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Cryo

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Can you please provide a diagram (of the interferometer)?

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https://commons.wikimedia.org/wiki/File:Beam_Split_and_fuse.svg

It's one of the setups used in the delayed choice experiments.

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Strilanc

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So, in the Mach–Zehnder interferometer we have a photon entering the system in the lower left corner. It encounters a beam splitter, which gives the photon a 50-50 chance of going either up or continue straight. Both upper left and lower right corners have a mirror deflecting the photon towards the upper right corner. In the open configuration there's nothing in the upper right corner, so the photon hits one of the detectors, which tell the observer which way the photon traveled through the system. In the closed configuration there's another beam splitter in the upper right corner, which has a 50-50 chance of sending the photon to one of the detectors. In this case the which-way information is not known, and the two detectors will exhibit interference effects.

So, the photons are still entangled, but each one will behave according to the path configuration they went through? I.e. photon entering the open system will hit only one detector, and the photon entering the closed system will cause interference at the detectors? If that's right, the delayed choice quantum eraser experiment makes no sense...

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bhobba

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http://quantum.phys.cmu.edu/CQT/chaps/cqt20.pdf

Probably a good idea to become acquainted with QM from a modern interpretation like Consistent Histories:

http://quantum.phys.cmu.edu/CQT/index.html

Thanks

Bill

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Strilanc

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So, the photons are still entangled, but each one will behave according to the path configuration they went through? I.e. photon entering the open system will hit only one detector, and the photon entering the closed system will cause interference at the detectors? If that's right, the delayed choice quantum eraser experiment makes no sense...

I don't see how what you asked is related to the delayed choice experiment. Delayed choice requires the ability to perform anti-commuting measurements related to the property the particles are entangled on, but a Mach-Zehnder interferometer won't do that for you unless you have additional details in mind that you haven't mentioned. That's why it doesn't matter that the particles are entangled.

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I'm not actually asking about the delayed choice experiment, I'm asking what would happen when instead of a single interferometer switching between open and closed, there are two interferometers (one always open and one always closed) and a photon enters each, both photons being entangled. Will one photon behave like a particle and hit only one detector in the open interferometer, while the other photon behave like a wave causing interference in the detectors of the closed interferometer?I don't see how what you asked is related to the delayed choice experiment. Delayed choice requires the ability to perform anti-commuting measurements related to the property the particles are entangled on, but a Mach-Zehnder interferometer won't do that for you unless you have additional details in mind that you haven't mentioned. That's why it doesn't matter that the particles are entangled.

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Strilanc

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I'm not actually asking about the delayed choice experiment, I'm asking what would happen when instead of a single interferometer switching between open and closed, there are two interferometers (one always open and one always closed) and a photon enters each, both photons being entangled. Will one photon behave like a particle and hit only one detector in the open interferometer, while the other photon behave like a wave causing interference in the detectors of the closed interferometer?

You can't approximate photons as a particle or as a wave when entanglement is involved. You won't be able to understand entanglement until you drop that terrible misleading outdated analogy. Photons are their own quantum thing.

For example, why do you think that the photon is "behaving like a particle" in the open interferometer? In what sense is it "not behaving like a wave"? The wave enters the beam splitters, gets evenly split over both paths, each paths lead to separate detectors, and so collapse distributes it randomly over the two.

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I think that I'm probably incorrectly applying the explanation behind the delayed choice quantum eraser. In that experiment, there are two entangled photons, and one of them goes directly to a detector, while the second one goes through a system that may determine the "which path" information. When the second photon produces the "which path" information, the first photon "behaves" like a particle, and if the second photon doesn't produce that information, the first photon acts like a wave producing interference pattern on the detector. The conclusion is that both photons behave either like a wave or like a particle at the same time.You can't approximate photons as a particle or as a wave when entanglement is involved. You won't be able to understand entanglement until you drop that terrible misleading outdated analogy. Photons are their own quantum thing.

The explanation I've seen says: because wave coming from the left doesn't interfere with the wave coming from the bottom, there must be only one wave coming from one direction, hence the photon behaves like a particle in the open interferometer.For example, why do you think that the photon is "behaving like a particle" in the open interferometer? In what sense is it "not behaving like a wave"? The wave enters the beam splitters, gets evenly split over both paths, each paths lead to separate detectors, and so collapse distributes it randomly over the two.

