swansont Posted January 7, 2017 Share Posted January 7, 2017 When one of the polarizers is set to 45 degrees the other is not set to 45. That would make it possible to determine the polarization. Also, if we were just trying to determine the polarization, we probably would not set the polarizer to 45 degrees. I didn't ask you to change the question, I asked you to answer it. When one of the polarizers is set to 45 degrees the other is not set to 45. That would make it possible to determine the polarization. The setting of the other polarizer depends on what you're trying to measure. From http://www.optics.rochester.edu/workgroups/lukishova/QuantumOpticsLab/2010/OPT253_reports/Josh_Lab1.pdf the coincidence counts are recorded for multiple polarizer positions. We used: a = b =0, a = b =45,a = b =90 and a = b =135degrees (Geller uses alpha and beta; I've substituted a and b) So the experiment you cite does, in fact, use both polarizers set to 45º, which is the opposite of your claim. Also, if we were just trying to determine the polarization, we probably would not set the polarizer to 45 degrees. But the goal of this is to test the Bell inequality, rather than to test Malus's law. It's important to understand why you want the polarizers at 45º for one of the measurements. Link to comment Share on other sites More sharing options...
Lazarus Posted January 7, 2017 Author Share Posted January 7, 2017 Swansont: Now, how about answering my question: When the polarizer is at 45º, how do you tell which BBO generated the photon? I didn't ask you to change the question, I asked you to answer it. Lazarus: As you suggest, if the polarizers are set to 45 degrees the result will not allow determination of the polarization of a photon or which BBO it came from. That is also true for the equivalent settings, 135, 225 and 315. All other settings allow for determining the polarization. Swansont: But the goal of this is to test the Bell inequality, rather than to test Malus's law. It's important to understand why you want the polarizers at 45º for one of the measurements. If the 45 degree settings are the only ones for which the polarization setting cannot be determined, the other setting should not be used in CHSH type experiments. Link to comment Share on other sites More sharing options...
swansont Posted January 7, 2017 Share Posted January 7, 2017 Swansont: Now, how about answering my question: When the polarizer is at 45º, how do you tell which BBO generated the photon? I didn't ask you to change the question, I asked you to answer it. Lazarus: As you suggest, if the polarizers are set to 45 degrees the result will not allow determination of the polarization of a photon or which BBO it came from. That is also true for the equivalent settings, 135, 225 and 315. All other settings allow for determining the polarization. OK. Set it at 30. Tell me conclusively which crystal. Swansont: But the goal of this is to test the Bell inequality, rather than to test Malus's law. It's important to understand why you want the polarizers at 45º for one of the measurements. If the 45 degree settings are the only ones for which the polarization setting cannot be determined, the other setting should not be used in CHSH type experiments. The ones used are most efficient, but they are not the only ones. The issue is what you are misunderstanding to make you think they are the only angles. Link to comment Share on other sites More sharing options...
Lazarus Posted January 7, 2017 Author Share Posted January 7, 2017 Swansont: OK. Set it at 30. Tell me conclusively which crystal. Lazarus: An easy way to do it is to set the other polarizer at 0 degrees. If it gets a + the 30 degree polarizer got a horizontally polarized photon. Swansont:The ones used are most efficient, but they are not the only ones. The issue is what you are misunderstanding to make you think they are the only angles. Lazarus: I ran all 360 degrees on a simulation. This is all moot if a BBO can’t reverse the beams. Link to comment Share on other sites More sharing options...
swansont Posted January 7, 2017 Share Posted January 7, 2017 Swansont: OK. Set it at 30. Tell me conclusively which crystal. Lazarus: An easy way to do it is to set the other polarizer at 0 degrees. If it gets a + the 30 degree polarizer got a horizontally polarized photon. Swansont: The ones used are most efficient, but they are not the only ones. The issue is what you are misunderstanding to make you think they are the only angles. Lazarus: I ran all 360 degrees on a simulation. This is all moot if a BBO can’t reverse the beams. Try filling in the gaps in your knowledge of the underlying physics. It's obvious you're missing something. Link to comment Share on other sites More sharing options...
Lazarus Posted January 7, 2017 Author Share Posted January 7, 2017 It appears the only real issue we have is whether A BBO can reverse beams. The rest of the differences rest on that. Link to comment Share on other sites More sharing options...
swansont Posted January 7, 2017 Share Posted January 7, 2017 It appears the only real issue we have is whether A BBO can reverse beams. The rest of the differences rest on that. Why does it have to reverse the beams? Link to comment Share on other sites More sharing options...
