Question about delayed choice quantum eraser

[sunjen]

I was reading about the Delayed Choice Quantum Eraser experiment here:

bottomlayer.com/bottom/kim-scully/kim-scully-web.htm

In this experiment the which-path info is erased or not at random AFTER the original (signal) photon hits the screen. The results of the screen are seen by the experimenter AFTER the choice to erase or not took place.

I wonder what happens If the experimenter looks at the screen after the photons hit the screen but BEFORE the choice to erase the which-path is made? That is to have the idler photons still traveling and not reach yet the mirror that has a 50-50 chance of erasing the info.

Any ideas?

 

[Hurkyl]

If you're describing what I think you're describing...


One of the main points (which, IMHO, is woefully underemphasized in popular accounts) is that the screen always looks 'normal'. The interference pattern only appears in post-processing, after you've separated the images on the screen into two sets based upon what the other detector saw. (And, of course, only appears if you retained the which-path info)

 

[Demystifier]

Whenever I have conceptual doubts about delayed choice and quantum eraser questions, I use a trick: I first analyze what would be the answer in the Bohmian interpretation. Then I simply apply the theorem that states that measurable predictions of the Bohmian interpretation are identical to those of the standard interpretation. The advantage of this trick is that, even if you do not favor the Bohmian interpretation, this approach is conceptually simpler in the sense that it does not involve the wave-function collapse.

To see in more detail how the Bohmian interpretation works for delayed choice, see:
D. Bohm, C. Dewdney, B.J. Hiley, Nature 315 (1985) 294-297.

 

[vanesch]

Whenever I have conceptual doubts about delayed choice and quantum eraser questions, I use a trick: I first analyze what would be the answer in the Bohmian interpretation. Then I simply apply the theorem that states that measurable predictions of the Bohmian interpretation are identical to those of the standard interpretation. The advantage of this trick is that, even if you do not favor the Bohmian interpretation, this approach is conceptually simpler in the sense that it does not involve the wave-function collapse.

I do exactly the same thing with MWI
It is my main - if not sole - justification for considering MWI.

In fact, the reason why as well Bohm as MWI give clear answers in this kind of cases, is that they don't have a "measurement ambiguity" built in their formalism, but have a universal dynamics. In other words, you can "close your eyes and think of England" and let the formalism crunch the numbers - while in every projection-based viewpoint, you have to decide when you project and when you don't, and that's the difficulty here.

 

[sunjen]

Thanks guys for the response.

... the screen always looks 'normal'. The interference pattern only appears in post-processing, after you've separated the images on the screen into two sets based upon what the other detector saw. (And, of course, only appears if you retained the which-path info)

Ok So let's see these 2 scenarios

1) I look at the screen AFTER completely erasing the which-path info of all photons.
I see the interference pattern. right?

2) I look at the screen BEFORE completely erasing the which-path info of all photons. The idler photons have not reached the detectors yet.
You say I see the the normal distribution. right?
So in this case it doesn't make a difference if I see or erase the which-path info, I already saw the normal dist. and that won't change. right?

 

[atyy]

Quantum Erasure: Quantum Interference Revisited
Stephen P. Walborn, Marcelo O. Terra Cunha, Sebastião Pádua, Carlos H. Monken
http://arxiv.org/abs/quant-ph/0503073

"Wait Bob, that wasn’t there before! How did you make the photons interfere after I already detected them and recorded it all in my lab book?!”

 

[sunjen]

Thanks. I just read the article, very good by the way.

So what I understand is that if you have any means to measure the which-path information (even if not measured yet), and you see the screen you will see the normal dist. If afterwards you erase the info from a subset of the photons, you can see interference only in that subset.

But what about if you erase the info on all photons?

 

[Autochthon]

Good article think I'll send it to a few non QM literate friends. Every time I read about Quantum Erasure I'm left with this odd niggling feeling that I'm peering through this strange lens called "entanglement" and that as I examine an object through the lens I can't decide if I'm changing my focus and thus seeing the a different aspect of the object or the object itself is changing.

 

[Cthugha]

So what I understand is that if you have any means to measure the which-path information (even if not measured yet), and you see the screen you will see the normal dist. If afterwards you erase the info from a subset of the photons, you can see interference only in that subset.

But what about if you erase the info on all photons?

If you look at the paper you linked in your first post, you will notice, that there are two subsets - or joint detection rates - which can show interference if you destroy which-way information (Fig. 3 and 4). If you take a close look, you will notice that they are out of phase.

If you erase which-way info on all photons and just look at the screen without choosing a subset by doing coincidence counting, you will now see both of these interference patterns superposed, which will again be a normal distribution as they are out of phase. So there is still no interference pattern without doing coincidence counting.

 

[sunjen]

Thanks for the response!

So there is still no interference pattern without doing coincidence counting.

Now this raises another question:

In this experiment there is no interference without doing coincidence counting, but in the original double slit experiments the interference pattern is shown.

So why is that, what is the difference between the two?

 

[Demystifier]

Now this raises another question:

In this experiment there is no interference without doing coincidence counting, but in the original double slit experiments the interference pattern is shown.

So why is that, what is the difference between the two?

An analogy from everyday life may also be helpfull:
https://www.physicsforums.com/blog.php?b=7

 

[Cthugha]

In this experiment there is no interference without doing coincidence counting, but in the original double slit experiments the interference pattern is shown.

So why is that, what is the difference between the two?

