Exploring the Possibility of FTL Communication with Entangled Particles

[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.

:rofl:

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? :tongue:

 

[Cane_Toad]

:rofl:

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.

I don't understand this in the context of a sequential stream of photons where the photons are "interfering with themselves".

Or does coherence length/time only apply to statistical samplings?

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.

Ack! How does this happen? Are you saying that beam B is no longer entangled because you are detecting inside the coherence length/time? I can't tell whether you are saying that the entanglement has been "broken", or that the entanglement is somehow no longer applicable.

Secondly, this doesn't make any sense to me unless the coherence length of beam A and beam B are both taken into account, i.e. you could reduce the coherence L/T of beam B until it again affects beam A?

I think we revisit this below...

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.

Let's see if I'm understanding anything...

To do this, you will take the idler samples collected from the "eraser" detectors, and correlate those with the signal beam.

The samples from signal beam, A, will be the same regardless of whether there is an idler beam, B, right? I'll try a better restatement of this below.

I'm really unsure as to what the which-path-erased samples from beam B are saying about beam A. They certainly aren't changing beam A. Right?

This gets to a point I have been trying to clear up about what is happening to the path integral for a given entangled photon pair. When the pair is still in flight, the path integral is fully indeterminate. Once the signal photon is collected, it's portion of the path integral collapses, right? However, the idler photon must use the path integral which includes the collapsed signal photon in order to get the [non]which-path information, so the collapsed signal photon wave function must be sufficient to calculate with the collapsing idler photon wave function to produce information which can be subsequently used to do a statistical correlation with the signal photon.

So, the which-path information in this scenario is unknown until the idler function has collapsed (after the signal has collapsed); the signal photon's detected position is invariant, meaning that if before collecting the idler photon, you removed the eraser setup when the idler is in mid-flight, the detected signal photon position is unaffected.

Presumably, this should work symmetrically if the idler photon is collected first.

Did I get any of this right?

Now, imagine we do this, using the "narrowed-down" slits which made beam A have an overall interference pattern.

By the way, I thought that the coherence length was the distance between the source and where the coherence breaks down, i.e. before or after the detection points. What's "narrowed-down"?

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.

Please explain, "FACTORED OUT". What is it about the coherence L/T that destroys the entanglement? This implies I've got something wrong with my previous descriptions.

 

[vanesch]

I don't understand this in the context of a sequential stream of photons where the photons are "interfering with themselves".

Or does coherence length/time only apply to statistical samplings?

Yes, coherence length/time etc... is a description of the (apparent) statistical mixture if you want. But the relationship between photons and classical electromagnetic field descriptions is quite involved if you want to do this entirely correctly, so in any case these explanations are always a bit approximate.

We take it here that we work in the low intensity limit, where we can associate a single photon with a single EM mode, preferentially of "pure" frequency. However, even that is not 100% correct: you can consider PURE photon states which contain different frequencies (a superposition of pure momentum states). These will not be statistical mixtures, but will nevertheless display finite "coherence time" in any interference experiment. Nevertheless, in "time-of-flight" experiments, they would turn out to be "pure" and not mixtures.

It's always the same:
a quantum state |a1> + |a2>, where a1 and a2 are two orthogonal quantum states, will be indistinguishable from a mixture of 50% |a1> and 50% |a2> when we only look upon the a1/a2 aspect, but will be able to be discriminated from such a mixture when looking in another basis, for instance the basis (|a1> + |a2>) and (|a1> - |a2>).

So, let us work in the basis of harmonic EM modes, and consider the time-dependent classical part as being exp(i w t), purely monochromatic. So infinite coherence time. We only look at spatial modes. In the low counting limit, and when we are not going to look at temporal correlations and so on, we can say that to each pure EM spatial mode, corresponds a pure quantum state of a photon.