Lazarus Posted January 7, 2017 Author Share Posted January 7, 2017 Swansont: Why does it have to reverse the beams? Lazarus: If the BBO’s do not reverse beams, the photons from the “left” BBO will always arrive in the same polarization and the photons from the “right” BBO will always arrive in the other polarization because we are only looking at coincidences. That means we can determine which BBO the photons came from. Link to comment Share on other sites More sharing options...
swansont Posted January 7, 2017 Share Posted January 7, 2017 Swansont: Why does it have to reverse the beams? Lazarus: If the BBO’s do not reverse beams, the photons from the “left” BBO will always arrive in the same polarization and the photons from the “right” BBO will always arrive in the other polarization because we are only looking at coincidences. That means we can determine which BBO the photons came from. No. You are wrong. Stop assuming you understand this. The BBOs are stacked. Photons go right and left from each. You can't tell which crystal generated the photons when the polarizers are placed at an arbitrary angle. Link to comment Share on other sites More sharing options...
Lazarus Posted January 8, 2017 Author Share Posted January 8, 2017 Here is my understanding of CHSH experiments: First is the Joshua Geller version. The BBO’s are Type I so all the generated pairs are parallel entangled. The 2 BBO’s are set at 45 degrees from 0 degrees. One is +45 and the other is -45 so there is 90 degrees between them. Any spot on their cones can be hit by generated photons. Both photons in the pairs from BBO #1 are all vertically polarized. From BBO #2 all photons are horizontally polarized. Following is a cross section of a BBO cone with the paths of some typical pairs. The 2 photons in an entangled pair are symmetrically aligned with the input beams path. The 2 cones overlap with the polarizers at the extremes of the overlap. The following is a cross section at the polarizers with some typical paths of pairs. The pairs from are centered on the axis of each BBO cone. Every spot of the overlap can be hit by both vertically and horizontally polarized photons. The problem is that the BBO source is apparent because all the photons from BBO #1 are vertically polarized and protons from BBO #2 are horizontally polarized. That doesn’t really show anything indeterminant. Next the same setup but using Type II BBO’s. That way both BBO’s generate both vertical and horizontal photons. Both polarizers see a mixture of vertical and horizontally polarized photons. That sounds like we now have indeterminate photons. But wait. As Swansont pointed out in post #59, one beam goes left of center and the other goes right of center. On a coincident event the A polarizer always gets the left photon and the B polarizer always gets the right photon. That means which BBO the photons came from can be determined. The solution is for the BBO’s to reverse the beams randomly. That also causes problems. Link to comment Share on other sites More sharing options...
swansont Posted January 8, 2017 Share Posted January 8, 2017 The cross-section of a cone is a circle. The problem is that the BBO source is apparent because all the photons from BBO #1 are vertically polarized and protons from BBO #2 are horizontally polarized. That doesnt really show anything indeterminant. I think I've covered this at least twice, so refer back to that for why you are wrong. But wait. As Swansont pointed out in post #59, one beam goes left of center and the other goes right of center. On a coincident event the A polarizer always gets the left photon and the B polarizer always gets the right photon. That means which BBO the photons came from can be determined. The solution is for the BBOs to reverse the beams randomly. That also causes problems. I've asked you to show this and you've failed to demonstrate it (45 and 30 degree orientations) You're going to assert it again? Seriously? Link to comment Share on other sites More sharing options...
Lazarus Posted January 9, 2017 Author Share Posted January 9, 2017 Swansont: I've asked you to show this and you've failed to demonstrate it (45 and 30 degree orientations) You're going to assert it again? Seriously? Lazarus: Closing your eyes does not make the color of the wall indeterminate. The detectors in the CHSH experiments certainly have the capability to determine the polarization of the photons and allow the determination of which BBO they came from. Setting the detectors at various angles, like 30 and 45, makes it difficult or impossible to determine the polarization. The detectors cannot determine the polarization, therefore the photons are entangled. The information is available, just being ignored Link to comment Share on other sites More sharing options...
swansont Posted January 9, 2017 Share Posted January 9, 2017 Setting the detectors at various angles, like 30 and 45, makes it difficult or impossible to determine the polarization. The detectors cannot determine the polarization, therefore the photons are entangled. BINGO! And since one is trying to establish whether or not the Bell inequality holds, we want entangled photons. The information is available, just being ignored Only if you arrange to not have entangled photons. Which is opposite of what the point of the experiment is. Link to comment Share on other sites More sharing options...
