Quantum Erasure 9903047 - Issues

[zekise]

The Quantum Erasure experiment by Scully et al http://arxiv.org/abs/quant-ph/9903047 appears to have generated more questions than it has settled. Please look at attached figure below, which I believe is isomorphic to their quantum erasure experiment Fig. 2. Photons A and B are entangled. Path A << Path B and registration on D1, D2, D3, and D4 is significantly delayed after that of D0.
The paper provides 4 patterns obtained on D0, each isolated by coincidence counting with registrations on D1, D2, D3, and D4. D0|D1 (pattern on D0 due to D1 coincidence counting) is a fringe. D0|D2 is an anti-fringe. D0|(D1+D2) is a Gaussian with mean 0. D0|D3 is a Gaussian with mean +d/2 where d is the distance between the two slits. D0|D4 is a Gaussian with mean -d/2.
Quesiton 1: Would the coincidence patterns change if Path A >> Path B ?
Question 2: what is the total pattern on D0, i.e. D0|(D1+D2+D3+D4), when the coincidence registrations are completely ignored.
Is D0|(D1+D2+D3+D4) = D0|(D1+D2) + D0|D3 + D0|D4 ?
If the total pattern on D0 is the sum of the other 4 patterns (i.e. Question 2 is positive), then I think it implies that D0 is NOT a Gaussian because d is not 0. (This will lead to contradictions )
I am sure this issue has been discussed before, and there is something fundamental that I am missing, but I could not find it here or elsewhere. Your insight will be highly appreciated, and thanks in advance.
Zekise

 

[zekise]

bump

ZapperZ and others, this is a basic erasure/entanglement question - please help !

Let me put the question in another way - Assume R0 is the total pattern on D0. What is the graph for R0? Is R0 independent of the setup on the delayed path B?

R0 should be independent of path B. If not, we can for example remove M1 and M2, and pass a message from M1, M2 to D0. This will violate locality if path B >> path A.

How can R0 be independent of the B setup, but at the same time delayed registrations on D1, D2, D3, and D4 can decompose R0 into components that are correlated with the B setup?

I must be missing some fundamental reality about this experiment. Pls. help. Thanks

 

[JesseM]

The Quantum Erasure experiment by Scully et al http://arxiv.org/abs/quant-ph/9903047 appears to have generated more questions than it has settled. Please look at attached figure below, which I believe is isomorphic to their quantum erasure experiment Fig. 2. Photons A and B are entangled. Path A << Path B and registration on D1, D2, D3, and D4 is significantly delayed after that of D0.
The paper provides 4 patterns obtained on D0, each isolated by coincidence counting with registrations on D1, D2, D3, and D4. D0|D1 (pattern on D0 due to D1 coincidence counting) is a fringe. D0|D2 is an anti-fringe. D0|(D1+D2) is a Gaussian with mean 0. D0|D3 is a Gaussian with mean +d/2 where d is the distance between the two slits. D0|D4 is a Gaussian with mean -d/2.
Quesiton 1: Would the coincidence patterns change if Path A >> Path B ?
Question 2: what is the total pattern on D0, i.e. D0|(D1+D2+D3+D4), when the coincidence registrations are completely ignored.
Is D0|(D1+D2+D3+D4) = D0|(D1+D2) + D0|D3 + D0|D4 ?
If the total pattern on D0 is the sum of the other 4 patterns (i.e. Question 2 is positive), then I think it implies that D0 is NOT a Gaussian because d is not 0. (This will lead to contradictions )
I am sure this issue has been discussed before, and there is something fundamental that I am missing, but I could not find it here or elsewhere. Your insight will be highly appreciated, and thanks in advance.
Zekise

I don't know how to actually do the math to derive the answers here, but just based on the principle that entanglement does not allow signalling faster than light, it seems clear that the total pattern at D0 must be a gaussian, because it would certainly be a gaussian if you removed the beam-splitters so that all the idler photons ended up at D3 or D4 and had their which-path information preserved, so if the total pattern at D0 would change with the beam-splitters in place, that would give you knowledge of whether a distant experimenter had removed them or not.

