sh asked in Science & MathematicsPhysics · 1 decade ago

John Bells Inequality?

What is Bell's Theorem and how does it resolve the EPR Paradox? What was the significance of the John Bell's contirbution to quantum physics?

2 Answers

Relevance
  • 1 decade ago
    Favorite Answer

    It was not until 1964 when John Bell published his theorem that finally after 30 some years, physicists were able to unshackle the EPR paradox and get out of Einstein's shadow.

    Today, no one takes Einstein's hidden variable theory seriously, and quantum entanglement is now a bedrock of quantum physics.

    So what is John Bell's Theorem?

    It is simply that "No physical theory of local hidden variables can ever reproduce all of the predictions of quantum mechanics." Thus, we must accept that on a QM level, violations of the "Principle of Locality" is an essential part of the laws of physics.

    One can prove this theorem by using either a thought experiment or perform actual experiments:

    The predictions of quantum mechanics in the EPR thought experiment are actually slightly different from the predictions of a very broad class of hidden variable theories. Quantum mechanics predicts a much stronger statistical correlations between the measurement results performed than the hidden variable theories. These differences, expressed using inequality relations now known as "Bell's inequalities", are also in fact experimentally detectable.

    After the publication of Bell's paper, a variety of experiments were devised to test Bell's inequalities. All the experiments conducted to date have found behavior in line with the predictions of standard quantum mechanics.

    So here we will describe Bell's Inequalities or a proof of Bell's Theorem:

    Given quantum spins a and b, the quantum mechanical correlation function is given by:

    Q(a,b) = <Ψ|σ·a ⊗ σ·|Ψ> = - a·b

    where a=(a1,a2,a3) and b=(b1,b2,b3) are unit 3-vectors

    σ·X = σiXi and σ is Pauli matrices, and

    |Ψ> = (1/sqrt(2)) (|1,-1> - |-1,1>) is a singlet state with spin 0

    (the notation above is just the quantum probability of finding the correlation of the 2 spin states, a and b)

    Given spins a and b, the classical correlation function is given by:

    C(a,b) = ∫ f¹(a,w) f²(b,w) dP(w)

    where Functions f¹ and f² should satisfy:

    | fⁿ(a, w) | ≤ 1, for n = 1, 2

    and dP(w) is a positive measure on some space Ω with ∫dP(w) = 1

    (the notation above is just the standard definition of a classical probability distribution function)

    However, Bell Theorem stipulates that we will not able to find the same results through probabilities measured via hidden classical variables, where

    C(a,b) = Q(a,b) = -a·b

    In this approach, quantum spin is represented by the random variable f(a, w) that takes values ±1, as defined earler. Therefore,

    | C(a,b) - C(a,b') + C(a',b) + C(a',b') | must be <= 2

    (here the sum adds to < or = 2 because the value of the correlation is ±1)

    But since singlet state |Ψ> is = (1/sqrt(2)) (|1,-1> - |-1,1>) for spin 0. So we see that the vector products:

    a·b = a'·b = a'·b' = -a·b' = 1/sqrt(2) or sqrt(2)/2

    and that

    | Q(a,b) - Q(a,b') + Q(a',b) + Q(a',b') | = 2sqrt(2)

    (here we sum sqrt(2)/2 4 times and get 2sqrt(2))

    So now we have proven that

    | C(a,b) - C(a,b') + C(a',b) + C(a',b') | is NOT equal to

    | Q(a,b) - Q(a,b') + Q(a',b) + Q(a',b') |

  • Anonymous
    1 decade ago

    It doesn't really resolve the EPR paradox. The paradox is resolved by rejecting one of the postulates of the EPR paradox--counterfactual definiteness.

    Bell's inequality mainly served to point out that there was a means of testing the paradox. These tests have been performed many times since the 70s and the results confirmed that you do in fact have to reject one of the postulates assumed in the EPR paper.

    Discussions of Bell's inequality and the "problems" it poses are still popular in philosophy classes, but most physicists have pretty much accepted the results and moved on.

Still have questions? Get your answers by asking now.