Slinkey Posted June 11, 2015 Share Posted June 11, 2015 (edited) Apologies if this has been asked before but as this thought experiment doesn't have a title that I'd know of it's hard to search for it to see if it has been asked. We have a pair of entangled particles. They fly apart toward a different experiment each. Particle A is flying towards the classic double slit experiment which reveals the wave nature of light and the classic interference pattern. Particle B is flying towards another double slit experiment but this one has a detector on one of the slits so we can see which slit the particle went through. There are three ways we can perform this experiment. 1. Particle A reaches its experiment first and we get interference. What happens with particle B? Do we get interference or no interference? Would particle A affect how particle B behaves in its experiment? 2. Particle B reaches its experiment first and we get no interference. What happens with particle A? Do we get interference or no interference? Would particle B affect how particle A behaves in its experiment? 3. They reach their respective experiment simultaneously. What happens? Thoughts? Edited June 11, 2015 by Slinkey Link to comment Share on other sites More sharing options...
imatfaal Posted June 11, 2015 Share Posted June 11, 2015 Sounds very like the (delayed choice) quantum eraser experiment. Yes it has been done and you should be able to get some great write ups. http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser Link to comment Share on other sites More sharing options...
swansont Posted June 11, 2015 Share Posted June 11, 2015 You aren't going to see wave behavior with one particle. Why are the wave and particle behaviors entangled? Link to comment Share on other sites More sharing options...
Slinkey Posted June 14, 2015 Author Share Posted June 14, 2015 I was under the impression that entangled particles are described by the same wave equation. I always took this to mean that the wave equation describes both particles and that if one particle collapses the wave equation then the collapse happens for the other particle as well. Thus in experiment 2 where we have the detected particle collapsing the wave equation first it will destroy the wave nature of the other particle. But as you have asked if these states are entangled it's made me think I have misunderstood something basic about entanglement. Link to comment Share on other sites More sharing options...
swansont Posted June 14, 2015 Share Posted June 14, 2015 The wave function describes the state of the particle. "Wave" vs "particle" is not something it describes. There is no "wave nature" or "particle" eigenfunction or eigenvalue. Link to comment Share on other sites More sharing options...
Slinkey Posted June 14, 2015 Author Share Posted June 14, 2015 The wave function describes the state of the particle. "Wave" vs "particle" is not something it describes. There is no "wave nature" or "particle" eigenfunction or eigenvalue. I don't recall saying that it did. I am under the impression that when we measure some property of a particle we collapse its wavefunction. This explains why we don't get interference in the double slit experiment when we look to see which slit it went through. We've asked it to be a particle so we can see where it went, and the wave function collapses. Is this not correct? So if the wave function is describing an entangled state, when it collapses doesn't its collapse result in two particles? Link to comment Share on other sites More sharing options...
swansont Posted June 15, 2015 Share Posted June 15, 2015 I don't recall saying that it did. I am under the impression that when we measure some property of a particle we collapse its wavefunction. This explains why we don't get interference in the double slit experiment when we look to see which slit it went through. We've asked it to be a particle so we can see where it went, and the wave function collapses. Is this not correct? So if the wave function is describing an entangled state, when it collapses doesn't its collapse result in two particles? When you detect a particle, you detect a particle. If there was wave behavior somewhere along the way, the results will reflect that. e.g. if you send a particle through a double slit, you will detect a particle. But a series of them will reflect an interference pattern. Do you have an example in mind where you detect a wave, rather than observe the effect of wave behavior prior to detection? Link to comment Share on other sites More sharing options...
Slinkey Posted June 15, 2015 Author Share Posted June 15, 2015 Yes, I understand how interference is created. If you recall my thought experiment called for the entangled particles to be sent to separate double slit experiments one of which has a detector on one of the slits to see which slit the particle went through, and the other does not. As I understand it in the DS experiment, if we let a particle pass through the apparatus undetected we reveal its wave behaviour and an interference pattern will build up on the screen. If we detect which slit the particle went through then we don't get an interference pattern. This implies that we have destroyed the wave property of the particle. Agreed? So, if we have a pair of entangled particles, which as I understand it is described by a single wave function, then if we collapse the wave function by detecting which slit the particle passed through in the DS experiment with the detector, will that make the other entangled particle behave like a particle or will it behave like a wave when it passes through its DS experiment without a detector (experiment 2)? Link to comment Share on other sites More sharing options...