Lazarus Posted January 9, 2017 Author Share Posted January 9, 2017 Swansont: BINGO! And since one is trying to establish whether or not the Bell inequality holds, we want entangled photons Only if you arrange to not have entangled photons. Which is opposite of what the point of the experiment is. Lazarus: This experiment is trying to prove Quantum entanglement. We can’t assume Quantum entanglement from the start. The strange part of the CHSH experiments is that 3 of the tests are really the same and one is different. Three of the tests are set with 22.5 degrees between the polarizers and one test set at 67.5 degrees. Even the Quantum Theory assumes that only the angle between the polarizers matters. Link to comment Share on other sites More sharing options...
swansont Posted January 10, 2017 Share Posted January 10, 2017 Lazarus: This experiment is trying to prove Quantum entanglement. We can’t assume Quantum entanglement from the start. The strange part of the CHSH experiments is that 3 of the tests are really the same and one is different. Three of the tests are set with 22.5 degrees between the polarizers and one test set at 67.5 degrees. Even the Quantum Theory assumes that only the angle between the polarizers matters. An experiment trying to show the Bell inequality is not trying to prove entanglement. (There's a pdf from my search that's a lab trying to demonstrate entanglement, but college lab courses are recreating known phenomenon) What does the abstract of your paper say? (You won't share information about it and said to Google it, so I don't know if I'm looking at the same paper as you) Your repeated objection is that H or V photons are produced and you could tell the difference, but have never explained how. The BBO crystal produces polarized photons orthogonal to the polarization of the input. But the input beam to the BBO crystal is polarized at 45%, so for either crystal you don't know if it produces H or V photons, since the input is in a superposition of the states. So, explain clearly how you would tell what the polarization is, and what crystal generated the photons. Link to comment Share on other sites More sharing options...
Lazarus Posted January 11, 2017 Author Share Posted January 11, 2017 Swansont: An experiment trying to show the Bell inequality is not trying to prove entanglement. (There's a pdf from my search that's a lab trying to demonstrate entanglement, but college lab courses are recreating known phenomenon) What does the abstract of your paper say? (You won't share information about it and said to Google it, so I don't know if I'm looking at the same paper as you) Lazarus: We are looking at the same paper if you have the Joshua Geller paper labeled Lab 1 from the Rochester University. Einstein’s EPR paper was objecting to Quantum Entanglement. The CHSH experiments claim to refute EPR. That sounds like CHSH was “proving” Quantum Entanglement. Swansont: Your repeated objection is that H or V photons are produced and you could tell the difference, but have never explained how. The BBO crystal produces polarized photons orthogonal to the polarization of the input. But the input beam to the BBO crystal is polarized at 45%, so for either crystal you don't know if it produces H or V photons, since the input is in a superposition of the states. So, explain clearly how you would tell what the polarization is, and what crystal generated the photons. Lazarus: We are in agreement that both H and V photons from both BBO’s can hit both polarizers. These experiments are restricted to cases when both detectors are activated by the photon pair. For perpendicular polarizations, the H generated photons are on one side of a center line and the V protons are always on the other side. With the setup: the A detector on the left side, the B detector on the right side, BBO’s set 45 degrees on each side of the center, BBO #1 sending the H protons to the right side and the V protons to the left and BBO #2 sending V protons right and H protons left the sequence of events can be described. When a V photon from BBO #2 hits the left detector (A), the H photon cannot hit the right detector (B) because it is left of the left detector. That event is not considered. The only pairs from BBO #2 that are considered cause the H proton to hit the left detector and the V proton to hit the right detector. With the polarizors both set at 0 degrees the polarization can be determined and which BBO generated it. BBP #1 is opposite so an H proton hitting the left detector means it came from BBO #2. Setting the polarizers at different angles defeats their ability to determine polarity but that does not change the polarization of the incoming photons any more than closing your eyes changes the color of the wall. Also, please comment on 3 of the CHSH tests being the same test of 22.5 degrees separation of the two polarizers. Link to comment Share on other sites More sharing options...
swansont Posted January 11, 2017 Share Posted January 11, 2017 We are in agreement that both H and V photons from both BBO’s can hit both polarizers. These experiments are restricted to cases when both detectors are activated by the photon pair. For perpendicular polarizations, the H generated photons are on one side of a center line and the V protons are always on the other side. What center line? And no, we are not in agreement.Photons go in either direction, with either polarization. With the setup: the A detector on the left side, the B detector on the right side, BBO’s set 45 degrees on each side of the center, BBO #1 sending the H protons to the right side and the V protons to the left and BBO #2 sending V protons right and H protons left the sequence of events can be described. Nope. "When vertically/horizontally polarized photons of wavelength (lambda) are incident on a type-I BBO crystal, two momentum and energy conserving horizontally/ vertically polarized photons of wavelength 2(lambda) are emitted from the BBO." (p.4) So the polarization of both emitted photons is the same, and are orthogonal to the photon that generated them. If H comes in, two V photons come out. If you think that H photons go to one side and V to the other, show specifically in the paper where it says that. Further, the incoming light is polarized at 45º to either crystal, so it's in a superposition of both H and V polarization. A crystal can thus generate H photons, or V photons. When a V photon from BBO #2 hits the left detector (A), the H photon cannot hit the right detector (B) because it is left of the left detector. There would not be a mix of polarizations, and you just said that photons can hit the detectors. Where do you think you read this in the paper? I see where they check to see that the system is aligned, so the light is hitting the detectors. Why do you think that photons will miss? Setting the polarizers at different angles defeats their ability to determine polarity but that does not change the polarization of the incoming photons any more than closing your eyes changes the color of the wall. You still have not explained how you would determine this, but part of that is that your description of the setup is wrong. Try reading the paper again. Ask questions about it, because it's obvious that you have misunderstood several things. Also, please comment on 3 of the CHSH tests being the same test of 22.5 degrees separation of the two polarizers. "We measure the sixteen combinations of polarization angles for polarizers A and B. We do this by fixing a value of the angle a and varying the angle b from 0 to 360 degrees. We do this for four fixed a values." (p.6) It's obvious from all the data points on the graph (Fig. 6) that your statement is flat-out wrong. Link to comment Share on other sites More sharing options...
