# In downconversion, can you get coherence between the signals from two sources?

by al onestone
Tags: coherence, downconversion, signals, sources
 P: 61 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
 Sci Advisor PF Gold P: 5,148 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
 P: 61 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.)?
P: 915

## In downconversion, can you get coherence between the signals from two sources?

 Quote by al onestone 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.)?
Assuming I understood your question correctly - the pump should not matter.

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.
PF Gold
P: 5,148
 Quote by al onestone 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.)?
I think from the references, you can deduce that by phase matching independent sources, you get indistinguishability and therefore interference (if there is interference to be had).
P: 1,563
 Quote by al onestone 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)?
I suppose adding the idlers at a beam splitter is a good first step, but not sufficient. The corresponding signal photons will show second-order interference if they are completely indistinguishable. That includes spatial, spectral and polarization properties. Depending on the kind of PDC you use, you will get a large spread in possible polarizations or emission angles or photon energies of the idler photons which will correspond to similar spreads on the signal side. Therefore, I suppose you need to use narrow filters for these properties on the idler side to see interference on the signal side. I think Mandel also needed to do that in his experiments. He might have used narrow slits/pinholes and interference filters on his idler photons, but I am not sure about that.
 P: 61 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.
P: 915
 Quote by al onestone 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.
Al Onestone,

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
P: 1,563
 Quote by al onestone Abra-Cadabra, we have a protocol for FTL!
Nope. As I noted before:

 Quote by Cthugha Therefore, I suppose you need to use narrow filters for these properties on the idler side to see interference on the signal side. I think Mandel also needed to do that in his experiments. He might have used narrow slits/pinholes and interference filters on his idler photons, but I am not sure about that.
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.

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.
PF Gold
P: 5,148
 Quote by al onestone 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.
If I interpret your setup as you intend, the answer is NO.

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.

 P: 61 Let's agree to disagree: San K's comment To get Bob's (or Alice's for that matter) interference patterns, we need to compare them with Alice's Cthuga's comment 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. DrChinese's comment As Cthugha points out, you must do coincidence comparison. You all seem dead set on there being no possibility of interference between the signals ALONE without comparison to the idlers, which is clearly what Mandel has achieved experimentally and with theoretical explaination. But then one of you cracks a little, Cthuga's comment 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. Correction cthuga, you can reduce entanglement with filtering but you absolutely cannot break entanglement. It is fundamentally wrong to state this. In the specific case of the ZWM you will get strong momentum entanglement due to the wave vector conservation in downconversion, and the positions will only be weakly entangled. To get coincidence interference you need to balance the two. But we only want interference between the signals (which are prepared IDENTICALLY) and Mandel did produce this, however with a different seup. We can optimise the setup to produce the strongest interference between signals, and forget about the coincidence interference. However, measuring the idler path will absolutely change the state description and negate the interference possibility. This entanglement (even if it is weak) is unbreakable and it is fundamental to the state description.
 Sci Advisor PF Gold P: 5,148 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?
P: 1,563
 Quote by al onestone Correction cthuga, you can reduce entanglement with filtering but you absolutely cannot break entanglement. It is fundamentally wrong to state this. In the specific case of the ZWM you will get strong momentum entanglement due to the wave vector conservation in downconversion, and the positions will only be weakly entangled. To get coincidence interference you need to balance the two. But we only want interference between the signals (which are prepared IDENTICALLY) and Mandel did produce this, however with a different seup. We can optimise the setup to produce the strongest interference between signals, and forget about the coincidence interference. However, measuring the idler path will absolutely change the state description and negate the interference possibility. This entanglement (even if it is weak) is unbreakable and it is fundamental to the state description.
That is incorrect. You also need to filter the signal in momentum space by using slits/pinholes/similar devices and the more you 'cut' your distribution, the more you reduce entanglement. If you perform a Bell test on the signal, you can also see that at some point you progress from violating Bell inequalities to not violating it anymore. This is quite obvious when going to the extreme case of having only one momentum state left after filtering. In that case, you always get the same measurement result. However, as there are no fluctuations around that value anymore, this is a purely classical result. In fact, interference and entanglement are complementary, see 'Demonstration of the complementarity of one- and two-photon interference' by Abouraddy et al., Phys. Rev. A 63, 063803 (2001) which exactly discusses the question here and even gives a duality relation for the two quantities in question. You can have either second-order interference, fourth-order interference (and therefore entanglement) or both at the same time with reduced visibility. This is the regime where Mandel operates. His second order interference pattern is at a visibility of about 30%, only. The visibility of fourth-order interference is not given. I suppose it is not larger than maybe 70-80 % under ideal conditions. For FTL communication you would need ideal visibility of both interference patterns at the same time, which is impossible.
PF Gold
P: 5,148
 Quote by Cthugha In fact, interference and entanglement are complementary, see 'Demonstration of the complementarity of one- and two-photon interference' by Abouraddy et al., Phys. Rev. A 63, 063803 (2001) which exactly discusses the question here and even gives a duality relation for the two quantities in question. You can have either second-order interference, fourth-order interference (and therefore entanglement) or both at the same time with reduced visibility.
http://arxiv.org/abs/quant-ph/0112065

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.
 P: 61 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.
PF Gold
P: 5,148
 Quote by al onestone ... I'm not an expert in non-linear optics but I think this leaves the idler and signal indistinguishable ...
As I see it (and may be wrong), I agree with your statement.

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?
P: 61

 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?
First of all it is not "pairs" that arrive in phase or out of phase, it is single systems that always arrive at the screen at point D. The other detector that Bob has is at point C which registers a classical optical beat, which has frequency equal to the difference in frequency of the two signals. When this optical beat frequency is low then the two are in phase. At this point the two being "in phase" also infers that they are coherent (because the phase was the only form of distinguishability). When they are coherent they interfere, meaning that they are now projected onto the state of a single particle state vector. An equivalent way of stating this would be to say that the two amplitudes combine before squaring to get the intensity at the detector at point D.

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.
P: 915
 Quote by al onestone 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.
Bob will get a blob.

 Quote by al onestone 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.
Again Bob will get a blob.

To get the interference pattern Bob needs to compare with Alice.

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