Now, the point is, that in the setup of the OP, we have a slit at L+(beam A) at +h and one at -h. We will call the two spatial beam modes that go through these respective slits, the SLITUPA and the SLITDOWNA mode, and with it correspond two orthogonal photon states. Now, the idea of an entangled beam with B is that the overall EM field is filled with 2-photon states of the kind:
|EM state> = |tagB1>|SLITUPA> + |tagB2>|SLITDOWNA>

It doesn't really matter what is tagB1 and tagB2, it are just two complementary (hence orthogonal) states of beam B that can be distinguished by the "tag" measurement on beam B.
THIS is what is meant with the correlation experiment:
IF we find a tagB1 photon in the B beam, then this means that the A beam was in the state SLITUPA, and hence that it was in the mode that went completely through slit A-up (and NOT through slit A-down).

When looking now purely at beam A by itself, this will be represented by a STATISTICAL MIXTURE of 50% photons in the state |SLITUPA> and 50% photons in the state |SLITDOWNA>. This means that this corresponds classically to the statistical mixture of the two modes which go each through a single slit. No interference will happen.

Ack! How does this happen? Are you saying that beam B is no longer entangled because you are detecting inside the coherence length/time? I can't tell whether you are saying that the entanglement has been "broken", or that the entanglement is somehow no longer applicable.

What follows will be highly symbolic and of course to be taken with a grain of salt.
But if we narrow down the slits at A, we have to re-write our modes:

|SLITUPA> = |slitabitupA> + |slitabitdownA> + |otherstuff>

which is the symbolic representation of the fact that the mode which was supposed to only go through the (original) slit UP will partly illuminate the two nearer slits (if not, you simply kill the whole beam!). So part of "slitupA" needs to illuminate the "center part" of the slit system at A, if we have not two totally separated beams at A (and then you cannot change the distance between the slits!).
If the slits are now much nearer to each other, the small part of SLITUPA that falls upon slitabitupA will be very similar to the equally small part of that same SLITUPA that falls upon slitabitdownA. This is why the two contributions are essentially equal.

In the same way, the |SLITDOWNA> will also be:

|SLITDOWNA> = |slitabitupA> + |slitabitdownA> + |otherstuffprim>

Indeed, the mode illuminating the original DOWN slit will also have to illuminate a bit the central part if we are going to be able to do some interference experiment with slits closer together.

And now you see what happens:
if you write the same state out in these terms, you find:

|EM state> = (|tagB1>+ |tagB2>) (|slitabitupA> + |slitabitdownA>) +
|tagB1> |otherstuff> + |tagB2> |otherstuffprim>

And now you see that, concerning those photons that get through the closer slits, only the first term matters, and you have a product state of beam B and beam A. Of course, most of the state is in the parts of the beam that would have gone through the original slits, and are still entangled, but in the "near slit" experiment, we don't look at those.

So the (small) part of the state we look at with our narrower slits is now in a product state: (|tagB1>+ |tagB2>) (|slitabitupA> + |slitabitdownA>)

So it is not entangled anymore. You can now measure on B what you want, this will not be correlated to the measurements on the A beam.
Moreover, the A beam is in a state |slitabitupA> + |slitabitdownA> which will of course give rise to an interference pattern, as this corresponds to a single mode of classical EM field with the two slits illuminated.

 

[JesseM]

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

Not if the equipment on the idler's side is set up after the signal and idler have been created (as could be true in Greene's hypothetical example where the idler traveled 10 light years). If the signal photon and the idler head off in opposite directions at the speed of light, then the only part of the idler's worldline that will lie in the past light cone of the signal photon at any moment along its path will be the event of the idler being created.

 

[Cane_Toad]

Not if the equipment on the idler's side is set up after the signal and idler have been created (as could be true in Greene's hypothetical example where the idler traveled 10 light years). If the signal photon and the idler head off in opposite directions at the speed of light, then the only part of the idler's worldline that will lie in the past light cone of the signal photon at any moment along its path will be the event of the idler being created.

