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I have asked the question “How do you know when pairs are entangled?” I have asked “What is the difference between entangled and correlated?” I have also asked several similar questions, but I keep missing the point. I have not articulated my thoughts properly and I apologies for that. If anyone comes close to understanding what I’m trying to ask, please feel free to contribute.

First of all, particles or objects can be physically correlated or anti-correlated so that they share properties such as polarization, spin or momentum etc. These systems can be created or explained and there is no mystery here. Roughly speaking the particles state always existed.

Entanglement on the other hand usually relates to an overall state where measuring pairs or systems show correlations that are said to be impossible to achieve in a classic system. Entanglement infers, because there is no physical explanation that the particle pairs somehow communicate or are connected through some field or wave function etc. Roughly speaking the particles state doesn’t exist until the wave function collapses.

The only reason (correct me if I’m wrong) that entanglement is preferred over classical correlation, is because of Bell’s inequity. You guys know the details so I won’t bore you but Bell also stated that “No physical theory of local hidden variables can ever reproduce all the predictions of quantum mechanics”. In other words, systems of entangled particles cannot satisfy the principle of local realism.

OK I get that, and I understand the math deriving those inequalities. It’s very simple and no matter how you work it, you can’t get around it.

So now I think I’ve narrowed my search and it does not question Bell’s inequality. I believe I am questioning Bell’s assumption. His assumption that particles have only one variable to consider. He also assumes that that single variable never varies itself.

If you consider other variables in addition to variables like polarization and calculate the probabilities of these correlated pairs of bi-variable particles you will find that Bell’s inequalities disappear.

I’m not asking you to imagine bi-variable particles even though it’s easy to do. I can describe a real system of correlated objects. Each object identical and carrying more than one variable. These correlated systems when tested later will reproduce the predictions of Q.M. and cos2theta.

Bell said no theory could ever reproduce this, but that’s not true. If I could describe a real system with real variables, that when correlated and tested will reproduce the prediction of QM, what would that mean for Bell’s theorem and inequalities?

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    $\begingroup$ "If you consider other variables in addition to variables like polarization and calculate the probabilities of these correlated pairs of bi-variable particles you will find that Bell’s inequalities disappear." If you post the calculation that led you to believe this, I will be happy to pinpoint your error for you. Of course if you just make the claim without showing the work that led to it, then nobody can help you. $\endgroup$ – WillO Jun 7 at 2:18
  • $\begingroup$ @WillO First answer my last question " If I could describe a real system with real variables, that when correlated and tested will reproduce the prediction of QM, what would that mean for Bell’s theorem and inequalities?" $\endgroup$ – Bill Alsept Jun 7 at 2:23
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    $\begingroup$ It would mean you had made a mistake, which, again, I'll be happy to find for you. $\endgroup$ – WillO Jun 7 at 2:29
  • $\begingroup$ Come on, you can do better than that. What would it mean? Also please answer the actual question if you can. Thanks $\endgroup$ – Bill Alsept Jun 7 at 2:35
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    $\begingroup$ If I told you that I knew of an even number that is not divisible by 2, what would it mean? What you're describing is equally mathematically impossible, so the answer to your question is the same as the answer to mine. $\endgroup$ – WillO Jun 7 at 2:38
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If I could describe a real system with real variables, that when correlated and tested will reproduce the prediction of QM, what would that mean for Bell’s theorem and inequalities?

Nothing at all. It would be as surprising as if you had claimed that you found an eight-legged animal that isn't an octopus. Which (given that spiders exist) is not surprising at all.

The argument of Bell-inequality violations is one of existence, not universality. There exist certain correlated observables (such as the left-hand side of the CHSH inequality) which are under certain constraints in local-hidden variables and for which the QM predictions (and experimental results) fall outside of those constraints. What happens to other observables is, to a large extent, immaterial. Bell does not claim that all observables exhibit nonlocal correlations, only that there exist some that do.

Coming up with QM observables that are explainable within LHV frameworks is not particularly surprising, and it is basically trivial to do with "single-variable" qubits. The extension to higher-dimensional systems is similarly trivial and similarly uninteresting.

His assumption that particles have only one variable to consider

Going to what you call "bi-variable" systems is at the same time unnecessarily broad (cause, as I said above, single variables are sufficient) and unnecessarily restrictive, because if you're going to work on nonlocality in larger systems then you might as well work in full generality. If you're working with finite-dimensional variables like spin, then no matter how many you have, it's still a finite-dimensional system, and it's both simpler and more general to work with arbitrary observables in arbitrary finite dimension $d$ (i.e. qudits). If you're working with continuous variables, then that's what continuous-variable quantum information is for.

