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In downconversion, can you get coherence between the signals from two sources? |
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| Jan11-13, 09:24 AM | #1 |
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In downconversion, can you get coherence between the signals from two sources?
In the ZWM study of 1991 Mandel et al get coherence between the signal photons of two distinct downconversion sources. This is shown with an interference collection between the two possible sources of signal photons (even though they are entangled with the idler). This is achievable because the two downconversions are pumped by a common source which is beamsplit, and because the idler from one downconversion crystal is transmitted through the other downconversion crystal and superimposed with the second idler (made indistinguishable). Thus the name of his paper "Induced coherence without induced emission".
Is it possible to get coherence between two distinct sources of signals (without the transmission and superimposing of the two idlers) if you simply add the two idlers at a beamsplitter to make them indistinguishable (which negates the which-path information)? You would have to look for interference between the two signals in the same way as you would look for interference between any two separate indistinguishable sources of photons (Hull 1949, Pfleegor & Mandel 1967). You can "postselect" the observations that are in phase or you can look for temporal interference between two detectors. A more thorough explaination of this thought experiment is at http://modifiedzwm.webs.com |
| Jan11-13, 11:48 AM | #2 |
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I think these may include some similar arrangements to what you describe:
Creating path entanglement and violating Bell inequalities by independent photon sources http://arxiv.org/abs/1001.3830 High-fidelity entanglement swapping with fully independent sources http://arxiv.org/abs/0809.3991 Entanglement Between Photons that have Never Coexisted http://arxiv.org/abs/1209.4191 |
| Jan12-13, 10:30 AM | #3 |
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Your first link to Weigner et al is about the creation of path entangled photons emitted by uncorrellated atoms, not similar to what I'm suggesting.
The second link is about entanglement swapping, which is a similar preparation to what I'm suggesting, but entanglement swapping is done with Bell state analysis which requires non-linear optical coupling between coherent photons of different entangled pairs. This non-linear coupling is not desired in my thought experiment. The third link is also entanglement swapping. In my thought experiment I wish to question whether you can get coherence between the two signals from independent (but indistinguishable) sources of downconverted photons. The type of coherence that gives you interference between the two. Mandel already achieved this with his ZWM apparatus(X. Y. Zou, L. J. Wang, and L. Mandel, Phys. Rev. Lett. 67, 318 1991), but this was with the two downconversions pumped by the same laser. What if they are separately pumped, can you still get interference the way you would get it in the two source interference effect (Hull 1949, etc.)? |
| Jan12-13, 10:48 AM | #4 |
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In downconversion, can you get coherence between the signals from two sources?however if you trying to distinguish between two photon and two-photon interference then in that case they are different. in two-photon interference the photons can be temporally (or spatially) separated and still interfere, but not so in two photon interference. |
| Jan12-13, 01:00 PM | #5 |
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| Jan12-13, 02:08 PM | #6 |
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| Jan13-13, 01:28 PM | #7 |
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DrChinese, San K and cthuga, you all seem to admit that indeed it is possible to get interference between the signals, even though they are from separate source emissions, and we may all explain in our own way how this is. And we must all admit that the interference at the signals (Bob's system) is contingent upon the combination of the idlers (Alice's system) with a beamsplitter.
Abra-Cadabra, we have a protocol for FTL! What if we do the same thing that Mandel did and modulate one of the idler pathways? Will Alice's choice of whether or not to add the idlers effect the outcome of Bob's interference collection. Remember, no addition of the idlers would allow for the "which-path" information of the system to be in principle knowable. Eureka! The great Al Onestone has done it! Instant communication. And if you want to get tied up with "no-signalling theorem", remember, no-signalling theorem always assumes a unitary OPERATION on Alice's system that will not disturb the measureable statistics of Bob's system. In this preparation we have "combination at a beam splitter" which is not a bonafied unitary operation, it is a projection onto a state where only the "which-path" information is different. It's the time reversal of a beam going into a beam splitter, but without detectors. And that is the type of preparation that Neils told us not to talk about. |
| Jan13-13, 02:19 PM | #8 |
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To get Bob's (or Alice's for that matter) interference patterns, we need to compare them with Alice's. Not sure how a difference pump source can make a difference re: FTL attempt Let's keep the below two things, in mind: 1. the interference pattern cannot be seen unless we compare Alice and Bob's photons (via a co-incidence counter). The following need to be separated ("un-embed"): a) interference due to coherence b) interference due to entanglement c) noise to separate them we need to compare Alice and Bob's photons 2. The is no coherence (hence no interference) when the photons are totally entangled. Coherence and entanglement are complimentary |
| Jan13-13, 02:37 PM | #9 |
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You can get a situation where interference is directly visible on the signal side alone if you strongly filter the signal photons. However, that breaks entanglement and the result does not depend on what happens at the idler side anymore. |
| Jan13-13, 06:02 PM | #10 |
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As I read it: Bob is looking for the pattern at the one side, while Alice makes the decision (or not) to entangle. Clever idea, no doubt. However, there is a minor detail (there always is). As Cthugha points out, you must do coincidence comparison. You probably think Bob is doing that, as he has both of the signal photons on his side. True enough. BUT... The coincidence counting must include a variable on Alice's side too! That is the type of entanglement which Bob's photons are cast into. There are multiple entangled states possible with entanglement swapping, not just one as there is from a single down conversion event. So the coincidences can be ++/-- or -+/+-. Bob's pattern looks the same until and unless he has Alice's information on which state of entanglement occurred.
