Can Entangled Particles Enable Faster-Than-Light Communication?

[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.