Why do you think that D0 being the sum of the other 4 patterns implies it is not gaussian? The sum of D1 and D2 would be a gaussian since the peaks of one line up with the valleys of the other, and D3 and D4 are already gaussian.

edit: OK, I think I see your question, I didn't catch the part about D0|D3 and D0|D4 having different means. I guess it might be that the total pattern would actually be the "double hump" pattern that's the sum of two gaussians with sufficiently different means (if the difference is small I believe you still get a single hump, but the curve may not qualify as 'gaussian' any more), like in the double-slit experiment when you measure which slit each particle went through. But in that case, presumably the sum of D0|D1 and D0|D2 would also have to be the same pattern.

 

[zekise]

JesseM - thanks for your reply. I am studying it. But another thing comes to mind.

What if D0|D1 fringe and D0|D2 anti-fringe are actually offset by d from one another? That is D0|(D1+D2) is a double-hump Gaussian (which is actually a non-Gaussian, so let's call it a semi-Gaussian) - so that it coincides with D0|D3 and D0|D4, exactly? Therefore, R0 is just one big double-humped semi-Gaussian, no matter what the setup is at delayed interferometer B?

But this still does not resolve the mystery that after registration of photon A, how can photon B, going through beamsplitters, "find the right detector D1, ..., D4", so that a fringe and anti-fringe emerges from coincidence selection of the total pattern R0?

 

[JesseM]

JesseM - thanks for your reply. I am studying it. But another thing comes to mind.

What if D0|D1 fringe and D0|D2 anti-fringe are actually offset by d from one another? That is D0|(D1+D2) is a double-hump Gaussian (which is actually a non-Gaussian, so let's call it a semi-Gaussian) - so that it coincides with D0|D3 and D0|D4, exactly? Therefore, R0 is just one big double-humped semi-Gaussian, no matter what the setup is at delayed interferometer B?

Yeah, whatever the pattern is for D0|(D1+D2+D3+D4), I think it'd have to be identical to D0|(D3+D4) and D0|(D1+D2), since looking at the screen shouldn't give you any information on whether a distant experimenter had removed the beam-splitters (insuring the idler would end up at D3 or D4) or replaced them with mirrors (insuring the idler would end up at D1 or D2).

But this still does not resolve the mystery that after registration of photon A, how can photon B, going through beamsplitters, "find the right detector D1, ..., D4", so that a fringe and anti-fringe emerges from coincidence selection of the total pattern R0?

It depends what kind of explanation you're looking for. Presumably an analysis of the setup using the laws of quantum mechanics would show that when the position of the signal photon at D0 is measured, it alters the wavefunction of the idler in such a way as to insure the correct probabilities (for example, if the signal photon is measured near a peak of the D0-D1 pattern and a valley of the D0-D2 pattern, then the idler should have a much higher probability of ending up at D1 then D2). But if you want a more conceptual interpretation it'd probably depend on which "interpretation" of quantum mechanics you prefer (the Copenhagen interpretation, Bohmian mechanics, the many-worlds interpretation, the transactional interpretation...take your pick).

 

[zekise]

Central Paradox of Entanglement

JesseM, so I think you are saying that the pattern R0 = D0|(D1+D2+D3+D4) is constant, and will not depend on the setup of the delayed interferometer B (DIB). This pattern is a triple-hump gaussian (a left hump due to D3, a higher center hump due to D1+D2, and a right hump due to D4). The distance between the left and right humps of course being d (the distance between the two slits, which could be arbitrarily large. Note - I have ruled out a double hump gaussian for R0, as suggested in a previous post, because if d can be arbitrarily large, I think we will be able to observe fringes in R0.)

But if you want a more conceptual interpretation it'd probably depend on which "interpretation" of quantum mechanics you prefer (the Copenhagen interpretation, Bohmian mechanics, the many-worlds interpretation, the transactional interpretation...take your pick).

I am sticking to Copenhagen as supplemented by the Epistomological (informatic) Interpretation of Zurek.

So are you saying that the idler photon 2 (p2), "knows ahead of time", which detector D1, ..., D4 it will end up with, so its entangled signal photon 1, registers in the proper gaussian/fringe of R0? Or alternately, p2 when arriving at delayed interferometer B will defeat the beamsplitters and go to the correct detector, depending on which gaussian/fringe its entangled twin has registered on D0? I know this is a crass attempt to conceptualize. But do I have any other choice? :grumpy:

Even this explanation has its problems. For example, we can do a delayed choice on p2, and reconfigure DIB. Let us remove M1 and M2. Then if P1 had fallen on the center gaussian which means its path has to be erased, then where is p2 going to fall? It can only fall on D3 and D4 and reveal its path.