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| Jan14-13, 08:22 AM | #11 |
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Let's agree to disagree:
San K's comment Code:
To get Bob's (or Alice's for that matter) interference patterns, we need to compare them with Alice's Code:
Only the coincidence counts with the strongly filtered detections on the idler side will give an interference in the coincidence counts, not the total number of the detected signal photons. Code:
As Cthugha points out, you must do coincidence comparison. Cthuga's comment Code:
You can get a situation where interference is directly visible on the signal side alone if you strongly filter the signal photons. However, that breaks entanglement and the result does not depend on what happens at the idler side anymore. |
| Jan14-13, 09:43 AM | #12 |
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In looking at the diagram you reference:
a) I see you have already considered the point I was making about the different Bell states: "It might be noted by some that the combined preparations of figures 1 and 2 make up the commonly practiced “entanglement swapping” protocol. Because Alice combines her two idler beams with a beam splitter, the beam splitter acts as a HOM filter which projects two idlers onto a symmetric Bell state basis and similarly projects Bob’s signals onto an entangled basis. This is a non-linear optical effect which is only relevant for a very small portion of the total system, so we disregard this effect when considering the whole of the system." Not sure why you are disregarding this. And it seems to me that this makes superfluous the detector you call C (Dc). b) It is also significant to note that the experimental results will NOT change regardless of when Alice makes a decision to entangle or not entangle Bob's photons. So this is irrelevant (emphasis added): "The preparation for Alice consists of nothing more than combining beams i1 & i2 at a beam splitter (BS), and specifically at identical optical path lengths that would have them combined PRIOR to the arrival of Bob’s beams at points A and B." c) I am not exactly following what pattern you expect to see at screen D. Some pairs coming through will be "in phase" and some will be "out of phase". So can you describe or point me to a picture of what a series of detections at D will look at depending on something Alice does or does not do? |
| Jan14-13, 10:08 AM | #13 |
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| Jan14-13, 12:08 PM | #14 |
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Demonstration of the Complementarity of One- and Two-Photon Interference A. F. Abouraddy, M. B. Nasr, B. E. A. Saleh, A. V. Sergienko, M. C. Teich (Submitted on 11 Dec 2001) The visibilities of second-order (single-photon) and fourth-order (two-photon) interference have been observed in a Young's double-slit experiment using light generated by spontaneous parametric down-conversion and a photon-counting intensified CCD camera. Coherence and entanglement underlie one-and two-photon interference, respectively. As the effective source size is increased, coherence is diminished while entanglement is enhanced, so that the visibility of single-photon interference decreases while that of two-photon interference increases. This is the first experimental demonstration of the complementarity between single- and two-photon interference (coherence and entanglement) in the spatial domain. |
| Jan14-13, 03:37 PM | #15 |
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Okay cthuga and DrChinese (thank you for your responses thus far), I've looked at your paper, which I think is not comparable to my thought experiment because the downconversion it uses is dissimilar, Abbouraddy sais "spontaneous parametric down-converted light in the degenerate collinear Type-I configuration" in a LiIO3 crystal (which is the same type of crystal that Mandel uses). I'm not an expert in non-linear optics but I think this leaves the idler and signal indistinguishable which is not the case for Mandel's preparation.
More importantly, even if the conclusions of Abbouraddy et al are correct and they do apply to this setup I have suggested, he and you are all making a drastic error as concerns your wording of his conclusion. In his paper he sais that the coincidence interference visibility is complementary to the one photon interference visibility. This absolutely does not mean that the entanglement is complementary to the one photon interference visibility. There is always entanglement. As long as you're using non-linear optics as your source, there is no theoretical possibility of exciting a signal photon in the absence of an idler. Coincidence interference may become less visible but there is still entanglement, there is still an idler. And this is the only requirement in my thought experiment. Just think about it in terms of single pumps exciting single pairs. The path information is fundamental to the state description. Forget about the visibility of coincidence interference. |
| Jan14-13, 04:00 PM | #16 |
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Can you answer this question I asked earlier today: c) I am not exactly following what pattern you expect to see at screen D. Some pairs coming through will be "in phase" and some will be "out of phase". So can you describe or point me to a picture of what a series of detections at D will look at depending on something Alice does or does not do? In other words, what pattern does Bob see when Alice transmits a 0, and what pattern does Bob see when Alice transmits a 1? |
| Jan15-13, 08:20 AM | #17 |
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DrChinese, your question
When Alice transmits a 0 (when she does not combine her idlers at the beamsplitter), Bob will see no interference at his detector (at point D), he will get the average intensity that you can calculate by adding the squares of the two amplitudes. When Alice transmits a 1 (by combining idlers at the beamsplitter which negates the "which-path" info), Bob will get interference at his detector (at point D) which can be calculated by adding amplitudes of the two signals prior to squaring. How does Bob actually measure interference? Easy, the same way anyone would do it, set up the appropriate aperture (pin hole or single slit) for each signal beam and choose a particular place on the screen at point D (CCD screen detector) where the two signals should produce an interference pattern. If you choose an intensity maxima then this will have a high reading for a 1 transmission from Alice, and will drop back to the average intensity when you have a 0 transmission. |
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