jason.p Posted January 9, 2012 Share Posted January 9, 2012 Experiments to demonstrate/prove entanglement or non locality seem to have been performed using photons, presumably because they are relatively easy to produce and measure. Is it reasonable to assume that this also applies to all elementary particles and their anti particles even though they may never be measured with any degree of accuracy? If this is the case then can we assume that all particles produced at the big bang share this non locality with their partners wherever they are in the universe? In other words could it be said that everything is interconnected by probability waves? Link to comment Share on other sites More sharing options...
questionposter Posted January 10, 2012 Share Posted January 10, 2012 What? Entanglement disappears when you measure it. I knew that bleep movie was no good. Link to comment Share on other sites More sharing options...
jason.p Posted January 10, 2012 Author Share Posted January 10, 2012 What? Entanglement disappears when you measure it. I knew that bleep movie was no good. Sorry QP. I don't understand your reply. Surely in that case it's only the particles that have been measured that may have lost their entanglement, all others as yet unmeasured would still have the potential "link". I don't see entaglement as a physical communication across the universe as such but if there is a theoretical "probability wave" mediating this phenomenon then everything has a connection whether real or mathematical. I accept that I probably have a very naive view of QM but I really am trying to get my head round it and appreciate any help I can get. ( Just Googled Bleep movie. Looks good ) Link to comment Share on other sites More sharing options...
swansont Posted January 10, 2012 Share Posted January 10, 2012 Interactions cause decoherence. It doesn't have to be a lab measurement. Link to comment Share on other sites More sharing options...
jason.p Posted January 11, 2012 Author Share Posted January 11, 2012 Interactions cause decoherence. It doesn't have to be a lab measurement. Thanks. I've just been re-reading Brian Greene's book. On first reading I had understood that interactions confuse the measurements rather than eliminate them, but now I see that all the interactions actually reduce the effect to nothing. I suppose the way to see it is that particles could be entangled at great distances, but in fact would not be because of the interactions they would have encountered. Link to comment Share on other sites More sharing options...
swansont Posted January 11, 2012 Share Posted January 11, 2012 Thanks. I've just been re-reading Brian Greene's book. On first reading I had understood that interactions confuse the measurements rather than eliminate them, but now I see that all the interactions actually reduce the effect to nothing. I suppose the way to see it is that particles could be entangled at great distances, but in fact would not be because of the interactions they would have encountered. This has been the difficulty in doing long-range experiments, and why they use photons. A photon's state can be preserved in polarization-maintaining optical fiber. Transporting a spin state of an electron is much more problematic, or doing a photon experiment in free space. Link to comment Share on other sites More sharing options...
Widdekind Posted January 11, 2012 Share Posted January 11, 2012 This has been the difficulty in doing long-range experiments, and why they use photons. A photon's state can be preserved in polarization-maintaining optical fiber. Transporting a spin state of an electron is much more problematic, or doing a photon experiment in free space. What does such fiber do, to "maintain polarization" ? Via what other mean(s), might you replicate that "maintenance" effect, i.e. "shepherding" of photons ? When you say "free space", I understand, that you are referring to ubiquitous space gas & dust, with which space-propagating photons would invariably "interact", and so, as you posted previously, "decohere" ? How do boson-like Cooper Pairs, in super-conductors, "maintain coherence", despite defects, in the super-conducting material, through which they propagate ? Might that make for some approximately-accurate analogy, to boson-like photons, propagating through "defect ridden" deep space ? Perhaps some "photon pair", combining a "spin-forward" right-handed photon, with a "spin-backwards" left-handed photon, super-posed with a suitable phase shift, e.g. [math]\theta = 0,\pi[/math] ?? Link to comment Share on other sites More sharing options...
