Can Entangled Particles Enable Faster-Than-Light Communication?

[Cane_Toad]

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 I can't tell whether you are saying that the entanglement has been "broken", or that the entanglement is somehow no longer applicable.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]

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.