Well the key to interference phenomena is indistinguishability of some sort. Comparing the simple double slit to quantum eraser experiments, you will notice, that we are talking about two different kinds of indistinguishability here.

The usual double slit uses the fact, that there is a fixed phase relationship of the incident light at both slits. If the phases at both slits were completely independent of each other, this would be some kind of which way information and the interference pattern would disappear. In order to avoid this, you need light, which is at least a bit coherent: The coherence length needs to be at least as large as the slit separation is. The light, which comes out of a BBO crystal used for spontaneous parametric down converion is rather incoherent. However one can increase the coherence length by putting the BBO crystal far away from the double slit. This equals choosing a small subset of wave vectors (or equivalently emission angles), which actually make it to the double slit, so the phase relationship at the double slit is better defined.

Quantum erasers and the like use indistinguishability of two-photon amplitudes. In this case the phase relationship of the two-photon state is well defined as the wave vectors of the two photons are correlated due to conservation of momentum. The detector D0 is positioned in the focal plane, so that each point inside the focal plane corresponds to exactly one wave vector. If the detector is small enough, this is a rather precise measurement of the photon wave vector.

Now the other entangled photon hits a double slit or some other kind of similar setup like in the paper you quoted. The light hitting this double slit is alone not coherent enough to show an interference pattern as there are plenty of different wave vectors arriving. However, if we detect a photon at the other detector, we measured the wave vector and therefore the wavevector of the other photon is pretty well defined due to conservation of momentum. So the subset of these joint detections has a clearly defined wave vector and therefore there will be some kind of interference effect in the coincidence counts.

However, to actually see an interference pattern in the coincidence counts at D0, you need a rather large spread of wave vectors, as every position in the plane corresponds to one certain wave vector. So the more wave vectors you include, the better will the visibility of your interference pattern be.

Now one sees that finding interference in the usual double slit needs a small spread in the wave vectors (which can be achieved by using a huge distance between light source and double slit) and finding interference effects in coincidence counting experiments needs a large spread in the wave vectors (which can be achieved by using a small distance between light source and double slit). As you can't have a small and a large distance simultaneously, both kinds of interference are complementary, so you can't have both at the same time with full visibility.

 

[blandrew]

Sorry in advance if this is an observation already covered but. In reference to the Delayed choice quantum eraser experiment http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser What would happen if BSa/BSb were moved so that the time taken for the idler photon to reach BSa/BSb was greater than the time taken for the signal photon to reach D0? i.e. the "choice" is made after the detection of the signal photon. (I'm also making the assumption that the data from D0 is not viewed until after the data from D1/D2/D3/D4)

 

[Cthugha]

What would happen if BSa/BSb were moved so that the time taken for the idler photon to reach BSa/BSb was greater than the time taken for the signal photon to reach D0? i.e. the "choice" is made after the detection of the signal photon. (I'm also making the assumption that the data from D0 is not viewed until after the data from D1/D2/D3/D4)

Although Wikipedia is usually not a really reliable source, your question is already answered in the article you quote. It does not matter, whether the path to D0 is shorter or longer than the other path. You will get the same results in both geometries. Additionally it does not even matter, whether you have a look at the data from D0 before the idler photon arrived at the other detector. You never get any information from looking at D0 alone.

 

[blandrew]

Although Wikipedia is usually not a really reliable source, your question is already answered in the article you quote. It does not matter, whether the path to D0 is shorter or longer than the other path. You will get the same results in both geometries. Additionally it does not even matter, whether you have a look at the data from D0 before the idler photon arrived at the other detector. You never get any information from looking at D0 alone.

If I've understood the experiment correctly, the idler photon is detected 8ns after the signal photon. However, there is no mention of the time at which the idler photon passes through BSa/BSb(the apparatus that give the idler a 50/50 which path/hidden path status). I have made the assumption that the transition through BSa/BSb is before the signal photon is detected. I may be wrong but I believe there is an important distinction between the idler passing through BSa/BSb before the signal detection and passing through after the signal detection. That is to say, if the idler photon passes through BSa/BSb after the signal photon is detected and D0 shows an interference pattern or a cluster pattern in keeping with the "which path" or "hidden path" results, then wouldn't this indicate something very strange happening? Again, apologies if I'm missing something obvious.

 

[Cthugha]

That is to say, if the idler photon passes through BSa/BSb after the signal photon is detected and D0 shows an interference pattern or a cluster pattern in keeping with the "which path" or "hidden path" results, then wouldn't this indicate something very strange happening? Again, apologies if I'm missing something obvious.

Well I don't know, how well you know the experiment, so let me start by asking you a question about the most important aspect:

Is it clear to you, that you will NEVER see an interference pattern on D0?

 

[blandrew]

Well I don't know, how well you know the experiment, so let me start by asking you a question about the most important aspect:

Is it clear to you, that you will NEVER see an interference pattern on D0?

If the signal photons that are paired with the idler photons that got diverted to the hidden path detector are plotted will they not show interference?

 

[JesseM]

If the signal photons that are paired with the idler photons that got diverted to the hidden path detector are plotted will they not show interference?

