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I Build Your Own Quantum Entanglement Experiment?

  1. May 21, 2017 #1
  2. jcsd
  3. May 22, 2017 #2

    Vanadium 50

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    I wouldn't call this entanglement.
     
  4. May 22, 2017 #3
    What would you call it?
     
  5. May 22, 2017 #4

    Vanadium 50

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    Not entanglement.
     
  6. May 22, 2017 #5

    stevendaryl

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    Why isn't it entanglement? Isn't the article talking about essentially a home-made version of the EPR experiment with twin photons?
     
  7. May 22, 2017 #6

    Nugatory

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    It's not doing the EPR thing in which entanglement is used in attempt to determine the value of two non-commuting observables... but it is a valid demonstration of entanglement (with a very high value*accessibility figure of merit).

    What it doesn't do is demonstrate that the observed correlations must be due to entanglement, as opposed to something equivalent to Bertlemann's socks.
     
  8. May 22, 2017 #7
    Do it yourself for just 20,000 dollars!!!
     
  9. May 23, 2017 #8

    zonde

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    Just to add.
    To tell apart simplest Bertlemann's socks type explanation from entanglement, results would have to be significant enough to tell apart ##\sin(\alpha-\beta)## correlation from ##\frac14+\frac12\sin(\alpha-\beta)## correlation. For ##\alpha-\beta=90^{\circ}## this would be telling apart correlations 1 and 0.75.
     
  10. May 23, 2017 #9

    vanhees71

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    The articles are, unfortunately a bit vague, indeed. E.g., you should always tell your readers which observables you are after that are supposed to be entangled. In the case of photons it cannot be position and momentum as in the original EPR, because there's no position observable of a photon. What's, entangled are the two momenta. The pair momentum is given by the total momentum of the electron and positron. In the center-momentum frame you have ##\vec{q}_1=-\vec{q}_2##, if ##\vec{q}_{1,2}## are the momenta of the photons in the considered pair-annihilation reaction ##e^+ + e^- \rightarrow \gamma+\gamma##. So there's an entanglement between the momenta of the photons, but the author doesn't demonstrate this entanglement but just doing coincidence measurements. As far as I can see, it's a correct measurement to demonstrate pair annihilation, but it's not a demonstration of entanglement.
     
  11. May 23, 2017 #10

    zonde

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    Author of the article was talking about polarization measurement.
     
  12. May 23, 2017 #11

    vanhees71

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    Argh! I should have read the 2nd part too. That's of course a nice demonstration of polarization entanglement. This is referring to the annihilation of positronium in matter. You have two classes of positronium states with regard to spin: Either positron and electron couple with total spin 0 (odd in the spin states) or total apin 1 (even in the spin states). The parity is even and odd, respectively, and the annihilation process is electromagnetic and thus parity is conserved. Thus only in the first case you can have an annihilation to 2 photons (in the 2nd case it must be an odd number of photons, and for kinematic reasons the annihilation can only be to 3 (or more)). So, if the decay is from positronium into 2 photons it must be from the spin-odd state, i.e., total spin 0, and thus the photon helicities must be opposite, i.e., the photon polarization are entangled to total helicity 0 (or written in the linear-polarization basis ##|HV \rangle-|VH \rangle##).
     
  13. May 23, 2017 #12

    Vanadium 50

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    I have a bit more time now, so...

    The article spends a good fraction of time on the angular correlation between the gammas, and sees an effect at 180 degrees - as expected. This is not entanglement any more than Bertlemann's socks. I hope we don't have to delve further into this.

    I don't know exactly what he's supposed to be measuring with respect to polarization, but he's not doing what he thinks he's doing. Compton polarimetry correlates the electron's spin with the photon's spin (and direction). An aluminum cube doesn't have its electrons spinning in any particular direction. Also, for a spin-0 state, the photons have the same helicities, so he should see a defiict, not an excess. So his polarimetry shouldn't work, and by the evidence that he has presented, it doesn't.

    Whatever it is, it's not a demonstration of entanglement.
     
  14. May 23, 2017 #13

    vanhees71

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    I think, it in fact is a demonstration of entanglement, because my argument above is also valid for free electrons and positrons annihilating: In the center-mass frame (which we can use to argue without loss of generality due to Lorentz invariance) the angular momentum (i.e., spin) of the pair is either ##S=0## or ##S=1##. Due to parity conservation under electromagnetic interactions, the exclusive channel ##e^+ + e^- \rightarrow 2 \gamma## (i.e., the annihilation into two photons and nothing else) can only occur for ##S=0##, and thus no matter, whether you have polarized or unpolarized electrons ans positrons (in the here discussed setup, they are of course unpolarized) the initial state must have ##S=0## in order to end up with two (and not three) photons. But then, due to angular-momentum conservation the helicities of the photons must be opposite, and thus you have polarization entanglement.
     
  15. May 23, 2017 #14

    zonde

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    Are you saying that there is no azimuthal modulation for linearly polarized gamma photons when they scatter from electrons with random spins?
    I don't know anything about Compton polarimetry but quick google search turned up this document: Developing a Compton Polarimeter to Measure Polarization of Hard X-Rays in the 50-300 keV Energy Range
    I couldn't however get access to the paper mentioned by author: Correlation between the States of Polarization of the Two Quanta of Annihilation Radiation
     
  16. May 23, 2017 #15

    zonde

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  17. May 23, 2017 #16

    Vanadium 50

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    The photons are not linearly polarized, but in any event, in what direction should it point? (Remember, the EM interaction is parity conserving, and that puts huge constraints on the answer)

    Normally Compton polarimetry looks at the incoming photon, the outgoing photon, and the scattered electron. Here you lose all information about the scattered electron because it is absorbed.
     
  18. May 23, 2017 #17

    vanhees71

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    The single photons are of course totally unpolarized but you have the entangled state ##|HV \rangle-|VH \rangle## (provided my parity argument above is right ;-)).
     
  19. May 23, 2017 #18

    Vanadium 50

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    Sure, the photons in the decay of a spin-0 states are entangled. But he's not measuring the entanglement if his polarimetry doesn't work.

    And the helicities are the same. The spins are opposite and the momenta are opposite.
     
  20. May 24, 2017 #19

    zonde

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    There are references to experiments that show it works. There are theoretical calculations that say it should work (Klein - Nishina differential cross section for polarized photons).
    What then is the basis for your doubt?
    Why does it matter?
     
  21. May 24, 2017 #20

    Vanadium 50

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    The JLAB Compton polarimeter has an analyzing power of about 6%. This is what you can do with the full resources of a national laboratory and a ton of money.

    Musser doesn't give his exact number, but he says "one per minute", "runs all day" and "consistently greater". There are 1440 minutes in a day, so you need a difference of about 100 or 8% to determine that one orientation is better. But this 8% requires both polarimeters to give the correct answer, which means that the cheap aluminum cubes need an analyzing power of at least 28%. For this to be physics, the cheap aluminum cubes have to be 4-5x better than the state of the art.

    With the colimator, he says the rate falls by a factor of 10, but gets a pair every 20 minutes. These numbers aren't completely consistent (perhaps by rounding), but you now get 72 counts per day, which means you need a difference of about 25, or a 30% effect. That requires an analyzing power of 54%.

    Whatever he is seeing, I don't think it's photon polarization.
     
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