Strange Posted June 15, 2015 Share Posted June 15, 2015 This implies that we have destroyed the wave property of the particle. Agreed? No. So, if we have a pair of entangled particles, which as I understand it is described by a single wave function, then if we collapse the wave function by detecting which slit the particle passed through in the DS experiment with the detector, will that make the other entangled particle behave like a particle or will it behave like a wave when it passes through its DS experiment without a detector (experiment 2)? This is how modern tests (such as the quantum eraser) are done. This shows that if you detect the "which path" information by using the "other one" of an entangled pair (so you do not directly affect the photon going through the slits) then you still destroy the interference pattern. Link to comment Share on other sites More sharing options...
swansont Posted June 15, 2015 Share Posted June 15, 2015 Yes, I understand how interference is created. If you recall my thought experiment called for the entangled particles to be sent to separate double slit experiments one of which has a detector on one of the slits to see which slit the particle went through, and the other does not. As I understand it in the DS experiment, if we let a particle pass through the apparatus undetected we reveal its wave behaviour and an interference pattern will build up on the screen. If we detect which slit the particle went through then we don't get an interference pattern. This implies that we have destroyed the wave property of the particle. Agreed? No, not the way you are portraying this. Wave vs particle is a matter of how you detect it. If you know which slit the particle went through but then passed it through another double slit, you would see interference. Link to comment Share on other sites More sharing options...
Enthalpy Posted July 20, 2015 Share Posted July 20, 2015 A few hints: - To observe an interference pattern, which is only a statistics, you need to detect many photons. One photon only gives one spot, in such an experiment. - "Knowing that the other particle etc" means that the experiment discards some observations: the ones when the other particle didn't behave as chosen. - There are uncertainties at entagled particles as well. That is, the observed property at one does not determine exactly the property at the other. - "Wave versus particle" is more a game for newspaper. A photon is both. Link to comment Share on other sites More sharing options...
Strange Posted July 20, 2015 Share Posted July 20, 2015 - "Knowing that the other particle etc" means that the experiment discards some observations: the ones when the other particle didn't behave as chosen. Does it? Link to comment Share on other sites More sharing options...
swansont Posted July 20, 2015 Share Posted July 20, 2015 A few hints: - There are uncertainties at entagled particles as well. That is, the observed property at one does not determine exactly the property at the other. . Yes, it does. That's part of the point in entanglement. You measure a property and know the other one. They're quantized, so there is no room for uncertainty. Link to comment Share on other sites More sharing options...
Enthalpy Posted February 4, 2016 Share Posted February 4, 2016 No, it doesn't. Some properties like the momentum or the position are not quantized for free particles. Entanglement leaves uncertainty in the relation between the particles, which can't be arbitrarily accurate. This has been measured experimentally and compared with Heisenberg's relations. Link to comment Share on other sites More sharing options...
imatfaal Posted February 4, 2016 Share Posted February 4, 2016 No, it doesn't. Some properties like the momentum or the position are not quantized for free particles. Entanglement leaves uncertainty in the relation between the particles, which can't be arbitrarily accurate. This has been measured experimentally and compared with Heisenberg's relations. Surely with position momentum entangled particles (spdc photons) if you measure the momentum to whatever accuracy you want of A you know the momentum to that accuracy of particle B - it is the position of particle B that is undefined within uncertainty parameters 1 Link to comment Share on other sites More sharing options...
swansont Posted February 4, 2016 Share Posted February 4, 2016 No, it doesn't. Some properties like the momentum or the position are not quantized for free particles. Entanglement leaves uncertainty in the relation between the particles, which can't be arbitrarily accurate. This has been measured experimentally and compared with Heisenberg's relations. Then you should have no trouble providing a citation and/or link Link to comment Share on other sites More sharing options...
Enthalpy Posted April 5, 2016 Share Posted April 5, 2016 Momentum and position not quantized for free particles shouldn't be a difficulty, so I guess you ask about entanglement having an uncertainty. Here's a paper about the relation between the precision of momentum entanglement and position precision:http://arxiv.org/ftp/arxiv/papers/0905/0905.4830.pdf the authors compare with Heisenberg's relation. In the situations where the process that creates entangled particles is known, the uncertainty appears logically from the properties (for instance the transverse dimensions) of the source. Link to comment Share on other sites More sharing options...
imatfaal Posted April 8, 2016 Share Posted April 8, 2016 Could you point out the bit - as it is all quite above my head - where the calculation/formula is along the lines of: (Variance in Position of X) and (Variance in Position of Y) are related to some function of Planck's Constant or other constant Cos that is what I would be looking for. What I see is lots of forms and extrapolations of: Variance^2(Difference in position of X1 and X2) and Variance^2(Difference in Momentum of X1 and X2) is related to some function of Planck's constant The first equation is what you have claimed above - the second is what everyone else has been saying. Link to comment Share on other sites More sharing options...
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