As the wikipedia article explains, the total pattern of signal photons never shows interference, it's only when you look at the subset of signal photons whose matching idlers were detected at a particular detector (one of the 'which-path erasing' detectors) that you find an interference pattern. Re-read these parts of the article:

However, it should be noted that an interference pattern can only be observed after the idlers have been detected, when the experimenter plots either the subset of signal photons at D0 that are entangled with idlers that went to the detector D1 (the D0/D1 coincidence count), or the subset of signal photons at D0 that are entangled with idlers that went to the detector D2 (the D0/D2 coincidence count). The total pattern of all signal photons at D0, whose entangled idlers went to multiple different detectors, will never show interference regardless of what happens to the idler photons;[2] one can get an idea of how this works by looking carefully at both the graph of the subset of signal photons whose idlers went to detector D1 (fig. 3 in the http://arxiv.org/pdf/quant-ph/9903047 ) and the graph of the subset of signal photons whose idlers went to detector D2 (fig. 4), and observing that the peaks of the first interference pattern line up with the troughs of the second and vice versa (noted in the paper as 'a π phase shift between the two interference fringes'), so that the sum of the two will not show interference.

and

In the delayed choice quantum eraser, the pattern reappears even if the which-path information is erased after the signal photons hit the primary detector. However, the interference pattern can only be seen retroactively once the idler photons have already been detected and the experimenter has obtained information about them, with the interference pattern being seen when the experimenter looks at particular subsets of signal photons that were matched with idlers that went to particular detectors. The total pattern of signal photons at the primary detector never shows interference, so it is not possible to deduce what will happens to the idler photons by observing the signal photons alone, which would open up the possibility of gaining information faster-than-light (since one might deduce this information before there had been time for a message moving at the speed of light to travel from the idler detector to the signal photon detector) or even gaining information about the future (since as noted above, the signal photons may be detected at an earlier time than the idlers), both of which would qualify as violations of causality in physics. In fact, a theorem proved by Phillippe Eberhard shows that if the accepted equations of quantum theory are correct, it should never be possible to experimentally violate causality using quantum effects,[4] although some physicists have speculated about the possibility that these equations might be changed in a way that would be consistent with previous experiments but which could allow for experimental causality violations.[5][6]

 

[blandrew]

"If the signal photons that are paired with the idler photons that got diverted to the hidden path detector are plotted will they not show interference?"

My conjecture with this question was that they(the D0/D1 and D0/D2 subsets plotted separately)would(was I right?). I was just curious as to what would happen if the experiment was repeated with a minor adjustment. The apparatus dealing with the idler photon being repositioned so that there would be a greater time for the idler photon to reach the beam splitters BSa/BSb than for the signal photon to reach D0.
As the experiment stands(sorry if I'm wrong about this and thankyou for your patience) there is a 8ns delay between the registry of the signal photon and its entangled idler(the idler being detected 8ns after the signal). However, for me, the crucial piece of the apparatus are the beam splitters BSa/BSb as it is here that the "fate" of the idler and its entangled signal photon are decided. So the fact that the idler photon is detected at D1/D2/D3/D4 8ns after the detection of the signal photon at D0 is irrelevant because BSa/BSb are encountered before the signal photon is detected at D0(this is an assumption on my part because I can't find any measurements on this). So, if this experiment was repeated in such a way that the signal photon was detected at D0 before the idler photon encountered BSa/BSb and the data was retrieved and plotted in the exact same way as before(data from D1/D2/D3/D4 was viewed then the data from D0 and the subsets were plotted) would an interference pattern in the D0,D1 and D0,D2 subsets still be seen?

 

[JesseM]

"If the signal photons that are paired with the idler photons that got diverted to the hidden path detector are plotted will they not show interference?"

My conjecture with this question was that they(the D0/D1 and D0/D2 subsets plotted separately)would(was I right?).

Yes, the D0/D1 and D0/D2 subsets individually show interference--I wasn't sure if you understood that the total pattern of photons at D0 does not show interference.

I was just curious as to what would happen if the experiment was repeated with a minor adjustment. The apparatus dealing with the idler photon being repositioned so that there would be a greater time for the idler photon to reach the beam splitters BSa/BSb than for the signal photon to reach D0.

I can't be certain that this is true of the experiments that have been done, but I'd imagine it would be--the diagrams of the setup show the detectors being close to the beam splitters, most of the longer path length for the idlers being prior to the point they reach the splitters. Anyway, I'm pretty sure the prediction of quantum theory would be that whether the signal photons reach D0 before or after the idlers reach the beam splitters shouldn't affect the outcome--either way you'd still see no interference in the total pattern of photons at D0, but if you plotted the subset of signal photons whose idlers when to D1 or D2, you'd see an interference pattern in this subset. Perhaps someone else can verify this?

 

[Cthugha]

Yes, the D0/D1 and D0/D2 subsets individually show interference--I wasn't sure if you understood that the total pattern of photons at D0 does not show interference.

Maybe one should add, that the Gaussian peak at D0 alone is in fact the superposition of both (D0/D1 and D0/D2) interference patterns. They are phase shifted by pi (as can be seen in the Kim paper) and therefore superpose to a simple peak.

I can't be certain that this is true of the experiments that have been done, but I'd imagine it would be--the diagrams of the setup show the detectors being close to the beam splitters, most of the longer path length for the idlers being prior to the point they reach the splitters. Anyway, I'm pretty sure the prediction of quantum theory would be that whether the signal photons reach D0 before or after the idlers reach the beam splitters shouldn't affect the outcome--either way you'd still see no interference in the total pattern of photons at D0, but if you plotted the subset of signal photons whose idlers when to D1 or D2, you'd see an interference pattern in this subset. Perhaps someone else can verify this?

There have been experiments showing this. I will have a look. Maybe I can find them.

However, for me, the crucial piece of the apparatus are the beam splitters BSa/BSb as it is here that the "fate" of the idler and its entangled signal photon are decided.