I'm having trouble visualizing what's happening with the light cones. Do the light cones of the two photons ever see each other at all after creation? The start out in each other's cones, and then as each photon moves away, it's light cone is expanding at C also, so why wouldn't they always remain intersected?

I was trying to use the light cone to help understand how the path integral evolves as the two photons arrive at the detectors, but this might not be valid.

 

[RandallB]

:rofl:
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.

Well as funny as it may seem, but I’m sorry you are wrong.
Our two positions are completely contradictory with no room for agreement. JesseM made it very clear that one beam by itself from a pair of entangled beams is not able to produce interference patterns when sent through a double slit. I am equally clear about that no beam can be independently tested to reveal if it is or ever was entangled, period! Such a beam will always produce a pattern! These two positions can never be anything but contradictory.

Your comment :
“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!”
Makes zero sense! That is exactly what each arm of light produced a PDC is; monochromatic & linearly polarized, one vertical the other horizontal. The idea the pattern might not been seen is wrong. The only way the pattern is “erased” is when only selected photons in the pattern are used, based on correlation or coordination with the “other arm” do those selected photon fail to show a pattern.

Yes I understand that is not Brain Greene says on page 198 in “Fabric of the Cosmos”, problem there is he is WRONG! What he describes is a “pattern creation” experiment which no one has done. The workable experiment is called an eraser experiment for a reason.
Yes I see Anton Zeilinger makes the same written claim by incorrectly using Dopfer results. If you actually look at those results you once again see that there is a pattern and the pattern is seen to be “erased” only when correlations are being made. Without that patterns are found.

So what if these guys are Profs with credibility – wrong is wrong, and these errors in print make it hard for those like Cane Toad trying to learn.

The proof is simple with the experiment DrC and Cane Toad referred to :

Anything <----PDC --->Double Slit ---> Detection

The “anything” on the left can even be a full collection of detailed 4 detector photon counts just as Greene describes in the book, just as long as none of that information is used on the right side. The right side detection across the screen of image created on the screen will always give an interference pattern.
This experiment is too simple to continue this debate without a direct reference to such an experiment showing otherwise with any type of entanglement.

Without that experiment those are my closing comments, good luck Cane Toad.

 

[JesseM]

Our two positions are completely contradictory with no room for agreement. JesseM made it very clear that one beam by itself from a pair of entangled beams is not able to produce interference patterns when sent through a double slit.

I only claim that if they are entangled in such a way that it would be possible in principle to determine which slit each photon went through by measuring its entangled twin in the right way, then you won't see interference in the total pattern of photons going through the double-slit. I've agreed with vanesch that there are situations where you can get an entangled beam to show interference, but these are exactly the situations where the details of the entanglement or the distance between slits are such that you could not use the entangled twins to determine which slit the first set of photons went through.

I am equally clear about that no beam can be independently tested to reveal if it is or ever was entangled, period! Such a beam will always produce a pattern!

So here you're pretty clearly making a blanket statement that you will never see interference if you send one beam of photons through a double-slit, even when it is possible to determine which slit each of these photons went through by measuring their entangled twins. This is the whole thing I've been disagreeing with you about, and vanesch has said that I am correct:

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.

Your comment :
“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!”
Makes zero sense! That is exactly what each arm of light produced a PDC is; monochromatic & linearly polarized, one vertical the other horizontal. The idea the pattern might not been seen is wrong.

I think vanesch may be using the word "ideal" to mean "in a pure quantum state rather than a mixed state", which would mean that by definition one half of a set of entangled photons could not be an "ideal" beam. But maybe he can elaborate on what he meant here...I'm pretty sure an entangled beam can be monochromatic and linearly polarized.

The only way the pattern is “erased” is when only selected photons in the pattern are used, based on correlation or coordination with the “other arm” do those selected photon fail to show a pattern.

