imatfaal Posted October 29, 2016 Posted October 29, 2016 I am pretty sure that I mentioned above that Bell's Inequality is purely mathematical and it is the experimentation which leverages a combination of the inequality with the results that will be obtained from a state of entanglement - ie the statistics / mathematics allows you to say what distributions of results can and cannot be produced from a single joint distribution. Are the pairs of pennies in an entangled unknowable state of superposition? (hint: No) Yet again you seem to have just cherry-picked an equation. You need to show that this actually happens. Or you need to accurately model what happens when a flying disc hits a bar and how that varies with the angle of discs axis with reference to the bar. 1
Lazarus Posted December 30, 2016 Author Posted December 30, 2016 I am pretty sure that I mentioned above that Bell's Inequality is purely mathematical and it is the experimentation which leverages a combination of the inequality with the results that will be obtained from a state of entanglement - ie the statistics / mathematics allows you to say what distributions of results can and cannot be produced from a single joint distribution. Are the pairs of pennies in an entangled unknowable state of superposition? (hint: No) Yet again you seem to have just cherry-picked an equation. You need to show that this actually happens. Or you need to accurately model what happens when a flying disc hits a bar and how that varies with the angle of discs axis with reference to the bar. While writing a CHSH type device simulator with pennies rather than photons is became apparent that there were inconsistencies in the CHSH experiments. Perhaps the biggest error is in the justification of the photons being indeterminate. From the overlapping cones of the paths of the photons it is clear that both the vertically polarized and horizontally polarized photons can reach either polarizer. But since we only see cases where there is detection in both the A and B sides, the left polarizer only sees vertically polarized photons and the right polarizer only sees horizontally polarized photons. When a left photon hits the left polarizer the right photon hits the right polarizer but when a right photon hits the left polarizer the left photon can not hit the right polarizer. The left photon is to the left of the left polarizer so is nowhere close to the right polarizer. No indeterminacy.
imatfaal Posted January 4, 2017 Posted January 4, 2017 OK CHSH is a mathematical inequality - your quibble is with Spontaneous Parametric Down Conversion. For Aspect Experiment etc you take your light from the intersection of the two cones of light; here you do not have horizontal and vertical polarisation - you have entangled photons in a state of superposition. If they were not entangled you would get different results - we don't get these different results
swansont Posted January 4, 2017 Posted January 4, 2017 You can't entangle pennies, so I'm thinking that it's why a simulation using them wouldn't work.
Lazarus Posted January 5, 2017 Author Posted January 5, 2017 1 A beam splitter produces 2 polarized beams. Here is a table showing the polarizations which are dependent on the angle of the input beam. 2 The only thing indeterminate is which detector the photons will hit. The pairs of photons have a known polarization and a definite spatial relation. The “proof” is that we can’t know which proton goes to which detector. By selecting only cases where both detectors are activated we are only counting cases where each detector is always getting the same photon polarization.
swansont Posted January 5, 2017 Posted January 5, 2017 The pairs of photons have a known polarization and a definite spatial relation. The “proof” is that we can’t know which proton goes to which detector. By selecting only cases where both detectors are activated we are only counting cases where each detector is always getting the same photon polarization. The "definite spatial relation" is that they are co-propagating. What is the experimental layout that corresponds to the down-conversion picture? You give one in post #10, but that's for counter-propgating photons.
Lazarus Posted January 5, 2017 Author Posted January 5, 2017 One of the better descriptions of a CHSH experiment is by Joshua Geller. You can download the full PDF description from the University of Rochester. In any CHSH experiment the SOURCE produces 2 polarized beams. The polarization in each beam always stays the same and should always go to the same detector. It does not make sense for the A beam to be going to the B detector.
imatfaal Posted January 5, 2017 Posted January 5, 2017 No the signal and idler photons are entangled - they exist in superposition the state should be [latex]\Bra{H}[/latex] \bra{\Phi}=\frac{1}{\sqrt{2}}\bra{V}_{signal} \bra{V}_{idler} + \frac{1}{\sqrt{2}}\bra{H}_{signal} \bra{H}_{idler} Getting something wrong in my latex [latex]|\phi \rangle_{type 1} = \frac{1}{\sqrt{2}} |V \rangle_{signal}|V \rangle_{idler} + \frac{1}{\sqrt{2}} |H \rangle_{signal}|H \rangle_{idler}[/latex] I think and for type two conversion which is what you had in your picture in earlier post [latex]|\phi \rangle_{type 2} = \frac{1}{\sqrt{2}} |V \rangle_{signal}|H \rangle_{idler} + \frac{1}{\sqrt{2}} |V \rangle_{idler}|H \rangle_{signal}[/latex]
Lazarus Posted January 5, 2017 Author Posted January 5, 2017 The math is wonderful but does not necessarily match the real world. The physical experiment must stand on its own feet. We have already shown that the three coins in the fountain proof is not valid. It simply says that adding a non-negative number to a positive number can not make the positive number smaller. Geller’s description of the experiment requires that A beam particles reach the B detector. That should never happen.
