Distinguishing classical physics vs. quantum physics

[PeterDonis]

Would you say that the usual CHSH or GHZ uses CFD?

Having looked at the Pan & Zeilinger paper you linked to, I would say that CFD is not necessary to prove the GHZ theorem, although that is not always made clear in statements of the theorem. The GHZ theorem can be stated, just like Bell's Theorem, entirely in terms of properties of actual measurements. In the GHZ case, the "inequality" is simpler, because it's a flat contradiction, not requiring statistics: prepare a large ensemble of sets of three photons in the GHZ state. Randomly choose which of four measurements to make on each set: xyy, yxy, yyx, or xxx. Compute the "result product" for each measurement. Quantum mechanics predicts that the "result product" will be -1 for the first three measurements and +1 for the fourth. Any "local realistic" model (by the GHZ definition) predicts that, if the result product is -1 for the first three measurements, it must also be -1 for the fourth.

 

[PeterDonis]

Seems I have ended up in someones ignore list

Oops, sorry, I lost track of who linked to what. You're right, the link to that paper was yours, not Zafa Pi's.

 

[Zafa Pi]

Randomly choose which of four measurements to make on each set: xyy, yxy, yyx, or xxx. Compute the "result product" for each measurement. Quantum mechanics predicts that the "result product" will be -1 for the first three measurements and +1 for the fourth. Any "local realistic" model (by the GHZ definition) predicts that, if the result product is -1 for the first three measurements, it must also be -1 for the fourth.

OK, now I have two questions:
1) What does your x represent?
2) What is your proof of your last sentence? I gave one in post #92 with totally different notation defined there.

 

[vanhees71]

I don't understand the confusion. You have a 2D Hilbert space and build Kronecker products of those. Physically realizable are these as polarization states of photons (corresponding to the two helicity states (circular polarized em. waves) or H/V states (linear polarized em. waves); it's just a change of bases between the two) or of spin 1/2 states (corresponding to two states of $\sigma_z = \pm 1/2$).

 

[PeterDonis]

What does your x represent?

I'm using the definitions in the paper zonde linked to. In that paper, H and V are the horizontal and vertical linear polarized states, x is a measurement of linear polarization in the 45 degree direction, and y is a measurement of circular polarization. So treating the polarization state space as a qubit, the eigenvectors of x are H + V and H - V, and the eigenvectors of y are H + iV and H - iV. That means the measurement operators, in the "z" basis (i.e., the H/V basis), are just the Pauli x and y spin matrices (and the measurement operator of horizontal/vertical linear polarization is the Pauli z spin matrix), which is the same as what you were using.

What is your proof of your last sentence?

It's just building a generic "local realistic" model that predicts -1 for the first three measurements, and then computing what it predicts for the fourth. The model requires assigning hidden variables to each photon in a particular way, so I would call it a "local hidden variable" model rather than a "local realistic" model. But the key aspect of the model is similar to Bell's model, that for each photon there is a function that is used to predict the results of measurements on it, and that function is a function only of the hidden variables assigned to that photon and the setting of the measurement device for that photon. It does not depend on any hidden variables or measurement device settings for other photons. That is the mathematical definition of "locality" in this model.

Some people would say that any model that includes hidden variables (i.e., variables that aren't measured) requires CFD (counterfactual definiteness), but that's a matter of words and definitions. The reasoning involved in constructing the predictions and testing them does not require any results to be assumed for measurements that are not actually made, so to me it does not require CFD. But we can always just eliminate the term "CFD" entirely and substitute whatever definition of it we are using, if there is any issue with definitions or clarity. The same is true of "locality", "realism", or any other such term. The point is to explore what kinds of properties can or cannot be possessed by models that are capable of reproducing the predictions of QM (and therefore of matching actual experiments).

 

[Zafa Pi]

I'm using the definitions in the paper zonde linked to. In that paper, H and V are the horizontal and vertical linear polarized states, x is a measurement of linear polarization in the 45 degree direction, and y is a measurement of circular polarization.

A measurement by either X or Y on a photon from |ψ⟩ (post #92) yields 1 or -1 with = probability. And I'm sure you know that different trials will not necessarily yield the same value.

