# FTL communication via delayed choice measurement

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• Alex Torres
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.

#### 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

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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)

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.

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.

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.

Alex Torres said:
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?

Alex Torres said:
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.

Strilanc said:
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...

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

This has been said multiple times now.

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...

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

Alex Torres said:
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.

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Alex Torres
PeroK said:
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!...

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

PeroK
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...

Alex Torres said:
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
Nugatory said:
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________________.

Alex Torres said:
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.

Nugatory said:
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________________.

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 ______________.

Alex Torres said:
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).

Nugatory said:
"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.

Alex Torres said:
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
Nugatory said:
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...

Alex Torres said:
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:

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Alex Torres
Alex Torres said:
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?

weirdoguy
mfb said:
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!

Alex Torres said:
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".

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Alex Torres
Nugatory said:
Then, since whatever happens or doesn't happen to the idlers is completely irrelevant to the overall pattern of signal photons at d0, we

Then, a definite interference pattern at d0 can be distinguished from an overall pattern at d0 resulting from different outputs being overlapped. Am i correct?

Alex Torres said:
Then, a definite interference pattern at d0 can be distinguished from an overall pattern at d0 resulting from different outputs being overlapped. Am i correct?

In one run of the experiment we accumulate a large number of position measurements, basically observations of the form "detector d0 detected a signal photon when it was at position x". We look at these and find a pattern: some areas get more detections than others. It's always the same pattern every time we run the experiment no matter what happens with the idlers, and that's all the information that we ever get from d0. There will never ever be but that one pattern at d0 (unless we change something else like the frequency of the pump laser, or the position and shape of the slits, or the characteristics of the BBO crystal).

Now suppose we take our large collection of position measurements and divide it into four groups. (We could put measurements 1, 5, 9, ... in the first group, 2, 6, 10, ... in the second group, 3, 7, 11, ... in the third group, and 4, 8, 12, ... in the fourth group. Or we could do the assignments randomly: for each measurement we generate a random number between 1 and 4 and put that measurement in that group. Or we could do something else). Then we look at the pattern for each group. Several things should be obvious:
- The total pattern will be the sum of the patterns for each group. It's nonsense to talk about the sum being somehow different from the total pattern (which is why the quick answer above is "no").
- The pattern for any single group doesn't have to look like the total pattern. For example, we could choose to put all the measurements in which x is less than .1 in one group; then we'll find that the pattern for that one group has a huge spike where the pattern for the other three groups has a huge trough, and neither the trough nor the spike appears in the pattern we actually measured. That doesn't mean that we've observed anything different; it's just an artifact of how we've manipulated our data.

There is one particularly interesting way of dividing our collection of position measurements into four groups. We are sitting around admiring the nice pattern we found at d0 when Alice walks up to us and suggests that we try putting this one, this one and this one into group one, then that one and that one into group two, and so forth. We're intrigued, so we ask her how she's choosing, and she explains that she's putting all the ones whose idler triggered d1 into one group, all the ones whose idler triggered d2 into another group, and so on. So we divide our collection of position measurements into four groups as Alice suggests, then we look at the pattern for each group in isolation... and we get something that looks like the patterns in Kim's paper, with interference fringes in groups 3 and 4 and not in groups 1 and 2.

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Alex Torres, Strilanc and Lord Crc
Nugatory said:
In one one run of the experiment we accumulate a large number of position measurements, basically observations of the form "detector d0 detected a signal photon when it was at position x". We look at these and find a pattern: some areas get more detections than others. It's always the same pattern every time we run the experiment no matter what happens with the idlers, and that's all the information that we ever get from d0. There will never ever be but that one pattern at d0 (unless we change something else like the frequency of the pump laser, or the position and shape of the slits, or the characteristics of the BBO crystal).
Then we get a definite interference pattern for each experiment run under the conditions proposed.

But there's still something i don't get from the rest of your answer,... if we run Kim's version exactly as it is, the overall patterns at d0 is always the same, which happens to be a bell shaped graph, see Strilanc's comment above under "what Alice sees" column.

So we can summarize all the above for the sake of simplicity as follows:

Setup A: is exactly as Kim's description.

Setup B: the idlers path is cleared from BS's and detectors, and has a single piece of paper as a unique target, so no measurement is done.

Overall results at d0, (without sorting the hits)

Setup A: a bell shaped graph
Setup B: a definite interference pattern

Or disclosing it in a more specific way. ...

The overall results at d0, (without sorting the hits) will show the following intensity across the screen:

Setup A: a single bell shaped pattern

Setup B: a definite interference pattern

Alex Torres said:
Or disclosing it in a more specific way. ...

The overall results at d0, (without sorting the hits) will show the following intensity across the screen:

Setup A: a single bell shaped pattern

Setup B: a definite interference pattern
No, both setups produce the same single-hump pattern. Strilanc's picture is correct and I didn't do the addition of the patterns properly the first time around. 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.