The beam splitters are not that crucial as you might think. There is no measurement done by splitting the beams. For example you could recombine the beams in a Mach-Zehnder-interferometer like way and thereby destroy the which-way info again. The important point is the measurement.

 

[blandrew]

There is no measurement done by splitting the beams. For example you could recombine the beams in a Mach-Zehnder-interferometer like way and thereby destroy the which-way info again. The important point is the measurement.

Although there is no measurement done at the BS, it is here that the ultimate arrival point of the idler is determined(D1/D2 or D3/D4). If you were to introduce a new piece of apparatus after the BS wouldn't you be merely moving the point of determination?(either forward or backward depending on what the new apparatus does).
I'll try to illustrate my point about the importance(just guess work really) of the position of the BS.

signal}----------------D0
idler }--------BS--------D1/D2 or D3/D4
In this scenario, the signal photon is "aware" of the destination of the idler photon before it(the signal photon) appears at D0. The possible detection points of the idler photon before arriving at BS are D1/D2 or D3/D4. That is to say, after passing BS the destination point of the idler has been decided at either a which-path detector(D3/D4) or a path-hidden detector(D1/D2) and so the signal photon can appear at D0 accordingly.

signal}----------------D0
idler }-------------------BS--------D1/D2 or D3/D4
In this scenario, the signal photon cannot be "aware" of the destination of the idler photon as it has yet to pass the BS. So I guess my questions are:
1) Do the D0/D1 and D0/D2 subsets in the second scenario still display interference?
2) HOW?

I'm using the word "aware" to replace intanglement phenomena which I am unfamiliar with. I'm also perposely ignoring the retro-active input of an observer(if it's relevant). Please pick me up on either of these.

 

[Cthugha]

Although there is no measurement done at the BS, it is here that the ultimate arrival point of the idler is determined(D1/D2 or D3/D4). If you were to introduce a new piece of apparatus after the BS wouldn't you be merely moving the point of determination?(either forward or backward depending on what the new apparatus does).

Not necessarily. The important point is, that there is the possibility to be unable to determine which-way information. For example:

signal}------------D0
idler}------BS-----------\
| |
| |
\------------BS----D1
|
|
D2

In this setup, you split the beam, but destroy the which-way information again. In this kind of setup there will be interference in D0/D1 and D0/D2 as long as you do not add another measurement, which determines, which way a photon took.

I'll try to illustrate my point about the importance(just guess work really) of the position of the BS.

signal}----------------D0
idler }--------BS--------D1/D2 or D3/D4
In this scenario, the signal photon is "aware" of the destination of the idler photon before it(the signal photon) appears at D0. The possible detection points of the idler photon before arriving at BS are D1/D2 or D3/D4. That is to say, after passing BS the destination point of the idler has been decided at either a which-path detector(D3/D4) or a path-hidden detector(D1/D2) and so the signal photon can appear at D0 accordingly.

signal}----------------D0
idler }-------------------BS--------D1/D2 or D3/D4
In this scenario, the signal photon cannot be "aware" of the destination of the idler photon as it has yet to pass the BS. So I guess my questions are:
1) Do the D0/D1 and D0/D2 subsets in the second scenario still display interference?
2) HOW?

1) Yes!
2) Well remember, that this is two-photon-interference. In usual interference, you have a superposition of two probability amplitudes, which lead to the same outcome, but cannot be distinguished experimentally. The key property here is coherence as only a fixed phase relationship over time spans, which are longer than the difference in travel time of the two alternative paths, assures that there is no possibility to determine which way information.

In two-photon-interference, the situation is more or less the same, but the coherence properties are different. Both beams are pretty incoherent for themselves. That is a reason, why there is no interference pattern looking just at D0. Single beam coherence is lost pretty fast, but - due to the entanglement - the two photon state is pretty coherent. So in this case you have also the interference of two probability amplitudes - the two paths the photon can take to D1 or D2, but only the coherence time of the two-photon state is long enough to show this effect.
The key is really, that you will not get right results, if you think of separate photons. As soon as you think of two-photon states, the situation gets much easier to handle.

 

[blandrew]

signal}------------D0
idler}------BS-----------\
| |
| |
\------------BS----D1
|
|
D2

sorry but I can't understand this schematic.

In two-photon-interference, the situation is more or less the same, but the coherence properties are different. Both beams are pretty incoherent for themselves. That is a reason, why there is no interference pattern looking just at D0. Single beam coherence is lost pretty fast, but - due to the entanglement - the two photon state is pretty coherent. So in this case you have also the interference of two probability amplitudes - the two paths the photon can take to D1 or D2, but only the coherence time of the two-photon state is long enough to show this effect.
The key is really, that you will not get right results, if you think of separate photons. As soon as you think of two-photon states, the situation gets much easier to handle.

So would it be true to say that both the signal and idler probability waves run through both sides of the aparatus?(the signal path and the idler path), is this what you mean by entanglement coherence? And that from the photons point of view it appears at the signal and idler destination at the same time? So from an observers point of view the coincidence counter would(in this case) show a 8ns delay between D0 and D1/D2/D3/D4, whereas from the photons point of view, no such delay ever existed?

 

[blandrew]

Can I ask what is essentially my main question about the delayed choice theory in a much more simplistic way?

If the decision to view or erase the which-way data is made after the photon, from the observers point of view, has arrived at the end location. Will the results be the same as if we were viewing(or not viewing) the which-way data live? That is to say, will an interference pattern be observed in the subsets that have had their which-path data erased without being viewed? I'm not asking how or why at this point.

