Understanding Quantum Eraser Experiments

[alexepascual]

I would like to start a discussion about the most basic principles behind quantum eraser experiments.
I understand this has been debated here many, many times and for some of you there is nothing to talk about, but I still find it interesting and would like to get some of your opinions.
I know there are many different experiments of this type that have been proposed and also realized.
Some of them try to show some new aspect of it and become quite complex, but I would like to focus on the most simple ones. It seems like the basic concept is that if we have two particles which are entangled with respect to some observable, an action on one of them that deletes information will have some non-local implication on the other particle. I understand that depending on your favorite interpretation you may deny non-locality, but at least that's what at first sight it looks like. If you have the two particles with some 2-state observable maximally entangled, which means this is a singlet and you take some action on the particle on the right that will make it impossible to distinguish between these tow states, then in principle it should be possible to observe interference (after sending a few particles of course) on the other side right?. Well, I know you will say "No", this only happens if coincidence counting is done. I wonder how we can model this concept if we send the singlets one at a time, don't loose any, and don't have any noise or non-entangled particles. In real-life we would have to choose a particular type of particle such as a pair of photons produced by down-conversion, some kind of fermion or even atoms, and choose the particular observable that we are entangling. But maybe we can think of a generalized example without getting into the details. My understanding is that in principle what we have is complementarity between "path" distinguishability and interference. But I think you might have different concepts and opinions. If you would prefer to talk about a particular simple model, I am willing to discuss that.

 

[StevieTNZ]

If you have the two particles with some 2-state observable maximally entangled, which means this is a singlet and you take some action on the particle on the right that will make it impossible to distinguish between these tow states, then in principle it should be possible to observe interference (after sending a few particles of course) on the other side right?

I don't understand this part -- action on one particle? What do you mean by that?

 

[bhobba]

It seems like the basic concept is that if we have two particles which are entangled with respect to some observable, an action on one of them that deletes information will have some non-local implication on the other particle.

It doesn't seem like that to me at all.

To me 'it seems' that in some simple cases decoherence can be undone - which of course is exactly what's going in.

Thanks
Bill

 

[zonde]

But maybe we can think of a generalized example without getting into the details.

You can think about generalized example but if you want to check your idea against reality you have to think about it in context of real experiments.

 

[alexepascual]

I don't understand this part -- action on one particle? What do you mean by that?

I didn't say what type of action because I wanted to keep it general, not specific to a particular experiment. But I'll try to explain in more detail what I meant. Interference between two (past) states can only be observed when there is no permanent recording that shows the particle having been being in one state or the other. This would apply to the double-slit experiment. During certain time there may be a way to make a measurement and obtain one of the two states as a result. But if instead of making the measurement we do one of two things 1) let it evolve until we can't distinguish anymore between the two states or 2) do something to erase any information that may be present and which would allow to distinguish in which of the two states the particle was before. Then interference between the two states can be seen.
In the case of the two entangled particles, we can say that each of them carries information about the other. One of the two particles could be manipulated in such a way that there is no way to distinguish between the two states, and then the other particle should be able to display interference. What kind of manipulation I am talking about? The one they do in all of these quantum eraser experiments. In the first thought experiment by Scully, the wall between two cavities was removed, in the one by Dopfer, the detector was placed at a position that would not be able to focus on the slits, in the one by Walborn, a polarizer is inserted. Then there is the one by Yoon-Ho Kim that is more complex as it uses two separate down-conversion sources. And of course there is Wheeler's initial thought experiment which did not involve entangled particles and in which I doubt we could talk about "erasure". We could get into one of these experiments, but I thought that if the principles behind all of them are basically the same, we could make an attempt to discuss those principles without getting into all the details of each experiment. But if you prefer to talk about one in particular, that's OK with me. I hope I made a little more clear what I had in mind during my initial post. If I didn't I apologize.

 

[alexepascual]

It doesn't seem like that to me at all.
To me 'it seems' that in some simple cases decoherence can be undone - which of course is exactly what's going in.
Thanks
Bill

According to the picture presented by Zurek, decoherence would imply an uncontrollable leak of information towards the environment, which would make it hard (perhaps not impossible) to "undo".
I would have to look into each of the experiments, but from what I remember, the only stage at which there was decoherence was when the photons were detected.
What you are saying makes some sense to me, specially when we talk about "erasure", but I don't recall any experiment where this could apply. Do you have a particular one in mind?

 

[bhobba]

According to the picture presented by Zurek, decoherence would imply an uncontrollable leak of information towards the environment, which would make it hard (perhaps not impossible) to "undo".

It has been discussed here before:
https://www.physicsforums.com/threa...and-the-delayed-choice-quantum-eraser.623648/

If you want to understand the exact mechanism in a particular set-up I will have to leave that up to an experimental type - but the principle is well known - in the quantum eraser all that's happening is you have undone decoherence.

Thanks
Bill

 

[alexepascual]

It has been discussed here before:
https://www.physicsforums.com/threa...and-the-delayed-choice-quantum-eraser.623648/

If you want to understand the exact mechanism in a particular set-up I will have to leave that up to an experimental type - but the principle is well known - in the quantum eraser all that's happening is you have undone decoherence.