swansont Posted January 11, 2012 Share Posted January 11, 2012 What does such fiber do, to "maintain polarization" ? Via what other mean(s), might you replicate that "maintenance" effect, i.e. "shepherding" of photons ? Stress-induced birefringence in the fiber. There are a number of ways to achieve this http://www.rp-photonics.com/polarization_maintaining_fibers.html http://en.wikipedia.org/wiki/Polarization-maintaining_optical_fiber When you say "free space", I understand, that you are referring to ubiquitous space gas & dust, with which space-propagating photons would invariably "interact", and so, as you posted previously, "decohere" ? How do boson-like Cooper Pairs, in super-conductors, "maintain coherence", despite defects, in the super-conducting material, through which they propagate ? Might that make for some approximately-accurate analogy, to boson-like photons, propagating through "defect ridden" deep space ? Perhaps some "photon pair", combining a "spin-forward" right-handed photon, with a "spin-backwards" left-handed photon, super-posed with a suitable phase shift, e.g. [math]\theta = 0,\pi[/math] ?? Cooper pairs are interacting electrons, maintaining a minimum energy state within the superconductor. AFAIK they are superconducting because they don't interact. I don't know to which photon pair you refer. Photons don't pair up like this. Link to comment Share on other sites More sharing options...
Widdekind Posted January 12, 2012 Share Posted January 12, 2012 Cooper pairs are interacting electrons, maintaining a minimum energy state within the superconductor. AFAIK they are superconducting because they don't interact. I don't know to which photon pair you refer. Photons don't pair up like this. Once 'mated together', CPs are bosons. And, in 'super-conductors', such bosons can encounter defects, without decohering. Perhaps such SC phenomena, could be applied, to space communications, so that photons (bosons) could encounter space dust ('defects') without decohering ?? For the former, 'decoherence' would be the breaking apart, of the CP, with each electron scattering off from some defect. For the latter, 'decoherence' would be ... the photons scattering of from some space dust. I.e. the behavior, of different bosons (CPs, photons), upon 'decoherence' is different. But, before decoherence, whilst coherent, in both cases S=1 bosons encounter 'defects', in their propagation media. So, if the former can "weather the rough ride", then perhaps the latter could too, with sufficient "QM magic" ?? Link to comment Share on other sites More sharing options...
swansont Posted January 12, 2012 Share Posted January 12, 2012 Once 'mated together', CPs are bosons. And, in 'super-conductors', such bosons can encounter defects, without decohering. Perhaps such SC phenomena, could be applied, to space communications, so that photons (bosons) could encounter space dust ('defects') without decohering ?? For the former, 'decoherence' would be the breaking apart, of the CP, with each electron scattering off from some defect. For the latter, 'decoherence' would be ... the photons scattering of from some space dust. I.e. the behavior, of different bosons (CPs, photons), upon 'decoherence' is different. But, before decoherence, whilst coherent, in both cases S=1 bosons encounter 'defects', in their propagation media. So, if the former can "weather the rough ride", then perhaps the latter could too, with sufficient "QM magic" ?? As I said before, electrons are interacting with each other in the superconducting medium (which makes them pretty much useless for entangled communication). I suspect that if they broke apart when encountering a defect, they would simply re-form afterwards, because that is the minimum-energy condition. Photons don't behave this way. Link to comment Share on other sites More sharing options...
Widdekind Posted January 13, 2012 Share Posted January 13, 2012 (edited) Stress-induced birefringence in the fiber. There are a number of ways to achieve this http://www.rp-photonics.com/polarization_maintaining_fibers.html http://en.wikipedia.org/wiki/Polarization-maintaining_optical_fiber Cooper pairs are interacting electrons, maintaining a minimum energy state within the superconductor. AFAIK they are superconducting because they don't interact. I don't know to which photon pair you refer. Photons don't pair up like this. Thanks for the links. Are you saying, that, being "composite bosons", formed from the combination of magnetically-attracting (spin-anti-parallel) electrons, CPs are "extra stable" vs. scattering / decohering, upon encountering defects, in their propagation media, i.e. "CPs are self-sealing" (very approximate analogy) ? At RF, space has negligible optical density, and negligible optical depth, i.e. RF photons rarely interact with "space gas & dust", even across cosmological distances. And, if RF photons do not interact with such "defects in their propagation medium", i.e. gas & dust in space, then they do not decohere; and they would not 'disentangle'. Er go, transmitting entangled RF photons, "through the water hole", seems feasible. Edited January 13, 2012 by Widdekind Link to comment Share on other sites More sharing options...