Again, it's the opposite--you only see interference in correlation patterns, not the total pattern. What's more, if you look at the correlation graphs in the paper you can see that the total pattern would just be the sum of the different correlation graphs, and the paper mentions that there is a phase shift of pi between the D0/D1 correlation graph and the D0/D2 correlation graph, meaning the peaks of one graph line up with the valleys of the other and vice versa (which you can see just by comparing fig. 3 and fig. 4), so that their sum would just be non-interference pattern like fig. 5, the D0/D3 pattern. During the course of a debate on the delayed choice quantum eraser last year I actually emailed one of the authors of the paper just to make sure the sum of the D0/D1 interference pattern and the D0/D2 interference pattern would not show any interference, and he confirmed it.

Yes I understand that is not Brain Greene says on page 198 in “Fabric of the Cosmos”, problem there is he is WRONG! What he describes is a “pattern creation” experiment which no one has done. The workable experiment is called an eraser experiment for a reason.
Yes I see Anton Zeilinger makes the same written claim by incorrectly using Dopfer results. If you actually look at those results you once again see that there is a pattern and the pattern is seen to be “erased” only when correlations are being made. Without that patterns are found.

So what if these guys are Profs with credibility – wrong is wrong, and these errors in print make it hard for those like Cane Toad trying to learn.

But why are you so sure they are wrong? Vanesch has also provided you with theoretical arguments for why one should not see interference in an entangled beams when it's possible to use the entangled twins to determine which slit each photon in the first beam has gone through. Have you actually done any calculations to see that you would see interference in these circumstances? Or are you claiming to be sure you'd see interference even if quantum theory predicts otherwise? If so, then you should be discussing your reasons in "Independent Research". If not, I don't understand how you can possibly be so confident that you're right and all these "Profs with credibility" are wrong when you've done no detailed calculations. Is there some principle of QM you're appealing to? What is the argument in your head that makes you so sure you couldn't be mistaken?

The proof is simple with the experiment DrC and Cane Toad referred to :

Anything <----PDC --->Double Slit ---> Detection

The “anything” on the left can even be a full collection of detailed 4 detector photon counts just as Greene describes in the book, just as long as none of that information is used on the right side. The right side detection across the screen of image created on the screen will always give an interference pattern.

That's not a proof at all, it's just an assertion. Why are you so sure that "anything" going through a double slit will show interference? Why why why?

This experiment is too simple to continue this debate without a direct reference to such an experiment showing otherwise with any type of entanglement.

Surely we can also discuss what quantum theory predicts about this type of experiment. Again, are you claiming to be a maverick disputing the predictions of QM, or are you claiming that a theoretical calculation in QM would support your statements?

In any case, I would still say the DCQE qualifies as the "experiment" you describe, since one can just graph the total pattern of signal photons at D0 and check if they show interference or not. The fact is that they don't.

 

[JesseM]

I'm having trouble visualizing what's happening with the light cones. Do the light cones of the two photons ever see each other at all after creation? The start out in each other's cones, and then as each photon moves away, it's light cone is expanding at C also, so why wouldn't they always remain intersected?

I was trying to use the light cone to help understand how the path integral evolves as the two photons arrive at the detectors, but this might not be valid.

The two past light cones just look like two triangles side-by-side (each one looking like the grey past light cone in the diagram here) with the right corner of the one on the left touching the left corner of the one on the right (that point is the event of the signal and idler being created), and the peaks being the events of the signal and idler being detected. Since both photons move at the speed of light, their worldlines are edges of the triangles representing the light cones--the right edge of the triangle on the left could be the idler's worldline and the left edge of the triangle on the right could be the signal photon's worldline. So then you can pick any point on the signal photon's worldline and draw the past light cone of that event, looking like a triangle with that point as the peak--it'll be a smaller triangle within the larger triangle (the larger triangle, again, is the past light cone of the signal photon being detected), and its left edge will coincide with a section of the left edge of the larger triangle. This means that the smaller triangle won't contain any more of the idler's worldline than the larger triangle--in both cases, only the event of the idler being created lies within the past light cone of the signal photon at any point on its worldline.

