Can Entangled Particles Enable Faster-Than-Light Communication?

[hellfire]

I am puzzled with the following gedankenexperiment.

Consider a pair of particles that are sent from x = 0 on different directions +x and -x with entangled position on the y axis.

Parallel to the y-axis we locate two double slit plates A and B. They plate A is located at x = -L and the plate B at x = +(L + e) ("e" is a very small distance). In the plate A one of the slits is closed. In the plate B both slits are open. The lower slit of the B plate is located at y = -h and the upper at y = +h.

When the particle 1 reaches x = -L its y-position will be measured, thus colapsing the whole 1+2 system and determining the y-position of particle 2. From this instant of time on, particle 2 will propagate with definite momentum spreading in y-position.

However, at that instant of time the particle 2 may be closer to one of the B-slits than to the other and there will be a greater probability that the particle 2 travels through one specific slit.

This will lead to a small interference pattern or to no interference pattern at all. Only in case that the y-position of particle 2 colapses at y = 0, the probability for superposition via both B-slits will be equal, leading to a clear interference.

However, if we open the second slit in plate A, the entanglement will not be destroyed (the y-position of the particle 1 will not be determined) and the y-position of particle 2 will remain uncertain. This should lead to a very noticeable interference pattern, more than in the first set-up.

At first sight this seams to be a way to have FTL comunication between A and B. Imagine that we send a series of particles one after the other and locate A and B at great distance. Observer A could set-up some kind of binary communication with B, opening the second slit in the A plate and allowing for interference between some hundred or thousand particles to be displayed at B. Afterwards, for the second set of hundred or thousand particles, A could close the second slit in the A plate and no interference (or practically no interference) would show up at B.

The problem in the usual EPR experiment with entangled spins is that the B observer has no way to know if collapse of the entangled system has taken place or not. His measurement looks always random for him and contains no information about whether A did measure or not. In this case, however, the double-slit experiment (interference or no interference) tells B if the collapse took place or not, knowing therefore if A did measure or not.

It is not clear to me whether it possible to have y-position entanglement to set up this experiment. if yes, will the results be as described? I assume it will not, but I cannot figure out the mistake.

 

[cesiumfrog]

Can you draw a diagram? It isn't completely clear what you mean the apparatus to look like. Also, if possible, could you replace the "y-position entanglement" with something more concrete, say a two-slit source that emits from "the same slit"/"a superposition of the two slits" (something that is clearly possible to construct)?

But without much idea what you're intending (and just comparing instead to the DCQE) it seems like you'll never get raw interference patterns. Even if you have two slits open on both sides, you can probably still figure out where the particles started by combining the time and position that they strike each screen.

 

[hellfire]

Can you draw a diagram? It isn't completely clear what you mean the apparatus to look like. Also, if possible, could you replace the "y-position entanglement" with something more concrete, say a two-slit source that emits from "the same slit"/"a superposition of the two slits" (something that is clearly possible to construct)?

Sorry I cannot explain a mechanism for this "y-position entanglement". Even I am not sure if this is possible. I just imagined that, same as spin entanglement, position entanglement should be possible. In such a case when determining the y-position of one particle, the other should be instantaneously determined.

Here I did my best to make some pictures. Consider A to the left and B to the right. The basic idea is that in the first case the y-position of the second particle should be determined in front of the B slits by measuring the y-position of the first particle at the A slit.

 

[cesiumfrog]

I understand this is your reasoning: in the first case, we know everything about the left particle, hence we effectively know which slit the right particle goes through, and there is no interference. In the second case, we do not know which slit the left particle goes through, thus do not know which slit the right particle goes through, therefore anticipate interference on the right. Yes?

You were thinking of FTL communication, that opening a slit on the left should "immediately" cause interference to appear on the right. FTL makes it easy to produce a paradox. Say the slit is open, and just after you've observed the interference, I'll alter the apparatus to insert a detector on the left (at x=-(L+2e)), to detect which slit the particle went through. If you prefer, then I can use some arrangement of mirrors and telescopes to make (or choose not to make) this "which path" measurement after you tell me whether the interference pattern appeared or not. (This is simply a convoluted form of Wheeler's delayed choice apparatus.)

The paradox is resolved if you never see an interference pattern on the right. This is to be expected because you can always, in principle, determine the "which path" information by measurements on the left. It's not enough to just not look at the measurement result on the left (which one might naively think is somehow different to not making the measurement at all), you would need to somehow quantum-mechanically "erase" the information (such that it cannot be reconstructed even in principle), and then (to preserve causality) you would have to somehow prove that you have done so (to the screen on the right) before seeing the interference pattern. This is exactly what occurs in the Delayed Choice Quantum Erasor experiment.

 

[hellfire]

You have correctly interpreted the experiment. So you are telling me that even if the four slits are open there will be no interference in any of both sides. I cannot figure out this because in such a case the entangled system will be still in a superposition state. Interference should show up.

I am not sure to follow your reasoning arguing that there exists the possibility to put a detector that could determine the path in principle. It seams to me that you are arguing that interference should not take place because causality should be preserved, but then you are cheating because this exactly what I want to prove. Sorry, but it seams that you have to be patient with me here...

 

[ueit]

Sorry I cannot explain a mechanism for this "y-position entanglement". Even I am not sure if this is possible. I just imagined that, same as spin entanglement, position entanglement should be possible.

A diatomic molecule like Hg2 can produce a pair of atoms entangled in the way you want.

In such a case when determining the y-position of one particle, the other should be instantaneously determined.

Only if you measure the other's particle position too.

 

[DrChinese]

You have correctly interpreted the experiment. So you are telling me that even if the four slits are open there will be no interference in any of both sides. I cannot figure out this because in such a case the entangled system will be still in a superposition state. Interference should show up.

I am not sure to follow your reasoning arguing that there exists the possibility to put a detector that could determine the path in principle. It seams to me that you are arguing that interference should not take place because causality should be preserved, but then you are cheating because this exactly what I want to prove. Sorry, but it seams that you have to be patient with me here...

cesiumfrog is correct. There is no interference in this case (all slits open) despite what you might expect. In fact, this idea has been proposed and shot down on this forum a number of times (count me as one of the victims). We have previously determined that an entangled photon does NOT yield an interference pattern when going through a double slit setup. As a result, you cannot use the wave-particle concept like a yes-no bit of information. You can see this in an enlightening article by Anton Zeilinger, p. 290, Figure 2.

Experiment and the foundations of quantum physics

 

[Tomsk]

I'm not sure I've understood this... Are you saying if we send two entangled photons in two different directions towards two double slit setups, and we have no way of determining the which-path information of either particle, we still don't get interference at either end purely because they are entangled? And this is because one of the photons doesn't know *that* the other one hasn't had it's which-path information measured (even if the two exprimenters agreed not to beforehand), because the photons are spatially separated?

Is that saying that this entanglement between the photons is somehow more important, or takes precedence over the passage through the double slits?

If we set up a normal double slit, and someone far away said, at any moment, I'll start sending entangled ones instead of normal ones, we could pinpoint exactly when they did, because the interference dissappears? (Obviously this doesn't violate causality, it just seems a bit weird... in fact it seems to say we can use wave/particle duality as yes/no information... although it's not on a photon-by-photon basis, only bunches of normal or entangled ones. Is that the catch?)

Interesting topic, this!

 

[DrChinese]

If we set up a normal double slit, and someone far away said, at any moment, I'll start sending entangled ones instead of normal ones, we could pinpoint exactly when they did, because the interference dissappears? (Obviously this doesn't violate causality, it just seems a bit weird... in fact it seems to say we can use wave/particle duality as yes/no information... although it's not on a photon-by-photon basis, only bunches of normal or entangled ones. Is that the catch?)

No catch: even if you look at the bunches, you still can't use this as a way to send yes/no information. You can determine if a bunch consists of an entangled source using this mechanism.