Thanks
Bill

I know this question is a little off-topic, but as you consider decoherence and it's undoing as the explanation for quantum eraser experiments, do you think decoherence solves the measurement problem?

 

[bhobba]

I know this question is a little off-topic, but as you consider decoherence and it's undoing as the explanation for quantum eraser experiments, do you think decoherence solves the measurement problem?

No.

Here is what it solves and does not solve:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

The issue is the difference between an improper and a proper mixed state.

Thanks
Bill

 

[alexepascual]

No.

Here is what it solves and does not solve:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

The issue is the difference between an improper and a proper mixed state.

Thanks
Bill

I did take a look at the article. So going back to two entangled particles running in opposite directions, how do you apply the concepts of improper and proper mixed states to quantum erasure? According to the article when describing one particle before the other has been measured it would be in an improper mixed state and after the other particle is measured, the first particle is in a proper mixed state if you haven't received information about the result of measurement. But when the other particle, even if it is far away and you can't know what the result of measurement is, you could conclude that it is in a definite eigenstate right?
Now if you describe this in the composite Hilbert space of both particles, it looks like measuring the other particle would "collapse" the wavefunction and not only reduce the density matrix but leave only one of the diagonal elements on it right?
I also don't see how all these ideas conflict with the picture I presented before.

 

[bhobba]

I did take a look at the article. So going back to two entangled particles running in opposite directions, how do you apply the concepts of improper and proper mixed states to quantum erasure.

You will need to go though an actual set-up with an experimental type - which I am not.

But in general when what you observe interacts with what you are observing it with, it becomes entangled with it, and a superposition gets converted to a mixed state. Its turns out in simple cases that entanglement can be undone. So what would happen is if you observe which slit it goes through it becomes entangled with what you are observing it with, and the superposition is now a mixed state. But it is possible to disentangle it so it becomes a superposition again.

Here is a complete analysis of the closely related delayed choice experiment:
http://quantum.phys.cmu.edu/CQT/chaps/cqt20.pdf

Thanks
Bill

 

[alexepascual]

You will need to go though an actual set-up with an experimental type - which I am not.

But in general when what you observe interacts with what you are observing it with, it becomes entangled with it, and a superposition gets converted to a mixed state. Its turns out in simple cases that entanglement can be undone. So what would happen is if you observe which slit it goes through it becomes entangled with what you are observing it with, and the superposition is now a mixed state. But it is possible to disentangle it so it becomes a superposition again.
Here is a complete analysis of the closely related delayed choice experiment:
http://quantum.phys.cmu.edu/CQT/chaps/cqt20.pdf

Thanks
Bill

That chapter you sent me the link to is part of Griffiths consistent histories book. I had read it before, and I always found it interesting.
I was just hoping to have a little debate on this topic. But that's OK. Thanks anyway Bill.

 

[Eye_in_the_Sky]

But maybe we can think of a generalized example without getting into the details.

Ok. So, here's a generalized example.

Consider the state

|ψ> = N ( |a>|b> + |a'>|b'> ) .

The probability density ρ(x) for finding system-1 at the point x, when no measurement is performed on system-2, is given by

ρ(x) = |N|² { |<x|a>|² + |<x|a'>|² + 2Re[<a|x><x|a'><b|b'>] } .

Observe that the interference term (the third term) vanishes when the two system-2 states |b> and |b'> are orthogonal; i.e. <b|b'>=0.

In the words of Zeilinger:

Formally speaking, the states |a>₁ and |a'>₁ ... cannot be coherently superposed because they are entangled with the two orthogonal states |b>₂ and |b'>₂.

Alternatively, one can put it this way:

With no measurement performed on system-2, the 'which-way' information for system-1 remains fully intact (via its entanglement with the two orthogonal states of system-2) and so there is no interference.
_____________________

Let us then set the apparatus up such that |b> and |b'> are orthogonal, in which case we can write

|ψ> = (1/√2) ( |a>|b> + |a'>|b'> ) .

... So how does one completely erase the 'which-way' information for system-1 encoded in system-2?

Perform a measurement of |e> on system-2, where |e> has the property |<e|b>|=|<e|b'>|; that is,

<e|b'> = α<e|b> , where α is a phase-factor.

In that case (whenever the result |e> is obtained), the state |ψ> → |φ> where

|φ> = N ( |a> + α|a'> ) |e> ,

for which the interference characteristics of system-1 are fully manifest.

This is what Zeilinger means by:

... the interference pattern can be obtained if one applies a so-called quantum eraser which completely erases the path information carried by particle 2. That is, one has to measure particle 2 in such a way that it is not possible, even in principle, to know from the measurement which path it took, b or b'.
_____________________

Here are the correspondences for the Walborn setup:

|a> ↔ slit-1

|a'> ↔ slit-2

|b> ↔ left-circular polarization

|b'> ↔ right-circular polarization

|e> ↔ horizontal polarization
_____________________

 

[Eye_in_the_Sky]

Here are the correspondences for the Walborn setup:

|a> ↔ slit-1

|a'> ↔ slit-2

|b> ↔ left-circular polarization

|b'> ↔ right-circular polarization

|e> ↔ horizontal polarization

I see here this correspondence is for a 'simplified' Walborn setup. You can see that setup at the bottom of page 341 [p. 6 in the PDF-viewer] at this link:

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

In this setup, though, one cannot argue for nonlocality.