I'm really sorry if this question crops up often and has been dealt with before. So again, to all, thanks for your patience.

 

[Cthugha]

signal}------------D0
idler}------BS-----------\
| |
| |
\------------BS----D1
|
|
D2

sorry but I can't understand this schematic.

Oh, sorry. I missed, that empty spaces are just deleted. Maybe this looks better:

signal}------------D0
idler}------BS-----------\
...|....|
...|....|
...\------------BS----D1
........|
........|
.......D2

The dots are just representing empty space. At D1, there is no which-path information present.

If the decision to view or erase the which-way data is made after the photon, from the observers point of view, has arrived at the end location. Will the results be the same as if we were viewing(or not viewing) the which-way data live? That is to say, will an interference pattern be observed in the subsets that have had their which-path data erased without being viewed? I'm not asking how or why at this point.

There will be an interference pattern, if you erased (for example like in the scheme above) the which-path information without measuring the which-path information even after the first of the two photons arrived at the end location. There will be no interference pattern, if you measure the which-path information.

 

[blandrew]

Thanks to everyone for the responses.

 

[ManyNames]

I was reading about the Delayed Choice Quantum Eraser experiment here:

bottomlayer.com/bottom/kim-scully/kim-scully-web.htm

In this experiment the which-path info is erased or not at random AFTER the original (signal) photon hits the screen. The results of the screen are seen by the experimenter AFTER the choice to erase or not took place.

I wonder what happens If the experimenter looks at the screen after the photons hit the screen but BEFORE the choice to erase the which-path is made? That is to have the idler photons still traveling and not reach yet the mirror that has a 50-50 chance of erasing the info.

Any ideas?

I may not know a lot, but i excell in my understanding of certain subjects.

The state of a system, unless collapsed into a value through decoherence, is found to be in a mixed state of probability. The best way to imagine this is by understanding that a photon doesn’t travel across the galaxy via one course or path. It actually travels through all of the potential paths, and out of which one upon measurement creates a defined history for that photon.

Wheelers Delayed-Choice Experiment was first proposed as a thought experiment by John Archibald Wheeler in 1978, and it was a variation of the famous Thomas Young’s Double Slit Experiment, in which the detector screen that ‘’picks up the presence of the photon can be removed at the very last moment according to the observer who measures the experiment.
It is a choice however that is made after the photon has passed the slit, and it could have traveled as a particle, or it could have traveled as a wave and we also find it could have traveled in many varied paths regardless of the instrument being used.
However, the instrument could be set up so that we couldn’t determine which path the photon had arrived, then the two paths are superpositioned together – and just as found in the Double Slit Experiment, the two waves are found to interfere with each other so that the final state is different to what would be expected without that slight change in the instrument being used.

In short, it is we who decide, by our choice of how the set-up is performed whether the photon traveled path A or path B or both A and B. It is the very last moment to decide whether to make that slight change that determines the past history of the quantum system… The experiment was verified in 1985 at College Park by three physicists by the names of Carroll Alley, William Wickes and Oleg Jakubowicz – and confirmed again later by Yoon-Ho Kim in 2000, using another advanced experiment called delayed-quantum eraser, verifying a delayed choice and proving backwards through time travel.

 

[Cthugha]

In short, it is we who decide, by our choice of how the set-up is performed whether the photon traveled path A or path B or both A and B. It is the very last moment to decide whether to make that slight change that determines the past history of the quantum system… The experiment was verified in 1985 at College Park by three physicists by the names of Carroll Alley, William Wickes and Oleg Jakubowicz – and confirmed again later by Yoon-Ho Kim in 2000, using another advanced experiment called delayed-quantum eraser, verifying a delayed choice and proving backwards through time travel.

If you think these experiments show time travel, you seriously misundersand them.

 

[blandrew]

Cthugha, not wanting to re-tread old ground but. Has an experiment actually been conducted where the decision to view or not of which-way path data and the plotting of a distribution graph have been done after all particles(entangled or otherwise) have(from the observers point of view) reached their targets? Is it even practical to try to conduct such an experiment? If I've understood the Delayed choice quantum eraser experiment correctly, the "choice" to have or not have the which-way path data is decided(by the aparatus) before the experiment has reached it's natural conclusion(i.e. before the idler is detected). I think what I'm trying to say is, in the DCQE it could be argued that the wave function is re-instated before the conclusion of the photons detection(if we assume that the photon is not fully detected until both entangled parts are). So isn't there a difference between that and an experiment where the photon, either as a whole or as an entangled pair, are/is detected before any "choice" and subsequent plotting of a graph is made?

 

[Cthugha]

Well, in fact, in almost any experiment you will only have a look at the distributions after the measurement has finished. But the choice, whether you look at the which-way information or not does not matter of course as you already measured this information.
As soon as both photons are detected, the experiment is done and you either have which-way information or you don't. Of course you can now have a look at the which-way information or you can put it aside. However at this moment, you do not have any choice left, whether there is which-way information or not as there was already a measurement of which-way information (or not). You can't undo this measurement.

The delayed choice experiment is not about the choice, which set of data you look at. It is about the kind of data you measure. And obviously therefore it is not possible to choose after both photons are detected.

 

[blandrew]

So if a basic double slit experiment was carried out ten times and data from the slits and the backplate was collected but not viewed. Then after the apparatus was dismantled the ten pairs of data were split into two sets. With the first set we view the slit data, the second set we erase the slit data. Then we look at the backplates, will the first set show non-interference clusters and the second set show interference clusters? Or is that an impossible(or even pointless) experiment to do?

