Exploring the Possibility of FTL Communication with Entangled Particles

[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.