 

[JesseM]

You have correctly interpreted the experiment. So you are telling me that even if the four slits are open there will be no interference in any of both sides. I cannot figure out this because in such a case the entangled system will be still in a superposition state. Interference should show up.

I am not sure to follow your reasoning arguing that there exists the possibility to put a detector that could determine the path in principle. It seams to me that you are arguing that interference should not take place because causality should be preserved, but then you are cheating because this exactly what I want to prove. Sorry, but it seams that you have to be patient with me here...

This is similar to the delayed choice quantum eraser, so although you won't see interference if you just look at the total pattern of photons on a given side, if you look at the subset of photons on side B whose entangled twins were detected in a particular location on side A (say, a particular value of y, or small interval on the y-axis), a location which prevents you from knowing which slit the photons on side B went through, then I think you could see an interference pattern in this subset. Again, look over the delayed choice quantum eraser example where something similar is true.

 

[Tomsk]

No catch: even if you look at the bunches, you still can't use this as a way to send yes/no information. You can determine if a bunch consists of an entangled source using this mechanism.

I think I'm not quite sure what you mean by yes/no information. If we set up two sources pointing in the same direction, one emitting normal photons, one emitting entangled ones, and we say that entangled=no interference=yes, not entangled=interference=no, and I send one or the other (not both together), and you receive them through a double slit, then you can tell whether I'm sending entangled or not entangled by looking at the interference, or lack of, and hence whether I'm sending yes or no.

 

[JesseM]

I think I'm not quite sure what you mean by yes/no information. If we set up two sources pointing in the same direction, one emitting normal photons, one emitting entangled ones, and we say that entangled=no interference=yes, not entangled=interference=no, and I send one or the other (not both together), and you receive them through a double slit, then you can tell whether I'm sending entangled or not entangled by looking at the interference, or lack of, and hence whether I'm sending yes or no.

But this isn't FTL, because the event of the entangled particles being created and the event of your measuring one of them do not have a spacelike separation.

 

[Cane_Toad]

... You can see this in an enlightening article by Anton Zeilinger, p. 290, Figure 2.

Experiment and the foundations of quantum physics

This was great, but I'm having troubling finding something similar now that it's been 8 years. Do you know of anything?

 

[Tomsk]

Ah, didn't realize it was supposed to be FTL, thanks for clearing that up.

 

[hellfire]

Thanks for the answers. The article by Zeilinger is great.

 

[DrChinese]

I think I'm not quite sure what you mean by yes/no information. If we set up two sources pointing in the same direction, one emitting normal photons, one emitting entangled ones, and we say that entangled=no interference=yes, not entangled=interference=no, and I send one or the other (not both together), and you receive them through a double slit, then you can tell whether I'm sending entangled or not entangled by looking at the interference, or lack of, and hence whether I'm sending yes or no.

True. Your signal will not exceed c, and I think you acknowledged this already.

 

[RandallB]

cesiumfrog is correct. There is no interference in this case (all slits open) despite what you might expect. In fact, this idea has been proposed and shot down on this forum a number of times (count me as one of the victims). We have previously determined that an entangled photon does NOT yield an interference pattern when going through a double slit setup. As a result, you cannot use the wave-particle concept like a yes-no bit of information. You can see this in an enlightening article by Anton Zeilinger, p. 290, Figure 2.

Experiment and the foundations of quantum physics

DrC your phrasing is a little misleading here.
First a single photon can never create an interference pattern. A beam of photons and a count of many photon location hits are required to determine if a pattern is being created.

Second, no one has ever demonstrated that a beam of light known to be entangled with a second beam of light will not create an interference pattern when sent through a double slit. By itself it is just a beam of light will always create a pattern. There is no objective test that can be preformed on a single beam of light alone that will reveal if it is entanglement with some other untested beam.

What does happen is if you log all the hits that created the pattern and correlate them with photons from the second beam in a method that would reveal “which way” AND you use only hit locations for those correlated photons to reconstruct a new screen display. Bingo – no more interference pattern

 

[JesseM]

Second, no one has ever demonstrated that a beam of light known to be entangled with a second beam of light will not create an interference pattern when sent through a double slit.

That is what quantum theory predicts, and the delayed choice quantum eraser experiment actually demonstrates it.

By itself it is just a beam of light will always create a pattern. There is no objective test that can be preformed on a single beam of light alone that will reveal if it is entanglement with some other untested beam.

If what you were saying was true, then in the DCQE experiment you'd see an interference pattern even if you preserved the which-path information of 100% of the entangled twins (the 'idlers'), a violation of complementarity. But this is not what happens--the total pattern of "signal" patterns going through the slits shows no interference, although if you correlate subsets of signal photons whose corresponding idlers went to a detector that erased their which-path information, you do see interference in this subset of signal photons. Have a look at the description of the experiment here, and in particular the illustration of the pattern at various detectors at the bottom (this image)--the "D0" pattern corresponds to the total pattern of signal photons going through the slits, and you can see that there's no interference.

 

[RandallB]

Originally Posted by RandallB
Second, no one has ever demonstrated that a beam of light known to be entangled with a second beam of light will not create an interference pattern when sent through a double slit

That is what quantum theory predicts, and the delayed choice quantum eraser experiment actually demonstrates it.

NOT TRUE
All "Delayed Choice" experiments require using both beams. No experiment has ever shown any beam of light to fail to produce interface patterns with a double slit unless something is introduced from a second beam “entangled” with the beam being tested. Either by correlation count coordination, or combining both beams somehow.

All the examples sited are using both beams.

 

[JesseM]

NOT TRUE
All "Delayed Choice" experiments require using both beams. No experiment has ever shown any beam of light to fail to produce interface patterns with a double slit unless something is introduced from a second beam “entangled” with the beam being tested. Either by correlation count coordination, or combining both beams somehow.

What is "not true"? You do use a coincidence count to see the interference pattern, and the diagram makes that clear by labeling the interference patterns "D0 correlated with D1" and "D0 correlated with D2", but the diagram also shows the pure "D0" pattern which you get when you only look at the beam of "signal photons" that goes through the slits, and this pattern shows no interference. And if the beam-splitters were removed so that all the entangled "idler photons" went to detectors D3 or D4 where their which-path information was preserved, there would be no interference anywhere, not in the total pattern of signal photons at D0 or in the D0/D3 and D0/D4 coincidence counts. In this case, for 100% of the signal photons you'd know which slit they went through, so you'd better not see interference in the total pattern of signal photons or that would be a violation of complementarity.

Brian Greene discusses the delayed choice quantum eraser in his book Fabric of the Cosmos, and on p. 197-198 he says:

Notice, too, perhaps the most dazzling result of all: the three additional beam splitters and the four idler-photon detectors can be on the other side of the laboratory or even on the other side of the universe, since nothing in our discussion depended at all on whether they receive a given idler photon before or after its signal photon partner has hit the screen. Imagine, then, that these devices are all far away, say ten light-years away, to be definite, and think about what this entails. You perform the experiment in Figure 7.5b today, recording–one after another–the impact locations of a huge number of signal photons, and you observe that they show no signs of interference. If someone asks you to explain the data, you might be tempted to say that because of the idler photons, which-path information is available and hence each signal photon definitely went along either the left or the right path, eliminating any possibility of interference. But, as above, this would be a hasty conclusion about what happened; it would be a thoroughly premature description of the past.

You see, ten years later, the four photon detectors will receive–one after another–the idler photons. If you are subsequently informed about which idlers wound up, say, in detector 2 (e.g., the first, seventh, eight, twelfth ... idlers to arrive), and if you then go back to the data you collected years earlier and highlight the corresponding signal photon locations on the screen (e.g., the first, seventh, eighth, twelfth ... signal photons that arrived), you will find that the highlighted data points fill out an interference pattern, thus revealing that those signal photons should be described as having traveled both paths. Alternatively, if 9 years, 364 days after you collected the signal photon data, a practical joker should sabotage the experiment by removing beam splitters a and b–ensuring that when the idler photons arrive the next day, they all go to either detector 1 or detector 4, thus preserving all which-path information–then, when you receive this information, you will conclude that every signal photon went along either the left path or the right path, and there will be no interference pattern to extract from the signal photon data.