swansont Posted January 5, 2017 Posted January 5, 2017 One of the better descriptions of a CHSH experiment is by Joshua Geller. You can download the full PDF description from the University of Rochester. You couldn't post a link so we're all looking at the same document? In any CHSH experiment the SOURCE produces 2 polarized beams. The polarization in each beam always stays the same and should always go to the same detector. It does not make sense for the A beam to be going to the B detector. What do you mean when you say the polarization of each beam is always the same? The polarization is unknown. It's the "same" in the sense that it's consistent — you have two photons of orthogonal polarization — but you don't know which path has which polarization. That's why they use two BBO crystals and pump with the polarization at 45º. So that the photons could come from either crystal, and you can't tell them apart. So saying that they always go to the same detector is wrong. You're not understanding the setup of the experiment.
imatfaal Posted January 5, 2017 Posted January 5, 2017 ...We have already shown that the three coins in the fountain proof is not valid... You have shown nothing conclusively in this thread. Where do you think you have proven something to anyone else's agreement?
Lazarus Posted January 5, 2017 Author Posted January 5, 2017 You have shown nothing conclusively in this thread. Where do you think you have proven something to anyone else's agreement? You are correct. It just wasn't refuted. You couldn't post a link so we're all looking at the same document? What do you mean when you say the polarization of each beam is always the same? The polarization is unknown. It's the "same" in the sense that it's consistent — you have two photons of orthogonal polarization — but you don't know which path has which polarization. That's why they use two BBO crystals and pump with the polarization at 45º. So that the photons could come from either crystal, and you can't tell them apart. So saying that they always go to the same detector is wrong. You're not understanding the setup of the experiment. Thank you for pointing out that there are 2 BBO’s in the source. I completely missed that. However, the problem is still there. The 45 degrees you mentioned does not cause the generated photons to be polarized randomly. The polarization is changed but each generated beam always has the same angle of polarization. There is a time separation between the generation of pairs between the 2 BBO’s so they are completely independent from each other. That means every thing about each incident can be calculated. There is no room for indeterminacy. The reason I didn't post the link to Geller's PDF is that the link is so long I had a problem with it. Searching for "entanglement bell joshua geller" will lead you to it.
imatfaal Posted January 5, 2017 Posted January 5, 2017 You are correct. It just wasn't refuted. I provided the correct formulation of Bell's inequality as it is used - please feel free to provide anything that casts doubt on either Bell's Inequality or CHSH Inequality. These are 'just' mathematics - so they can be proved or disproved; feel free
Lazarus Posted January 5, 2017 Author Posted January 5, 2017 I provided the correct formulation of Bell's inequality as it is used - please feel free to provide anything that casts doubt on either Bell's Inequality or CHSH Inequality. These are 'just' mathematics - so they can be proved or disproved; feel free It is highly improbable that the mathematics is flawed. I appreciate your patience.
swansont Posted January 5, 2017 Posted January 5, 2017 You are correct. It just wasn't refuted. Thank you for pointing out that there are 2 BBOs in the source. I completely missed that. However, the problem is still there. The 45 degrees you mentioned does not cause the generated photons to be polarized randomly. The polarization is changed but each generated beam always has the same angle of polarization. There is a time separation between the generation of pairs between the 2 BBOs so they are completely independent from each other. That means every thing about each incident can be calculated. There is no room for indeterminacy.The reason I didn't post the link to Geller's PDF is that the link is so long I had a problem with it. Searching for "entanglement bell joshua geller" will lead you to it. You aren't measuring the time, and if the crystals are e.g. 3 mm thick that's an average of 10 picoseconds of difference. You aren't going to measure that. Plenty of room for indeterminacy. Since the two crystals give opposite polarizations, you do not have the same polarization in each detector. A pretty good rule to follow is that if you think you've found an obvious flaw in well-established physics papers, it means you missed something. Assume the paper is right and you are wrong, and look for your mistake.