I guess we are touching on the central paradox of entanglement. And I am sure there is an explanation. ZapperZ, could you please confirm the identity of pattern R0, and provide us with the necessary Copenhagen hints to this paradox. TIA.

 

[JesseM]

JesseM, so I think you are saying that the pattern R0 = D0|(D1+D2+D3+D4) is constant, and will not depend on the setup of the delayed interferometer B (DIB). This pattern is a triple-hump gaussian (a left hump due to D3, a higher center hump due to D1+D2, and a right hump due to D4).

Why would D1+D2 be a center hump? I think D1+D2 must be the same pattern as D3+D4, because again, you are free to either remove the beam splitters (insuring that every signal photon is paired with an idler that ends up in D3 or D4) or replace them with mirrors (insuring that every signal photon is paired with idlers that end up at D1 or D2) after you have already observed the location of the signal photons. So if the D3+D4 pattern was different than the D1+D2 pattern, this would create the possibility of FTL or backwards-in-time signalling.

I am sticking to Copenhagen as supplemented by the Epistomological (informatic) Interpretation of Zurek.

So are you saying that the idler photon 2 (p2), "knows ahead of time", which detector D1, ..., D4 it will end up with, so its entangled signal photon 1, registers in the proper gaussian/fringe of R0? Or alternately, p2 when arriving at delayed interferometer B will defeat the beamsplitters and go to the correct detector, depending on which gaussian/fringe its entangled twin has registered on D0? I know this is a crass attempt to conceptualize. But do I have any other choice? :grumpy:

Where did you get either of these ideas from what I wrote? What I said was that if the signal photon is detected first, this alters the wavefunction of the idler so that it has the correct probabilities for ending up at D1, ..., D4. It wouldn't be hard to design a simulation of this on an ordinary classical computer, with the program picking the position of the signal randomly with no advance plan for where the idler would end up, and then once the signal photon's position has been selected, the program would take this into account in determining the probabilities that the idler would end up at the different possible detectors.

I don't really understand what you're suggesting in the second version, what does it mean to "defeat the beamsplitter", and why would this be necessary?

Even this explanation has its problems. For example, we can do a delayed choice on p2, and reconfigure DIB. Let us remove M1 and M2.

If you remove M1 and M2, then the idler is guaranteed to go to D3 or D4. But as I said, the pattern for D3+D4 must be identical to the pattern for D1+D2+D3+D4, as well as to the pattern for D1+D2.

Then if P1 had fallen on the center gaussian which means its path has to be erased, then where is p2 going to fall? It can only fall on D3 and D4 and reveal its path.

If by "center guassian" you are suggesting that D1+D2 looks different from the two-hump pattern of D3+D4, I'm confident this assumption is wrong, because it would mean that observing the signal photons could give you precognition about whether M1 and M2 were going to be removed in the future (or at a point with a spacelike separation from your observation).

 

[zekise]

Predestination?

JesseM, thanks for the clarification. I was under the impression that you were saying D0 is a triple-hump. But now I understand you are saying that D1+D2 = D3+D4. Then please confirm for me each of the following a1 thru a6 (a1 already confirmed):

a1) D1+D2 = D3+D4 is a double hump gaussian (DHG)

a2) Pattern D0 is invariant to the setup of delayed interferometer B (DIB), and always registers the same DHG. This is true even for delayed choice experiments.

a3) Pattern D0 remains invariant, even if we remove DIB altogether, and idler p2 never registers.

pa1) Problem with a1) - d the distance between the two slits can be arbitrarily large. Imagine d is increased to be about the size of the standard deviation of the gaussian. Then how can the anti-fringe D2 cancel out the fringe D1, so that we get a DHG and there is no hint of a fringe remaining in D1+D2. When d increases, the small "tail" of the D2 anti-fringe cannot cancel the large center body of the D1 fringe, and there will be a hint of a fringe. (note that if there is any hint of a fringe, then we can measure the frequency, and at the same time remove M1 and M2 in a delayed choice, and also obtain which-way information (WWI) - and thus violate the Principle of Complementarity.) Therefore, D1+D2 is a single center hump.