 

[DrChinese]

Second, no one has ever demonstrated that a beam of light known to be entangled with a second beam of light will not create an interference pattern when sent through a double slit. By itself it is just a beam of light will always create a pattern. There is no objective test that can be preformed on a single beam of light alone that will reveal if it is entanglement with some other untested beam.

I would disagree with this. My reference above states my position clearly, and he should know. It is true that I have not seen a separate experiment to test this because it doesn't prove much. But it would be interesting to see it in print anyway. If I can't find a reference, I will see if it is possible to get someone to write it up and publish it.

All that is needed is to put a double slit apparatus on both sides in place of the beamsplitters. When the entangled photons go through, they should not (or should depending on who you believe) make an interference pattern. This would demonstrate that entangled photons objectively act differently than unentangled photons.

 

[Cane_Toad]

Brian Greene discusses the delayed choice quantum eraser in his book Fabric of the Cosmos, and on p. 197-198 he says:

Alternatively, if 9 years, 364 days after you collected the signal photon data, a practical joker should sabotage the experiment by removing beam splitters a and b–ensuring that when the idler photons arrive the next day, they all go to either detector 1 or detector 4, thus preserving all which-path information–then, when you receive this information, you will conclude that every signal photon went along either the left path or the right path, and there will be no interference pattern to extract from the signal photon data.

I'm confused. Is this a deduction on his part, or a restatement of a real experiment? Re:

I would disagree with this. My reference above states my position clearly, and he should know. It is true that I have not seen a separate experiment to test this because it doesn't prove much. But it would be interesting to see it in print anyway. If I can't find a reference, I will see if it is possible to get someone to write it up and publish it.

Brian Green's story seems to be a misinterpretation (or extrapolation) of the DCQE, which is based on the situation where detectors D3,D4 collect photons *after* D0, the "after" part being the "delayed choice"? On the other hand, maybe it works that way after all?

 

[Cane_Toad]

All that is needed is to put a double slit apparatus on both sides in place of the beamsplitters. When the entangled photons go through, they should not (or should depending on who you believe) make an interference pattern. This would demonstrate that entangled photons objectively act differently than unentangled photons.

Dammit, why can't I buy one of these from Edmund Scientific, and be done with it!

 

[JesseM]

I'm confused. Is this a deduction on his part, or a restatement of a real experiment? Re:

Is which part a deduction? Obviously the part about having the detectors 10 light years apart is an extrapolation, if that's what you mean.

Brian Green's story seems to be a misinterpretation (or extrapolation) of the DCQE, which is based on the situation where detectors D3,D4 collect photons *after* D0, the "after" part being the "delayed choice"?

That's what Brian Greene describes too. He does use a different notation in his diagram if that's what's bothering you, he labels the two idler detectors which preserve the which-path information 1 and 4 rather than D3 and D4.

 

[RandallB]

I would disagree with this. My reference above states my position clearly, and he should know. It is true that I have not seen a separate experiment to test this because it doesn't prove much. But it would be interesting to see it in print anyway. If I can't find a reference, I will see if it is possible to get someone to write it up and publish it.

All that is needed is to put a double slit apparatus on both sides in place of the beamsplitters. When the entangled photons go through, they should not (or should depending on who you believe) make an interference pattern. This would demonstrate that entangled photons objectively act differently than unentangled photons.

I disagree with JesseM and your assessment of your reference in the first paragraph. It does not demonstrate your position clearly. If it real does state your position, it is providing absolutely no justification for such a statement as the only tests used there involve correlations.

I totally agree with your second paragraph; that is all anyone would need to do to prove the point that a single beam of light evolved is unique from a normal single beam of light that by it not capable of producing an interference pattern.
PLUS your experiment producing two independent beams of light both failing to produce interface patterns is so amazingly simple one has to ask.
WHY hasn’t some one done it! Where are the results! They don’t even need a coincidence counter to do the experiment – they don’t get much simpler!

It is because the interference pattern is seen all the time as labs they set up for more complex experiments. The interference patterns that will in fact show up in your example, will only disappear when some form of correlation between the A side and B side is taken into account. And the pattern will certainly disappear if the joint testing between A & B could reveal “which way” on either side.

 

[JesseM]

I disagree with JesseM

What do you disagree with, specifically? Have you actually looked at the delayed choice quantum eraser experiment, or are you just making statements based on your own preconceptions? Please tell me which of these statements you disagree with:

1. Do you disagree that the D0 detector corresponds to the "screen" in the double-slit experiment?

2. Do you disagree that the diagram I linked to shows the pattern of signal photon hits at the D0 detector alone (i.e. no correlation with the idler photons), and that it shows no interference?

3. Do you disagree that Brian Greene was also saying that the total pattern of signal photons would shown no interference when he said "You perform the experiment in Figure 7.5b today, recording–one after another–the impact locations of a huge number of signal photons, and you observe that they show no signs of interference"?

4. Do you disagree that if you remove the beam-splitters from the path of the idler photons, all the idlers will go to detectors that preserve their which-path information (D3 and D4 in the paper on the double-slit experiment, or detectors 1 and 4 in Brian Greene's diagram)?

4a. Do you disagree that if you look at the correlation graph of idlers at one of these detectors with signal photons, they will show no interference?

4b. Do you disagree that in the case where all the idlers go to one of these two which-path-preserving detectors, the total pattern of signal photons at D0 will just be the sum of the D0/D3 correlation graph and the D0/D4 correlation graph?

4c. Do you disagree that the sum of two non-interference patterns cannot itself be an interference pattern?

If it real does state your position, it is providing absolutely no justification for such a statement as the only tests used there involve correlations.

False. The diagram I linked to showed the uncorrelated pattern of signal photon hits at D0 along with the the correlation patterns between D0 hits and idler hits at other detectors--did you actually look at the diagram? What do you think the pattern just labeled "D0" is showing, exactly? And what do you think Brian Greene meant when he said "You perform the experiment in Figure 7.5b today, recording–one after another–the impact locations of a huge number of signal photons, and you observe that they show no signs of interference"? (keep in mind the context, where he was discussing a case where the idler photons would not reach their detectors until 10 years after the signal photons were detected--it's obvious that he couldn't have been talking about any correlation pattern in this sentence!)

It is because the interference pattern is seen all the time as labs they set up for more complex experiments. The interference patterns that will in fact show up in your example, will only disappear when some form of correlation between the A side and B side is taken into account.

If you will actually read about the details of the delayed choice quantum eraser instead of just making assumptions, you will see that you have it backwards--non-interference is what you see in the total pattern of signal photons, interference patterns only appear when "some form of correlation between the A side and B side is taken into account". Specifically, you see an interference pattern when you look at the subset of signal photons at D0 whose corresponding idlers went to a detector that erased their which-path information--that's what the diagram I linked to shows in the "D0 correlated with D1" and "D0 correlated with D2" picture (just click the link and look at it!), and that's what Brian Greene meant when he said "If you are subsequently informed about which idlers wound up, say, in detector 2 (e.g., the first, seventh, eight, twelfth ... idlers to arrive), and if you then go back to the data you collected years earlier and highlight the corresponding signal photon locations on the screen (e.g., the first, seventh, eighth, twelfth ... signal photons that arrived), you will find that the highlighted data points fill out an interference pattern, thus revealing that those signal photons should be described as having traveled both paths."

And the pattern will certainly disappear if the joint testing between A & B could reveal “which way” on either side.

