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A delayed choice experiment with a paradox. What did I do wrong? |
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| Feb19-05, 09:37 PM | #1 |
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A delayed choice experiment with a paradox. What did I do wrong?
Some time ago I read some stuff about delayed choice and quantum eraser experiments, but rather than reading the articles thorouhgly, I just glanced at them quickly, picked up a few ideas, and then tried to design a gedanken experiment of my own. Now I seem to have run into a bit of a problem. My gedanken experiment seems to either contain a paradox or violate the principles of quantum mechanics. I'm sure I'm making some sort of mistake, but so far I haven't been able to find it. Perhaps someone here can can help me figure this out.
Look at the attached drawing. A "down converter" is a crystal that that emits two photons, each with half the energy of the original, when it's hit with just one. According to what I read, such crystals actually exist. I imagine that it would be difficult, if not practically impossible to build a down converter that always emits the two photons in the directions I've drawn in the picture, but I don't think it's impossible in principle. Imagine that we have carefully adjusted the lengths of the paths in this experiment so that the lower right detector never clicks, because of destructive interference. All the photons that reach the half-silvered mirror below the detectors will go to the left detector. If we imagine the same setup, but without the half-silvered mirror at the upper right, there wouldn't be any interference at the lower detectors. They would both detect the same number of photons. This is because the upper detectors would give us "which-way information" about the original photon. For example, the detector at the top of the picture will only click if the original photon was reflected at the first half-silvered mirror. The purpose of the half-silvered mirror at the upper right (and the mirrors at the upper left and lower right), is to eliminate the possibility to obtain which-way information that would destroy the interference. There's no way, even in principle, to determine if a photon that's been detected by one of the detectors at the upper right went through the last half-silvered mirror or was reflected by it. Let's move on to the paradox. Suppose that we sometimes remove the half-silvered mirror at the upper right, after one of the lower detectors has clicked. This will make it possible to determine which way the original photon went at the first half-silvered mirror, and this should destroy the interference at the lower two detectors. If the interference really is destroyed, then the lower right detector should detect a photon 50% of the times we do this (instead of 0%). This is retro-causation. The effect precedes the cause. The detector will click because someone at a later time removed a half-silvered mirror. If that isn't strange enough, suppose that we immediately after the lower right detector has detected a photon decide not to remove the half-silvered mirror. Then we have a paradox. If we never remove the half-silvered mirror, that detector should never click. Maybe it's just impossible to destroy the interference by removing the half-silvered mirror after one of the lower detectors has already clicked. But in that case, we can obtain which-way information about the original photon and still get interference. Doesn't this violate the principles of quantum mechanics? |
| Feb20-05, 10:45 AM | #2 |
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I had to shrink the image to upload it. That made it rather unpleasant to look at. If you'd like to see a version of that picture that's a bit easier on the eyes, you can use this link.
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| Feb20-05, 10:47 PM | #3 |
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I spent a little time today trying to get my head around this. I need to spend some more.
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| Feb21-05, 03:30 AM | #4 |
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A delayed choice experiment with a paradox. What did I do wrong?
Hello Fredrik,
I tried to follow your gedanken-experiment. It would be nice if you could give a link to the article you read, so we'd have background knowledge and understand your idea better. and know which way the photon took at the first H-mirror, then, in my opinion, you would still get interference at the middle detectors. For example: Suppose we have no H-mirror (right top) and detect a photon at the right top detector. This means: We know the photon was NOT reflected at the first H-mirror, but it went through it. So, the photon goes straight through the first H-mirror, is downconverted, and then is incident on the H-mirror in the center. Because this mirror in the center is half-silvered, you don't know which way it takes (in the center) and hence still get interference. I think what you are considering has something to do with two-photon-interference, where you look if there's COINCIDENCE between the detectors in the center and on the top. Polarization plays a role there. http://www.bu.edu/qil/pdf/PRL-09-02-96.pdf |
| Feb21-05, 06:15 AM | #5 |
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Note that after the first half-silvered mirror, every line drawn in the diagram represents a photon that's in a superposition of existence and non-existence. In the language that's often used in discussions about similar experiments, each line represents a "half" photon. The interference I'm talking about is interference between two "halves" that have the last part of their paths to the detector in common. If we know that the original photon went through the first half-silvered mirror, then there was never any "half" photons around that could interfere with each other. |
| Feb21-05, 06:16 AM | #6 |
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| Feb21-05, 07:33 AM | #7 |
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I have to admit that I've understood for the first time, why this 100% and 0% situation occurs, the site above really gives a good explanation. I have to think about your gedankenexperiment. |
| Feb21-05, 08:05 AM | #8 |
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If you don't remove the H-mirror in the right top, only detector 1 in the center will click due to interference (like in the Mach-Zehnder-interferometer). I don't think that you can destroy the interference afterwards. I can't imagine how that would look like. So I agree with your comment: and test it in the lab.