 

[Cthugha]

I think, I do not understand exactly, what exeriment you are trying to do.

Do you mean, that you want to repeat a double slit experiment 10 times and you will 5 times block one of the slits (giving you which-path information) and you will have no block in front of the slits the other 5 times (giving you an interference pattern)?

 

[blandrew]

I think, I do not understand exactly, what exeriment you are trying to do.

Do you mean, that you want to repeat a double slit experiment 10 times and you will 5 times block one of the slits (giving you which-path information) and you will have no block in front of the slits the other 5 times (giving you an interference pattern)?

No, here is a link to what I am talking about. I'm just trying to fathom if what this lecturer is talking about has actually been varified or if he's misunderstood the DCQE experiments.

http://uk.youtube.com/watch?v=QBOaXcG3sJ0

 

[blandrew]

http://www.bottomlayer.com/bottom/kim-scully/kim-scully-web.htm

This article has the name Ross Rhodes associated with it, he is also in the next link giving a lecture about the double slit experiment and specifically about delayed choice. Can anyone verify his commentary of the Delayed choice quantum eraser experiment? Do the article and the lecture on youtube amount to a consensus about what happened in the DCQE?

http://uk.youtube.com/watch?v=QBOaXcG3sJ0

 

[JesseM]

http://www.bottomlayer.com/bottom/kim-scully/kim-scully-web.htm

This article has the name Ross Rhodes associated with it, he is also in the next link giving a lecture about the double slit experiment and specifically about delayed choice. Can anyone verify his commentary of the Delayed choice quantum eraser experiment? Do the article and the lecture on youtube amount to a consensus about what happened in the DCQE?

http://uk.youtube.com/watch?v=QBOaXcG3sJ0

I'm not sure the guy on the video is Ross Rhodes--go to around 4:06 in the video, he says "If Mr. Rhodes is right about his theory, his crazy theory, what should we see at the back wall?" Maybe he's just talking about himself in the third person though, I don't know. (edit: never mind, I see from his website that it is him) Anyway, he seems to be saying that the total pattern of hits on the back wall would depend on whether or not you chose to look at or erase information from detectors at the slits, which is definitely not correct. If there's any time when the photons interacted with a device that could have told you which slit they went through, then even if you later make it so this information is unrecoverable--"erase" it--the total pattern of photons on the back wall won't show interference, although you may be able to find interference patterns in subsets of these photons when you do some kind of coincidence count.

 

[blandrew]

No, the guy on the video isn't Ross Rhodes, he's just describing his understanding of the Ross Rhodes article. Go to around 4:06 in the video, he says "If Mr. Rhodes is right about his theory, his crazy theory, what should we see at the back wall?" Anyway, he seems to be saying that the total pattern of hits on the back wall would depend on whether or not you chose to look at or erase information from detectors at the slits, which is definitely not correct. If there's any time when the photons interacted with a device that could have told you which slit they went through, then even if you later make it so this information is unrecoverable--"erase" it--the total pattern of photons on the back wall won't show interference, although you may be able to find interference patterns in subsets of these photons when you do some kind of coincidence count.

My mistake, I assumed the guy speaking was Rhodes because at the beginning of each clip it says "presentation by Ross Rhodes" I just thought he was talking about himself in the third person.

So the measurement itself collapses the wave function and not observation of the measurement?

 

[JesseM]

My mistake, I assumed the guy speaking was Rhodes because at the beginning of each clip it says "presentation by Ross Rhodes" I just thought he was talking about himself in the third person.

See the edit I made to my post above, it is him after all.

So the measurement itself collapses the wave function and not observation of the measurement?

You should be able to model the interaction with the measuring device in terms of entanglement, without any "collapse" of the wavefunction happening at the moment the device interacts with the particle. But the entanglement changes the probability distribution for the particles hitting the back wall when you do measure them, in such a way that the probability distribution for these particles shows no overall interference pattern (but there is still interference in the joint probabilities for combinations of outcomes when you measure all the parts of the entangled system, like with coincidence counts of both signal photons and idlers in the delayed choice quantum eraser).

Of course, the fact that entanglement creates a sort of pseudo-collapse is one of the reasons its tempting to think that all measurements are just creating entanglements between measuring-device and system being measured, with no "collapses" at all--this would lead you to the many-worlds interpretation. There are conceptual problems with explaining how this picture of an eternally-evolving universal wavefunction is connected to the subjective appearance of distinct probabilistic outcomes of measurements, though.

 

[Cthugha]

Do the article and the lecture on youtube amount to a consensus about what happened in the DCQE?

http://uk.youtube.com/watch?v=QBOaXcG3sJ0

This is what I supposed, when I read your last post. This guy is either a crackpot or wants to be completely misunderstood. The choice in the delayed choice is not about us looking at the data (or part of the data) or not. The choice happens between the first and the second detection of a photon. After both photons are detected, the measurement is done and all choices have been made.

edit: Wow, I am typing really slow today. JesseM was faster.

 

[Cthugha]

As this experiment seems to be the source of quite some confusion let me explain a simple model I sometimes present students, who want to know, what happens in DCQE experiments.

