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B Proving entanglement using polarisation & Bell's Inequality

  1. Feb 26, 2016 #1
    I've seen some articles using particle spin experiments to 'prove' that the results violate Bell's inequality and consequently local reality.
    I've also seen stated that the same experiments can be done using other particle attributes such as polarisation. I can see how with polarisation, you can use Malus's law to 'prove' that the results are non-local, however, I can't seem to connect Bell's Inequality to experiments with polarised particles.
    Can anyone help me with this?
     
  2. jcsd
  3. Feb 28, 2016 #2

    jfizzix

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    The traditional Bell inequality deals with measurements of variables that have one of two outcomes given one's measurement setting.

    In the case of spin, the two outcomes are spin-up or spin down, relative to the orientation defined by your measurement setting

    In the case of polarization, the two outcomes are horizontally polarized, or vertically polarized, relative to the orientation defined by your measurement setting.

    There are more general forms of the Bell inequality, where you can have variables with a continuum of outcomes, though the math requires that that set be bounded. One neat example that's been proposed is to use the position correlations in beams of entangled photons

    http://www.pas.rochester.edu/~jschneel/Schneeloch_CVBellIneq_PRA_2016.pdf
    Yes, it's my paper, but the point stands that you can violate a Bell inequality with more than just spin (or the photon equivalent, polarization) as seen there and in its contained references (It can be really really hard, though).
     
  4. Feb 28, 2016 #3

    Mentz114

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    Thanks for posting the reference. I look forward to reading ( and maybe partly) understanding it.
     
  5. Feb 28, 2016 #4
    Thanks for taking the time to reply "jfizzix", very helpful.

    Though I can't open your link at the moment (timeout error), maybe the server is down?

    So when using spin, the particles can only have 2 states, while with polarisation, the photons have an infinite number of states?
    I'm working on a little article (mainly to help me understand) that gives the following example:
    A twin photon generator emits photon trains to opposite site polarising filters (30 degrees and -30 degrees). I then show that when you compare both results, the local condition shows 50% mismatch (or less), while the entangled result reveals 75% mismatch (using Malus's law).
    Can I regard the -30 and 30 degrees as 2 states and apply Bell's inequality? If so, what would the equation look like?
     
  6. Feb 28, 2016 #5

    Nugatory

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    The photons only have two possible polarization states (parallel and perpendicular) for any given angle of the polarization detector, and the spin-1/2 particles only have two possible spin states (up and down) for any given angle of the spin detector. Thus, in both cases there are an infinite number of angles you can set the detector at but only two possible states once you've set the detector angle; these are the conditions in which Bell's inequality applies.

    Be aware that there is a closely related inequality, the CHSH inequality, that is more often used in practical experiments.
     
  7. Feb 28, 2016 #6
    Thanks for that! That clears things up a lot. It is the binary result state of the particle's attribute that counts (after having passed through the filter). Can you please confirm the following Bell's Equation for polarisation of photons:

    The polarisation filters can be set to 3 angles: -30 degrees, 0 degrees, 30 degrees

    Now I have the following states:

    A = polarisation of photon after passing a -30 degree filter in the left detector
    !A = ... -30 degree filter in the right detector
    B = ... 0 degree filter in the left detector
    !B = ... 0 degree filter in the right detector
    C = ... 30 degreee filter in the left detector
    !C = ... 30 degree filter in the right detector

    For N being the number of photons passing, Bell's Inequality states:
    N(A !B) + N(B !C) >= N(A !C)

    this means for the number of photons NOT passing (mismatches):
    N(A !B) + N(B !C) <= N(A !C) [ is this true?]