 

[alexepascual]

I see here this correspondence is for a 'simplified' Walborn setup. You can see that setup at the bottom of page 341 [p. 6 in the PDF-viewer] at this link:
http://www.mat.ufmg.br/~tcunha/2003-07WalbornF.pdf
In this setup, though, one cannot argue for nonlocality.

Hello Eye_in_the_Sky. Thanks for your interesting posts. This morning I sent you an email.
After looking at your post, I grabbed my (hard) copy of Zeilinger's "Experiment and Quantum Physics" which I think is the article you were looking at.
I'll have to read it again. (I have read it many times already but that was a few years ago).
However, what I see is that Zeilinger in page S290 talks about one pair of photons. His interpretation of quantum erasure is very clear. Now, I just read that page and I haven't read yet sections IV and V. But what I get from section III is that his argument, being based on just one pair of photons, would not (at least in principle) require coincidence counting for interference to appear. Of course in the real experiment due to technical reasons, you only see interference when you consider the photons that are entangled and discard all the others. This has been discussed here a lot many years ago and the discussion eventually died down. The arguments that were presented in favor of the idea that coincidence counting is needed in order to observe interference were not convincing to me. Tonight I'll take another look at the other sections of Zeilinger's paper so that I can see if he expresses (at least in that paper) some argument in favor of the need for coincidence counting. I think I have read before that he does think it is necessary.
Now, going to Walborn's experiment, I don't see that one as much different than a typical EPR-Bohm experiment. I think describing a hypothetical experiment of that nature could clear things up. Let's say we have this pair of entangled particles (fermions) where like always one goes to the left (A) and the other to the right (B) and we record the arrival time for each photon. Let's say we measure spin in the vertical direction on A and in the horizontal direction on B. When we look at our measurements on A, we find that half of the time the particle has spin up and half of the time spin down. If we look for interference by recombining the paths in the Stern-Gerlach we don't find any. Same thing happens for particle B being measured in the horizontal direction. But if after the experiment (which was ideal and did only contain entangled photons) we select only those pairs (which we can do because we recorded the time of detection) where the photon B was detected as having horizontal spin pointing to the right (not looking at the paper from the top but looking along the direction of motion of the particle) we do see interference on particle A. Of course , to see interference on A we have to set it up to measure interference and not the up-down value on that particle. Now, if we select the pairs where the spin of B was pointing left, then we also see interference but the interference fringes are displaced with respect to the previous ones. So here there is no demonstration of non-locality (as you pointed out). Now Walborn seems to be saying that the idea that quantum-eraser experiments could be used for faster-than-light (he actually says back-in-time) communication is false and that his experiment confirms it. I think his experiment shows a particular setup in which faster-than-light communication is clearly non-feasible. But this does not prove that in a different type of setup it could not be done. Of course there is Eberhard's theorem, but that's a different story.
As I said before, I will read Zeilinger's paper in more detail and then I'll post again.

 

[Eye_in_the_Sky]

But what I get from section III is that his argument, being based on just one pair of photons, would not (at least in principle) require coincidence counting for interference to appear.

Oh, but we do need coincidence counting.

Let's see what's going on.

At the A-arm of the apparatus, the dots begin to fill the screen one, by one, by one. We keep track of the sequence of dots: dot-1, dot-2, ... dot-N. Once the dots have filled-in sufficiently, we find a bell-shaped distribution which shows no interference at all. This will be the case no matter what actions are performed at the B-arm of the apparatus.

However, if we are told which of the dots on the screen (at the A-arm) correspond to a measurement of |e> (at the B-arm) for which the outcome was "yes", then we do indeed find that those dots taken alone will show an interference pattern.

... Coincidence counting required.

 

[alexepascual]

Oh, but we do need coincidence counting.

Let's see what's going on.

At the A-arm of the apparatus, the dots begin to fill the screen one, by one, by one. We keep track of the sequence of dots: dot-1, dot-2, ... dot-N. Once the dots have filled-in sufficiently, we find a bell-shaped distribution which shows no interference at all. This will be the case no matter what actions are performed at the B-arm of the apparatus.

However, if we are told which of the dots on the screen (at the A-arm) correspond to a measurement of |e> (at the B-arm) for which the outcome was "yes", then we do indeed find that those dots taken alone will show an interference pattern.

... Coincidence counting required.

The picture you are presenting seems to be similar to the one I mentioned in my previous post which was a simplified version of Walborn's experiment. Well, I think it was simplified at least at the conceptual level. Experimentally it would be harder to implement as it uses spin 1/2 particles instead of photons. Using your example, if you select those dots where the outcome is "no" when measuring |e> at the B-arm, do you see interference at the A arm?
So we could keep track of all particles, assign a number to each pair and put them in a database right?
Are you the same Eye I sent the email yesterday? I am wondering because each eye is connected to a different hemisphere in the brain. Perhaps one of your hemispheres is into classical physics and the other into quantum physics. Oh, changing the subject, I just read Zeilinger's section IV. I didn't find new evidence of his denial of the possibility of interference without coincidence-counting (I know that's his position, but at least he is not saying it in that section). What I did see is a train of thought much in sync with the Copenhagen interpretation and against realism. (no mention of non-locality in that section but it may be implied).