Both of these conceptual frameworks are perfectly well explored, and any claim from your side that they aren't boils down to ignorance on your part. This is technical material and it gets boring really fast, so it's not something that's discussed often in introductory material and other public-facing text (as I said above, single variables and qubits are perfectly sufficient to exhibit the nontrivial behaviour), but that does not mean that it does not exist. It just means that you do need to go to the primary technical literature to get it.

And on that note: claims of the form "there is no result of the type X in the technical literature", like the ones you've made here, are extraordinarily hard to make in a justified fashion. It takes about two years' worth of PhD study in the specific area in question before one is in a position to make those claims (and before those claims are really believable by others). Technical literature searches are hard stuff, and you make statements of absence of results at your peril. If you haven't found it, it's because you don't know how to look for it, not because it's not there.

He also assumes that that single variable never varies itself.

To be clear: "that single variable never varies itself" is basically meaningless. Provide a clear mathematical framework for the type of process you mean by this, and it will be easy to point out how that framework falls under its own weight, but "variables varying themselves" is not a thing in physics.

If you consider other variables in addition to variables like polarization and calculate the probabilities of these correlated pairs of bi-variable particles you will find that Bell’s inequalities disappear.

That's a staggering claim (which is also completely hollow). Taken literally, you claim to have a theorem that proves that no Bell-type inequalities are at all possible in "bi-variable particles", but you are not providing the proof for it. Extraordinary claims like this one require extraordinary evidence, so: show your proof. Or back down from the claim.

In reality, there is nothing of the sort. There are plenty of Bell-type inequalities for higher-dimensional systems. You haven't seen them because they're boring and technical and they only appear in the technical literature where you haven't looked.

And, to be honest, the leap of going from that willful ignorance of the literature to the hubris of the claim that "you will find that Bell's inequalities disappear" is, frankly, extremely offensive. Your text implies that you think that you know better than all of the physicists working on quantum information, but you cannot be bothered to do a proper literature search? Some humility is in order there.

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You can find plenty of examples in which Bell's inequalities are not violated, to prove Bell's claim ("no local hidden variable theory can reproduce all the prediction of quantum mechanics"), it is enough to find one prediction of quantum mechanics and show it cannot be reproduced by any hidden variable theory. The CHSH game provides an example with just one variable to measure, why complicate things?

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I know your question states that you already understand the mathematics of Bell inequalities, but perhaps a closer look at the classical prediction will make matters clearer. Also, I have been unable to find a coherent response for Bells theorem on physics SE, so I'm including most of the details here.

Consider the bell state (equation 1) which can be used to describe two entangled and spatially separated particles with spin $\frac{1}{2}$. Here, $| \uparrow \rangle_{k}$ i s the spin up eigensate $| s, m \rangle$ for particles $k\in\{A, B\}$.

\begin{align} \tag{1} | \psi \rangle=\frac{1}{\sqrt{2}}[ | \uparrow\rangle_{A} | \downarrow \rangle_{B}-| \downarrow \rangle_{A} | \uparrow \rangle_{B} ] \end{align}

We consider three spin measurements $S_{x}, S_{y}, S_{z}$ which are performed on each particle, and introduce the Pauli spins $\sigma_z$,$\sigma_y$, and $\sigma_x$ with corresponding eigenvalues of $\pm 1$. We also define the spin observable with equation (2). It describes a measurement of (1) in a lab frame at any angle to the entangled pairs frame.

\begin{align} \tag{2} \sigma_{\theta}=\cos \theta \sigma_{z}+\sin \theta \sigma_{x} \end{align}

Our first task is to work out the expectation value of the spin product $E(\theta, \phi) \equiv\left\langle\sigma_{\theta}^{A} \sigma_{\phi}^{B}\right\rangle$. To start this, we choose a basis set corresponding to $\sigma_z$ which places the Pauli spin matrices into a familiar form (equation 3).

\begin{align} | \uparrow \rangle=\left( \begin{array}{c}{1} \\ {0}\end{array}\right), \quad | \downarrow \rangle=\left( \begin{array}{c}{0} \\ {1}\end{array}\right) \end{align} \begin{align} \tag{3} \begin{aligned} \sigma_{x}&=\left( \begin{array}{cc}{0} & {1} \\ {1} & {0}\end{array}\right) \\ \sigma_{y}&=\left( \begin{array}{cc}{0} & {-i} \\ {i} & {0}\end{array}\right) \\ \sigma_{z} &=\left( \begin{array}{cc}{1} & {0} \\ {0} & {-1}\end{array}\right) \end{aligned} \end{align}