The reasoning involved in constructing the predictions and testing them does not require any results to be assumed for measurements that are not actually made, so to me it does not require CFD.

Well you didn't offer a classical proof that if Alice, Bob, and Carol all chose to measure with X the product of the results would be -1. For three different runs/trials not requiring CFD give, for example,
X_A•Y_B1•Y_C1 = -1, Y_A1•X_B•Y_C2 = -1 and Y_A2•Y_B2•X_C = -1. Where Y_A2, for example, is the 2nd time Alice measures with Y

Now what? I also notice you didn't answer my question at the end of post #92.

 

[PeterDonis]

you didn't offer a classical proof that if Alice, Bob, and Carol all chose to measure with X the product of the results would be -1.

I don't see the point of rehashing the entire model described in the paper. The predictions of that model are clear.

I also notice you didn't answer my question at the end of post #92.

I'm not sure what the question is. You ask "now what?" as if there's some problem I'm supposed to solve, but I don't see what it is. I have never denied that QM gives the correct predictions. I personally don't care what ordinary language terms you use, or whether you think "CFD" is "required" or not. The mathematical requirements for any theory that matches the QM predictions are clear; what ordinary language labels you want to paste on them is irrelevant.

 

[Zafa Pi]

This is not correct. P(a, b) and P(a, c) in the formula at the top of p. 406 refer to probability distributions obtained from two sets of runs--one set with settings a, b for the two measuring devices, the other with settings a, c for the two measuring devices.

If they are obtained from two sets of runs Why does is A(a,λ) the same in two different runs. It's the same difficulty you have with GHZ.

 

[PeterDonis]

Why does is A(a,λ) the same in two different runs.

Because that's what the model says. The model is just math; it's a mathematical engine for making predictions. You can take some property of the math, such as the fact that the same function A(a,λ) is used in the model to generate predictions for different runs, and label it with a name like "CFD" or "hidden variables" or whatever. But that has nothing to do with the math or the predictions or comparing the predictions with the results. So I have no interest in arguing about it. Sure, call that property "CFD" if you want, I don't care. It makes no difference to the physics.

 

[Zafa Pi]

Because that's what the model says. The model is just math; it's a mathematical engine for making predictions. You can take some property of the math, such as the fact that the same function A(a,λ) is used in the model to generate predictions for different runs, and label it with a name like "CFD" or "hidden variables" or whatever. But that has nothing to do with the math or the predictions or comparing the predictions with the results. So I have no interest in arguing about it. Sure, call that property "CFD" if you want, I don't care. It makes no difference to the physics.

Just like when you said that P(a,b) was a probability distribution and it wasn't, you are now saying A(a,λ) is a function. It is not, A(a,λ) is a fixed number either 1 or -1 and it's not necessarily the same number in two different runs. How does one get away with using the same value, CFD = "element of reality" is the answer.

Similarly for Zeilinger: On page 228:
"Calling these elements of reality, of photon i, Xi with values +1(−1) for H′(V ′) polarizations and Yi with values +1(−1) for R(L) polarizations we obtain the relations Y1Y2X3 = −1, Y1X2Y3 = −1 and X1Y2Y3 in order to be able to reproduce the quantum predictions of (16.4) and its permutations [12]."

Notice that the same symbol for a number, Y1, appears in two different equations.

I will honor your statement: "So I have no interest in arguing about it. Sure, call that property "CFD" if you want, I don't care. It makes no difference to the physics."
And I thank you once again for getting me to read Bell's paper, it was long over due.

 

[PeterDonis]

A(a,λ) is a fixed number either 1 or -1 and it's not necessarily the same number in two different runs

In other words, A(a,λ) is a function that gets evaluated for each run; when it is evaluated it can give either 1 or -1. That's the underlying model, which is what I was talking about.

How does one get away with using the same value, CFD = "element of reality" is the answer.

No, the answer is that in the underlying model there is a mathematical function. "CFD = element of reality" is just some ordinary language verbiage you choose to use to label that function. It has nothing to do with the physics.