Well, yes, that's the whole point--in the delayed choice quantum eraser experiment, joint testing of both members of a signal photon/idler photon entangled pair can tell you which way the signal photons went through the slits, and that's exactly why the total pattern of signal photons doesn't show interference.

edit: I also notice that p. 290 of the reference DrChinese provided says very specifically that entanglement will not be observed in particles going through a double slit if they are members of entangled pairs whose total momentum is zero:

The situation is strikingly illustrated if one employs pairs of particles which are strongly correlated (‘‘entangled’’) such that either particle carries information about the other (Horne and Zeilinger, 1985; Greenberger, Horne, and Zeilinger, 1993). Consider a setup where a source emits two particles with antiparallel momenta (Fig. 2). Then, whenever particle 1 is found in beam a, particle 2 is found in beam b and whenever particle 1 is found in beam a', particle 2 is found in beam b'. The quantum state is

| \psi > = \frac{1}{\sqrt{2}} (|a>_1 |b>_2 + |a'>_1 |b'>_2 )

Will we now observe an interference pattern for particle 1 behind its double slit? The answer has again to be negative because by simply placing detectors in the beams b and b' of particle 2 we can determine which path particle 1 took. Formally speaking, the states |a>_1 and |a'>_1 again cannot be coherently superposed because they are entangled with the two orthogonal states|b>_2 and |b'>_2

And on http://www.advancedphysics.org/forum/showthread.php?t=6632&page=2 from another forum, the user "smilodon" gives some explanation of why no interference is seen if you measure just one of the particles:

The problem is that to demonstrate "interference" you need to have a statistical sample. You cannot, with one single event, determine whether or not there was interference. Now, what we find, when we look at the local reduced density matrix, is that the "photon which is part of an entangled system" corresponds to a reduced density matrix which is a mixture and not a pure state, and this mixture is such, that it is a perfect, well, mixture, of the two complementary interference patterns, which makes you see no interference at all over a statistical ensemble. So a "photon beam" full of photons entangled with something else, shows up as a non-pure beam which is a statistical mixture. The degree of freedom that shows up as "non-pure" is exactly the degree of freedom that is entangled: this can be: angular direction, wavelength, polarization, or a combination of all of them.

So, if you want to, you can still conceptually think of each photon in the beam as "interfering with itself" ; only, because of the non-pure character, there will be no interference pattern (if the pattern depends upon the degree of freedom of the photon that is entangled) by the entire beam.

 

[Cane_Toad]

edit: I also notice that p. 290 of the http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf" DrChinese provided says very specifically that entanglement will not be observed in particles going through a double slit if they are members of entangled pairs whose total momentum is zero:

The document I see for this only has 10 pages. Where is the one you refer to that has >= 290 pages?

 

[vanesch]

I disagree with JesseM and your assessment of your reference in the first paragraph. It does not demonstrate your position clearly. If it real does state your position, it is providing absolutely no justification for such a statement as the only tests used there involve correlations.

I think we've gone over this discussion already a few times. The point is the following. Of course it is possible to get an interference pattern using "one arm" of an entangled beam ; only, at that point, the entangled quantities do not have anything to do anymore with the interference pattern.

"One arm" of an entangled beam is always a statistical mixture (when looked at locally). Now, it is perfectly possible to make an interference pattern with a statistically mixed beam! You can even obtain interference patterns with white light for that matter. The point is, however, that the "interference" is then not an interference anymore between the different statistical compounds of the mixture (in other words, a superposition).

Let us get back to the OP's experiment. The point was of course that the entanglement was such, that if you would place photon counters between all the slits, that each time you get a hit on the (x=-L,y=+h), you'd also have a hit on (x=+L,y=-h), and not on the other one (at x=+L,y=+h).
THIS is what is meant by having the "y" coordinates entangled.

Now, when seen from a single side, this means that the light impinging on the +h slit on the +L side is TOTALLY UNCORRELATED with the light impinging on the -h slit on the +L side. That's the statistical mixture I am talking about.
As such, there will be no interference in this case.

But we can of course narrow down the slits by diminishing h, say. From a certain point on, there WILL be interference. But at that moment, the hits on the -L side will not be strictly correlated anymore with the hits on the +L side, in other words, the interference pattern tested is not between entangled degrees of freedom.

 

[JesseM]

The document I see for this only has 10 pages. Where is the one you refer to that has >= 290 pages?

I wasn't going by the number of pages in the PDF document, but by the page-numbers in the corners of each page--the first page has "S288" in the lower-left corner, the next page has "S289" in the upper-right corner, and the third has "S290" in the upper-left corner, that's the page the quote is from (it also has a diagram and some additional discussion of the experiment).

 

[RandallB]

I think we've gone over this discussion already a few times. The point is the following. Of course it is possible to get an interference pattern using "one arm" of an entangled beam ; only, at that point, the entangled quantities do not have anything to do anymore with the interference pattern.

"One arm" of an entangled beam is always a statistical mixture (when looked at locally). Now, it is perfectly possible to make an interference pattern with a statistically mixed beam! ...

Vanesch You’re preaching to the choir here, I understand. What you explain here is exactly what JesseM is saying is not true. Give him a little help he obviously doesn’t want to here it from me.

Of course anytime you use the other beam in a manner that could reveal which way information interference will disappear.

 

[RandallB]

What do you disagree with, specifically? Have you actually looked at the delayed choice quantum eraser experiment, ...
...
2. Do you disagree that the diagram I linked to shows the pattern of signal photon hits at the D0 detector alone (i.e. no correlation with the idler photons), and that it shows no interference? ...

Of course I’ve looked at DCQE before, have you looked at your diagrams?? You can refer to a picture (with incorrect information) 3 or 4 times and call it a diagram and it is still just a photo.

Did you actually look at the http://strangepaths.com/category/physics/en/" they put up in the web page that shows where all 5 detectors 0 through 4 are ?? One thing is sure these guys did not actually run any experiment based on that “diagram” !
Care to point out where the source of photons is in the real diagram.
At least find an example that make some sense.

 

[JesseM]

Vanesch You’re preaching to the choir here, I understand. What you explain here is exactly what JesseM is saying is not true. Give him a little help he obviously doesn’t want to here it from me.

Uh, no he isn't. Read the bolded section of his post:

Let us get back to the OP's experiment. The point was of course that the entanglement was such, that if you would place photon counters between all the slits, that each time you get a hit on the (x=-L,y=+h), you'd also have a hit on (x=+L,y=-h), and not on the other one (at x=+L,y=+h).
THIS is what is meant by having the "y" coordinates entangled.

Now, when seen from a single side, this means that the light impinging on the +h slit on the +L side is TOTALLY UNCORRELATED with the light impinging on the -h slit on the +L side. That's the statistical mixture I am talking about.
As such, there will be no interference in this case.

But we can of course narrow down the slits by diminishing h, say. From a certain point on, there WILL be interference. But at that moment, the hits on the -L side will not be strictly correlated anymore with the hits on the +L side, in other words, the interference pattern tested is not between entangled degrees of freedom.

The point is that you can only get interference if you alter the distance between the slits in such a way that there is no longer any way to tell which slit the photon on one side went through by measuring its entangled twin on the other side. I was not claiming that arbitrary forms of entanglement would destroy interference, just claiming that if the entangled twin is entangled in such a way that there would be any potential to deduce which slit the first photon went though, you will see no interference pattern in the photons going through the slits.

Of course anytime you use the other beam in a manner that could reveal which way information interference will disappear.

It does not depend on how you actually use the other beam, only on whether they are entangled in such a way that there was the potential to learn which slits the photons went through by measuring their entangled twins. In the delayed choice quantum eraser you are free to measure all the entangled idler photons in such a way that the which-path information is lost forever, but this won't create an interference pattern in the total pattern of signal photons (although it will create one in the correlation graphs between idlers that went to a particular detector and the subset of signal photons which were entangled with those idlers), the total pattern of signal photons will still show no interference because they were entangled with the idlers in such a way that you could have learned which slit the signal photon went through if you had measured the idler in the right way, and of course the experimenter at the double-slit apparatus cannot get any FTL information about whether the idlers actually were measured in this way or whether they were measured in a way that erased their which-path information.