Note: However, if the standard deviation of D1+D2 increases along with d, then it conceivably can be the case that the anti-fringe tail will actually cancel out the center body of the fringe. The math needs to be done to ascertain this.

a4) assume we remove D1, D2 altogether, and replace mirror Me with detector D5. I can see that we will get a fringe at D0 with half of the registrations. But where is the anti-fringe that we need to keep D0 invariant? How can D5, which is nothing but a screen, generate both a fringe and an anti-fringe?

a5) But there still remains the central paradox. Assume all 3 beamsplitters are 50%. Therefore the probability of p2 registering on D1, D2, D3, and D4 is 25% each and they are independent random variables. Now if you randomly select 25% of registration of the signal p1 on D0, you will get a double hump with 1/4 of the amplitude of D0. You will NOT get a fringe or anti-fringe.

The collapse of p1 will alter the wavefunction of p2. But there is no way that it can tell p2 to go to a certain detector (by defeating the beamsplitter - meaning the photon telling the beamsplitter which way it wants to go).

So how is it that when we do a coincidence of D1 with D0, we get a fringe in the experiment? If p2 is going to randomly choose D1 25% of the time, we will not get a fringe for D0|D1. When p1 registers, neither p1 nor p2 know that this registration belongs to let's say D1. Even if they did know, there is no way to guide p2 to go and hit D1.

JesseM: What I said was that if the signal photon is detected first, this alters the wavefunction of the idler so that it has the correct probabilities for ending up at D1, ..., D4. ... and then once the signal photon's position has been selected, the program would take this into account in determining the probabilities that the idler would end up at the different possible detectors.

You surely are not suggesting that p2 can be guided to hit a desired detector. But that is how I read this. How can you guide p2 to go through a maze of beamsplitters and hit a predetermined detector?

a6) Are you saying that the wavefunction for p2 is pointing to D1, ..., D4 with equal probability. Upon the collapse of p1, depending on where it has registered on D0, the 4 branches of p2 wavefunction coalesce onto one of the branches - namely the branch that probabilistically suits the p1 registration? Please explain. If this is the case, I think I can come up with some delayed choice setup that would result in a contradiction.

 

[JesseM]

JesseM, thanks for the clarification. I was under the impression that you were saying D0 is a triple-hump. But now I understand you are saying that D1+D2 = D3+D4. Then please confirm for me each of the following a1 thru a6 (a1 already confirmed):

a1) D1+D2 = D3+D4 is a double hump gaussian (DHG)

Well, I agree D1+D2 = D3+D4. I don't think D3+D4 will always have a double hump--the sum of two guassians sufficiently close will only have a single hump, I believe--but whatever D3+D4 looks like in a particular setup, D1+D2 should look identical.

a2) Pattern D0 is invariant to the setup of delayed interferometer B (DIB), and always registers the same DHG. This is true even for delayed choice experiments.

Yes, as long as the atoms A and B emitting the photons have the same position relative to each other and relative to the screen, I believe the total pattern on the screen should always be the same, just because the alternative would seem to imply FTL or backwards-in-time signalling.

a3) Pattern D0 remains invariant, even if we remove DIB altogether, and idler p2 never registers.

Sure.

pa1) Problem with a1) - d the distance between the two slits can be arbitrarily large. Imagine d is increased to be about the size of the standard deviation of the gaussian. Then how can the anti-fringe D2 cancel out the fringe D1, so that we get a DHG and there is no hint of a fringe remaining in D1+D2. When d increases, the small "tail" of the D2 anti-fringe cannot cancel the large center body of the D1 fringe, and there will be a hint of a fringe.

How can you be sure of this, without doing any math? For example, perhaps it could be that the the further apart the distance between the emitting atoms A and B, the weaker the interference pattern, and the more D1 will resemble D3 and D2 will resemble D4.

Note: However, if the standard deviation of D1+D2 increases along with d, then it conceivably can be the case that the anti-fringe tail will actually cancel out the center body of the fringe. The math needs to be done to ascertain this.