 

[Eye_in_the_Sky]

Using your example, if you select those dots where the outcome is "no" when measuring |e> at the B-arm, do you see interference at the A arm?

Yes, you do.

In general, you will see the combined intensities of two patterns:

1) The anti-interference pattern ,

plus

2) A bell-shaped distribution which shows no interference .

The first aspect will always be present.

However, in an experiment for which P("yes")=P("no") (and, hence, both are ½), the second aspect vanishes.

Note that, for any erasing experiment which can be 'mapped' to our generalized example, we must necessarily have P("yes")≤½. This follows from the conditions:

|b> and |b'> are orthogonal ,

|<e|b>| = |<e|b'>| ,

and the original state is

(1/√2) ( |a>|b> + |a'>|b'> ) .
__________________________________

So we could keep track of all particles, assign a number to each pair and put them in a database right?

In the idealized scenario, yes.

 

[alexepascual]

Yes, you do.
In general, you will see the combined intensities of two patterns:
1) The anti-interference pattern ,
2) A bell-shaped distribution which shows no interference .
Note that, for any erasing experiment which can be 'mapped' to our generalized example, we must necessarily have P("yes")≤½. This follows from the conditions:
In the idealized scenario, yes.

Did you really mean P("Yes")≤1/2 or you meant P(Yes)≠1/2?
It looks to me that this scenario is exactly the same as both the Walborn experiment and the one I mentioned with spin 1/2 particles. Basically there are fringes and anti-fringes and when you add them up you get the Bell-shaped curve. In this case it is very clear that coincidence-counting is necessary.
Now, let me make it clear. I am not saying that I am in favor of the idea that a quantum-eraser experiment can be done without coincidence-counting. All that I am saying is that the explanations I have seen (in my opinion) do not prove this concept in a simple and intuitive way. If the need (in a setup such as Dopfer's) for coincidence-counting is true, there should be a simple way to explain it. It should be an explanation as simple as the one you showed or the one I mentioned. But as I said, I think these both refer to a different setup and don't necessarily apply to something like Dopfer's experiment.
I'll try to think about a convincing argument about the fundamental difference between these two experiments and explain it in my next post.

 

[alexepascual]

I'll try to think about a convincing argument about the fundamental difference between these two experiments and explain it in my next post.

OK, here it is. The main difference between the experiments we discussed above and Dopfer's is that in the latter the number of photons that form the interference pattern is the same as the ones that form the Bell shape curve. So we can't recover after-the-fact an interference pattern on the lower arm by selecting a sub-group of the photons according to the results of the measurement done on the upper arm (the Heisenberg lens arm) We can make the detector in the upper arm large enough that no photons will be lost. Now, there is something that complicates the picture. Zeilinger mentions (referring to Klishko) that this can be seen as if the source were the double-slit and the down-conversion crystal acted as a mirror. Now, when the (entangled) photons reach the double-slit, some pass through and the ones that don't go through get absorbed by the material of the double-slit. The "partners" of those photons that didn't go though will still be detected at the upper branch. So when we place the detector at the imaging plane we don't "see" the slits (but I guess if the detector was a very efficient CCD camera we could for each photon know if it's partner went through the slits by seeing at what position in the imaging plane it landed, and even determining through which slit it went. So this should destroy interference at the lower arm (behind the slits). Now if we put the detector at a distance of 2f, at which the image gets totally blurred, we are destroying which-way information and the interference pattern should show behind the slits. Each photon that reaches the detector in this position will be detected at a random position and there can be no inference from this as to which slit the other one went through. Even with the additional non-entangled (I mean non-entangled with the ones that went through the slits) photons in the upper branch, this should not provide any additional information that would allow us to find which-way information. So interference at the lower branch should be seen.
Another thing that distinguishes this experiment from Walborn's or the one you showed is that we don't have an |e> measurement that gives a particular result (in your example it was "yes" or "no". In this case it could be argued that the detector placed at the 2f (blurring) distance is the one that does the |e> measurement because it is at that moment that which-way information is erased. OK, but you don't get "yes/no" or anything like that. The only thing you can measure at that detector is position, and it happens to be random, without providing any additional information (of course the other thing you measure is the time of arrival). And you get the interference pattern at the lower arm without having to select a sub-group of the photons based on the result of the upper branch measurement. Now, you may say that coincidence-counting is what does that selection, but besides filtering-out any "noise" detection events, coincidence counting does not seem to select a particular sub-set of the entangled photons. Unless you say that the arrival time at the upper branch detector is what is in this case represented by your |e>. That could at first sight be a feasible argument, but I doubt it could be made to work.

 

[Cthugha]

The only thing you can measure at that detector is position, and it happens to be random, without providing any additional information (of course the other thing you measure is the time of arrival). And you get the interference pattern at the lower arm without having to select a sub-group of the photons based on the result of the upper branch measurement. Now, you may say that coincidence-counting is what does that selection, but besides filtering-out any "noise" detection events, coincidence counting does not seem to select a particular sub-set of the entangled photons.

