Question about delayed choice quantum eraser

[Cthugha]

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

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

 

[blandrew]

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

 

[Cthugha]

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

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

 

[blandrew]

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

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

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

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

 

[blandrew]

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

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

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

 

[JesseM]

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

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

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

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

 

[blandrew]

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

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

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

 

[JesseM]

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

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

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

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

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

 

[Cthugha]

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

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

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

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

 

[Cthugha]

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

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

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

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

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

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

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

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

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

 

[blandrew]

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

 

[San K]

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

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

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

 

[guillefix]

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

 

[ijdavis]

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

 

[msumm21]

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

 

[DrChinese]

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

Welcome to PhysicsForums, ijdavis!

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

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

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

 

[ijdavis]

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

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

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

 

[ijdavis]

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

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

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

 

[DrChinese]

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

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

 

[DrChinese]

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

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

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

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

 

[ijdavis]

Same number of photons yes. But the emergent distribution of where the photons at D0 strike a screen will if it cannot be determined which slit these photons passed through be to produce a diffraction pattern at D0, while to not produce this diffraction pattern if the red and blue paths differ as to where they arrive consequent of it then being possible to say which slit the photons went through. Or am I misunderstanding how the diffraction pattern is to be observed/derived when it exists at D0.

 

[DrChinese]

Same number of photons yes. But the emergent distribution of where the photons at D0 strike a screen will if it cannot be determined which slit these photons passed through be to produce a diffraction pattern at D0, while to not produce this diffraction pattern if the red and blue paths differ as to where they arrive consequent of it then being possible to say which slit the photons went through. Or am I misunderstanding how the diffraction pattern is to be observed/derived when it exists at D0.

You can see from the diagram that all D0 photons are focused by a lens to a single point. Therefore, there is no pattern to observe. Again, nothing changes at D0 based on any action on the other side.

And if you remove that lens, I am not really sure what you would expect to see either other than maybe a couple of blobs. One for the red stream and one for the blue stream.

 

[San K]

However, if one can force all entangled twins to arrive at D3 or D4 it seems one can force a result at D0, which no longer requires any correlation to be performed.

D0-D3 (alone) forms an interference pattern.
D0-D4 (alone) forms an interference pattern.

However D0 contains both D3 and D4.

The two patterns overlap such that the crest of one aligns with the trough of the other.

Thus no interference pattern is seen, only a "blob" is seen.

 

[kaplan]

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

I don't understand that attitude. If the MWI makes it easier and conceptually simpler to understand QM - and it does - then why not accept it as the best interpretation of QM?