You can treat this experimet as if every photon here originates from either point A or B (following the pictures in the DCQE paper by Kim). Now the setting is as following:

a) In entangled photon experiments each photon on its own behaves like incoherent light. This means, there is no or at least not much correlation between the phases (\phi_1 and \phi_2) of the fields originating from points A and B at the same moment. For the sake of simplicity I will now treat the phase of these fields as completely random and the amplitude as constant and equal.

b) The two-photon state has a well defined phase. This means that the fields of both paths (signal and idler), which originate from the same point (A or B) have a fixed phase relationship. For the sake of simplicity I will now assume, that the initial phase is the same for signal and idler.

Let's first consider, what happens at the detector D0, which scans the x axis. Just like in an usual double slit experiment you will not detect any photons, if there is destructive interference and will detect a large number of photons, if there is constructive interference. Each point P on the x axis, which the detector scans, corresponds to a certain path difference between the paths from A to P and B to P, which can be expressed in terms of an additional phase difference \Delta \phi. So you will have constructive interference at a point if \Delta \phi +(\phi_2 -\phi_1) = 2 \pi. As \Delta \phi is a constant for each point this means that a detection at a certain point P means, that the phase difference between fields at point A and B had a fixed value \phi_2 -\phi_1 at a certain time short before. So in fact, scanning the x-axis means scanning the phase difference of the fields. As the fields change completely random and independent of each other, each phase difference will be realized and there will be no interference pattern at D0 alone.

Now let's have a look at the other side. There are two detectors, which both fields can reach and two detectors, which can only be reached by one field. Let me explain D1 first. I will assume that the distances from A and B to each of the detectors are equal. Before the field originating from A reaches the detector, it crosses 2 beam splitters (no reflection) and a mirror. This influences the phase of the field. Assuming that the beam splitters are 50-50 each transmission changes the phase by \frac{\pi}{2} and each reflection changes the phase by \pi. So summarizing the phase of the field originating will be \phi_1+\pi(reflection at the mirror) + 2*\frac{\pi}{2} (transmission at the beamsplitters. The field reaching D1 from point B is reflected twice and transmitted once, so the phase will be \phi_2 + 2*\pi+\frac{\pi}{2}, so the phase difference at the detector will be \Delta \phi=\phi_2+2*\pi+\frac{\pi}{2}-(\phi_1+\pi+2*\frac{\pi}{2})=\phi_2-\phi_1+\frac{\pi}{2}. Of course a detection implies again, that there is no destructive interference and most detections will occur, if \phi_2-\phi_1-\frac{\pi}{2}=2 \pi.

Now let's check the other detector. Here the field originating from A is reflected twice and transmitted once and the field originating from B is transmitted twice and reflected once, which leads to a phase relationship of \Delta \phi = \phi_2 + \pi + 2* \frac{\pi}{2}-(\phi_1+ \frac{\pi}{2}+2*\pi)=\phi_2-\phi_1-\frac{\pi}{2}. So the \Delta \phi at the two detectors are exactly \pi out of phase. This means that constructive interference on one detector at some certain phase difference automatically implies destructive interference at the other. So each detector selects a set of phase differences. Let me once again stress that the phases are completely random, so there will be no interference on these detectors either. The detectors D§ and D4 are simpler. As there is only one field present, there will be no interference and the phase does not matter. The detections will be independent of \phi_1 and \phi_2.

Now we're almost done. Now we have to consider the two-photon state, where the relative phases are not random anymore. As I stated before, a certain spot on the x-axis of D0 corresponds to a certain phase difference \phi_2 - \phi_1. This very same phase difference will also correspond to a certain amount of constructive (or destructive) interference on D1 and consequently also (due to the different geometries concerning transmission and reflection mentioned above) to an equivalent amount of destructive (or constructive) interference on D2. So you will see this interference pattern in the coincidence counts of D0/D1 and D0/D2 due to the fixed phase difference of the two photon state. Now it is also clear, why there is no interference pattern if you have which-way information. If you have which-way information, there is just one field present, which has a random phase. There is no interference pattern present, which corresponds to the phase difference, which is present at D0 and therefore no interference pattern can show up in the coincidence counts.

I hope this simplified scheme shows, why the choice between the interference pattern and the which-way information can be done after the signal photon has already been detected, why it does not depend on whether we have a look at the data or not and that there are no problems with causality.

 

[blandrew]

Thankyou Cthugha, I'll go away and digest this :)

 

[San K]

In two-photon-interference, the situation is more or less the same, but the coherence properties are different. Both beams are pretty incoherent for themselves. That is a reason, why there is no interference pattern looking just at D0. .

A single photon beam is coherent, but a two-photon (entangled) beam, for themselves, is not?

can we make the two-photon beam, for themselves, coherent? and then repeat the experiment? would we then get interference pattern at D0?

 

[guillefix]

So basically the coincidence counter is needed because the two beams have a phase difference of pi, resulting in no interference. Is it impossible to find a configuration in which both paths have a phase difference of 2*pi or a multiple of it?. Because if there is such a configuration we would find interference when the photon is detected D1 or D2 without a coincidence counter. Wouldn't then be possible to have FTL communication if we find a way to decide when the photon is detected at D3/D4 or D1/D2?