swansont Posted January 13, 2012 Share Posted January 13, 2012 Thanks for the links. Are you saying, that, being "composite bosons", formed from the combination of magnetically-attracting (spin-anti-parallel) electrons, CPs are "extra stable" vs. scattering / decohering, upon encountering defects, in their propagation media, i.e. "CPs are self-sealing" (very approximate analogy) ? Cooper pairs form readily in a superconductor. At RF, space has negligible optical density, and negligible optical depth, i.e. RF photons rarely interact with "space gas & dust", even across cosmological distances. And, if RF photons do not interact with such "defects in their propagation medium", i.e. gas & dust in space, then they do not decohere; and they would not 'disentangle'. Er go, transmitting entangled RF photons, "through the water hole", seems feasible. OK. Now you have to figure out how to generate them and detect them above background. Link to comment Share on other sites More sharing options...
swansont Posted January 16, 2012 Share Posted January 16, 2012 ! Moderator Note Cooper pair/superconductivity question splithttp://www.scienceforums.net/topic/63093-cooper-pairs/page__pid__651570#entry651570 Link to comment Share on other sites More sharing options...
Widdekind Posted February 4, 2012 Share Posted February 4, 2012 (edited) I understand, that multiple particles are only "entangled", when they form a single "system", with a single quantum state, which must be a super-position, of several "pure" product states. Now, that is not to say, that super-positions -- specifically factorable super-positions -- must represent entanglements. For counter example, consider two separate electrons, each in the ground state, of two separate hydrogen atoms. Each of their individual states, can be written in any basis, e.g. position basis [math]\psi_0(x) = \psi_0(x_1) \delta(x-x_1) + \psi_0(x_2) \delta(x-x_2) + ...[/math]. And, if so, then their composite state would involve cross terms: [math]\psi_{0,1} \otimes \psi_{0,2} = \left( \psi_0(x_1) \delta(x-x_1) + \psi_0(x_2) \delta(x-x_2) + ... \right)_1 \otimes \left( \psi_0(x_1) \delta(x-x_1) + \psi_0(x_2) \delta(x-x_2) + ... \right)_2[/math] However, upon a position measurement, the location "chosen" by one particle, would in no way constrain or limit, the position "chosen" by the other particle, i.e. "each could appear 'here' or 'there', without care, as to what the other was doing". Such "care free attitude" defines non-entangled states. Intuitively, the particles are non-entangled, b/c they behave, upon measurement, as if they "never noticed each other". Conversely, entangled particles must be represented, by non-factorable super-positions, of pure product states, embodying some "mutually exclusive behaviors", i.e. some "either-or" quality, according to which the "behavior" of one particle, upon measurement, necessarily limits & constrains, the "behavior" of their partner particle, as if "there is only so much, of some limited quantity, to be shared between them". Er go, a system state such as [math]|0\rangle_1 |1\rangle_2 + |1\rangle_1 |0\rangle_2[/math], embodying the possible "behaviors" of "either the first particle is in the ground state, and the second particle is in an excited state; or vice versa". Such entangled states are due to interactions, which impose the familiar Conservation Laws, for Conserved Quantities, e.g. "conservation of energy". I.e. particles are entangled, when they behave, upon measurement, as if "they negotiated with the other, on how to divvy up the earnings" of their interaction, e.g. "particle one takes 'spin up', particle two takes 'spin down'". I understand all of this, from Compendium of Quantum Physics (Springer, 2009); and from 101 Quantum Questions by Kenneth Ford. Edited February 4, 2012 by Widdekind Link to comment Share on other sites More sharing options...
jason.p Posted February 4, 2012 Author Share Posted February 4, 2012 Thanks for all the responses. I'm afraid I got lost at around post #7. It seems I've got a hell of a lot to learn Thanks anyway. Link to comment Share on other sites More sharing options...
Widdekind Posted February 4, 2012 Share Posted February 4, 2012 Scientific American has published a simplified explanation, with 10 minute video, of QE. Link to comment Share on other sites More sharing options...
IM Egdall Posted February 5, 2012 Share Posted February 5, 2012 (edited) Scientific American has published a simplified explanation, with 10 minute video, of QE. Clever video! Or I should say "clever John Bell".I'll give the link to my students in my quantum emtangement class. Thanks. Edited February 5, 2012 by IM Egdall Link to comment Share on other sites More sharing options...
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