Yes, I think it's necessary to do the math, but just based on principles like "no FTL signalling in QM" I feel confident that D1+D2 would always look like D3+D4.

a4) assume we remove D1, D2 altogether, and replace mirror Me with detector D5. I can see that we will get a fringe at D0 with half of the registrations.

Even if Me is replaced with a detector D5, that detector could still in principle measure the momentum of incoming photons, and distinguish between ones coming from one beam-splitter vs. another; therefore the which path-information is preserved here, so I would predict no interference in this case. Even if the detector is not designed to specifically measure momentum, my guess is it would in some sense be registered by the environment (including the physical structure of the detector itself, even if it does not record this information in a form we can read); as an analogy, note that if you try to perform the double-slit experiment with electrons and you fail to create an adequate vacuum, the interaction between the electron and the air will destroy interference, even if you don't have the technical ability to reconstruct which slit the electron went through by measuring air molecules.

a5) But there still remains the central paradox. Assume all 3 beamsplitters are 50%. Therefore the probability of p2 registering on D1, D2, D3, and D4 is 25% each and they are independent random variables. Now if you randomly select 25% of registration of the signal p1 on D0, you will get a double hump with 1/4 of the amplitude of D0. You will NOT get a fringe or anti-fringe.

Agreed, if you choose 25% of the signal photons at random, you will simply get a diminished version of the total pattern of signal photons.

The collapse of p1 will alter the wavefunction of p2. But there is no way that it can tell p2 to go to a certain detector (by defeating the beamsplitter - meaning the photon telling the beamsplitter which way it wants to go).

The collapse of each individual signal photon wavefunction can certainly alter the wavefunction of its corresponding idler in such a way as to make it more likely that particular idler will end up at a particular detector, although if you average over many randomly-selected signal photons, the average probability that a randomly-selected member of the corresponding group of idlers will end up at a particular detector should be 25%. I don't see any contradiction here.

If p2 is going to randomly choose D1 25% of the time, we will not get a fringe for D0|D1.

You're not distinguishing between total probability and conditional probability. if you don't include any information about the location of p1, then p2 will choose D1 25% of the time, ie P(p2 ends up at D1) = 0.25; but P(p2 ends up at D1 | p1 was detected at position x on the screen) will not necessarily be 0.25. Only when you average over all possible positions x that p1 could be detected at, weighted by the probability p1 will be detected there, do you know you'll get a probability of 0.25 [in other words, P(p2 ends up at D1 | p1 was detected at position x1 the screen)*P(p1 was detected at position x1) + P(p2 ends up at D1 | p1 was detected at position x2 the screen)*P(p1 was detected at position x2) + P(p2 ends up at D1 | p1 was detected at position x3 the screen)*P(p1 was detected at position x3) + ... = 0.25].

When p1 registers, neither p1 nor p2 know that this registration belongs to let's say D1. Even if they did know, there is no way to guide p2 to go and hit D1.

Well, I don't think there'd be any position on the screen where there'd be a 100% chance the idler ended up at D1--for that to happen, the D0/D2 and D0/D3 and D0/D4 patterns would have to be zero at that position while D0/D1 was nonzero, I don't think that happens anywhere. But certainly for certain positions, the probability p2 hits D2 may be higher than 25%--if you repeat the experiment many times and look only at the subset of trials where p1 hit that position on the screen, you will find that p2 is detected at D1 in more than 25% of of this subset of trials. Are you just confused about some sort of causal explanation of how this could be true, or are you disagreeing that this is what would actually be predicted by the theory of quantum mechanics?

You surely are not suggesting that p2 can be guided to hit a desired detector. But that is how I read this. How can you guide p2 to go through a maze of beamsplitters and hit a predetermined detector?

This question seems too classical to me, you can't just assume p2 has a single definite path while it's traveling through the apparatus, this is equivalent to assuming that every photon in the ordinary double-slit experiment must have definitely traveled through one slit or the other, even when no detector is placed at the slits. In both cases, it's better to think of a wave moving along all paths which can interfere with itself.

a6) Are you saying that the wavefunction for p2 is pointing to D1, ..., D4 with equal probability.

