

You get the same pattern at D0 no matter what you do with the idler. Read @Strilanc's post in #3 again.

 

[Alex Torres]

You need to split the measurements of d0 into two groups in order to see the interference or not, but in order to do the splitting you need Alice's measurement results.

I understand, that completely applies as long as Alice is measuring both a particle and a wave in the same experiment run. My simple question is, what Bob will get at d0 if he just let all the idlers get absorbed at a piece of paper, since you are not making a definite measurement of anything at the idler, the most reasonable answer is that he gets a definite interference pattern at d0...but that's just my guess... don't know of any experiment done this way...

 

[mfb]

What Alice does has no impact on the result Bob gets.

This has been said multiple times now.

 

[Alex Torres]

Ok, take Alice out of this...Bob is doing the experiment at home, he takes out all detectors and Beam splitters in the idlers path and put a piece of paper instead...my question is, what pattern, if any, will the signals photons form at d0 after all hits are recorded...

 

[mfb]

The same as with detectors, but now a sketch of the setup would help.

 

[PeroK]

Given Kim's version of DCQE is it possible FTL communication ...

In order to send a message, you have to change something under your control. In the case of entanglement, all you can do is observe the state. You cannot control the result of your observation (although you can choose what to observe).

It's fundamentally no different in this respect from the right shoe/left shoe scenario. If Alice receives a box with one shoe in it and Bob receives the other shoe, then they can observe what is in their box. That determines, but doesn't change, what's in the other box. Each has no control over what shoe they observe. And, if one of them replaces the shoe with an opposite shoe, that does not change the shoe the other one has.

Fundamentally, you cannot send a message by observation.

 

[Alex Torres]

If Alice receives a box with one shoe in it and Bob receives the other shoe, then they can observe what is in their box.

Again, let's take Alice out of this context ...Bob is doing the experiment at home, he takes out all detectors and Beam splitters in the idlers path and put a piece of paper instead...my simple question is, what pattern, if any, will the signals photons form at d0 after all hits are recorded... I am not asking here anything about FTL communication either!...

 

[weirdoguy]

You received an answer:

The same as with detectors, but now a sketch of the setup would help.

 

[Alex Torres]

In the proposal Bob is deciding himself no to measure position, my guess is that he should get a definite fringes pattern at d0, in Kim's papers it's pretty well explained why every single detector contributes to the single solid band at d0, but doesn't tell if all detectors are removed from the idlers path...

 

[Nugatory]

Again, let's take Alice out of this context ...Bob is doing the experiment at home, he takes out all detectors and Beam splitters in the idlers path and put a piece of paper instead...my simple question is, what pattern, if any, will the signals photons form at d0 after all hits are recorded.

That experiment is easier said than done, because you need some way of picking the signal photons out from the much larger number of of incident photons that aren't part of signal/idler pairs. However, the downconverted photons do have a different frequency, so it should be possible in principle.

So what you're really asking is: Suppose we were to identify every detection at D0 that would have been included if we had our D1-D4 detectors and coincidence counter set up; what pattern do they form? Phrased that way, the answer should be clear: it's the sum of the four individual patterns.

 

[Alex Torres]

if we had our D1-D4 detectors and coincidence counter set up; what pattern do they form? Phrased that way, the answer should be clear: it's the sum of the four individual patterns.

Agree...then if we remove d1-d4 detectors the pattern at d0 will be________________.

 

[Nugatory]

Agree...then if we remove d1-d4 detectors the pattern at d0 will be________________.

Unchanged from if the detectors were there, so we can use the distributions in the paper to see what the pattern will be. Qualitatively, the rate-position graph is a hill with bumps on it, or you can visualize the pattern by overlapping the R₀₁, R₀₂, R₀₃, and R₀₄ illustrations from the wikipedia article.

 

[Alex Torres]

Unchanged from if the detectors were there, so we can

sorry for no being specific, should read: if we remove d1-d4 detectors and replace them with a piece of paper then run the experiment again, when the experiment is done the pattern at d0 will be________________.

 

[Alex Torres]

In other words, if you remove all BS's and detectors in the idlers path an just let it make its run toward a single piece of paper, after the experiment is done, the pattern at d0 will be ______________.

