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QM prediction violating Bell’s inequality

  1. Nov 30, 2014 #1
    I have read many explanations of Bell’s proof that mention in passing something like “According to QM, the correlation between measurements of spin at different angles should be given by the cosine of the angle between them.” Sometimes they talk about 1-cos(x)/2. Sometimes they talk about cos^2(x). I suppose that the cos is in some wavefunction, whose amplitude squared is the probability. What is that wavefunction and where does it come from?

    Some explanations involve Pauli matrices, which I understand somewhat. But I don’t understand the rules for using them in this calculation. Can you explain them in terms of some Hamiltonian and Schrodinger’s equation? Or is it just a matter of an expectation-value calculation from state vectors?
     
  2. jcsd
  3. Nov 30, 2014 #2
    Schrodinger's equation doesn't have anything to do with it because time evolution isn't a factor in the violations of Bell inequalities. They're derived from considering a particular fixed entangled state. Basically, your last sentence has it right.

    The inequalities you usually see these days are the CHSH inequalities for spin-1/2 particles, so there's really no getting around the Pauli matrices altogether since they represent the spin observables that we want to measure. I think the quantum information wiki (http://www.quantiki.org/wiki/Bell%27s_theorem [Broken]) has a pretty nice, simple presentation of it.
     
    Last edited by a moderator: May 7, 2017
  4. Nov 30, 2014 #3

    atyy

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    LastOneStanding is right, but just in case the OP is asking for nitty gritty detail, it's usually only the spin part of the wave function that is considered, and the coordinate part of the wave function omitted. But if one wants to include that, the particles are free, so the wave function just evolves under the free Schroedinger equation. Another detail that is usually left out is that in real experiments the particles are identical so one should symmetrize or anti-symmetrize the wave function. Also for photons, one has to use the relativistic theory, and it's usually easier to work in the Heisenberg picture. However, none of these additional details change the picture in which only a fixed state - one of the Bell maximally entangled states - is considered.

    Actually, the relativistic picture might change things a little bit: http://arxiv.org/abs/1204.6220.
     
  5. Nov 30, 2014 #4
    Well, so long as you pick a Bell state with the right parity for your particles, you don't have to do anything more. In the quantiki article, the derivation is for electrons so they use the ##|\Psi^-\rangle## state which is antisymmetric under particle exchange.
     
  6. Dec 1, 2014 #5
    Thanks folks, I read the quantiki. Can you walk me through it just a bit more slowly…I don’t quite understand where the definitions of B(b) & B(b’) come from. B(b) makes sense for 45 degrees, but I don’t understand the minus sign; nor the minus sign in B(b’). And I can’t quite see how the products come out to 1/sqrt(2).
     
  7. Dec 1, 2014 #6

    DrChinese

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    There are several levels to your question, not sure if LastOneStanding and atyy have addressed them all.

    Q. What is the relationship of 2 entangled particle spins?

    For PDC Type I entangled photon pairs, the match ratio is cos^2(theta). Note: This is slightly different from the true *correlation* ratio which is cos^2(theta)-sin^2(theta) - ie matches less mismatches (and varies from 1 to -1). Most of the time, this difference is ignored for the sake of discussion. For PDC Type II entangled photons, the match ratio is 1-cos^2(theta).

    For typical entangled electron pairs, which have opposite spins, the match ratio is 1-cos^2(theta/2).

    Are you looking for the derivation of these?
     
  8. Dec 1, 2014 #7
    Dr Chinese yes I was looking for derivations (that I can understand). (Our last posts probably crossed.) Also I don't know what PDC types are.
     
  9. Dec 1, 2014 #8

    DrChinese

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    Yup, they crossed. :-)

    I will see if I can cobble something together to show this.

    PDC stands for Parametric Down Conversion. PDC crystals are used to create most entangled photon pairs. There are 2 types. Type I pairs have the same polarization, and Type II have orthogonal (crossed) polarization. So you adjust your formula accordingly.
     
  10. Dec 1, 2014 #9

    atyy

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    The notation seems to be inconsistent in the quantiki, which is confusing. Anyway you can take B(b) and B(b') as given, because these are the measurements that Bob can choose to make. A different choice of measurements by Bob will give a different result. For some choices the quantum mechanical prediction will not violate the Bell inequality, and for others there will be a violation but by a different amount.

    The term ##\langle A(a) B(b)\rangle## is the expectation value of the observable A(a)B(b), so you have to sandwich the observable between the bra and ket of the state ##|\phi\rangle##
     
    Last edited: Dec 1, 2014
  11. Dec 1, 2014 #10
    Thanks atyy, I should have known that because it says <ABphi | phi>, but I’m still a beginner.

    ((And by the way I really appreciate how you guys try to feed beginners the right information without knowing their backgrounds.))

    But now I’m stuck on how to turn |+x> into <+x| (I guess I don’t know what these vectors are exactly—are they {1 0} and {0 -1}?), and am also shaky on calculations with tensor products. Would anyone be willing to spell it out for me?
     
  12. Dec 1, 2014 #11

    atyy

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  13. Dec 1, 2014 #12
    I think I got it. This was a learning experience for me, as I wasn’t sure about some of the vectors and haven’t done much with tensor products. Here’s how I got <Φ|A(a)B(b) |Φ> = 1/√2 (in example from quantiki):

    |Φ> = 1/√2 (|+x> ⊗ |-x>) - (|-x> ⊗ |+x>)

    |+x>= [1 0]
    |-x>= [0 -1]
    (writing column vectors horizontally)

    |+x> ⊗ |-x> = [0 -1 0 0]

    |-x> ⊗ |+x> = [0 0 -1 0]

    so
    |Φ> = [0 -1/√2 1/√2 0]

    Sz+Sx =
    1 1
    1 -1

    I ⊗ (Sz +Sx) =
    1 1 0 0
    1-1 0 0
    0 0 1 1
    0 0 1-1

    B(b) = (-1/√2) I ⊗ (Sz +Sx)

    B(b)|Φ> = [1/2 -1/2 -1/2 -1/2]

    A(a) = Sz ⊗ I =
    1 0 0 0
    0 1 0 0
    0 0 -1 0
    0 0 0 -1

    A(a)B(b) |Φ> = [1/2 -1/2 1/2 1/2]

    <Φ| = [0 -1/√2 1/√2 0]

    <Φ|A(a)B(b) |Φ> = 0 + 1/2√2 + 1/2√2 + 0 = 1/√2
     
  14. Dec 1, 2014 #13

    atyy

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    If you choose the representation of Sz to be

    Sz =
    1 0
    0 -1

    Sx =
    1 0
    0 1

    then we will get

    |+z> = [1 0]
    |-z> = [0 1]
    |+x> = [1/sqrt(2) 1/sqrt(2)]
    |-x> = [1/sqrt(2) -1/sqrt(2)]

    This is because |+z> and |-z> are eigenvectors of Sz, meaning that Sz |+z> produces a multiple of |+z>.

    Similarly, Sx|+x> should produce a multiple of |+x>.


     
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