I'm not sure what it means to say the wavefunction is "pointing at" a particular detector prior to a measurement (before the photon has had time to reach any detector, the probability should be close to zero it will be detected at any of them), I would think you can only look at the amplitude squared of the wavefunction at the moment a measurement is made to find the probability the photon will be detected at different detectors at that moment. But I suppose we could say that if the measurement of the idler is made before the measurement of the signal photon, in this case there should be an equal probability it will be detected at any of the 4 detectors (in this case it would be the wavefunction of p1 that is altered by this measurement).

Upon the collapse of p1, depending on where it has registered on D0, the 4 branches of p2 wavefunction coalesce onto one of the branches

I don't think that any measurement of p1 would lead to a 100% chance that p2 would be measured at a particular detector, but what I'm saying is the probabilities of p2 would be altered by the measurement of p1, so it would not necessarily be 25% for each detector any more.

 

[zekise]

Again thanks for the explanation JesseM. What you say makes a lot of sense. You also touch upon the issue of decoherence, so I will take that up in another subthread in this thread as a separate matter. BTW, I find it odd that nobody else has shown an interest in this thread, when the question is so central. I am sure it can't be because the issue is too pedestrian or too uninteresting. Also, I noticed I have been confusing the signal and idler photons as p1 and p2, when I should be designating them as pA and pB (or if you prefer, pS and pI).

Even if Me is replaced with a detector D5, that detector could still in principle measure the momentum of incoming photons, and distinguish between ones coming from one beam-splitter vs. another; therefore the which path-information is preserved here, so I would predict no interference in this case.

a4) But the momentum is complementary to the WWI. So how can this reveal both momentum and WWI at the same time? The D5 detector here acts as an erasure and should not give us WWI. Am I missing something here? Or is it the way it is drawn?

a4.2) Pls. look at diagram 2 enclosed. This shows DIB being replaced by D5, and M1, M2 being replaced by full mirrors. Path B >> Path A. D0 and D5 are both erasures and there is no WWI at D0 or D5. I understand you are saying that in this situation, we will still get a DHG on D0 consisting of a fringe and anti-fringe (and no which-way gaussians).

But what gives rise to the anti-fringe? There is no symmetry here like in diagram 1. However, we must get anti-fringes, or D0 will not be invariant. So where do the anti-fringes come from? Is it that half the time we get fringe, and the other half anti-fringe randomly? Then what accounts for the phase shift in one of the wavefunctions necessary for the anti-fringe? And why don't we get anti-fringes in a standard non-entangled double slit experiment?

[Aside: Is this symmetric interference the preferred way for the wavefunction to collapse in diagram 2 and is this what is meant by einselection (or superselection?) where certain states of the system is preferred and emerges out of the setting to create the classical world?]

a7) Back in diagram 1, we turn that into a Mach-Zehnder interferometer. All we have to do is tweak M1 or M2. What will happen is all pB going to Me, due to interference registered by pA, will end up in D1 and none will end up in D2 (or vice versa). Therefore there can't be any anti-fringes, and D0 will cease to be DHG. And coincidence correlation will show the fringe, but no anti-fringe. What gives?

 

[JesseM]

a4) But the momentum is complementary to the WWI. So how can this reveal both momentum and WWI at the same time? The D5 detector here acts as an erasure and should not give us WWI. Am I missing something here? Or is it the way it is drawn?

What does WWI stand for, besides World War I? ;) since you talk about complementarity I assume you're talking about position, and you're right, I hadn't thought of it before but it seems that if D5 could make its position measurement arbitrarily accurate, at some point the uncertainty in momentum would be large enough that all which-path information would be lost. Hmm, this is a good question...if my argument earlier about no backwards-in-time or FTL signalling makes sense then it still seems impossible that you'd see an interference pattern in the total collection of signal photons at D0, but with only one detector D5 here I don't see how that non-interference pattern could be understood as the sum of two or more subsets which do show interference. But if the which-path information is lost in this setup, doesn't there have to be some kind of interference phenomena in this situation? It's possible that our intuitive notion that "no which-path information -> interference" just breaks down here, I suppose. But I confess I don't know how to resolve this, I wish I knew how to actually analyze these sorts of setups mathematically to see what sort of patterns would be predicted...

a7) Back in diagram 1, we turn that into a Mach-Zehnder interferometer. All we have to do is tweak M1 or M2. What will happen is all pB going to Me, due to interference registered by pA, will end up in D1 and none will end up in D2 (or vice versa). Therefore there can't be any anti-fringes, and D0 will cease to be DHG. And coincidence correlation will show the fringe, but no anti-fringe. What gives?

