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How can we measure entanglement?

  1. Sep 27, 2013 #1

    naima

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    Let us begin with a state belonging to the four dimensional tensor product of two particles.
    [tex] \sqrt {p_1}|+_x>|+_{x'}> + \sqrt {p_2}|+_x>|-_{x'}> + \sqrt {p_3}|-_x>|+_{x'}> + \sqrt {p_4}|-_x>|-_{x'}> [/tex]
    We can compute the Von Neumann entropy by tracing out and taking the log of the matrix.
    Suppose now than Bob and Alice receive a lot of pairs of particles prepared in this state.
    They ignore the directions x and x'. Can they get the entropy out of theirs results when they compare them?
     
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  3. Sep 27, 2013 #2

    mfb

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    What do they measure, if they ignore x and x'?
    I you measure them (-> distinguish +x from -x and the same for x'), you get access to the squared value of all amplitudes. This is sufficient to calculate the entropy. Getting the phases would need more work.
     
  4. Sep 27, 2013 #3

    naima

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    Bob and Alice ignore what are x and x' but can do all the measurements they want as offten as they want.
    suppose that p1 = 1 and p2 = p3 = p4 =0
    they wil see that among their results x and x' are the only directions with absolute correlation and
    that the S von neumann entropy is null (tensor product of two vectors.
    With Bell states (without knowing (x and x') they will find S = 1.
    My question is: can they find the degree of correlation in the general case?
    My english is not very good have i to rephrase the question?
     
  5. Sep 27, 2013 #4

    kith

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    If Alice and Bob know that the whole system is in a pure state, they needn't even compare their results. They could simply calculate the entropy of their subsystems from the reduced density matrices which are completely determined by local observables.

    If they don't know whether the whole state is pure or mixed, the von Neumann entropy of the subsystems is not enough to specify the degree of entanglement. You have to relate it somehow to the entropy of the whole system. I don't know how this is done in practice. In principle, you could reconstruct the density matrix of the whole system by performing lots of measurements but there's probably an easier way.
     
  6. Sep 27, 2013 #5

    mfb

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    They see even more, they get the same result every time, if I understand your notation correctly.

    If your measurement can distinguish between your four cases, you can simply count.
     
  7. Sep 27, 2013 #6

    naima

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    the experimentalist has only told them that the whole system is in a pure state.
     
  8. Sep 30, 2013 #7

    naima

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    Thank you mfb i think i found the correct answer:
    The unknown state is defined by coeffs in a peculiar basis.
    If Bob and Alice choose two orthogonal directions this defines another 4 dimentional basis in the hilbert space. the state has coeffs in the new basis that can be measured an give the Von Neumann entropy.
    I realized that this entropy must be basis independent (to be proved mathematically).
     
  9. Sep 30, 2013 #8

    Zafa Pi

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  10. Sep 30, 2013 #9

    naima

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    the state belongs to a space generated by tensor products. This pace is not a space of tensor products. A basis may be made with 4 tensor products.
     
  11. Sep 30, 2013 #10

    Zafa Pi

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    OK, I get it, thanks. I was misled by the phrase "a state belonging to the four dimensional tensor product of two particles", rather than the 4-D space generated by ...
     
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