 

[RandallB]

So here you're pretty clearly making a blanket statement that you will never see interference if you send one beam of photons through a double-slit, even when it is possible to determine which slit each of these photons went through by measuring their entangled twins. This is the whole thing I've been disagreeing with you about, and vanesch has said that I am correct:

NO I don’t say that, you claim you will NOT see interference – I’m say you WILL see interference! And yes even if data exists that can select which photons went through which slit.

In any case, I would still say the DCQE qualifies as the "experiment" you describe, since one can just graph the total pattern of signal photons at D0 and check if they show interference or not. The fact is that they don't.

No you’re still using correlations – the part of the DCQE that applies to the experiment DrC and Cane Toad suggested is without correlations. . You’re still missing why the experiment is called ERASER!. You get test results for 1,000,000 photons at D0 that give you interference. With each photon numbered 1 – through 1,000,000 and where it landed. Even if you have a list of 150,000 photon IDs that you know WhichWay about, when you look at all 1,000,000 events it still shows a pattern!
Only when you pick out only the 150,000 photons that you do know WhichWay info do you see the pattern for that smaller group of photon go away or ERASE! This is why I know Greene is wrong

What Greene is claiming is the group of 1,000,000 photons will not showing a pattern just because he has in hand a list of 150,000 of them that he knows WhichWay info. And only if he selects for another list of photons 200,000 that he is sure he does not know WhichWay will he get the interference pattern to appear that group. That is not erasing a pattern that is creating a pattern and no one has done that. It is so illogical I don’t understand how anyone buys into it. You atually have to use the correlations as they do in every DCQE to get the pattern to ERASE!

The large photon count unreduced by selective correlations will always show interference from a single beam of light. No matter the source or “how much” entanglement” for that single beam. A PDC can give you a horizontal or vertical beam, or the “sweet spot” used by most experiments, mixing both to appear as circular polarization, doesn’t matter. The only way a pattern is not going to show is if you use a horizontal beam on vertical slits. A good vertical slit or slits should act as a polar filter and block all horizontal light.

 

[JesseM]

So here you're pretty clearly making a blanket statement that you will NEVER see interference if you send one beam of photons through a double-slit, even when it is possible to determine which slit each of these photons went through by measuring their entangled twins. This is the whole thing I've been disagreeing with you about, and vanesch has said that I am correct:

NO I don’t say that, you claim you will NOT see interference – I’m say you WILL see interference! And yes even if data exists that can select which photons went through which slit.

That's exactly what I just said you were saying--see the part in bold if it wasn't clear.

No you’re still using correlations

No I'm not, I'm talking about the total pattern of signal photons at the D0 detector, which is the sum of 4 different correlation patterns D0/D1, D0/D2, D0/D3, and D0/D4. Since these correlation patterns cover all possibilities, every signal photon should be present in one of them, so their sum should be the total pattern of all signal photons.

You’re still missing why the experiment is called ERASER!. You get test results for 1,000,000 photons at D0 that give you interference. With each photon numbered 1 – through 1,000,000 and where it landed. Even if you have a list of 150,000 photon IDs that you know WhichWay about, when you look at all 1,000,000 events it still shows a pattern!
Only when you pick out only the 150,000 photons that you do know WhichWay info do you see the pattern for that smaller group of photon go away or ERASE!

No, you've got it wrong, and if you actually read the paper you would see this. If you have 1,000,000 signal photons at D0, these photons show no interference. If 250,000 of these signal photons had idlers that went to D3, you can graph just this subset of signal photons (shown in fig. 5 of the paper), and since D3 preserved the which-path information, this subset won't show inteference either (same with D4). But if you look at 250,000 signal photons whose corresponding idlers went to D1, then since D1 has erased the which-path information, if you just graph this subset of signal photons (shown in fig. 3) you will see an interference pattern (same with D2, shown in fig. 4). This is the meaning of "eraser".