That is incorrect, if I understand you correctly. The position at the detector in the Heisenberg lens arm corresponds to a well defined momentum value (if placed at a distance of f). Moving the detector to another position will result in a different momentum value. In coincidence counting, the interference pattern seen will correspond to the photons with this fixed momentum (or rather with a fixed momentum that depends deterministically on the momentum picked in the Heisenberg lens arm). For every momentum you may choose, you will see a different pattern in coincidence counting. Without coincidence counting (or when placing the Heisenberg lens detector at 2f where you have no momentum info), you just get a mixture of many interference patterns, which corresponds just to no pattern at all.

You can get an interference pattern without coincidence counting (see Dopfer's thesis, unfortunately in German only) by placing the non-linear crystal far away from the double slit. However, in that case you have already broken entanglement because the few photons that make it to the double slit are already filtered so strongly in momentum that you cannot violate Bell's inequality anymore.

 

[alexepascual]

That is incorrect, if I understand you correctly. The position at the detector in the Heisenberg lens arm corresponds to a well defined momentum value (if placed at a distance of f). Moving the detector to another position will result in a different momentum value. In coincidence counting, the interference pattern seen will correspond to the photons with this fixed momentum (or rather with a fixed momentum that depends deterministically on the momentum picked in the Heisenberg lens arm). For every momentum you may choose, you will see a different pattern in coincidence counting. Without coincidence counting (or when placing the Heisenberg lens detector at 2f where you have no momentum info), you just get a mixture of many interference patterns, which corresponds just to no pattern at all.
You can get an interference pattern without coincidence counting (see Dopfer's thesis, unfortunately in German only) by placing the non-linear crystal far away from the double slit. However, in that case you have already broken entanglement because the few photons that make it to the double slit are already filtered so strongly in momentum that you cannot violate Bell's inequality anymore.

Hi Cthugha, I recall that we discussed this several years ago. Since then I was busy with other problems in my life and didn't have a chance to look into this in more detail as I would have liked to do. Well, now I am back to this. Your explanation from an abstract point of view seems very reasonable. You are using measurement of momentum as the "eraser" of position information. That makes sense being that position and momentum are complementary. Actually, as I was writing the post you just responded to, that picture crossed my mind and I was trying to remember if I had seen a similar explanation in the original Dopfer thesis. As you said, Dopfer's thesis is in German and as far as I know nobody has translated it. I tried to translate it using the Google translator and another one but only got a small part of it translated. I still have the original as hard copy in my bookshelf and I am going to take another look at it. From what you said I get the impression that you understand German. I even though about studying German at that time when I was translating Dopfer's thesis but I found it hard and it was taking too much time so I didn't continue. If what you are saying is correct as an explanation of this experiment, it was not explained like that by Zeilinger. As I did not have an English version of Dopfer's paper, I had to rely on Zeilinger's explanation which at least in my mind did not provide any convincing proof about the need for coincidence-counting.
Now, even if I think that your explanation makes a lot of sense from a theoretical point of view, I have trouble fitting it in the actual experiment. I will see if I can find something in the original thesis and maybe I'll get some further explanation from you that can help (which I'll appreciate). You may wonder why I have trouble making this concept work in the actual experiment. The first thing that I notice is that you are assuming that placing the detector at a particular position near 1f in the Heisenberg lens arm will select certain fraction of the photons. Let's say that 1/2 of the photons on that arm are selected. So my first reaction would be: Where did the other photons go? Now, as I am writing this I am thinking and I realize you may say that the photons on the upper arm are always the same but that we are only looking at the ones that are in coincidence with the ones at the lower arm. That could make things a little more reasonable, but I still would have to think about it to get a more complete picture. I think we could imagine a situation in which all the entangled pairs where the one in the lower arm hits the material of the double-slit (and does not go through) get discarded. So what we have left is certain number of entangled pairs where we detect (assuming 100% detection efficiency) all of them. We could imagine let's say 1000 photons going in each arm, but I guess we could equally think about just one pair (as Zeilinger does) and think that the statistical features that we see with an ensemble of photons reflect something also present in each entangled pair. In that sense we could talk about interference even if we are looking at just one entangled pair. Mmmmm... maybe you don't like this (it would not allow "selecting" certain photons because you have only two). Just to make sure, when you say that the detector in the upper arm is selecting photons with certain momentum, are you thinking about momentum in the direction of the arm, perpendicular or just a particular angle of the momentum vector?.

 

[Cthugha]

Hi Cthugha, I recall that we discussed this several years ago. Since then I was busy with other problems in my life and didn't have a chance to look into this in more detail as I would have liked to do. Well, now I am back to this.

Oh, sorry, I am sometimes forgetting stuff and am not as active here as I used to be.

Your explanation from an abstract point of view seems very reasonable. You are using measurement of momentum as the "eraser" of position information. That makes sense being that position and momentum are complementary. Actually, as I was writing the post you just responded to, that picture crossed my mind and I was trying to remember if I had seen a similar explanation in the original Dopfer thesis. As you said, Dopfer's thesis is in German and as far as I know nobody has translated it. I tried to translate it using the Google translator and another one but only got a small part of it translated. I still have the original as hard copy in my bookshelf and I am going to take another look at it. From what you said I get the impression that you understand German. I even though about studying German at that time when I was translating Dopfer's thesis but I found it hard and it was taking too much time so I didn't continue. If what you are saying is correct as an explanation of this experiment, it was not explained like that by Zeilinger. As I did not have an English version of Dopfer's paper, I had to rely on Zeilinger's explanation which at least in my mind did not provide any convincing proof about the need for coincidence-counting.