A measurement $\sigma^A_\theta$ on $A$ and $\sigma_\phi^B$ on $B$ is represented by

\begin{align} \left\langle\sigma_{\theta}^{A} \sigma_{\phi}^{B}\right\rangle&=\left\langle\psi|\sigma_{\theta}^{A} \sigma_{\phi}^{B}| \psi\right\rangle \end{align}

An abbreviation of the trigonometric functions has been made, where $\cos(x)$ and $\sin(x)$ are represented as $C_x$ and $S_x$ respectively. The spin product is then writen as

\begin{align} \left\langle\sigma_{\theta}^{A} \sigma_{\phi}^{B}\right\rangle&=C_\theta C_\Phi \left\langle\sigma_{z}^{A} \sigma_{z}^{B}\right\rangle+ S_\theta S_\Phi \left\langle\sigma_{x}^{A} \sigma_{x}^{B}\right\rangle\\ &+S_\theta C_\Phi \left\langle\sigma_{x}^{A} \sigma_{z}^{B}\right\rangle +C_\theta S_\Phi \left\langle\sigma_{z}^{A} \sigma_{x}^{B}\right\rangle \end{align}

The problem has thus been reduced to finding the expectation values of four spin products. The first is in the $z$ direction for both particles.

\begin{align} \left\langle\sigma_{z}^{A} \sigma_{z}^{B}\right\rangle&=\left\langle\psi|\sigma_{z}^{A} \sigma_{z}^{B}| \psi\right\rangle\\ \because~\left.|\sigma_{z}^{A} \sigma_{z}^{B}| \psi\right\rangle&=\frac{1}{\sqrt{2}}\sigma_{z}^{A} \sigma_{z}^{B}\left(| \uparrow\rangle_{A} | \downarrow \rangle_{B}-| \downarrow \rangle_{A} | \uparrow \rangle_{B}\right)\\ &=\frac{1}{\sqrt{2}}\left(\sigma_{z}^{A} \sigma_{z}^{B}| \uparrow\rangle_{A} | \downarrow \rangle_{B}-\sigma_{z}^{A} \sigma_{z}^{B}| \downarrow \rangle_{A} | \uparrow \rangle_{B}\right)\\ &=\frac{1}{\sqrt{2}}\left(\sigma_{z}^{A} | \uparrow\rangle_{A} \sigma_{z}^{B}| \downarrow \rangle_{B}-\sigma_{z}^{A} | \downarrow \rangle_{A} \sigma_{z}^{B}| \uparrow \rangle_{B}\right)\\ \end{align}

\begin{align} \therefore~\sqrt{2}\left.|\sigma_{z}^{A} \sigma_{z}^{B}| \psi\right\rangle&= \begin{aligned}\left( \begin{array}{cc}{1} & {0} \\ {0} & {-1}\end{array}\right)\left( \begin{array}{c}{1}\\ {0} \end{array}\right)\left( \begin{array}{cc}{1} & {0} \\ {0} & {-1}\end{array}\right)\left( \begin{array}{c}{0}\\ {1}\end{array}\right) \end{aligned}\\ &-\begin{aligned}\left( \begin{array}{cc}{1} & {0} \\ {0} & {-1}\end{array}\right)\left( \begin{array}{c}{0}\\ {1} \end{array}\right)\left( \begin{array}{cc}{1} & {0} \\ {0} & {-1}\end{array}\right)\left( \begin{array}{c}{1}\\ {0}\end{array}\right) \end{aligned}\\ &=\begin{aligned}\left( \begin{array}{c}{1}\\ {0}\end{array}\right)\left( \begin{array}{c}{0}\\ {-1} \end{array}\right)-\left( \begin{array}{c}{0}\\ {-1}\end{array}\right)\left( \begin{array}{c}{1}\\ {0}\end{array}\right) \end{aligned}\\ &=| \uparrow\rangle_{A} (-| \downarrow \rangle_{B})-(-| \downarrow \rangle_{A}) | \uparrow \rangle_{B}\\ \therefore~\left.|\sigma_{z}^{A} \sigma_{z}^{B}| \psi\right\rangle&=-\left.| \psi\right\rangle \end{align} \begin{align} \therefore~\left\langle\sigma_{z}^{A} \sigma_{z}^{B}\right\rangle&=1 \end{align}