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Strilanc

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When the second photon produces the "which path" information, the first photon "behaves" like a particle

It doesn't. Even the single slit experiment doesn't produce a pattern that looks like particles (e.g. narrowing the slit widens the area being hit, but if photons were particles you'd expect narrowing the slit to narrow the area being hit).

When the second photon produces the "which path" information, the first photon "behaves" like a particle, and if the second photon doesn't produce that information, the first photon acts like a wave producing interference pattern on the detector. The conclusion is that both photons behave either like a wave or like a particle at the same time.

There's never a visible interference pattern on the detector. There's a correlation between where the signal photon hits on the detector and the measurement result of the idler photon, and that correlation has a wave pattern.

The explanation I've seen says: because wave coming from the left doesn't interfere with the wave coming from the bottom, there must be only one wave coming from one direction, hence the photon behaves like a particle in the open interferometer.

That's not a correct explanation of the experiment.

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There's never a visible interference pattern on the detector. There's a correlation between where the signal photon hits on the detector and the measurement result of the idler photon, and that correlation has a wave pattern

.... in simple words, you are pointing out that you need the results from

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https://arxiv.org/abs/quant-ph/0106078

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This is for me the most fascinating of all quantum erasure measurements

Please...can you tell some characteristics making it to outstand, let's say, Kim's version?...thanks!

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A nice explanation of this beautiful experiment can be found here:

http://laser.physics.sunysb.edu/~amarch/eraser/

although, as I said before, the original paper is very well understandable with basic knowledge about the two-level quantum system.

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There's no spooky action at distance and also no influence in the past.

How about an experiment where Alice's delayed choice "forces" a quantum correlation at Bobs?....(for this one Bob will be receiving entangled pair or photons and as his measurement always is e.g. spin up or spin down his entangled pair will always correlates)

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https://en.wikipedia.org/wiki/Quantum_teleportation

There you need "classical communication" and thus there's no FTL communication.

It is very clear mathematically that according to QED there is no possibility for FTL communication, no matter which states you try to use to do that. This is the linked cluster theorem valid for local microcausal relativistic QFTs.

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https://en.wikipedia.org/wiki/Quantum_teleportation

There you need "classical communication" and thus there's no FTL communication.

It is very clear mathematically that according to QED there is no possibility for FTL communication, no matter which states you try to use to do that. This is the linked cluster theorem valid for local microcausal relativistic QFTs.

....it's he realization of Asher Peres thought experiment by Zeilinger in 2012....

https://arxiv.org/abs/1203.4834

he places Alice and Bob performing

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DrChinese

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....it's he realization of Asher Peres thought experiment by Zeilinger in 2012....

https://arxiv.org/abs/1203.4834

he places Alice and Bob performingwhile Victor projects a delayed choice of quantum correlation (entanglement) or classical correlations into Alice's and Bob's pair of photons...see that Zeilinger did that on purpose, but he could well have let Bob and Alice toindependent measurementstheir measurements and limit their outcomes, for lets say, spin up or spin down....in this way Alice and Bob, receiving the photons at the same station will be 100% certain that Victor projected a quantum correlation for a single experiment run without the classical information channel.....that's given Victor projected quantum correlation for the whole experiment run.... see that in the actual experiment Victor projects entanglement at random during a single experiment run...coordinate

Just a reminder that when Victor chooses to project what Alice and Bob see into an entangled state: there are 2 possible resulting entangled Bell states. One is + entanglement, the other is - entanglement. I.e. they either have same polarization or opposite. The selection of which occurs is outside of Victor's control. But he can see which one results. He can communicate that to Alice and Bob via classical channels.

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One is + entanglement, the other is - entanglement. I.e. they either have same polarization or opposite.

It means that even thou Alice's and Bob's pairs are entangled, if they make the same measurement, lets say....spin up/spin down, they still might find both photons spinning up or spinning down? ..just like if Victor projected them with classical correlations...?

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DrChinese

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It means that even thou Alice's and Bob's pairs are entangled, if they make the same measurement, lets say....spin up or spin down, they still might find both photons spinning up or spinning down? ..just like if Victor projected them with classical correlations...?

Here is one type of entanglement:

1. Alice/Bob:

H H

H H

V V

H H

V V

etc.

I.e. the spins are the same.

2. Here is another:

H V

V H

V H

H V

V H

etc.

I.e. the spins are opposite.

