EPR and FTL

1. May 17, 2007

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.

Last edited: May 17, 2007
2. May 17, 2007

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.

Last edited: May 17, 2007
3. May 17, 2007

hellfire

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.

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4. May 17, 2007

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.

Last edited: May 17, 2007
5. May 17, 2007

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...

Last edited: May 17, 2007
6. May 17, 2007

ueit

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

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

7. May 17, 2007

DrChinese

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

Last edited: May 17, 2007
8. May 17, 2007

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!

9. May 17, 2007

DrChinese

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.

10. May 17, 2007

JesseM

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.

11. May 17, 2007

Tomsk

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.

12. May 17, 2007

JesseM

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.

13. May 18, 2007

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

14. May 18, 2007

Tomsk

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

15. May 18, 2007

hellfire

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

16. May 18, 2007

DrChinese

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

17. May 18, 2007

RandallB

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 its self 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

18. May 18, 2007

JesseM

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

19. May 19, 2007

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

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.

20. May 19, 2007

JesseM

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:

21. May 19, 2007

DrChinese

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.

22. May 20, 2007

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

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?

23. May 20, 2007

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

24. May 20, 2007

JesseM

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.
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.

25. May 20, 2007

RandallB

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.