 

[Nugatory]

sorry for no being specific, should read: if we remove d1-d4 detectors and replace them with a piece of paper then run the experiment again, when the experiment is done the pattern at d0 will be________________.

In all all questions of the form "If <some condition involving the idlers> and then we run the experiment again, the pattern formed by the signal photons at d0 will be _________?" we can fill in the blank with the same words: "What you get when you overlap the R₀₁, R₀₂, R₀₃ and R₀₄ illustrations from the wikipedia article; the rate-position graph is a hill with bumps".

(As a practical matter, we need some way of recognizing the signal photons; for every one that arrives at d0 we will get hundreds of thousands or millions of unrelated photons that will totally overwhelm the pattern formed by the signal photons. The idler detectors and the coincidence counter circuity take care of that when they're present so those are conditions under which it's easiest to actually observe the pattern. But once we've observed it, it's our answer no matter what we do with the idlers).

 

[Alex Torres]

"If <some condition involving the idlers

the proposal is that no condition is forced upon the idler, we let it fly freely all along until absorbed by a piece of paper, as someone stated before, the pattern at d0 will be the sum of the patterns provided by d1-d4, given they are in place, so it will still be a definite and distinctive pattern if compared with something else.

 

[Nugatory]

the proposal is that no condition is forced upon the idler, we let it fly freely all along until absorbed by a piece of paper,

That's a condition (as would also be "we pay no attention to the idler" or "we have no idea what is done with the idler" or "the idler falls into a black hole" or ...).

 

[Alex Torres]

That's a condition (as would also be "we pay no attention to the idler" or "we have no idea what is done with the idler" or "the idler falls into a black hole" or ...).

yep.. you are right...so let me use another term: you are not measuring the idlers in anyways..still take into account that it just passed thru the double slit...

 

[Strilanc]

In other words, if you remove all BS's and detectors in the idlers path an just let it make its run toward a single piece of paper, after the experiment is done, the pattern at d0 will be ______________.

See also: "A Classical Delayed Choice Experiment" http://algassert.com/post/1720 , which makes it a bit clearer what the mistake you're making is:

 

[Alex Torres]

View attachment 229561See also: "A Classical Delayed Choice Experiment" http://algassert.com/post/1720 , which makes it a bit clearer what the mistake you're making is:

View attachment 229563
View attachment 229562

don't really know what mistake you are talking about...im just asking a question given a proposal...besides, the pictures you brought in are really good to explain what you get at d0 and why, given d1-d4 are in place...

 

[mfb]

im just asking a question given a proposal

An you are getting the same correct answer over and over again. Do you expect the answer to change if you ask yet another time?

 

[Alex Torres]

An you are getting the same correct answer over and over again. Do you expect the answer to change if you ask yet another time?

Believe me, it's not my intention to be redundant, but haven't get a definite answer yet, so will make this a final question like a poll, a yes or no ...will do it... Here is it: ...After the entangled photons are deflected by the Glan-Thompson prism and the one named the idler takes the longer path and gets absorbed at a piece of paper without being measured...once the experiment is done and given both the signal and the idler went thru the double slits as a single photon, and no measurement was made, will d0 show a definite difraction pattern?...yes or no...thanks!

 

[Nugatory]

once the experiment is done and given both the signal and the idler went thru the double slits as a single photon, and no measurement was made, will d0 show a definite difraction pattern?...yes or no...thanks!

We can't answer that question with a yes or no because it's based on a false premise - neither the signal nor the idler ever go through the double slit. They are created together in the BBO crystal after the pump photon has passed through the double slit and is down-converted.
(Another problem is is that you're asking about one photon, and of course one photon never yields any sort of pattern, just a single detection somewhere. I'm assuming that's just careless wording and you meant to say something along the lines of "a statistically significant number of signal photons have reached d0 and d0 has made a full sweep across its range of motion").

We can remove that false premise from your question and reword it as "Once the experiment is done and no measurement of the idlers was made, will d0 show a definite interference patttern?". Then, since whatever happens or doesn't happen to the idlers is completely irrelevant to the overall pattern of signal photons at d0, we can further simplify the question to "Once the experiment is done, will d0 show a definite interference pattern?" The answer to that question is "yes, and we've told you what that pattern looks like several times already".