 

[ijdavis]

A simple question I have never got a reply to from any quantum physicist I have ever asked. Take the standard delayed choice quantum eraser experiment. Send entangled photons in a stream through it. Interrupt the stream with pauses in transmission at regular intervals for synchronisation between sender and receiver. Receiver converts the experiment to have mirrors instead of glass which electronically can be raised to reflect light or lowered so that light goes through the now "virtual" glass. The mirrors are positioned in the pauses between the stream of entangled photons. Connect the governance of the mirrors to a computer which causes path indeterminate to signal a 0 and path known to signal a 1. Given enough photons within each burst why can one then not read a message sent from the future as in this burst I observe diffraction (a 0) and in this burst I do not observe diffraction (a 1)? If it is impossible to send a message back in time, how is the thought experiment broken? The original experiment used mirrors, a stream of photons, and I can certainly block this stream to achieve synchronization ensuring sender and receiver are dealing with the same bit logic. Is it the movement of the mirrors in the pauses which somehow breaks the entanglement? or what gives here?

 

[msumm21]

Could you provide a link or something to the particular eraser experiment you are referring to and specify where you want to place mirrors, ... within that setup?

 

[DrChinese]

A simple question I have never got a reply to from any quantum physicist I have ever asked. Take the standard delayed choice quantum eraser experiment. Send entangled photons in a stream through it. Interrupt the stream with pauses in transmission at regular intervals for synchronisation between sender and receiver. Receiver converts the experiment to have mirrors instead of glass which electronically can be raised to reflect light or lowered so that light goes through the now "virtual" glass. The mirrors are positioned in the pauses between the stream of entangled photons. Connect the governance of the mirrors to a computer which causes path indeterminate to signal a 0 and path known to signal a 1. Given enough photons within each burst why can one then not read a message sent from the future as in this burst I observe diffraction (a 0) and in this burst I do not observe diffraction (a 1)? If it is impossible to send a message back in time, how is the thought experiment broken? The original experiment used mirrors, a stream of photons, and I can certainly block this stream to achieve synchronization ensuring sender and receiver are dealing with the same bit logic. Is it the movement of the mirrors in the pauses which somehow breaks the entanglement? or what gives here?

Welcome to PhysicsForums, ijdavis!

Just to be clear: Nothing Alice does will change the pattern Bob sees. Period. The changes are only evident when Alice and Bob's results (one the message sender, the other the receiver) are brought together.

http://www.fsc.ufsc.br/~lucio/2003-07WalbornF.pdf

See p. 342 for a discussion, which pretty much mimics the above.

 

[ijdavis]

http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser
http://en.wikipedia.org/wiki/File:Kim_EtAl_Quantum_Eraser.svg

The mirrors I am going to have mechanically raised or lowered under a computers control, itself doing so only in the pauses between transmission of entangled photons are BSa, BSb, and BSc (all local to me). When disengaged from the incoming ray of entangled photons the light path will be straight as shown. When engaged the light path will be as reflected in the diagram by the partially reflective glass.

To send a non-diffraction pattern backwards in time to D0 I will raise mirrors BSa and BSb to reflect light from both the red and blue path so that I can detect which-path information. To produce a diffraction pattern I will lower BSa and BSb and coerce the red and blue paths to arrive in phase synchronisation having traveled exactly the same distance at detectors D1 or D2. Even if I am unable to replace the half reflective glass at BSc I should still produce a very different signal at D0.

 

[ijdavis]

Thanks Dr Chinese. I will read the paper with interest. But my understanding (perhaps wrong) of the Delayed Choice Quantum Eraser experiment is that the coincidence counter is used to divide into two inputs at D0 those entangled photons whose twin arrives at D3 or D4, from those that arrive at D1 or D2. When so correlated the two now distinct input streams at D0 differ in that one produces a diffraction pattern when sent to a screen while the other does not.

However, if one can force all entangled twins to arrive at D3 or D4 it seems one can force a result at D0, which no longer requires any correlation to be performed. So another way of asking my question is to ask what is to stop us forcing the two potential paths to arrive either at one of D3 or D4, or alternatively to necessarily arrive at one of D1 or D2.

The randomness in the constructed experiment which requires the use of the coincidence counter seems to be itself erased if mirrors are used in place of partially reflective glass.

 

[DrChinese]

To send a non-diffraction pattern backwards in time to D0 I will raise mirrors BSa and BSb to reflect light from both the red and blue path so that I can detect which-path information. To produce a diffraction pattern I will lower BSa and BSb and coerce the red and blue paths to arrive in phase synchronisation having traveled exactly the same distance at detectors D1 or D2. Even if I am unable to replace the half reflective glass at BSc I should still produce a very different signal at D0.

Nothing ever changes at D0! By experimental design, every entangled pair produced yields a click at D0. That click is used to form a coincidence with D1, D2, D3 or D4. Not a lot of message capability in that.

 

[DrChinese]

Thanks Dr Chinese. I will read the paper with interest. But my understanding (perhaps wrong) of the Delayed Choice Quantum Eraser experiment is that the coincidence counter is used to divide into two inputs at D0 those entangled photons whose twin arrives at D3 or D4, from those that arrive at D1 or D2. When so correlated the two now distinct input streams at D0 differ in that one produces a diffraction pattern when sent to a screen while the other does not.

However, if one can force all entangled twins to arrive at D3 or D4 it seems one can force a result at D0, which no longer requires any correlation to be performed. So another way of asking my question is to ask what is to stop us forcing the two potential paths to arrive either at one of D3 or D4, or alternatively to necessarily arrive at one of D1 or D2.

The randomness in the constructed experiment which requires the use of the coincidence counter seems to be itself erased if mirrors are used in place of partially reflective glass.

Yes, we can force all photons to arrive at D3 or D4. How does that change what arrives at D0? The diagram indicates D0 gets everything. So nothing ever changes there. Same number of photons per second regardless of what you do on the other side.