Yes, I am a native speaker. If you are interested in some small parts of the thesis, I may be able to translate it, but translating all of it would be way too much work. I am not sure about Zeilinger's own explanation. He gives very different arguments for audiences on different levels, so I tend to be a bit careful with his statements because some of them are a bit simplified. My explanation is not really about momentum as a "eraser". This whole erasing stuff is overemphasized a bit in my opinion. I will give a somewhat naive and rough picture of what happens below.

The first thing that I notice is that you are assuming that placing the detector at a particular position near 1f in the Heisenberg lens arm will select certain fraction of the photons. Let's say that 1/2 of the photons on that arm are selected. So my first reaction would be: Where did the other photons go? Now, as I am writing this I am thinking and I realize you may say that the photons on the upper arm are always the same but that we are only looking at the ones that are in coincidence with the ones at the lower arm. That could make things a little more reasonable, but I still would have to think about it to get a more complete picture.

One important thing to note is that the detector in this arm is small. It is smaller in space than the spread of photons and too small to detect all photons. This is a point that is often not explained properly, but it is very important. This will not work with a bucket detector that detects all photons.

Just to make sure, when you say that the detector in the upper arm is selecting photons with certain momentum, are you thinking about momentum in the direction of the arm, perpendicular or just a particular angle of the momentum vector?.

Momentum as in wave vector. This is more or less the angle of the vector.

Now to the simplified explanation. Take a large white light source and place it next to a double slit. You will not notice an interference pattern. Now place a narrow pinhole in between, quite far from the slit. You will notice that now you will see an interference pattern. Young also had to introduce a pinhole for his first interference experiment. Now move that pinhole perpendicular to the slit. You will see that the interference pattern will change. If you move it to every possible position, record each possible pattern and superpose all these patterns, you will end up with the no-interference double slit pattern. The pinhole acts as a momentum filter.

Now that was simple and classical stuff. In the quantum eraser stuff, you more or less play a different trick with the pinhole. Instead of putting it in front of the double slit, you use entangled photons and put it in the other arm. If you are able to do some imaging, which allows you to identify the momentum of the entangled particles, putting a small detector there is the equivalent of a narrow pinhole. All of the photons ending up on the detector will come from a narrow range of momenta. Thus, when doing coincidence counting, all the "partner" photons will also share a narrow range of momenta and this filtered set of photons will show interference just as if you had placed a pinhole in front of the double slit at the corresponding position. The detector acts like a non-local pinhole. Therefore using a large detector would not help (unless of course it is position sensitive). If you used a large position sensitive CCD (which is not practical because they do not show single-photon sensitivity), you could indeed detect all the photons in the Heisenberg lens arm. However, that alone does not give you interference just because the which-way info is erased. You really need to have the momentum information, so you could do coincidence counting for every pixel of that CCD and would get many different interference patterns on the other side for each pixel. Their sum, however, does not look like one, even if you use all photons because there is not "the" interference pattern. There are many.

 

[Eye_in_the_Sky]

Did you really mean P("Yes")≤1/2 ... ?

Yes.

With the state given by

(1/√2) ( |a>|b> + |a'>|b'> )

where |b> and |b'> are orthogonal, it follows that

P("Yes") = ½ ( |<e|b>|² + |<e|b'>|² ) ≤ ½ .
______________________________________

Now, let me make it clear. I am not saying that I am in favor of the idea that a quantum-eraser experiment can be done without coincidence-counting. All that I am saying is that the explanations I have seen (in my opinion) do not prove this concept in a simple and intuitive way. If the need (in a setup such as Dopfer's) for coincidence-counting is true, there should be a simple way to explain it. It should be an explanation as simple as the one you showed or the one I mentioned. But as I said, I think these both refer to a different setup and don't necessarily apply to something like Dopfer's experiment.

For the Walborn scenario, P("Yes")=P("No")=½.

In 'simplified'-Walborn, the horizontal polarizer (i.e. |e>-detector) is stationed between the slits and screen.

The |e>-"No" outcomes are, thus, physically removed from the outgoing beam, while the |e>-"Yes" outcomes can proceed to the screen ... where interference is observed.

Compare this to 'simplified'-Dopfer, as explained by Cthugha:

Take a large white light source and place it next to a double slit. You will not notice an interference pattern. Now place a narrow pinhole in between, quite far from the slit. You will notice that now you will see an interference pattern.

In 'simplified'-Dopfer, the pinholed-blocker (i.e. |p_o>-detector) is stationed between the source and slits.

The |p_o>-"No" outcomes are, thus, physically removed from the outgoing beam, while the |p_o>-"Yes" outcomes can proceed to the slits ... from which interference will be observed.

In this experiment, P("Yes") << P("No").