To find the spin product for in the x-direction, we note that $\sigma_x$ is simply a bit flip or XOR operator

\begin{align} \because~\left.|\sigma_{x}^{A} \sigma_{x}^{B}| \psi\right\rangle&=\frac{1}{\sqrt{2}}\sigma_{x}^{A} \sigma_{x}^{B}\left(| \uparrow\rangle_{A} | \downarrow \rangle_{B}-| \downarrow \rangle_{A} | \uparrow \rangle_{B}\right)\\ &=\frac{1}{\sqrt{2}}\left(| \downarrow\rangle_{A} | \uparrow \rangle_{B}-| \uparrow \rangle_{A} | \downarrow \rangle_{B}\right)\\ &=-\left.| \psi\right\rangle\\ \therefore~\left\langle\sigma_{x}^{A} \sigma_{x}^{B}\right\rangle&=1 \end{align}

We already know the result of various operators applying to each state, so we can shorten the derivation for the effect of different spin products.

\begin{align} \because~\left.|\sigma_{x}^{A} \sigma_{z}^{B}| \psi\right\rangle&=\frac{1}{\sqrt{2}}\left(\sigma_{x}^{A} | \uparrow\rangle_{A} \sigma_{z}^{B}| \downarrow \rangle_{B}-\sigma_{x}^{A} | \downarrow \rangle_{A} \sigma_{z}^{B}| \uparrow \rangle_{B}\right)\\ &=\frac{1}{\sqrt{2}}\left(| \downarrow\rangle_{A} (-| \downarrow \rangle_{B})-| \uparrow \rangle_{A} | \uparrow \rangle_{B}\right)\\ \therefore~\left\langle\sigma_{x}^{A} \sigma_{z}^{B}\right\rangle&=0 \end{align} \begin{align} \left.|\sigma_{z}^{A} \sigma_{x}^{B}| \psi\right\rangle&=\frac{1}{\sqrt{2}}\left(\sigma_{z}^{A} | \uparrow\rangle_{A} \sigma_{x}^{B}| \downarrow \rangle_{B}-\sigma_{z}^{A} | \downarrow \rangle_{A} \sigma_{x}^{B}| \uparrow \rangle_{B}\right)\\ &=\frac{1}{\sqrt{2}}\left(| \uparrow\rangle_{A} | \uparrow \rangle_{B}+| \downarrow \rangle_{A} | \downarrow \rangle_{B}\right)\\ \therefore~\left\langle\sigma_{z}^{A} \sigma_{x}^{B}\right\rangle&=0 \end{align}

And so the spin product from earlier yields

\begin{align} \left\langle\sigma_{\theta}^{A} \sigma_{\phi}^{B}\right\rangle&=C_\theta C_\Phi + S_\theta S_\Phi\\ &=\cos(\theta) \cos(\Phi) + \sin(\theta) \sin(\Phi) \end{align}

We now consider the predictions made by all local hidden variable (LHV) theories. These predict the spin products result to be constrained by a Bell inequality (equation 4).

\begin{align} \tag{4} B&=E(\theta, \phi)-E\left(\theta, \phi^{\prime}\right)+E\left(\theta^{\prime}, \phi\right)+E\left(\theta^{\prime}, \phi^{\prime}\right)\\ |B|&=\left|E(\theta, \phi)-E\left(\theta, \phi^{\prime}\right)+E\left(\theta^{\prime}, \phi\right)+E\left(\theta^{\prime}, \phi^{\prime}\right)\right| \leq 2 \end{align}

If the spins are predetermined and exist independently of the measurement at each location, then the prediction would be

\begin{align} B&=\left\langle\lambda_{\theta} \lambda_{\phi}\right\rangle-\left\langle\lambda_{\theta} \lambda_{\phi^{\prime}}\right\rangle+\left\langle\lambda_{\theta^{\prime}} \lambda_{\phi}\right\rangle+\left\langle\lambda_{\theta^{\prime}} \lambda_{\phi^{\prime}}\right\rangle\\ &=\left\langle\lambda_{\theta} \lambda_{\phi}-\lambda_{\theta} \lambda_{\phi^{\prime}}+\lambda_{\theta^{\prime}} \lambda_{\phi}+\lambda_{\theta^{\prime}} \lambda_{\phi^{\prime}}\right\rangle\\ &\equiv\left\langle B_{\lambda}\right\rangle\\ \text{where }B_{\lambda}&=\lambda_{\theta} \lambda_{\phi}-\lambda_{\theta} \lambda_{\phi^{\prime}}+\lambda_{\theta^{\prime}} \lambda_{\phi}+\lambda_{\theta^{\prime}} \lambda_{\phi} \end{align}

