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completing the good Frank answer : Bell designed an experiment that cuts between forecasts of Einstein and Bohr. In a particular situation, Bohr's theory provides a result that other theories could not foresee. Technically, Bohr theory predicts $cos²(angle1-angle2)$ correlations where other theories could not expect better than a line sloping at 0.5. ...


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It all goes back to EPR experiment. In a paper published by Einstein, Podolsky and Rosen the authors argued that quantum mechanics is "incomplete". They argued that principles (such as the principle of locality) needed to be restored in order for it be a complete theory. The problem is, ultimately quantum theory is a "nonlocal" theory. What this means is ...


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It is perhaps illuminating to think in terms of polarisation-entangled photons, where this result is sometimes recognised as the quantum Malus' law. A reference can be found here. As you will see from Eq. (3), this simply derives from transmission probability $\left|\mathcal{A}\right|^{2}=\left|\langle\Omega|\vec{a}'\rangle\right|^{2}$ (consistent with your ...


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This is a very complex question you are asking and very much depends on the notion of "operational" you consider. This paper, for instance, shows that any entangled state can help to do some task better than without it. Regarding PPT vs. NPT, e.g. PPT states cannot be distilled, while most (?) NPT states can be distilled. (Whether all NPT states can be ...


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There can be no blanket answer about what properties are shared between two entangled particles because it depends entirely on the type of entanglement. Moreover, the entanglement of a single property does not necessitate the entanglement of the others (see this question). My understanding is that if particles A and B are say, spin-entangled, a kick applied ...


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You can use the teleportation protocol to teleport and part of a larger quantum state (which can be arbitrarily entangled), and it will work the way it should: I.e., if initially A+C hold $\vert\psi\rangle_{AC}$, after the protocol B+C hold $\vert\psi\rangle_{BC}$. The same is true if the initially shared state is mixed. This follows from the linearity of ...


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This is an area rife with potential misunderstanding, so we need to be absolutely clear what we mean. Suppose I take a ruler and a clock and I use rulers to mark out $x, y, z$ axes in space and the clock to note the positions of events in time. Assuming spacetime is flat, I now have a universal coordinate frame that everyone who is stationary relative to me ...


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I was under the impression that the standard teleportation protocol preserves teleportation. That is to say that if Alice shares some entangled state with Bob, and then teleports her part of this state to Carol in the usual way, Carol will now share that entangled state with Bob. Using this teleportation protocol requires Alice and Carol to share a ...


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The entanglement does not depend on the velocity vectors of the particles. The SG apparatus will determine the spin state of the measured particle, not necessarily the whole system. By performing a measurement on one particle in a known entangled state you would know what the state of the other particle is as well. You do not need to measure it.


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$\newcommand{\ket}[1]{\left| #1 \right>}$A state $ \ket \psi \in H_1 \otimes H_2 \otimes H_3$ is said to be entangled if there exist no coefficients $a_i,b_i,c_i$ such that: $$\ket \psi = \sum_{ijk} d_{ijk} \ket{e^1_i} \otimes \ket{e^2_j} \otimes \ket{e^3_j} = \sum_i a_i \ket{e^1_i} \otimes \sum_j b_j \ket{e^2_j} \sum_k c_k \ket{e^3_k} \tag{1}$$ ...


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You can calculate the density matrix of the state $$\rho = |\varphi\rangle\langle\varphi|$$ And then, the reduced density matrix for one of the particles taking the partial trace $$ \rho_A = \textrm{tr}_{B,C}(\rho)$$ The state is entangled if and only if the reduced matrix is a mixed state. This can be checked, for example, calculating the von Neumann ...


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I've done a lot more work on this since asking, and now have the answers to my questions. First, I mentioned that it seemed that with single qubit gates you can always change the state given by c-z-z-z... (from the equal superposition) to one with this 'bit-flip' property, but that I hadn't proven it. Turns out this is true for any circuit made up of c-z ...


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It seems that you're confusing, perhaps rightfully, the difference between exhibiting entangled behavior and describing entangled behavior. It's no surprise that a classical system can describe entanglement. As you point out, a digital computer can simulate an entagled system. However, there is no meaningful entanglement happening in the computing process; ...


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I will try my own answer, which is mostly thinking out loud, and am ready to delete it if it is too non-sensical. For starter, we need to agree on what we mean by theory; in particular a theory of what and with which logical rules. Let's roughly say that a theory is set of true statements about objects of interest for the theory with rules that enables ...


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If the emission of two entangled photons from nonlinear optical crystals does satisfy your definition of any QM model, then yes, it is possible to simulate it on a digital computer to any degree of precision. And I believe that it is more a question about how good undetermined are two randomization processes on the computer, first for the polarization of one ...


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Here are some facts: As others have said, the evolution of a quantum state, including entanglement, can be simulated arbitrarily well classically with sufficient resources. Actually, modelling the evolution of a quantum system is not even (believed to be) NP hard- if it was, a quantum computer could solve NP problems! That said, it does generally require ...


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No, entanglement is not a classical phenomenon. That is because its very definition is quantum: Let $H_1, H_2$ be quantum spaces of states of two system. A quantum state $\chi\in H_1\otimes H_2$ of the combined system is called separable if it is a simple tensor, i.e. if there are $\phi\in H_1,\psi\in H_2$ such that $\chi = \phi\otimes\psi$. A state is ...


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Quite inefficiently compared to "direct implementation" in quantum hardware, but yes, entanglement defintely can be simulated to arbitrary degree of precision give adequate memory and time resources in a conventional, entirely classical digital computer! Just go and solve multi-particle Schrodinger equation. And this fact has nothing to do with Bohmian ...


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My problem with a positive answer would be that some local rule cellular automaton are Turing universal, which would imply that entanglement could be simulated by a model that uses a classical local rule. This seems wrong, doesn't it? There is a classical, local, deterministic and realistic equivalent of QM, called bohmian mechanics. And no, it is NOT ...


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yes it is wrong, see EPR experiment conclusion ! Quantum experiment reach an efficiency of 92% on each arm ( Anton Zeilinger 2013 http://arxiv.org/abs/1212.0533 ) In the best case , a perfect polarizer emulation in the EPR experiment will reach an efficiency ratio of 75%. If you modify the parameters of the simulation to increase the efficiency, you will ...


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As previously said (e.g. in this answer), you’re right, any possible physical theory can be described by hidden variables. The whole point of Bell inequality is to look at some properties of these hidden variables. The bell tests rule out local hidden variables. But non-local hidden variable are still allowed, and quantum physics can be seen as a non-local ...


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It's not uncommon that popular science articles will refer to paradoxes in physics. However it is extremely important to understand that there are no paradoxes in physics. Our current theories of physics are self consistent and do not contain paradoxes (though there are some conditions not covered by any of our existing theories). Non-physicists tend to use ...


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(Just adding one more answer to the clear ones given above) The "weirdness" of the entanglement of two spin 1/2 particles is the following: If you have two observers Alice and Bob, separated by any distance, and tell them to measure the spins of the entangled pair, in such a way that the analyzers are rotated randomically so that when a measurement on one ...


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When someone says that spin measured about different axis can't both be known, they mean that whatever state you pick will have variability in at least one of the possible spin measurements you can do. So that is what you will get when measure the spin, you will get variable results. This happens even with entanglement with even just one particle. With ...



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