 

[JesseM]

Of course I’ve looked at DCQE before, have you looked at your diagrams?? You can refer to a picture (with incorrect information)

What information do you think is incorrect?

3 or 4 times and call it a diagram and it is still just a photo.

I don't think that's an actual photo, no. How would you photograph correlation patterns like "D0 correlated with D1", where you only want to depict a certain subset of the photons that arrive at D0?

Did you actually look at the http://strangepaths.com/category/physics/en/" , with the D4 detector added in red).

Care to point out where the source of photons is in the real diagram.

In fig. 1, instead of a traditional double-slit the photons are being emitted from one of two possible atoms A and B which are in the position of the slits. The paper explains this in the opening:

One proposed quantum eraser experiment very close to the 1982 proposal is illustrated in Fig.1. Two atoms labeled by A and B are excited by a laser pulse. A pair of entangled photons, photon 1 and photon 2, is then emitted from either atom A or atom B by atomic cascade decay.

P. 2 of the paper discusses the actual experiment they performed which is illustrated in fig. 2:

We wish to report a realization of the above quantum eraser experiment. The schematic diagram of the experimental setup is shown in Fig.2. Instead of atomic cascade decay, spontaneous parametric down conversion (SPDC) is used to prepare the entangled two-photon state. SPDC is a spontaneous nonlinear optical process from which a pair of signal-idler photons is generated when a pump laser beam is incident onto a nonlinear optical crystal [6]. In this experiment, the 351.1 nm Argon ion pump laser beam is divided by a double-slit and incident onto a type-II phase matching [7] nonlinear optical crystal BBO (? ? BaB2 O4 ) at two regions A and B. A pair of 702.2nm orthogonally polarized signal-idler photon is generated either from A or B region.

The signal photons then pass through the slits and are focused by a lens to detector D0, while the idlers go through a prism and various beam-splitters to end up at either D1 or D2 (which-path info erased) or D3 or D4 (which-path info preserved).

 

[Cane_Toad]

...

the total pattern of signal photons will still show no interference because they were entangled with the idlers in such a way that you could have learned which slit the signal photon went through if you had measured the idler in the right way, a...

.

I thought you got no interference pattern because you have to add the two correlations, D0:D1, and D0:D2. That sure looks like what would happen if you add up the bars as shown in:

http://strangepaths.com/wp-content/uploads/2007/03/patterns-01.jpg"

 

[RandallB]

Uh, no he isn't. Read the bolded section of his post:

Oh for crying out loud those sections are talking about correlations -- did you read his sentence:
[Of course it is possible to get an interference pattern using "one arm" of an entangled beam]

All I’m saying is any beam of light I don’t care how entangled it may be, when put to the test as described in post 25 will always create a pattern.
If not where is the experiment to proof it! One where no part of the one arm is split off and combined or measured against the other arm. And especially no correlations of detection left between left arm and right arm before after of during their trip through the double slits.

Treated locally by itself one beam from a pair of entangled beams will test the same as a non entangled beam of the same polarization. Period.

Come on Vanesch help this guy out, he’s digging himself into hole.

 

[JesseM]

I thought you got no interference pattern because you have to add the two correlations, D0:D1, and D0:D2.

Well, that is a way of understanding why you don't see interference in the total pattern of signal photons at D0 even if you replace some beam-splitters with mirrors so all the idlers end up at D1 or D2 and have their which-path information erased. I discussed this point in the last paragraph of my earlier post on the DCQE from this thread:

Even in the case of the normal delayed choice quantum eraser setup where the which-path information is erased, the total pattern of photons on the screen does not show any interference, it's only when you look at the subset of signal photons matched with idler photons that ended up in a particular detector that you see an interference pattern. For reference, look at the diagram of the setup in fig. 1 of this paper:

http://xxx.lanl.gov/PS_cache/quant-ph/pdf/9903/9903047.pdf

In this figure, pairs of entangled photons are emitted by one of two atoms at different positions, A and B. The signal photons move to the right on the diagram, and are detected at D0--you can think of the two atoms as corresponding to the two slits in the double-slit experiment, while D0 corresponds to the screen. Meanwhile, the idler photons move to the left on the diagram. If the idler is detected at D3, then you know that it came from atom A, and thus that the signal photon came from there also; so when you look at the subset of trials where the idler was detected at D3, you will not see any interference in the distribution of positions where the signal photon was detected at D0, just as you see no interference on the screen in the double-slit experiment when you measure which slit the particle went through. Likewise, if the idler is detected at D4, then you know both it and the signal photon came from atom B, and you won't see any interference in the signal photon's distribution. But if the idler is detected at either D1 or D2, then this is equally consistent with a path where it came from atom A and was reflected by the beam-splitter BSA or a path where it came from atom B and was reflected from beam-splitter BSB, thus you have no information about which atom the signal photon came from and will get interference in the signal photon's distribution, just like in the double-slit experiment when you don't measure which slit the particle came through. Note that if you removed the beam-splitters BSA and BSB you could guarantee that the idler would be detected at D3 or D4 and thus that the path of the signal photon would be known; likewise, if you replaced the beam-splitters BSA and BSB with mirrors, then you could guarantee that the idler would be detected at D1 or D2 and thus that the path of the signal photon would be unknown. By making the distances large enough you could even choose whether to make sure the idlers go to D3&D4 or to go to D1&D2 after you have already observed the position that the signal photon was detected, so in this sense you have the choice whether or not to retroactively "erase" your opportunity to know which atom the signal photon came from, after the signal photon's position has already been detected.

This confused me for a while since it seemed like this would imply your later choice determines whether or not you observe interference in the signal photons earlier, until I got into a discussion about it online and someone showed me the "trick". In the same paper, look at the graphs in Fig. 3 and Fig. 4, Fig. 3 showing the interference pattern in the signal photons in the subset of cases where the idler was detected at D1, and Fig. 4 showing the interference pattern in the signal photons in the subset of cases where the idler was detected at D2 (the two cases where the idler's 'which-path' information is lost). They do both show interference, but if you line the graphs up you see that the peaks of one interference pattern line up with the troughs of the other--so the "trick" here is that if you add the two patterns together, you get a non-interference pattern just like if the idlers had ended up at D3 or D4. This means that even if you did replace the beam-splitters BSA and BSB with mirrors, guaranteeing that the idlers would always be detected at D1 or D2 and that their which-path information would always be erased, you still wouldn't see any interference in the total pattern of the signal photons; only after the idlers have been detected at D1 or D2, and you look at the subset of signal photons whose corresponding idlers were detected at one or the other, do you see any kind of interference.

However, you will also see exactly the same non-interference pattern at D0 if you arrange things so all the idlers go to D3 or D4 and have their which path-information preserved (in which case the D0 pattern will be the sum of the D0:D3 correlation pattern and the D0:D4 correlation pattern) or if you run the experiment normally so the idlers go to a combination of D1, D2, D3 and D4.

 

[JesseM]

Oh for crying out loud those sections are talking about correlations -- did you read his sentence:
[Of course it is possible to get an interference pattern using "one arm" of an entangled beam]

He said it's possible (by changing the distance between slits so that no measurement of the other members of each entangled pair can help you figure out which slit the first set of photons went through), not that you would always see interference if you just look at the photons going through the slits. In the case where you don't change h, you will not see interference in the photons that go through the slits, because in this case a measurement of the entangled photons could tell you which slit each photon went through.

Vanesch, can you confirm?

Treated locally by itself one beam from a pair of entangled beams will test the same as a non entangled beam of the same polarization. Period.

Once again, the DCQE proves you wrong. Please read the paper or at least the description on the wikipedia page, then address my questions to you above in post #26.

 

[Cane_Toad]

Sorry, but I think I'm losing track.

All that is needed is to put a double slit apparatus on both sides in place of the beamsplitters. When the entangled photons go through, they should not (or should depending on who you believe) make an interference pattern. This would demonstrate that entangled photons objectively act differently than unentangled photons.