 

[alexepascual]

Now that was simple and classical stuff. In the quantum eraser stuff, you more or less play a different trick with the pinhole. Instead of putting it in front of the double slit, you use entangled photons and put it in the other arm. If you are able to do some imaging, which allows you to identify the momentum of the entangled particles, putting a small detector there is the equivalent of a narrow pinhole. All of the photons ending up on the detector will come from a narrow range of momenta. Thus, when doing coincidence counting, all the "partner" photons will also share a narrow range of momenta and this filtered set of photons will show interference just as if you had placed a pinhole in front of the double slit at the corresponding position. The detector acts like a non-local pinhole. Therefore using a large detector would not help (unless of course it is position sensitive). If you used a large position sensitive CCD (which is not practical because they do not show single-photon sensitivity), you could indeed detect all the photons in the Heisenberg lens arm. However, that alone does not give you interference just because the which-way info is erased. You really need to have the momentum information, so you could do coincidence counting for every pixel of that CCD and would get many different interference patterns on the other side for each pixel. Their sum, however, does not look like one, even if you use all photons because there is not "the" interference pattern. There are many.

Your explanation seems very consistent. Now I have to take the time to look at it in more detail and that will take some time. The troubling aspect of this is that at first sight, as you say, the "eraser" part of it seems to loose meaning. Although Zelinger was not using that word, reading his paper it appears to me that he was implying it. Now I'll focus more on Dopfer's paper.

 

[alexepascual]

Yes. With the state given by
(1/√2) ( |a>|b> + |a'>|b'> )
where |b> and |b'> are orthogonal, it follows that
P("Yes") = ½ ( |<e|b>|² + |<e|b'>|² ) ≤ ½ .
______________________________________

For the Walborn scenario, P("Yes")=P("No")=½.
In 'simplified'-Walborn, the horizontal polarizer (i.e. |e>-detector) is stationed between the slits and screen.
The |e>-"No" outcomes are, thus, physically removed from the outgoing beam, while the |e>-"Yes" outcomes can proceed to the screen ... where interference is observed.
Compare this to 'simplified'-Dopfer, as explained by Cthugha:
In 'simplified'-Dopfer, the pinholed-blocker (i.e. |p_o>-detector) is stationed between the source and slits.
The |p_o>-"No" outcomes are, thus, physically removed from the outgoing beam, while the |p_o>-"Yes" outcomes can proceed to the slits ... from which interference will be observed.
In this experiment, P("Yes") << P("No").

Thanks a lot Eye_in_The_Sky, I'll have to think about it and compare those steps with what Cthugha says.

 

[Gerinski]

I wrote this in another thread but it might be applicable also here:

An 'observation' is anything that produces an irreversible change in our universe's course of history.

In terms of the double slit experiment, when an electron is not 'observed' it is a wave, and thus it can pass through 2 slits simultaneously. This means that both possibilities coexist in our universe as a superposition, they both contribute 50/50 to our current 'now' reality, because a universe in which nobody will ever be able to know which slit it went through is just one universe, not two. The universe did not have to make a choice. Nothing has changed in a universe where that information is and will never be known, it is one and only one universe reality.

You might imagine that there could still be 2 different universes, one in which it passed through slit A but nobody will ever know, and another where it passed through slit B but also nobody will ever know. But this is how the quantum works, in practice those 2 imaginary universes would be identical, nothing would be irreversibly changed in them making them ever distinguishable from each other, their futures are identical until the end of times, so nature is economical and merges those 2 options as one single universe reality which is in fact a 50/50 superposition of both.

On the other hand if the electron passes through slit A or through slit B leaving any irreversible change in the fabric of the universe, even if that change (whether it passed through A or B) is not observed now but it will only become apparent within billions of years and in a very distant place, these are necessarily 2 different realities, two different futures.
If there is any way in which it will ever be possible to know that once an experiment was conducted on Earth in which the electron passed through slit A and not through B, then the universe needs to make a choice and become that reality and discard the option where it will be possible to know that it passed through slit B and not through A, because that would be a different universe, something irreversible was changed in the course of the experiment changing the future (Many Worlds says that when we put the detector the universe does actually split in 2 versions of itself, one in which the detector will detect slit A and the other where it will detect slit B).

Anything we do which will allow, now or in the distant future, to know which slit the electron went through does necessarily 'force the universe to make the choice' because they would be 2 different futures, they can not coexist. Anything that has the potential to alter the future is 'an observation'.

When you think about it, it is actually completely logical, it could not be otherwise. What is actually amazing is that as long as the futures are identical, all the possible options leading to that future do actually coexist in superposition in the form of waves creating an interference among them, a kind of blurred reality consisting in a haze of all the possible realities, all the possible 'nows' which while being different are still completely consistent with that future. In the simple double slit experiment (when unobserved) we have only 2 different realities in superposition with 50/50 contribution, but in real life complex situations all the 'unobserved' phenomena, i.e. those which do not represent any difference in their future histories, the different possible 'nows' can be many and are superposed with different contributions according to their probability of resulting in that specific future.

 

[alexepascual]

I wrote this in another thread but it might be applicable also here:
An 'observation' is anything that produces an irreversible change in our universe's course of history.
In terms of the double slit experiment, when an electron is not 'observed' it is a wave, and thus it can pass through 2 slits simultaneously. This means that both possibilities coexist in our universe as a superposition, they both contribute 50/50 to our current 'now' reality, because a universe in which nobody will ever be able to know which slit it went through is just one universe, not two. The universe did not have to make a choice. Nothing has changed in a universe where that information is and will never be known, it is one and only one universe reality.