Another good point. I'm sorry to say your cleverness has got me stumped, hopefully someone on this board knows how to do an explicit quantitative analysis of the delayed choice quantum eraser and interferometer experiments, and can address this...

 

[JesseM]

OK, I just had a new idea about a solution to the problems above. Just as we can imagine replacing D0 with a screen, which is equivalent to varying the position of D0 over different trials and seeing the frequency the signal photon hits it at different locations, so we could also imagine replacing D5 with a screen. In this case perhaps even though the total pattern of signal photons at D0 shows no interference, if you looked at just the subset of signal photons whose idlers ended up at a particular location on the screen replacing D5, then you would see interference in that subset. But what does that mean in the situation where we just have a single detector D5 and no screen? Perhaps in this case you do see interference when you look at the D0/D5 coincidence graph, but this doesn't account for all the signal photons, because many of the signal photons are paired with idlers that simply missed D5 altogether (they would have landed at a different location on the screen if we had put one in the region of D5). Thus the total pattern of signal photons at D0's location could still show no interference, even though the D0/D5 coincidence graph does show interference.

I'm not sure how this would apply to the case of an interferometer, but it strikes me that the photons in an interferometer aren't always going to travel along straight lines, ie if you replaced the detectors in an interferometer with screens you might detect photons at a variety of locations. So again, it might be that if you look at a coincidence graph between idlers that have gone through the interferometer and signal photons detected at D0, you see interference, but the total pattern of signal photons includes many whose idlers didn't hit any of the discretely-placed detectors at all, and thus the total pattern can show no interference.

This is all just speculation, but it seems plausible enough to me. Again, if anyone knows how to analyze these setups quantitatively, any help would be appreciated.

 

[zekise]

WWI == which-way information (i.e. did pB come from C1 or C2).

Note D5 in diagram 2, which like D0 is a CCD array - equivalent to a screen, when it registers a photon, it only obtains momentum (statistically when the fringe emerges, wavelength being momentum). The position on the array D5 signifies the momentum, and not WWI, because WWI is erased. If WWI had not been erased (by D0 let's say), then the position on D5 could signify WWI (i.e. which hump did pB register).

The way I understand diagram 2, is that since WWI is lost, interference will arise in the form of the fringe and anti-fringe. So both D0 and D5 will identically show the usual double hump gaussian, and will be invariant, and will not have any gaussian component (like the old D3 and D4 used to provide). That is, D5 is purely made of a fringe and an anti-fringe (the old D1 and D2) and is a DHG.

However, we cannot descipher the fringes, because our interferometer is not providing us the coincidences to distinguish fringe from anti-fringe. Its as if you performed the Scully Erasure experiment, and then you deleted the file containing the D1 and D2 registrations. So you will lose the ability to decipher the fringes.

The real question (a4) I think is how does anti-fringe arise in this case where there is no symmetry? As you recall, the anti-fringe must arise to make sure that D0 remains DHG and invariant.

... if you looked at just the subset of signal photons whose idlers ended up at a particular location on the screen replacing D5, then you would see interference in that subset. ... Thus the total pattern of signal photons at D0's location could still show no interference, even though the D0/D5 coincidence graph does show interference.

You can assume all detectors are CCD arrays (screens). It should not make a difference in the outcome of the experiment.

When you say we can retrieve the fringe on D0, by coinciding a subset of the registrations on D5 - how do you know which subset? D5 is showing the same DHG as D0. You do not know the frequency of the photon. How are you going to separate the fringe from the anti-fringe? I think this information is lost.

 

[zekise]

JesseM, let me ask you about decoherence. If you want, I can make another thread out of this. Any pointers to reading material is highly appreciated.

Assume we are doing a Zeilinger matter wave experminent shooting flouridated buckyballs C60F48, the largest object yet to be interfered, to a double slit.

Now before the FBB can pass through the double slit, it has to be cohered, otherwise it will not be in a coherent superposition.

So how do you do that? If you are shooting FBBs, you can tell from the recoil of the gun what the momentum of the FBB is, and you know where the gun is, and you can predict the path of the particle. You would know exactly which slit it went through. How do you cohere the beam in the first place, and what does it mean for a single particle to be coherent?