Ok this means:

backdrop --- double slit --- light source --- double slit --- backdrop

where the light source will emit entangled photons with antiparallel momentum?

I totally agree with your second paragraph; that is all anyone would need to do to prove the point that a single beam of light evolved is unique from a normal single beam of light that by it not capable of producing an interference pattern.

I'm getting lost in the grammer of this statement.

But anyway...

It is because the interference pattern is seen all the time as labs they set up for more complex experiments.

All this was in response to:

Originally Posted by RandallB View Post
First a single photon can never create an interference pattern. A beam of photons and a count of many photon location hits are required to determine if a pattern is being created.

Second, no one has ever demonstrated that a beam of light known to be entangled with a second beam of light will not create an interference pattern when sent through a double slit. By itself it is just a beam of light will always create a pattern. There is no objective test that can be preformed on a single beam of light alone that will reveal if it is entanglement with some other untested beam.

What does happen is if you log all the hits that created the pattern and correlate them with photons from the second beam in a method that would reveal “which way” AND you use only hit locations for those correlated photons to reconstruct a new screen display. Bingo – no more interference pattern

As I see it, there was a disjunct in understanding between Jesse and Randall because Randall was saying quite literally, that a *single* beam of light will always create a pattern, implying that the other untested beam is not going through the D1,2,3,4 system, but Jesse was still considering the ramifications of D1,2,3,4 because he thought you were still discussing the DCQE? Right?

A la, Jesse's response:

That is what quantum theory predicts, and the delayed choice quantum eraser experiment actually demonstrates it.

Thus Randall replies:

NOT TRUE
All "Delayed Choice" experiments require using both beams. No experiment has ever shown any beam of light to fail to produce interface patterns with a double slit unless something is introduced from a second beam “entangled” with the beam being tested. Either by correlation count coordination, or combining both beams somehow.

Still thinking about the situation of a single beam without all the DCQE stuff.

 

[Cane_Toad]

Back to the question that was weirding me out, that I didn't state clearly earlier.

Brian Green:
...[ explains DCQE with idler detectors 10 light years away ]

You see, ten years later, the four photon detectors will receive–one after another–the idler photons. If you are subsequently informed about which idlers wound up, say, in detector 2 (e.g., the first, seventh, eight, twelfth ... idlers to arrive), and if you then go back to the data you collected years earlier and highlight the corresponding signal photon locations on the screen (e.g., the first, seventh, eighth, twelfth ... signal photons that arrived), you will find that the highlighted data points fill out an interference pattern, thus revealing that those signal photons should be described as having traveled both paths.

Alternatively, if 9 years, 364 days after you collected the signal photon data, a practical joker should sabotage the experiment by removing beam splitters a and b–ensuring that when the idler photons arrive the next day, they all go to either detector 1 or detector 4, thus preserving all which-path information–then, when you receive this information, you will conclude that every signal photon went along either the left path or the right path, and there will be no interference pattern to extract from the signal photon data.

So, the situation is that the signal photons have been detected and recorded when all the idler detectors are in place.

Situation 1: the idlers are detected and correlated, you the the expected patterns.

Situation 2: you remove the eraser detectors, i.e. D1,2, before detecting the idlers, leaving D3,4 alone, then you will not be able to correlate any patterns. Has this been done, removing the detectors while the idlers are in flight?

He says that *all* the photons will appear to have gone through D3,4. Presumably this is because the wave fronts were always available to go through D3,4 , even if they would have gone through D1,2 had those detectors not been removed?

Is it saying simply that your correlation process is broken because your signal and idler detections are no longer consistent and valid?

Or is it saying something wiggy about the path integral being rewritten to accommodate the missing detectors?

I can't figure out what this is saying about the signal photons.

The correlation process is to create data sets of photons by matching signal to idler by order received. So in the case where you remove D1,2 then of course you will end up with only two data sets instead of four.

It seems like the reason you get no pattern is because you have merged two data sets which would have been for D1,2 into the two data sets for D3,4 so it should come as no surprise that nothing good happens.

So, were the signal photons that were recorded, when the D1,2 were still in place, affected by the presence of D1,2 , or not? They should have been, right? At that time, the path integral would have included D1,2 when the position was afixed for the signal photon?

I dunno, I still come to the conclusion that what you get is corrupted data that can't be resolved because the D1,2 data was not recorded, and you have a mash of signal points, which would still form a pattern if you magically knew how to separate properly, unlike the situation where D1,2 never existed at all.

But I have the nagging feeling that some weirder is implied, or else he wouldn't gone through the trouble to state something obvious.

 

[JesseM]

So, the situation is that the signal photons have been detected and recorded when all the idler detectors are in place.

Situation 1: the idlers are detected and correlated, you the the expected patterns.

Situation 2: you remove the eraser detectors, i.e. D1,2, before detecting the idlers, leaving D3,4 alone, then you will not be able to correlate any patterns. Has this been done, removing the detectors while the idlers are in flight?

What do you mean by "you will not be able to correlate any patterns"? If all the idlers go to D3 or D4, then you'll know which slit all the signal photons went through. But to make all the idlers go to D3 and D4, you remove the beam-splitters BSA and BSB in the setup shown here with mirrors, rather than just removing D1 and D2.

He says that *all* the photons will appear to have gone through D3,4. Presumably this is because the wave fronts were always available to go through D3,4 , even if they would have gone through D1,2 had those detectors not been removed?

Like I said, you need to remove the beam-splitters to make them go to D3 and D4, if you just remove D1 and D2 the idlers that would have gone there will just end up missing. Greene specifies this when he says:

Alternatively, if 9 years, 364 days after you collected the signal photon data, a practical joker should sabotage the experiment by removing beam splitters a and b–ensuring that when the idler photons arrive the next day, they all go to either detector 1 or detector 4, thus preserving all which-path information–then, when you receive this information, you will conclude that every signal photon went along either the left path or the right path, and there will be no interference pattern to extract from the signal photon data.

Is it saying simply that your correlation process is broken because your signal and idler detections are no longer consistent and valid?

What do you mean "your correlation process is broken"? Again, measuring the idlers at D3 or D4 (1 and 4 in Greene's notation) means that you know which slit the corresponding signal photons went through, which is why the D3/D0 correlation graph and the D4/D0 correlation graph show no interference.

It seems like the reason you get no pattern is because you have merged two data sets which would have been for D1,2 into the two data sets for D3,4 so it should come as no surprise that nothing good happens.

The D3/D0 and D4/D0 correlation graphs never show interference, regardless of whether D1 and D2 are present or not (fig. 5 of the paper shows the D0/D3 graph). They are the non-erasing detectors, so if you have the which-path information, you don't expect interference, just like you don't expect interference in the ordinary non-entangled double-slit experiment if you measure which slit each particle went through.

So, were the signal photons that were recorded, when the D1,2 were still in place, affected by the presence of D1,2 , or not? They should have been, right? At that time, the path integral would have included D1,2 when the position was afixed for the signal photon?

The calculation of the joint probabilities would be different, but the end result for just the signal photons always ends up being the same--no interference, regardless of what apparatus you send the idlers through. All that matters is that you could have measured the idlers in a way that would have revealed which slit the signal photons went through.

 

[JesseM]

Ok this means:

backdrop --- double slit --- light source --- double slit --- backdrop

where the light source will emit entangled photons with antiparallel momentum?

Yes, and if you look on p. 290 of the Zeilinger paper DrChinese linked to (p. 3 of the PDF), you can see an experiment much like this in fig. 2, although they don't actually place a double slit on the right side. As I pointed out earlier, Zeilinger says of the photons on the left that go through the double slit:

Will we now observe an interference pattern for particle 1 behind its double slit? The answer has again to be negative because by simply placing detectors in the beams b and b' of particle 2 we can determine which path particle 1 took

As I see it, there was a disjunct in understanding between Jesse and Randall because Randall was saying quite literally, that a *single* beam of light will always create a pattern, implying that the other untested beam is not going through the D1,2,3,4 system, but Jesse was still considering the ramifications of D1,2,3,4 because he thought you were still discussing the DCQE? Right?