    So for local reality we have:
    delta 30 degree mismatch + delta 30 degree mismatch <= 2 times delta 30 degree mismatch, or
    1/4 + 1/4 <= 1/2 (because in local reality, the 2 filters -30, 30 act as a 0,30 + 0,-30 = 1/4 + 1/4 = 1/2 )

    But for entangled photons, the results show that the 2 filters -30, 30 act as one angle of 60 degrees, which according to Malus should have a mismatch of 75 % i.e.
    1/4 + 1/4 !<= 3/4
     
  8. Feb 29, 2016 #7
    When a beam of photons goes through 2 polarizers with orientations [itex]\theta_{1}[/itex] and [itex]\theta_{2}[/itex], the resulting intensity is proportional to [itex]cos(\theta_{1}-\theta_{2})^2[/itex]. This is Malus law and it can be demonstrated in classical optics.

    Now, when a pair of entangled photons, going into opposite directions, go through 2 polarizers (one polarizer for each photon), the probability for the 2 photons to have the same outcome (transmitted or absorbed) is also [itex]cos(\theta_{1}-\theta_{2})^2[/itex]. But this is not Malus law, this is quantum mechanics.
     
  9. Feb 29, 2016 #8
    @mbond But couldn't you say that the entanglement (which is quantum mechanics) observes Malus's law here?
     
  10. Feb 29, 2016 #9

    Nugatory

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    Despite the appearance of ##\cos^2\theta## in both Malus's law and the quantum mechanics of polarization-entangled photons, they're different phenomena.
     
  11. Feb 29, 2016 #10
    Interesting, thanks. I suppose it is erroneous to mix up both classical and quantum mechanics like that. Do you think that the explanation I gave above regarding linking photon polarisation to Bell's Equation has any validity (barring the connection of Malus's law to entanglement)? In particular:
    1) Did I allocate the states (A, !A, etc) correctly?
    2) Is it ok to switch signs for Bell's equation from >= to <= when regarding mismatches? This is the part I feel least confident about since I couldn't make the equation work any other way...
     
  12. Mar 1, 2016 #11
    Guy Blaylock wrote a nice simple and very informative treatment of the two photon example.
     
  13. Mar 1, 2016 #12
    Thanks sheaf, looks like a great article which I will read through into more depth. At first glance, even though Bell's Inequality is covered, I can't see (even a remote resemblance) to the version of Bell's Inequality that I derived in the 6th post in this thread. I still would like to know if the reasoning in that post is valid. My aim is to represent this phenomenon in the simplest way possible, so that a non-physicist can get some basic understanding of this.
     
  14. Mar 1, 2016 #13

    Nugatory

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    For that you might want to familiarize yourself with the web page maintained by our own @DrChinese and especially the "Bell's Theorem with Easy Math" section. I also recommend this Scientific American article: https://www.scientificamerican.com/media/pdf/197911_0158.pdf

    You've written the inequality correctly when you said N(A !B) + N(B !C) >= N(A !C). To get an intuitive sense of what that inequality is saying, imagine that A is the statement "smokes cigarettes", B is the statement "is left-handed", and C is the statement "is a woman". The inequality is saying that N(A !C), the number of male smokers in a room full of people, must be less than or equal to the sum of the number of right-handed smokers and the number of left-handed men.
     
  15. Mar 1, 2016 #14
    Yes, it is entirely plausible that adding all RH smokers and LH men is very likely higher than all male smokers.

    I got that equation from http://www.upscale.utoronto.ca/GeneralInterest/Harrison/BellsTheorem/BellsTheorem.html
    who gives a simple proof as well. The purpose of my question actually involves using the same argument for photon polarisation.

    My main question now is, did I do right by reversing the equality sign when considering mismatches?
     
    Last edited: Mar 1, 2016
  16. Mar 2, 2016 #15
    I'm sorry, there's a mistake in the said reasoning. I don't have to reverse the inequality sign, but I do have to consider mismatches to make the inequality work, i.e,
    num mismatches of (-30, 0) + number of mismatches (0,30) >= number of mismatches (-30, 30)
    0.25 + 0.25 >= 0.5 (2 x 0.25) in local reality
    0.25 + 0.25 >= 0.75 when entangled, which obviously is not the case..

    So I wonder, given the assumptions were correct, why I had to consider mismatches instead of matches?
     
  17. Mar 2, 2016 #16
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