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Yes, it is possible to conceive theories with "stronger correlations" than those given by quantum mechanics. One way to make this statement precise is to consider some kind of "measurement apparatus" (you can think of it as simply a black a box with some buttons that you can push and different LEDs that correspond to different possible outputs), and analyse ...

6

Basically, it means that the correlations could be used to send a message. Or simply that Bob’s measurement outcomes can reveal some details of Alice’s actions. This is impossible when Alice and Bob each hold one qubit of a Bell pair. Despite the entanglement present, as well as contextuality, signaling in this case would result faster than light ...

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I'll describe the case of two party correlations but this can straightforwardly extended to more parties. Let's give a box to Alice and a box to Bob. Alice and Bob can interact with their boxes by providing them classical inputs $x \in \mathcal{X}$ and $y \in \mathcal{Y}$ respectively. They can also receive outputs from their boxes, $a \in \mathcal{A}$ and $... 5 EPR pairs are a particular case of entangled pairs of qubits. From Wikipedia: "Quantum entanglement is a physical phenomenon which occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently of the state of the other." More to the point regarding your question, ... 5 I think that I can explain the definition through the following simple example: Suppose that you perform two experiments in the same house in two separate rooms. In the first you measure the observable$A$and instantaneously in the second you measure the observable$B$. The measurements are afflicted with noise, so you do not get a definite answer every ... 4 I think you're doing things a little bit backwards. You probably shouldn't be calculating$P(a|x)$or$P(b|y)$in advance, because you're simply trying to ask: Given a set of$\{P(ab|xy)\}$, do there exist assignments to$P(a|x)$and$P(b|y)$that satisfy$P(ab|xy)=P(a|x)P(b|y)$for all$a,b,x,y$? So, how do you evaluate the probability of getting ... 4 It perhaps helps to express$P(ab|xy)$in words: the probability that Alice gets answer A and Bob gets answer B given that choices x and y were made Now independence in classical probability holds if and only if $$P(e_1\text{ and }e_2)=P(e_1)P(e_2)$$ where$e_1$and$e_2$are events, and practically, you can see what it means through Bayes' theorem $$... 3 The state |00\rangle + i|11\rangle has deterministic Z parity and random X parity. Basically the trick is to rotate |00\rangle + |11\rangle around the Z axis to mess up the agreement of the X observables. Instead of ZZ, YY, and XX being deterministic; ZZ, XY, and YX are. 3 If you were to try and imagine the simplest form of correlation, you might think of two bits that were randomly either both 0, or both 1. Bell states are just this, but quantum. We have bits instead of qubits, and the randomness is due to superposition. Since they are most conceptually simple form of entanglement, and the easiest to describe using ... 3 The equation P(ab|xy) = P(a|x)P(b|y) would imply that any dependence that the output ab has on the inputs xy (expressed by the lhs) is solely due to a depending on x alone, and b depending on y alone. This is expressed by the rhs by treating the value of a and its dependence on x as an independent event from the value of b and its ... 3 It is the tensor product of the Pauli X with itself. Preskill is specifying the kind of correlation by giving you a matrix with a +1 eigenspace corresponding to the desired set of states. States where the simultaneous application of an X gate to each qubit has no effect, including phase kickback when conditioned on an ancilla qubit. This notation is very ... 3 ... and the question changes again. Re Update 4: I'm not sure what you intended the variable a to be in the question (I think part of the problem is that you've got muddled between the chosen questions and the given answers), but I have used it very carefully, to always be the \pm1 answer given. In a quantum setting, a is the answer, a value \pm 1 ... 3 Yes; in fact, \rho is both separable and pure. We can start by writing any state \rho in its eigenbasis$$\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|,$$where p_i are probabilities (i.e., positive and sum to unity) and |\psi_i\rangle are pure states that may or may not be entangled. If \rho is bipartite, each eigenstate |\psi_i\rangle is ... 3 Yes, they do. If it were true that$$\operatorname{Tr}[(\Pi^A_a\otimes\Pi^B_b)\rho] = \operatorname{Tr}(\Pi^A_a\rho_A) \operatorname{Tr}(\Pi^B_b\rho_B)$$for all POVMs (or projective measurements, it doesn't make a difference), then we would just have$$\rho = \rho_A \otimes \rho_B,$$because you'd be in particular doing tomography of your quantum state. Now ... 3 Coupling is a dynamic concept that characterizes the evolution of a composite system. It means that the evolution involves an interaction between subsystems. In a quantum circuit, coupling corresponds to multi-qubit gates. Entanglement is a static concept that characterizes the state of a system. It is related to coupling in that it arises due to coupled ... 2 Consider two copies of a d-dimensional system, \mathcal X_1,\mathcal X_2, and take two mutually unbiased bases (MUBs) in each space. Denote these bases with$$\newcommand{\ket}{\lvert #1\rangle} U\equiv \{\ket{u_k}\}_k,\,V\equiv \{\ket{v_k}\}_k.$$Consider the state \mathcal X_1\otimes\mathcal X_2\ni\ket\psi\equiv\sum_k \ket{u_k}\otimes\ket{u_k} (... 2 I do understand that the sum of these three probabilities is greater than one because there are some constraints already involved; like if we uncover all three coins at least two have to be the same. So naturally, there's some redundancy leading to a sum of probabilities that is greater than one! I would say the explanation is simpler than that. ... 2 The other answer already covered most of the bases. I'd just add an explicit example of a no-signalling, non-quantum distribution, because I think it's useful to have some in mind when discussing these things. Consider a two-two behaviour, that is, a conditional probability distribution p(ab|xy) with a,b,x,y\in\{0,1\}, such that p(a,a|x,y)=1/2 for xy=... 2 I'm not sure exactly what the question is, but I can expand a bit about these states. The states you mention are sometimes referred to as "one-way quantum-classical correlated states" (eg here and arxiv version) to refect the properties you describe. They differ from "strictly classical-classical states" of the form \sum_j p_j|j\rangle \... 2 No, it is not possible without communication. To see why, consider B and C, and just ignore A -- since they cannot communicate, for anything B and C can do A's presence is irrelevant. Then, B and C's measurement outcomes are completely uncorrelated (since they don't share any entanglement, or correlations, just each of them holds a maximally mixed state). ... 2 It's fundamentally similar to/the same as Baker–Campbell–Hausdorff (BCH). Generally, in quantum physics, this is most often used (or at least taught) with commuting Hamiltonians:$$e^{-i\left(H_1+H_2\right)t} = \sum_{n=1}^\infty\frac{\left(-it\right)^n}{n!}\left(H_1+H_2\right)^n = e^{-iH_1t}e^{-iH_2t}e^{\frac{1}{2}\left[H_1, H_2\right]t^2}\cdots$$where the ... 1 Given the quantum discord is always minimized for a 1 dimensional projector:$$D_{A}(A:BC)=I(A:BC) - J_{A}(A:BC)$$where$$J_{A}(A:BC)=max_{\{\Pi_{i}^{A}\}}(S(BC)-\sum_{i}p_{i}S(BC_{i}))$$In this case,$S(BC_{i})=Tr_{A}((\Pi_{i}^{A}\otimes I_{BC})\rho_{ABC}(\Pi_{i}^{A}\otimes I_{BC}))/(Tr((\Pi_{i}^{A}\otimes I_{BC})\rho_{ABC}(\Pi_{i}^{A}\otimes I_{BC})))\$ ...

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