Lazarus Posted January 6, 2017 Author Posted January 6, 2017 You aren't measuring the time, and if the crystals are e.g. 3 mm thick that's an average of 10 picoseconds of difference. You aren't going to measure that. Plenty of room for indeterminacy. Since the two crystals give opposite polarizations, you do not have the same polarization in each detector. A pretty good rule to follow is that if you think you've found an obvious flaw in well-established physics papers, it means you missed something. Assume the paper is right and you are wrong, and look for your mistake. You seem to be talking about the time difference in the 2 beams from one BBO. The time separation of concern is the time between the output one BBO and the other. The time increment that is allowed for one event is about 25ns. The coincident detections have to come from one of the 2 BBO’s. The BBO’s take turns. They very rarely both react in the same 25ns.
swansont Posted January 6, 2017 Posted January 6, 2017 You seem to be talking about the time difference in the 2 beams from one BBO. The time separation of concern is the time between the output one BBO and the other. The time increment that is allowed for one event is about 25ns. The coincident detections have to come from one of the 2 BBO’s. The BBO’s take turns. They very rarely both react in the same 25ns. No, I'm talking about the time delay between coming from one BBO or the other. If you can't tell, or don't know, you can't tell which crystal it came from. They put in a quartz plate to compensate for the phase delay for the longer path length. Hence, you don't know which polarization pair you are getting. The BBOs take turns? Where do think you read that? What I read in the paper I got was "spontaneous parametric down- conversion, a nonlinear X(2) process of very low probability – on the order of 10-10 " IOW, it doesn't take turns
imatfaal Posted January 6, 2017 Posted January 6, 2017 It is highly improbable that the mathematics is flawed. I appreciate your patience. Who said any maths was flawed? You stated this "We have already shown that the three coins in the fountain proof is not valid" - Where? You made a few comments about a simplified version of Bell's Inequality (never claiming it was wrong) and I provided the inequality that was actually used. You cannot just come with a claim like that - you need to specify/answer 1 . where is the three coins in a fountain proof? 2 . how is it shown to be wrong? 3 . does this impact on Bell?
Lazarus Posted January 6, 2017 Author Posted January 6, 2017 Who said any maths was flawed? You stated this "We have already shown that the three coins in the fountain proof is not valid" - Where? You made a few comments about a simplified version of Bell's Inequality (never claiming it was wrong) and I provided the inequality that was actually used. You cannot just come with a claim like that - you need to specify/answer 1 . where is the three coins in a fountain proof? At the start of this thread. 2 . how is it shown to be wrong? The net result is the statement, adding a non-negative number to a positive number doesn't make the positive number smaller. 3 . does this impact on Bell? No, since this version may not actually match Bell's theory. No, I'm talking about the time delay between coming from one BBO or the other. If you can't tell, or don't know, you can't tell which crystal it came from. They put in a quartz plate to compensate for the phase delay for the longer path length. Hence, you don't know which polarization pair you are getting. The BBOs take turns? Where do think you read that? What I read in the paper I got was "spontaneous parametric down- conversion, a nonlinear X(2) process of very low probability – on the order of 10-10 " IOW, it doesn't take turns Here is typical time line so you can point out the error. A couple billion polarizer photons hit each BBO. One of the photon that hit the A BBO split into a pair of entangled photons which are recorded by the detectors. None of the photons that hit the B BBO split. Another couple billion photons hit each BBO some nanoseconds later. Same thing happens again, one A BBO photons splits and none of the B one do. A third set of a couple billion photons hit each BBO. None of the A photons split but one of the photons that hit the B BBO splits into a pair of entangled photons which are recorded by the detectors. The fourth time that a couple billion photons hit each BBO something rare happens. Both BBO’s have a photon split into two entangled photons. All 4 photons hit the detectors within 25 ns. What happens is dependent on the design of the Coincidence Monitor. It should ignore them because that is not in the playbook. None of the next batch of photons split in either BBO so no detections. Where is any indeterminacy involved?
