Question on Bell's Inequality?

Question

One version of Bell's theorem analysis is given as =2sqrt(2) where A,B,C,D are quantum variables(observable) and <> means expected (average) value. So if A,B,C,D are classical random variables let S=.

Find possible values of S if these calssical random variables can only take values of +1 or -1.

Answer 1

In demonstrating a proof of John Bell's Theorem, I have calculated the values of the classical random variables for quantum spins.

Recall that for the 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

What this means is that the classical variables representing spin can take on the expected valus of ±1.

So in your notation of S=, we have

AC = BC = BD = AD = 1

so that S <= 2

Now you might ask, "why is it that the QM expected value turned out to be 1/sqrt(2)?" Since, it's obvious now that it's this definition that made all the difference in proving Bell's Theorem.

The answer involves a bit of quantum mechanics as well as some linear algebra using matrices. So you will need to have some background knowledge to fully understand this. Of course, knowing something about classical probability distribution functions would help alot too. Unfortunately, one must understand the mathematical formulation of quantum spin in order to fully understand Bell's Theorem

Also, we use spin as the quantum variable in this calculation mainly because there is NO classical equivalent of quantum spin. So for example, we can look at the spin degree of freedom for an electron, which is associated with a two-dimensional Hilbert space H. So that the operators corresponding to the spin along the x, y, and z direction are Sx, Sy, and Sz respectively, can be represented using the Pauli matrices (http://en.wikipedia.org/wiki/Pauli_matrices).

Then the eigenstates of Sz are represented by:

|+z> or [1,0] and |-z> or [0,1]

And then the eigenstates of Sx and Sy are represented by:

|+x> or (1/sqrt(2)) [1,1] and |-x> or (1/sqrt(2)) [1,-1]

|+y> or (1/sqrt(2)) [1,1] and |-y> or (1/sqrt(2)) [1,-1]

Here you see the appearance of the factor of (1/sqrt(2)). The reason it's here is because the electron has 3-degrees of freedom when we measure its spin. So when we look at the probability of finding its spin, we must account for the fact that we may see the spin along any of the three of the axis x, y, and z.

So when the spin is up, it's +z. When the spin is down, it's -z. But we may also find the spin of the electron aligned somewhere in between along the x and y axis. But the spins are quantized. It's not like a top where you can orient the top in any angle you like. So to indicate this mathematically, we normalize the probability of finding the spin some combination of x and y axis.

This is where the factor (1/sqrt(2)) comes in. This factor will normalize the probability to 1 or 0. Of course, this factor is only true for the quantum variable spin with 2 states, up and down. Had the quantum variable been something else, say with 3 dinstinct states, then the normalizing factor would have been different. But Bell only needed to demonstrate one example where hidden variables can't work to prove his point.

Answer 2

S=0