If you take the sum of all the correlation graphs--the graph of the 250,000 then went to D1 + the graph of the 250,000 that went to D2 + the graph of the 250,000 that went to D3 + the graph of the 250,000 that went to D4--then naturally what you have is a graph of the total pattern of all 1,000,000 signal photons. And as I said before, if you look at the graphs in the paper and read what they said about a pi phase shift between the D0/D1 interference pattern and the D0/D2 interference pattern (which you can see visually by looking at fig. 3 and fig. 4 and noting the peaks of one line up with the valleys of the other and vice versa), you can see that in their sum the interference is canceled out. And naturally since D0/D3 and D0/D4 don't show interference, adding those two won't recreate any interference, so the sum of all 4 correlation graphs shows no interference.

Please actually read through the paper to make sure you understand the setup, and then look at the diagrams, you'll see that what I'm telling you is correct.

What Greene is claiming is the group of 1,000,000 photons will not showing a pattern just because he has in hand a list of 150,000 of them that he knows WhichWay info.

No, Greene didn't make that sort of argument, in fact he didn't explain the reasons for his prediction about the total pattern of signal photons at all (it's a popular book, so he's free to just explain what the results would be according to orthodox QM without explaining the derivation). He just said, flat out, that the total pattern of signal photons on the screen won't show interference. If you want reasons why this is the prediction of orthodox QM, you could start by reading vanesch's posts.

That is not erasing a pattern that is creating a pattern and no one has done that. It is so illogical I don’t understand how anyone buys into it. You atually have to use the correlations as they do in every DCQE to get the pattern to ERASE!

You're misunderstanding the meaning of the term "delayed choice quantum eraser". "Erasing" refers not to erasing an interference pattern, but to erasing the which-way information for the idlers, which creates an interference pattern (I hope you're not basing your whole case on this misunderstanding of the etymology--if you like I can dig up quotes from physicists where they talk about 'erasing' which-path info in the context of the DCQE experiment, whereas I don't think you'll find quotes where they talk about 'erasing' interference patterns in this context). Again, please read the paper and look at the graphs.

The large photon count unreduced by selective correlations will always show interference from a single beam of light. No matter the source or “how much” entanglement” for that single beam.

You just keep repeating this claim over and over, but you never provide any calculations or even refer to principles of QM in order to justify it. Why are you so confident this is correct? I would think that having big-name physicists like Greene and Zeilinger (along with knowledgeable posters on this board like vanesch) disagreeing with you would at least give you pause if you don't have an airtight case for believing this.

And again, it would help if you would answer my earlier question: are you claiming that orthodox QM would make the same prediction that you are making here, or are you just saying you're sure this is what would be seen experimentally, and if orthodox QM predicts something different then QM is wrong?

 

[RandallB]

That's exactly what I just said you were saying--see the part in bold if it wasn't clear.

? How is "you will NEVER see interference" the same as "you WILL see interference!"

No I'm not, I'm talking about the total pattern of signal photons at the D0 detector, which is the sum of 4 different correlation patterns D0/D1, D0/D2, D0/D3, and D0/D4.

Well of course that is not true D1+D2+D3+D4 will not give the total in D0. Nothing prevents two or more detectors side by side in D1 area and others for D1a, D1b, D1c etc.
The only thing that counts is measure D0 alone WITHOUT CORRALATION COUNTS to anything, the patttern will be there.
One arm only.

 

[JesseM]

? How is "you will NEVER see interference" the same as "you WILL see interference!"

Sorry, that was my mistake, I got confused when I was contrasting your argument with mine, since I had just said you can sometimes see interference, I was looking for the opposite and it came out "never see interference" when I really should have said "always see interference". I understood what you meant though, if you look at the rest of that post you can see I was arguing against the position that you'll see interference in cases like the DCQE.