swansont Posted January 6, 2017 Posted January 6, 2017 Here is typical time line so you can point out the error. A couple billion polarizer photons hit each BBO. One of the photon that hit the A BBO split into a pair of entangled photons which are recorded by the detectors. None of the photons that hit the B BBO split. Another couple billion photons hit each BBO some nanoseconds later. Same thing happens again, one A BBO photons splits and none of the B one do. A third set of a couple billion photons hit each BBO. None of the A photons split but one of the photons that hit the B BBO splits into a pair of entangled photons which are recorded by the detectors. The fourth time that a couple billion photons hit each BBO something rare happens. Both BBO’s have a photon split into two entangled photons. All 4 photons hit the detectors within 25 ns. What happens is dependent on the design of the Coincidence Monitor. It should ignore them because that is not in the playbook. None of the next batch of photons split in either BBO so no detections. Where is any indeterminacy involved? You don't know if the photons came from A or B. Their source is undetermined, as far as the experiment is concerned. As far as the "rare" event goes, you should be able to tell how often that's expected and to what extent it skews the results. Rare implies a very small effect.
Lazarus Posted January 7, 2017 Author Posted January 7, 2017 You don't know if the photons came from A or B. Their source is undetermined, as far as the experiment is concerned. As far as the "rare" event goes, you should be able to tell how often that's expected and to what extent it skews the results. Rare implies a very small effect. The rate of generation of pairs is intentionally slowed so that there is normally one generation in the 25ns window. Whether the double generations are rare or not shouldn’t be an issue because they should not be counted. The detectors know which BBO the photon came from because the polarization is different when it comes from A than it is when it comes from B and the detector can tell us.
swansont Posted January 7, 2017 Posted January 7, 2017 The rate of generation of pairs is intentionally slowed so that there is normally one generation in the 25ns window. Whether the double generations are rare or not shouldn’t be an issue because they should not be counted. The detectors know which BBO the photon came from because the polarization is different when it comes from A than it is when it comes from B and the detector can tell us. The 25 ns is the coincidence window. I don't know what other significance you are attaching to this. The detections have to occur within the window, or they will be rejected. The detectors do not know if the polarizer is at an arbitrary angle. Let's say it's at 45 degrees. The probability of an H or V photon getting through is 0.5. You detect a photon. Which crystal generated it?
Lazarus Posted January 7, 2017 Author Posted January 7, 2017 The 25 ns is the coincidence window. I don't know what other significance you are attaching to this. The detections have to occur within the window, or they will be rejected. The detectors do not know if the polarizer is at an arbitrary angle. Let's say it's at 45 degrees. The probability of an H or V photon getting through is 0.5. You detect a photon. Which crystal generated it? The 25ns is the window that is to catch both photons from one BBO, not to catch photons from both BBO’s in one window. The experiment is designed to space out the down conversions so that they happen one at a time. There is no requirement or expectation that both BBO’s will generate pairs at the same time. The output of a BBO is 2 polarized beams. All the photons in one beam have the same angle of polarization. The BBO’s are 90 degrees apart. When the left BBO is set so the left beam is vertically polarized and the right beam horizontal the vertical photons will always hit the left detector and the horizontal photons will always hit the right detector when both detectors are activated. The right BBO has the reverse effect.
swansont Posted January 7, 2017 Posted January 7, 2017 The 25ns is the window that is to catch both photons from one BBO, not to catch photons from both BBO’s in one window. The experiment is designed to space out the down conversions so that they happen one at a time. There is no requirement or expectation that both BBO’s will generate pairs at the same time. Yes. That's what "the coincidence window is 25 ns" means. The output of a BBO is 2 polarized beams. All the photons in one beam have the same angle of polarization. Correct. The BBO’s are 90 degrees apart. When the left BBO is set so the left beam is vertically polarized and the right beam horizontal the vertical photons will always hit the left detector and the horizontal photons will always hit the right detector when both detectors are activated. The right BBO has the reverse effect. It's not left and right, it's front and back. The beams from the two nearly overlap, so that both beams headed toward a detector will hit it. Now, how about answering my question: When the polarizer is at 45º, how do you tell which BBO generated the photon?
Lazarus Posted January 7, 2017 Author Posted January 7, 2017 It's not left and right, it's front and back. The beams from the two nearly overlap, so that both beams headed toward a detector will hit it. That is a key point. If the 2 generated photons can hit A to A and B to B but also can hit A to B and B to A that implies that neither generated beam is polarized. ------------------------------------------------ Now, how about answering my question: When the polarizer is at 45º, how do you tell which BBO generated the photon? 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.
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