No, RandallB was saying that you would always see interference if you send a beam of light through a double slit, regardless of what happens to their entangled twins. He makes this clear in his most recent post:

All I’m saying is any beam of light I don’t care how entangled it may be, when put to the test as described in post 25 will always create a pattern.
If not where is the experiment to proof it! One where no part of the one arm is split off and combined or measured against the other arm. And especially no correlations of detection left between left arm and right arm before after of during their trip through the double slits.

Treated locally by itself one beam from a pair of entangled beams will test the same as a non entangled beam of the same polarization. Period.

Meanwhile, what I am saying is that it doesn't actually matter what you do to the idlers in the DCQE, you could send them into space if you wanted, the total pattern of signal photons going through the double slit still wouldn't show interference as long as they were entangled in such a way that there was the potential to have found the which-path information by measuring the idlers in the right way. If a scientist could look at how the signal photon/idler photon pairs were emitted, and could see the double-slit apparatus the signal photon went through but had no idea what happened to the idlers, then as long as he would say "yes, with the right experimental setup on the idler side you'd be able to find out which slit the signal photons went through", that's enough to guarantee you won't see interference in the signal photons alone, regardless of whether anyone actually does measure the idlers in this way. After all, if the pattern of signal photons depended on what actually happened to the idlers rather than what could have happened, then by deciding whether or not to measure the which-path information of the idlers you could send an FTL message to a person watching the signal photons come through the slit!

 

[Cane_Toad]

What do you mean by "you will not be able to correlate any patterns"?

Short for, you won't be able to make any correlations which produce interference patterns.

Like I said, you need to remove the beam-splitters to make them go to D3 and D4, if you just remove D1 and D2 the idlers that would have gone there will just end up missing.

That's what I meant to say. Please substitute for the same mistake later.

What do you mean "your correlation process is broken"? Again, measuring

Some, roughly half, of the data collected for the idlers, no longer has consistency with the corresponding signal photons.

the idlers at D3 or D4 (1 and 4 in Greene's notation) means that you know which slit the corresponding signal photons went through, which is why the D3/D0 correlation graph and the D4/D0 correlation graph show no interference.
...

All that matters is that you could have measured the idlers in a way that would have revealed which slit the signal photons went through.

This is ambiguous. I'm not convinced that you know which slit the signal photons passed through for the idlers which would have gone to D1,2 , since you would get the same result from the correlations of those data points, which have been mapped from D1,2 to D3,4, if they are simply corrupt.

In both cases (all D3,4 readings show which-path, or half D3,4 readings meaningless), the D0/3 and D0/4 correlations would show no interference patterns.

The difference is significant to understanding what's happening with the path integral, for me at least.

 

[Cane_Toad]

Yes, and if you look on p. 290 of the Zeilinger paper DrChinese linked to (p. 3 of the PDF), you can see an experiment much like this in fig. 2, although they don't actually place a double slit on the right side. As I pointed out earlier, Zeilinger says of the photons on the left that go through the double slit:

Will we now observe an interference pattern for particle
1 behind its double slit? The answer has again to be
negative because by simply placing detectors in the
beams b and b8 of particle 2 we can determine which
path particle 1 took.

This example doesn't fill me with as much confidence as the others, as it seems to be a thought experiment, i.e. "The answer [has] again to be", and what happens to photon 2 isn't well defined, though he says that you "can" put detectors along b or b'.

Lastly, the example seems to contain an error in the description:

Obviously, 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, a' or b'.

Since, of course, it's stated earlier that 2 can take paths b or b'.

No, RandallB was saying that you would always see interference if you send a beam of light through a double slit, regardless of what happens to their entangled twins. He makes this clear in his most recent post:

All I’m saying is any beam of light I don’t care how entangled it may be, when put to the test as described in post 25 will always create a pattern.
If not where is the experiment to proof it! One where no part of the one arm is split off and combined or measured against the other arm. And especially no correlations of detection left between left arm and right arm before after of during their trip through the double slits.

Randall is talking about "arms" and correlation, which makes me think DCQE. He also doesn't say "regardless of what happens to their entangled twins", he says "I don't care how entangled", which isn't the same, but doesn't help any either.

The waters are muddy. Post #25 doesn't suggest a clear experiment either, but is relative to another.

I don't know about y'all, but it would help me if this stuff was restated more clearly.

After all, if the pattern of signal photons depended on what actually happened to the idlers rather than what could have happened, then by deciding whether or not to measure the which-path information of the idlers you could send an FTL message to a person watching the signal photons come through the slit!

As a side issue, I'm still suggesting that the pattern of the signal photons depends on what potential the path integral has at the time when the signal photons are collected. In either case, though no FTL is possible.

 

[JesseM]

This is ambiguous. I'm not convinced that you know which slit the signal photons passed through for the idlers which would have gone to D1,2 , since you would get the same result from the correlations of those data points, which have been mapped from D1,2 to D3,4, if they are simply corrupt.

What does "corrupt" mean physically? Are you suggesting that the photons somehow know the beam-splitters BSA and BSB in the diagram of the setup here were "supposed" to be there and that something is "wrong" if they're missing? Physically, if you measure a photon at D3, then in terms of the diagram of the setup it could only have come from the atom at A rather than the atom at B (regardless of whether BSA is present or missing), right?

 

[JesseM]

This example doesn't fill me with as much confidence as the others, as it seems to be a thought experiment, i.e. "The answer [has] again to be", and what happens to photon 2 isn't well defined, though he says that you "can" put detectors along b or b'.

Perhaps, but it at least shows that RandallB is wrong if he thinks that his claims agree with QM's predictions. And if he admits that QM predicts no interference in this case, then if he thinks something different will happen he should be posting in the "Independent Research" forum for novel theories, not here

Lastly, the example seems to contain an error in the description

True, although from the context it's pretty clear what he meant to type.

Randall is talking about "arms" and correlation, which makes me think DCQE. He also doesn't say "regardless of what happens to their entangled twins", he says "I don't care how entangled", which isn't the same, but doesn't help any either.

Either way, he says "All I’m saying is any beam of light ... when put to the test as described in post 25 will always create a pattern." The beam of signal photons going through the slit seems to match the quote from DrChinese in post 25:

All that is needed is to put a double slit apparatus on both sides in place of the beamsplitters. When the entangled photons go through, they should not (or should depending on who you believe) make an interference pattern.

Now, it's true that in the DCQE you don't put a double slit apparatus on both sides, just one. But then RandallB says any beam of light will create an interference pattern, I'm pretty sure he wouldn't admit that even a single beam of light can fail to produce interference when sent through a double-slit apparatus, regardless of what happens to another entangled beam. But if I'm misunderstanding, then hopefully he'll clarify.

As a side issue, I'm still suggesting that the pattern of the signal photons depends on what potential the path integral has at the time when the signal photons are collected. In either case, though no FTL is possible.

What do you mean by the "potential" of the path integral? If the photons are moving in opposite directions at the speed of light, then nothing that happens to an idler after the idler/signal photon pair is created will be in the past light cone of the event of the signal photon hitting the screen (or D0), so if the probability that the signal photon would be detected in a particular spot depended on anything that happened to the idler after the moment it was created, this would imply the possibility of FTL messages.

 

[Cane_Toad]

What do you mean by the "potential" of the path integral? If the photons are moving in opposite directions at the speed of light, then nothing that happens to an idler after the idler/signal photon pair is created will be in the past light cone of the event of the signal photon hitting the screen (or D0), so if the probability that the signal photon would be detected in a particular spot depended on anything that happened to the idler after the moment it was created, this would imply the possibility of FTL messages.

I don't know if "potential" is a good word to use, but I mean that the instant before the signal photon is collected, the path integral should still contain all possibilities, regardless of the light years involved, but after the signal photon is collected, the path integral only contains the idler photon and its possibilities.