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| Feb21-05, 08:17 AM | #9 |
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I think for the detectors in the right top the probabilities
should also be 100% and 0%, because the outer mirrors also represent a Mach-Zehnder interferometer. I just noticed that your experimental setup consists of two Mach-Zehnders, an inner and an outer interferometer. |
| Feb21-05, 09:30 AM | #10 |
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| Feb21-05, 09:36 AM | #11 |
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It's an interesting addition to the experiments to have as Edgardo says, and I also noticed, essentially two interferometers in inner and outer loops.
I've been drawing the experiment as a superposition of two alternative histories which interfere. After the first H mirror, one of my drawings is the history which follows the path of photon up and then the paths of the two photons created this way, while the other drawing is the history which follows path of the photon right and then the paths of the two photons created that way. I'm imagining the drawings as then interfering and seeing if that gives me some clues. I'd guess the photon goes around the inner loop by two paths in superposition has inteference effects with itself whatever else happens. What is happening to the photon in the outer loop is what I'm wondering now. |
| Feb21-05, 10:30 PM | #12 |
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Fredrik-- the quote below greatly puzzles me.
If we imagine the same setup, but without the half-silvered mirror at the upper right, there wouldn't be any interference at the lower detectors. They would both detect the same number of photons. This is because the upper detectors would give us "which-way information" about the original photon. For example, the detector at the top of the picture will only click if the original photon was reflected at the first half-silvered mirror. My instinct is to say "higher detectors" should replace "lower detectors" in the first sentence, there's a typo in other words. But, perhaps not. In which case note that you have a 'sort-of" standard interferometer with the lower detectors. That is to say, interference will be alive and well at the lower detectors no matter the state of the upper apparatus. This is certainly true for classical E&M, and for QM as well. the Basic idea can be shown by looking at thr quantum fields at the measurement points. Both classical and quantum fields share the same spatial form. So at Lower Detector 1 (left), or LD1, have for 2 photons, with momentum k, a wave function (poor nomenclature) EXP (-k{x11 + x21}) where x11 and x21 are the photon coordinates at LD1, fully symmetric in the two photons. Interference is demonstrated by varying x11 + x21. If the sum is pi/2 there is full destructive interference, and so on. Two channels feeding a detector guarantees interference. After the fact causality? I don't think so. Regards, Reilly Atkinson |
| Feb22-05, 01:58 AM | #13 |
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| Feb23-05, 03:31 PM | #14 |
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Hello Fredrik,
I've read about quantum erasers. And read these websites: http://www.lns.cornell.edu/spr/1999-03/msg0015208.html and especialy this one: http://grad.physics.sunysb.edu/~amarch/ It mentions, that the which way information is erased AFTER it made click. (at least that's how I understood it) In your gedankenexperiment I now think that the interference is destroyed, that is you don't get any interference although you take away the mirror AFTER it made click (at the center-detectors). However, that's my first thought after having read about these quantum erasure experiments. I will let you know, if I get any new insight into this. Regards Edgardo P.S. Check out this program about quantum erasure. http://www.physik.uni-muenchen.de/di...interfere.html Note that you can rotate the polarizers, which I at first didn't get. |
| Feb23-05, 06:40 PM | #15 |
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Talking of erasing the past, in Roland Omnes's Understanding Quantum Theory, he mentions the ideal von Neumann experiment.
A particle is emitted and can go down one of two channels. There is a detector in one channel. If, after a certain time, the particle is detected then you know it went along the channel with the detector and if the particle is not detected then you know it went along the channel without the detector. So far, so simple. Now here comes the interesting part as, if you arrange it so that the two channels curve and later join up into the same direction, then when the particle reaches the point at which the channels join up then the detector result will go from "detected" or "not detected" into a superposition and "don't know" Schrodinger Cat state even if the particle is now a huge distance away from the detector! It's EPR and Schrodinger's Cat in one simple, fun package. Decoherence keeps this lunacy from happening in the real world, thankfully.
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| Feb24-05, 02:10 AM | #16 |
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| Feb24-05, 04:33 AM | #17 |
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This thought experiment has captured my interest, and I have been thinking about it for some days now. I do not know much about "quantum erasure" and ((S)P)DCs, and so I have only been able to take the setup at face value on the basis of heuristic considerations.
The following "rule" appears to remove all elements of inconsistency: Interference effects will occur in the signal only if the 'decision' to erase the "which-way" information occurs a spacetime event which lies in the causal past of the signal detection event. If one would like to argue that this "rule" does not apply, namely that interference effects in the signal exist even when (i) the 'decision' to erase and signal detection are spacelike separated or, much more drastically, (ii) the 'decision' to erase lies in the causal future of signal detection , then, in case (i) we can arrange for faster-than-light signaling, and in case (ii) we can arrange (with some abuse of the term 'decision') for an action performed now to 'undo' an event which has already taken place (... whatever that could mean). I think that this "rule" and the ramifications of its violation for cases (i) and (ii) above apply quite generally to all erasure-type scenarios, and not just to the thought experiment at hand. |
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