Lazarus Posted January 12, 2017 Author Share Posted January 12, 2017 Swansont: What center line? And no, we are not in agreement.Photons go in either direction, with either polarization. Lazarus: If both vertical and horizontal photons of a Barium Borate crystal can go in either generated beam, then the output of the BBO is unpolarized. We could use flashlights instead. If we can’t agree on this none of the rest of it matters. Link to comment Share on other sites More sharing options...
swansont Posted January 12, 2017 Share Posted January 12, 2017 Lazarus: If both vertical and horizontal photons of a Barium Borate crystal can go in either generated beam, then the output of the BBO is unpolarized. We could use flashlights instead. They look at the overall output and confirm it's unpolarized as part of the testing. (2) in procedure on p.6 "We next check that the two SPDC cones coming from the BBOs are completely overlapping by checking the polarization of the light incident on our detectors; if the cones are overlapped properly the total SPDC light cone is unpolarized. " If you think that a flashlight will work then you really need to take a giant step back and learn about entanglement before trying to attack Bell tests. Why do you think flashlight photons would have the same polarization correlation that these entangled photons would? Link to comment Share on other sites More sharing options...
Lazarus Posted January 13, 2017 Author Share Posted January 13, 2017 Swansont: They look at the overall output and confirm it's unpolarized as part of the testing. (2) in procedure on p.6 "We next check that the two SPDC cones coming from the BBOs are completely overlapping by checking the polarization of the light incident on our detectors; if the cones are overlapped properly the total SPDC light cone is unpolarized.“ Lazarus: The only reason that is unpolarized is that there are 2 BBO’s. Since each detection comes from 1 BBO, both of the generated photons are polarized. Link to comment Share on other sites More sharing options...
swansont Posted January 13, 2017 Share Posted January 13, 2017 You tell me what the output will be for input light polarized at 45 degrees. Link to comment Share on other sites More sharing options...
Lazarus Posted January 13, 2017 Author Share Posted January 13, 2017 You tell me what the output will be for input light polarized at 45 degrees. Link to comment Share on other sites More sharing options...
swansont Posted January 13, 2017 Share Posted January 13, 2017 polarize2.jpg Did you not understand the question? (There is nothing in my query about angle of incidence) Link to comment Share on other sites More sharing options...
Lazarus Posted January 13, 2017 Author Share Posted January 13, 2017 Swansont: Did you not understand the question? (There is nothing in my query about angle of incidence) Lazarus: The output of an absorptive polarizer set at 45 degrees is half of the light, polarized at the polarizer setting. An arriving photon has a 50-50 chance of making it through. The photons making it through have their polarization changed. The output of a beam-splitting polarizer is 2 polarized beams. At 45 degrees an arriving photon has a 50-50 chance of which beam it follows. The polarization of the incoming photons can be determined. Blinding the polarizer by changing its alignment does not change the polarization of the incoming photon. Where is anything indeterminate? Link to comment Share on other sites More sharing options...
swansont Posted January 13, 2017 Share Posted January 13, 2017 Swansont: Did you not understand the question? (There is nothing in my query about angle of incidence) Lazarus: The output of an absorptive polarizer set at 45 degrees is half of the light, polarized at the polarizer setting. An arriving photon has a 50-50 chance of making it through. The photons making it through have their polarization changed. The output of a beam-splitting polarizer is 2 polarized beams. At 45 degrees an arriving photon has a 50-50 chance of which beam it follows. The polarization of the incoming photons can be determined. Blinding the polarizer by changing its alignment does not change the polarization of the incoming photon. Where is anything indeterminate? I asked about the output of the crystal, but even here you haven't said how the polarization could be determined, and just claimed you could do it. The output photon takes the horizontal polarization path. What was the incoming polarization? Link to comment Share on other sites More sharing options...
Recommended Posts