Second, if a coherent FBB hits a molecule of N2 on its way to the slits, but the N2 does not interact with its environment, would the N2 be entangled with the FBB, or are they "independent"? Would FBB be decohered if N2 does not interact with the environment? Does it matter if N2 was coherent or decoherent before FBB hit it?

 

[JesseM]

WWI == which-way information (i.e. did pB come from C1 or C2).

Note D5 in diagram 2, which like D0 is a CCD array - equivalent to a screen, when it registers a photon, it only obtains momentum (statistically when the fringe emerges, wavelength being momentum). The position on the array D5 signifies the momentum, and not WWI, because WWI is erased. If WWI had not been erased (by D0 let's say), then the position on D5 could signify WWI (i.e. which hump did pB register).

OK, I don't know much about the way this sort of equipment works, I was assuming each detector just detects whether the photon arrives at its location or not without any more fine-grained positional information, and that the coincidence graphs were obtained by varying the position of D0 on different trials. Is this incorrect, and are you saying that each detector can measure a range of positions, like a screen? Or does your statement "it only obtains momentum" mean it does not obtain any position information besides the one-bit statement "the photon set off the detector" or "the photon was not registered by the detector"? And even if so, is it possible the the environment would in some sense measure the particle's position in a more fine-grained way through decoherence, or does the detector measure the momentum accurately enough that the uncertainty principle rules out any more detailed information about the photon's position? If you can clear up my confusion about how these types of detectors work, I think I can explain better how the suggestion I made in my last post might resolve the problems you bring up.

 

[zekise]

Well, I am not the expert. But this is how I understand it. The detectors can be CCD arrays that are a 2 dimensional array of cells. They can tell which cell the photon got registered. D0 in the QE experiment is a mechanical contraption that essentially simulates a CCD array. You can assume all detectors are CCD arrays. I don't think this will gain you any extra information, just make your life easier.

The x,y position you get from an array just tells you where on the graph you are for that detector. A single detection does not tell you anything about the fringe yet. However, if the detector is D3 or D4 for example, then a single registration obviously gives you WWI, (and you don't need an array). Using a Mach-Zehnder, with a single photon you can get information if there has been interference or not. Probably a photo multiplier can give you wavelength with a single photon by figuring the energy of the photon.

But let's say you want to measure the fringe. This has to be done statistically. So in a standard double slit experiment, the array receives the stream of self-interfering photons, and the fringe appears on the 2D array. You can then calculate the momentum from the period of the fringe.

What I was saying is that the x,y position of the registration on the Array is not related to the WWI. However, if the photon is not interfering, i.e. somebody knows its WWI, and a stream from a double-slit experiment hits the array, then technically you can figure out the WWI with a degree of uncertainty by looking at the double-hump. Each hump is coming from its corresponding slit, and the more distance between the two slits, the higher the certainty.

Did you see the post I made for decoherence? Any insight on the questions?

 

[zekise]

JesseM, I think I got the answer to the a4 riddle!

The anti-fringe (on D0 and D5) is created by the entangled twin (because its properties are symmetric). That is why in regular double slit experiment (without entangled twin), there is no anti-fringe.

What do you think?

What happens then with triple-entanglement? The local particle creates the fringe, the non-local creates the anti-fringe, and what does the 3rd particle create?

 

[ganstaman]

Even if the detector is not designed to specifically measure momentum, my guess is it would in some sense be registered by the environment (including the physical structure of the detector itself, even if it does not record this information in a form we can read); as an analogy, note that if you try to perform the double-slit experiment with electrons and you fail to create an adequate vacuum, the interaction between the electron and the air will destroy interference, even if you don't have the technical ability to reconstruct which slit the electron went through by measuring air molecules.

I realize that this quote/thread is very old, but I'm curious about this point. Has it actually been verified that performing the double-slit experiment in a non-vacuum leaves you with no interference pattern?

If so, then I have a question about the Kim/Scully delayed-choice quantum eraser set-up that I believe is discussed at the beginning of this thread. If the air molecules 'detect' the electrons as stated above, then why do the beam splitters (that send the photons to either D3 or D4 or D1/D2) not also serve to 'detect' the photons, thus providing which path information for all photons?