Once all the signal photons have been collected, you should be able shut it all down, take your signal data and run a program to figure out what the idlers should have been in order to get the interference pattern you want. The idlers are obsolete.

 

[JesseM]

I don't know if "potential" is a good word to use, but I mean that the instant before the signal photon is collected, the path integral should still contain all possibilities, regardless of the light years involved, but after the signal photon is collected, the path integral only contains the idler photon and its possibilities.

But wouldn't the path integral for both particles be different given different experimental setups (different placements of detectors/beam-splitters/etc.)? And yet the choice of which setup to use on the idler side can be outside the past light cone of the event of the signal photon's detection. So if you're saying that the choice of setup to use on the idler's side can affect the probabilities that the signal photon will be detected at different positions on the screen--and I'm not clear if you are--then this would imply the possibility of FTL communication.

Once all the signal photons have been collected, you should be able shut it all down, take your signal data and run a program to figure out what the idlers should have been in order to get the interference pattern you want. The idlers are obsolete.

You mean, just highlight a subset of the signal photon detection positions such that the highlighted dots will form an interference pattern, and then imagine that the corresponding idlers went to a which-path-erasing detector? You could do this, but you aren't saying you could use this to actually predict which idlers go to a particular detector, are you?

 

[Cane_Toad]

But wouldn't the path integral for both particles be different given different experimental setups (different placements of detectors/beam-splitters/etc.)? And yet the choice of which setup to use on the idler side can be outside the past light cone of the event of the signal photon's detection. So if you're saying that the choice of setup to use on the idler's side can affect the probabilities that the signal photon will be detected at different positions on the screen--and I'm not clear if you are--then this would imply the possibility of FTL communication.

Yes. When signal and idler are in flight, the idler setup is still in the light cone of the signal, right?

However, I'm saying that the effects of the idler detector setup ceases to affect the signal photon at the moment the signal photon is collected. Thus FTL is excluded.

You mean, just highlight a subset of the signal photon detection positions such that the highlighted dots will form an interference pattern, and then imagine that the corresponding idlers went to a which-path-erasing detector? You could do this, but you aren't saying you could use this to actually predict which idlers go to a particular detector, are you?

This was a joke.

 

[vanesch]

Come on Vanesch help this guy out, he’s digging himself into hole.

Well, I'm sorry but I'm rather in agreement with what JesseM writes, only, it seems that you guys are talking next to each other. That said, I am also in agreement with certain points you make ; however, both are not contradictory.

There are two totally different issues here, which I hoped I was making clear. There is one point, which is: "making interference patterns" of beams of a certain quality. And then there is another point, and that is: doing measurements in such a way that the entangled quality of two beams is used (and eventually, how this plays out in trying to make FTL signaling devices).

The confusing issue here is the term "making interference patterns", which can mean different things. Clearly, an ideal, monochromatic, linearly polarized beam of light will ALWAYS make an interference pattern in a 2-slit experiment ; also, such a beam can NEVER be the "one arm" of an entangled set of beams!

In quantum speak, if our beam is "in a pure state", then it factors out in the overall wavefunction, and hence, by definition, is not entangled.

Now, in order to have interference, it is not necessary to have a pure state. You can also have interference in mixtures, but not always: only when the two components that are to interfere are sufficiently correlated. This is what is classically described by coherence lengths and times. This is why even "noisy" light can give interference patterns, if only we work with small enough coherence lengths and times. But once we go beyond these small coherence lengths and times, there is no interference anymore.
So a statistical mixture of pure states will give rise to a limited capability to give rise to interference.

Now, what with entanglement ? The whole point by using interference in entangled states is to try to have "one slit" of beam A to correspond with a measurable property of beam B, and "the other slit" of beam A to correspond with the complementary property of beam B. This is interesting because it gives us the idea that we might "cheat" on the interference mechanism: by using the measureable property on beam B, we might find out (potentially) through which slit beam A went, and nevertheless have an interference pattern. THIS is what is impossible, for the following reason.
AS LONG AS IT IS POTENTIALLY POSSIBLE (I'm with JesseM here) to do so, no interference pattern can be obtained by beam A.
But this is not due to some magic "weird behaviour of beam A", rather, it is because beam A, when looked at locally, will correspond exactly to such a statistical mixture, that the desired interference experiment will lie outside of the coherence lengths and times that are necessary for obtaining an interference pattern in these conditions.
Now, somebody who only has access to beam A, would be totally in agreement with this find, because he would, after doing some spectral analysis and so on, find out that indeed, beam A is statistically so mixed, that its coherence length is too short to see an interference pattern.

But of course you can now do ANOTHER interference experiment with beam A, which is within the coherence length of that beam, and then you WILL of course find a pattern. Only, THIS specific interference experiment hasn't gotten anything to do anymore with the original aim of the entangled beams, which was, to be able to "cheat" on the interference mechanism. Indeed, now you will find out that the entangled beams are such, that beam B hasn't gotten anything to say anymore about which slit beam A might go through. In other words, the two slits of beam A are now such, that the states corresponding to the new slits are not entangled to orthogonal states in beam B. No measurable quantity on beam B can tell you now through which slit you went at A.

Let us go back to our original interference setup at A, so that there is a potential measurable property of beam B that tells us through which slit beam A went. No interference pattern should occur in this case.
Now, what if you "destroy" beam B, or whatever measurement you do on it that will make it impossible for you to restore the "which slit" information ?
It won't change anything: beam A, as seen just as a single beam, hasn't changed, and is still the statistical mixture it was before B got destroyed/measured/whatever. As such, its coherence length is not good enough to produce an interference pattern. It is not by doing something with beam B, that something will change on the A side.

However (and these are those famous DQE experiments), you could do a measurement on beam B, which makes it impossible to restore the which-slit information, simply because it is an incompatible measurement.
Now you can USE this information obtained by this measurement on beam B, to go and SUBSAMPLE the hits you found on beam A. And THEN, it is possible, USING THIS SUBSAMPLING, to find in the selected dataset an "interference pattern" on the A-side. But this pattern is a *subsample* of the total pattern on side A, which didn't show any interference overall.

Now, imagine we do this, using the "narrowed-down" slits which made beam A have an overall interference pattern. You can now do on beam B what you want, and use this information to subsample the data on the A side as much as you like, you will ALWAYS find the same interference pattern on the A side. Why ? Because the interference pattern on the A side is now due to a state which FACTORED OUT and is hence not entangled with any property on the B side. So there is no specific correlation that will appear.

EDIT: upon re-reading some of the posts here, I would like to make something a bit more clear, and I'm not sure whether it is RandallB or JesseM who is (mis)understanding something, or whether there's an empty dispute.
It is true that locally, one cannot see any difference between an "entangled" and a "non-entangled" beam. However, a non-entangled beam might be a pure beam, while an entangled beam will ALWAYS (APPEAR TO) BE A STATISTICAL MIXTURE OF SOME KIND.
But when these statistical properties are accounted for, there is indeed, locally, no way of discriminating, by no experiment, between a "truely statistical mixture of non-entangled beams" and "one arm of an entangled beam".

 

[DrChinese]

It is true that locally, one cannot see any difference between an "entangled" and a "non-entangled" beam. However, a non-entangled beam might be a pure beam, while an entangled beam will ALWAYS (APPEAR TO) BE A STATISTICAL MIXTURE OF SOME KIND.
But when these statistical properties are accounted for, there is indeed, locally, no way of discriminating, by no experiment, between a "truely statistical mixture of "un-entangled beams" and "one arm of an entangled beam".

I think what you have said is similar to the following:

An entangled beam is always a statistical mixture, say of H and V. Therefore self-interference effects will cancel in a double slit setup. There will be no interference pattern, just the 2 bars.

An un-entangled beam can be either in a pure state - which leads to an intereference pattern after the double slit - or a mixed state, which will not lead to a pattern.

Am I close?