Well of course that is not true D1+D2+D3+D4 will not give the total in D0. Nothing prevents two or more detectors side by side in D1 area and others for D1a, D1b, D1c etc.

I had thought that virtually all the idlers would end up at one of the four detectors--remember that we're dealing with lasers whose paths are very close to the perfect straight lines depicted in the diagram with minimal spreading, and I think it'd also be true that the reason the photons at the D0 detector have a wider range of possible positions is that they go through a double-slit which increases their momentum uncertainty by narrowing their position. Still, you could be right that some significant fraction of signal photons at D0 will not have their entangled idlers detected by any of the four detectors, so it's true that my argument about adding the four is not airtight. On the other hand, if the total pattern of signal photons at D0 did show interference as you imagine, I can't see why the subset of signal photons at D0 whose idlers happened to end up at any of the 4 detectors (not just the which-path preserving ones) would show non-interference. I emailed one of the authors of the paper a question about it in the past, maybe one of us should email them again to see if they recorded the total pattern of photons at D0 and checked what type of pattern they made?

Anyway, regardless of what you think of my argument, the claims of Greene/Zeilinger/vanesch about not seeing interference in the total pattern of photons on the screen were not justified in the same way that I justified it (Zeilinger and vanesch were not even talking about the DCQE experiment), so again, why are you so confident that they are all wrong and you are right? What principle are you appealing to that makes you confident the total pattern must show interference if you haven't even done the calculations to find the probability distribution?

 

[vanesch]

I think vanesch may be using the word "ideal" to mean "in a pure quantum state rather than a mixed state", which would mean that by definition one half of a set of entangled photons could not be an "ideal" beam. But maybe he can elaborate on what he meant here...I'm pretty sure an entangled beam can be monochromatic and linearly polarized.

Well, there must be some stochastic element to a beam which is "half an entangled" beam: limited coherence, or lack of polarization. The reason is the following:

if the entangled state is of the kind: |a1>|b1> + |a2>|b2>, where a1 and a2 are two pure quantum states, which we take (in the low intensity limit) to correspond to two perfectly monochromatic (pure k vector) and/or polarized beams, BUT WHICH ARE OF COURSE PERFECTLY DISTINGUISHABLE (that means: different direction, or different frequency, or different polarization), then the quantum description of the a-beam, by a reduced density matrix, is given by a mixture of 50% |a1> and 50% |a2>. As such, this beam cannot be purely monochromatic, plane (single k) and perfectly polarized, because then there aren't any degrees of freedom left to make the mixture. So at least one of these degrees of freedom (or a combination of them) must be stochastic: it could be the directivity (direction of k, hence finite spatial coherence), it could be the frequency (hence finite temporal coherence), or it could be the polarization (hence unpolarized light). But for sure, such a beam cannot be in a pure plane monochromatic and perfectly polarized state, as that would correspond to a single pure quantum state of the photons in this beam, which would factor out, and we wouldn't have any entanglement.

It is the quality (the degree of freedom, be it frequency, or direction, or polarization, or a combination) which is entangled, which will show stochastic mixing if one looks only on one side.

EDIT: as to the PDC, the two entangled beams are NOT perfectly plane, monochromatic and polarized! Indeed, for instance in:

http://scotty.quantum.physik.uni-muenchen.de/publ/achtbild.pdf

in PDC type II conversion, we need to MIX THE TWO CONES, so that there are two beams of overlapping polarization cones ; the two beams are then UNPOLARIZED.
Here, you can keep the monochromaticity (given by the momentum conservation) and the direction, but you "sacrifice" the polarization. You could however, also play on the "rainbow" of a PDC X-tal, where you change the wavelength as a function of angle, and work with a purely polarized beam. But in any case, you need to leave SOME degree of freedom "free" for the entanglement, and it is exactly this degree of freedom which will appear as "stochastic" on one side.