Well Gerinski, if we get into the interpretational side, it can be a long story. I have my own ideas when it comes to interpretation. But I am not dogmatic about them and I am willing to change if I find something that makes more sense. In this thread I was trying to focus more on the way these experiments work. With respect to interpretations, my view is that the most important classification of them is One-World versus Many-Worlds. There are a few that consider "other worlds" but don't assign to them the qualification of "reality" or try not to say anything about it. Consistent histories and Relational interpretation would fall in that blurry category. Some people consider that they are actually Many-worlds in disguise. If we go with the Copenhagen interpretation, as soon as se see collapse, we take the other "histories" or "branches" out of the way and consider only one. That may simplify things a little, but being that there are still things that appear paradoxical, I would not discard these other "histories" and get them "out of the picture". So my approach would be closer to the many-worlds one but even then there are things that remain unexplained, such as Born's rule. (I know Deutsch would not agree).
Then there is the concept of "reality" that is hard-wired in our brain. I would try not to consider "reality" as something fundamental that I can use to classify everything else but just a useful concept that at some point may need modification.

 

[bhobba]

In terms of the double slit experiment, when an electron is not 'observed' it is a wave, and thus it can pass through 2 slits simultaneously. This means that both possibilities coexist in our universe as a superposition,

I don't know where you are getting this from, but the formalism, by which I mean the theory devoid of actual interpretation, except some very basic things such as the meaning of probability, doesn't say anything like that. The electrons are never a wave, possibilities do not coexist (whatever that is supposed to mean). Its not passing through both slits simultaneously (that's part of Feynmans sum over histories interpretation which is actually a hidden variable theory - but of a very novel and non-trivial type). All that's happening is QM is silent about what's going on when not observed. QM is a theory about observations. I have posted it many times but will again. This wave particle stuff is wrong - here is the 'correct' explanation of the double slit:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

Thanks
Bill

 

[alexepascual]

I don't know where you are getting this from, but the formalism, by which I mean the theory devoid of actual interpretation, except some very basic things such as the meaning of probability, doesn't say anything like that. The electrons are never a wave, possibilities do not coexist (whatever that is supposed to mean). Its not passing through both slits simultaneously (that's part of Feynmans sum over histories interpretation which is actually a hidden variable theory - but of a very novel and non-trivial type). All that's happening is QM is silent about what's going on when not observed. QM is a theory about observations. I have posted it many times but will again. This wave particle stuff is wrong - here is the 'correct' explanation of the double slit:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

Thanks
Bill

I don't want to get into this fight, but I think that interpretations so far are just that. For the same reason that you can't prove that one is the best one, you can't either prove that one of them is wrong unless it conflicts with the results of some experiment. Maybe some time in the future one interpretation will prove to better represent the underlying phenomena, but so far I think it is wise to be humble, respect other people's ideas and even try sometimes to look at things from their point of view.

 

[bhobba]

I don't want to get into this fight, but I think that interpretations so far are just that

Indeed they are.

Well said.

Thanks
Bill

 

[alexepascual]

Indeed they are. Well said. Thanks
Bill

But when I said that they are just interpretations I did not mean to diminish their importance.
And even I know you will disagree on this, I consider the "ensenble interpretation" also as a "interpretation"
But I am not saying that you are wrong and I am right. I am just saying that we have different points of view.

 

[bhobba]

And even I know you will disagree on this, I consider the "ensenble interpretation" also as a "interpretation"
But I am not saying that you are wrong and I am right. I am just saying that we have different points of view.

The ensemble interpretation is just an interpretation like any other - it's what I hold to - but that means nothing.

No interpretation is better than another - they are all simply different viewpoints.

Thanks
Bill

 

[Gerinski]

In my humble opinion, an electron passing through the slits unobserved is like a qbit, it is both 0 and 1 simultaneously. An 'observation' is anything which turns that qbit into either a 0 or a 1 (= slit A or B). It doesn't matter what, when, where or who will 'notice' that change. A bit in the universe's information has been changed. The 'superposed' 0+1 was one state. A 0 is a different state, and a 1 is yet another different state. If we setup the apparatus so that it is not 'observed', the state will remain as it was, 0+1 simultaneously, we will not have caused a change in that bit of the universe's information. If we cause the qbit to change to either 0 or 1, (even if we do not 'look' at that change), that constitutes an 'observation'.

 

[Swamp Thing]

This wave particle stuff is wrong - here is the 'correct' explanation of the double slit:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

If I understand Marcella correctly, the slits prepare a state in terms of position, and the detection screen measures the momentum.

Would it be ok to say instead, that:
(a) The slits put the system in a certain position state, which is equivalent to the sum of many momentum states
(b) The system evolves in time from this initial state, such that the corresponding projection back onto the position basis also evolves in time
(c) The evolution thus takes the system through various states which are the sums of many position states
(c) Finally, the screen measures the position again at some point in its evolution.

Is this a valid, and mathematically equivalent, description to Marcella's? If so, then - although more complicated when described in words - it is perhaps a more acceptable picture. After all, a screen actually measures position and not momentum!