\begin{array}{||c | c | c | c | c | c | c | c | c||} \lambda_\theta & \lambda_{\theta'} & \lambda_{\phi} & \lambda_{\phi'} & \lambda_{\theta}\lambda_{\phi} & \lambda_{\theta'}\lambda_{\phi} & \lambda_{\theta}\lambda_{\phi'} & \lambda_{\theta'}\lambda_{\phi'} & B_\lambda \\ +1&+1&+1&+1&+1&+1&+1&+1&2\\ +1&+1&+1&-1&+1&+1&-1&-1&2\\ +1&+1&-1&+1&-1&-1&+1&+1&-2\\ +1&+1&-1&-1&-1&-1&-1&-1&-2\\ +1&-1&+1&+1&+1&-1&+1&+1&0\\ +1&-1&+1&-1&+1&-1&-1&+1&2\\ +1&-1&-1&+1&-1&+1&+1&-1&-2\\ +1&-1&-1&-1&-1&+1&-1&+1&2\\ -1&+1&+1&+1&-1&+1&-1&+1&2\\ -1&+1&+1&-1&-1&+1&+1&-1&-2\\ -1&+1&-1&+1&+1&-1&-1&+1&2\\ -1&+1&-1&-1&+1&-1&+1&-1&-2\\ -1&-1&+1&+1&-1&-1&-1&-1&-2\\ -1&-1&+1&-1&-1&-1&-1&+1&0\\ -1&-1&-1&+1&+1&+1&+1&-1&0\\ -1&-1&-1&-1&+1&+1&+1&+1&2\\ \end{array}

From the table we arrive at the Bell inequality (equation 4). This proof can be generalized to cases that are not perfectly predetermined, so that there exists a hidden variable state $\{\lambda\}$ for each location that gives an average prediction for the spin.

If we consider the specific case that $\theta=0, \phi=\pi / 4, \theta^{\prime}=\pi / 2, \phi^{\prime}=3 \pi / 4$, we find

\begin{align} \left\langle\sigma_{\theta}^{A} \sigma_{\phi}^{B}\right\rangle&=\cos(\theta) \cos(\Phi) + \sin(\theta) \sin(\Phi) \\ &=\cos(0) \cos(\pi / 4) + \sin(0) \sin(\pi / 4) \\ &=\frac{\sqrt{2}}{2} \end{align} \begin{align} \left\langle\sigma_{\theta}^{A} \sigma_{\phi'}^{B}\right\rangle&=\cos(0) \cos(3\pi / 4) + \sin(0) \sin(3\pi / 4)\\ &=-\frac{\sqrt{2}}{2} \end{align} \begin{align} \left\langle\sigma_{\theta'}^{A} \sigma_{\phi}^{B}\right\rangle&=\cos(\pi / 2) \cos(\pi / 4)+\sin(\pi / 2) \sin(\pi / 4)\\ &=\frac{\sqrt{2}}{2}\\ \end{align} \begin{align} \left\langle\sigma_{\theta'}^{A} \sigma_{\phi'}^{B}\right\rangle&=\frac{\sqrt{2}}{2}\\ \end{align}

Therefore, $B=E(\theta, \phi)-E\left(\theta, \phi^{\prime}\right)+E\left(\theta^{\prime}, \phi\right)+E\left(\theta^{\prime}, \phi^{\prime}\right)=4\left(\frac{\sqrt{2}}{2}\right)=2\sqrt{2}$.

The conclusion from this is that QM and local hidden variables give different predictions for a simple entangled state. Experiments have recently been conducted contradicting the local hidden variable outcome.

Now on to your question. It is stated in your topic that the Bell inequality will disappear when there is more than one hidden variable or when the variables are able to vary themselves.

To answer the first part, we note that even if there were extra variables, the experiment only measures the spin of both particles and computes there spin product. We treat the experiment like a black box and consider every outcome of this product. It does not matter how the variables vary, or whether there are extra variables that can also be measured, all that is required for the proof to work is that the spin product of the classical prediction falls within $-2$ and $2$.

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  • $\begingroup$ Thanks for your answer. What I mean by a bi-variable could be both polarization and a rotating cycle or frequency. For a crude example picture a knife rotating vertically as it moves toward a target. Of course the vertical polarization is important but so is the rotating frequency. Place a vertical slot along the way and the knife has no problem getting through. Place a horizontal slot and it's stopped. Rotate the slot to different angles and the knife may or may not make it through, depending on where the rotation is when it it gets to the edge of the slot. The results match QM and cos2theta. $\endgroup$ – Bill Alsept Jun 7 at 8:32

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