When Victor casts into an entangled state, either one of the above occurs randomly. Victor can see if 1 or 2 is the result, but cannot influence which occurs. When Alice and Bob compare results, they need to have information from Victor so they can confirm the entanglement. Because all they see are random patterns otherwise, no hint at all of entanglement.

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morrobay

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Case 2 the detectors are perpendicular and both photons are always opposite, H V , V H at both detectors.

So if Alice and Bob whether the polarizers are aligned or perpendicular, then each should know what the other got based on what they recorded. And hence have entangled photons.

Above is for special cases 1 & 2.

When polarizers are at different angles θ

then the probability of photon 1 being also Horizontal at θ

* By the way are Horizontal/Vertical interchangeable with absorbed/transmitted. ( for probabilities only ) ?

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DrChinese

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It looks like cases above are for entangled twin state photons. Case 1 the the polarizers at Alice and Bob are aligned so either both photons are Horizontal at A & B or both photons are Vertical at A & B.

Case 2 the detectors are perpendicular and both photons are always opposite, H V , V H at both detectors.

So if Alice and Bob whether the polarizers are aligned or perpendicular, then each should know what the other got based on what they recorded. And hence have entangled photons.

The cases I presented were for the same measurement being performed on both photons. Notice that when you know which of my 2 cases is occurring, the entanglement is obvious. On the other hand, without information from Victor, you can't tell one from the other or distinguish the cases (which also occur often) where there is no Bell entanglement.

This situation occurs in entanglement swapping as a key characteristic. This is different than ordinary entanglement via BBo crystals (parametric down conversion).

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This is different than ordinary entanglement via BBo crystals (parametric down conversion).

But check onto figure 2 of Zeilinger's article, he actually used BBO crystals for the SPDC, the entanglement is the same type, he only swapped it.....

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DrChinese

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But check onto figure 2 of Zeilinger's article, he actually used BBO crystals for the SPDC, the entanglement is the same type, he only swapped it.....

They are NOT entangled by PDC, even though the source photons were themselves created by PDC. They are entangled by swapping. The mechanism for causing the swap is the Bell State Analyzer device that Alice oversees (see Figure 1 on page 8 of the below Zeilinger reference, so we can all be on the same page). When Alice successfully entangles 0 & 3, the swapping produces multiple resulting entangled states (each roughly equal probability). We are interested primarily in the psi+ and psi- cases, which I abbreviate as + and - entanglement.

https://arxiv.org/pdf/quant-ph/0201134.pdf

In the experiment itself, for technical reasons, only the psi- state is considered (even though psi+ is produced too). But Alice has to announce which ones that represents, so that the results make sense for Bob.

The point is: the entangled pairs, when measured on the same basis, do not show the HH+VV pattern you might expect. Only those with psi+ entangled yield that. The psi- pairs will show HV+VH. And we ignore the phi+ and phi- cases completely, as they cannot be recognized by Bob when performing linear polarization measurements.

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That reference is from 2008....but check the previous reference is from a more advanced version done in 2012...

https://arxiv.org/abs/1203.4834

In this case the bipartite state analyzer is at Victors who's location is delayed in both time/distance and there are current versions with this being delayed by over several kilometers....

What it means is that by the time Victor receives his pair of photons....the other pair received by Alice and Bob "*have already been measured and no longer exists"....*

In the experiment Alice and Bob are making linear polarization measurements or what is called sinusoidal plane of the electromagnetic wave....so their results will always be hh-vv or hv-vh (while quantum correlation exists)....so all cleared up to this point here's the question....

can we have Alice and Bob measuring the*same direction* of the photon's total angular momentum (or spin)....instead?

https://arxiv.org/abs/1203.4834

In this case the bipartite state analyzer is at Victors who's location is delayed in both time/distance and there are current versions with this being delayed by over several kilometers....

What it means is that by the time Victor receives his pair of photons....the other pair received by Alice and Bob "

In the experiment Alice and Bob are making linear polarization measurements or what is called sinusoidal plane of the electromagnetic wave....so their results will always be hh-vv or hv-vh (while quantum correlation exists)....so all cleared up to this point here's the question....

can we have Alice and Bob measuring the

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DrChinese

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That reference is from 2008....but check the previous reference is from a more advanced version done in 2012...

https://arxiv.org/abs/1203.4834

Yes this is good, the actors are relabeled but it is functionally much the same. As you point out, the entangled pair is cast into an entangled state after the entangled pair no longer exists. There are also versions of this where the entangled photons never existed at the same time. So some pretty interesting situations to consider.

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