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

San K

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.

*al onestone*

You know San K, I'm getting tierd of having to tell people that it is not true that

Code:

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

I would ask you to simply look at the original ZWM and tell me that Mandel had to "compare the signal detections to the idler detections" in order to get second order interference at the signal detector. It simply is not true. You only need to "compare the signals to the idlers" when you collect fourth order coincidence interference. Take a closer look at the original ZWM.

X. Y. Zou, L. J. Wang, and L. Mandel, Phys. Rev. Lett. 67, 318 (1991).

and

Phys. Rev. A 44, 4614–4622 (1991)
http://pra.aps.org/abstract/PRA/v44/i7/p4614_1

DrChinese

Quote by al onestone

DrChinese, your question

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.

I need to ask some additional questions.

a) So Bob's signal photons are directed to separate screens. And the interference (or non-interference pattern) he sees is essentially visible by looking at either of the 2 patterns. Is that correct?

b) What is the purpose of the detector C which is determining whether or not the signal photons are coherent? Are you doing coincidence counting using only those that are coherent? And how do you not see that this detection of their relative phase is occurring without disturbing the system in some fashion?

c) I guess you have read all the information that says that entangled photons are not coherent and reject that conclusion, is that correct?

--------------------------------------

Entangled photons are not coherent. In your setup, either Bob's 2 signal photons are entangled with each other OR they are entangled with their idler mates. So all they will ever produce is a blob on the detector screens, as San K says.

If all of the other references weren't enough, here is yet another which explicitly says that you cannot get an interference pattern from such photons:

See fig. 2, p 290.
http://www.hep.yorku.ca/menary/cours...oundations.pdf

And from the abstract of the article previously referenced:

"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."

Cthugha

Quote by al onestone

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.

It does not really matter whether for this question whether you have them at the same or different energies. This is basically a question of phase matching. It does not really change the physics involved.

Quote by al onestone

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.

This is the central point where you are wrong. Having a signal and idler at a constant sum of total energy/momentum/whatever is NOT sufficient for entanglement. To distinguish whether you really have entanglement or it is just plain classical correlation, you need to demonstrate a violation of Bell inequalities (or some similar scheme like CHSH in other geometries). Filtering your signal/idler strongly prevents exactly this. When going to the extreme of vanishing two-photon interference, you can replace signal and idler by two completely independent beams having the same characteristics and will get the same results as for strongly filtered signal/idler beams exactly because entanglement becomes meaningless in that case.

Quote by al onestone

Forget about the visibility of coincidence interference.

Why should I? This is the important physics involved here.

*al onestone*

Hey cthuga, have you forgotten your own quote,

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.

This is all that I demand in my thought experiment. Second order interference between the signal beams while there is still meaningfull entanglement with the idler beams. This leaves Bob's system of signal beams capable of producing an interference that is contingent upon the "which-path" information of the entangled partners.

I can accept that the meaningfullness of the entanglement disipates when I prepare conditions for the signal beams to interfere, but as long as there is entanglement then the knowability of the idler path information COMPLETELY destroys the signal interference. Remember, Mandel already did it, the ZWM, and I'm only proposing a different method of interference collection. The collection I propose is the same as in the link below, the good old fashioned way of collecting interference between separate sources.

DrChinese, this link below is exactly how I wish for Bob to create interference,

Phys. Rev. 159, 1084–1088 (1967)

http://prola.aps.org/abstract/PR/v159/i5/p1084_1

or a free version

https://docs.google.com/viewer?a=v&q...yC4HO2KG_AOpBQ

Cthugha

Quote by al onestone

I can accept that the meaningfullness of the entanglement disipates when I prepare conditions for the signal beams to interfere, but as long as there is entanglement then the knowability of the idler path information COMPLETELY destroys the signal interference. Remember, Mandel already did it, the ZWM, and I'm only proposing a different method of interference collection.

It completely destroys the interference pattern which had a gorgeous visibility of at most 30%. That increases the possibility of having useless noise for any single detection from 70% to 100%. Also as entanglement is not 100% anymore, you also lose something. I leave it up to you to calculate what the necessary amounts of first order visibility and entanglement purity are to realize a meaningful communication protocol. Hint: You pretty much need to exactly violate the duality relation between interference visibility and entanglement to realize it.

edit: I also do not see why you are so interested in having two different pump source instead of just a single one split using a beamsplitter. That does not change the physics involved at all (besides that is more complicated to keep the different pump sources mutually coherent).

*al onestone*

Cthuga, DrChinese, San K and others, thank you for your responses to my thread. This discussion has been enlightening for me.

Cthuga, your final comment,

I also do not see why you are so interested in having two different pump source instead of just a single one split using a beamsplitter. That does not change the physics involved at all (besides that is more complicated to keep the different pump sources mutually coherent).

Simply put, the coherence of common pumps is lost due to a combination of two factors, downconversion is highly improbable and spontaneous. If the pump goes to one crystal and downconverts there is no promise that if it had gone to the other crystal it certainly would have downconverted there also. In the ZWM the transmission of one idler through the other crystal locks the coherence again, and makes the common pump meaningful. In my setup the common pump is useless, and coherence is only gained by brute force, comparison in the form of an optical beat.

The link to the thought experiment has been appropriately updated and will continue to be:
http://modifiedzwm.webs.com

Cthugha

Oh, I see. Ok.

Final question, just out of curiosity. How exactly do you intend to make the idlers indistinguishable? Hong-Ou-Mandel interference? I am just asking because just placing both idlers on a beamsplitter somehow, obviously will not work.

*al onestone*

Cthuga your question

How exactly do you intend to make the idlers indistinguishable? Hong-Ou-Mandel interference?

The possibility of HOM projection onto a Bell state is very negligible therefore it is not a significant consideration.

The idlers are already indistinguishable in every way other than phase and path of origin. Bob only postselects observations that are "in phase" and rejects all others. And the path distinguishability of idlers goes away when they are added at the beamsplitter by Alice.

Cthugha

I do not get your point. Either the idlers are indistinguishable with the exception of their origin: then they definitely will undergo HOM interference, thus wiping out the information about their origin, or they are somewhat distinguishable and will not undergo HOM interference. But in the latter case, you do not erase the information about their path of origin. Just placing two photons on a beamsplitter somehow does not achieve that.

DrChinese

Quote by al onestone

Bob only postselects observations that are "in phase" and rejects all others.

And the path distinguishability of idlers goes away when they are added at the beamsplitter by Alice.

As I have said before: Bob cannot observe the relative phase of a pair of photons without causing a type of collapse on those photons. Otherwise you would be able to violate the HUP.

And although nothing Alice does will ever change anything Bob sees in the sense that he can detect that without coincidence counting: even the coincidence counting will be meaningless after Bob checks the relative phase of the photons.

Just saying...

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Cthugha Recognitions: Science Advisor	Oh, I see. Ok. Final question, just out of curiosity. How exactly do you intend to make the idlers indistinguishable? Hong-Ou-Mandel interference? I am just asking because just placing both idlers on a beamsplitter somehow, obviously will not work.