7 votes
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Are correlations stronger than those allowed by quantum mechanics possible?

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" ...
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6 votes
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If the partial traces $\rho_A,\rho_B$ are pure, does it imply that $\rho$ is a product state?

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., ...
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6 votes
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What is the difference between signaling and non-signaling quantum correlations, and what is a signaling channel?

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 ...
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5 votes
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What is the no-signaling set and how can it be related to other types of correlations?

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 ...
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How are EPR Pairs used in quantum computing?

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 ...
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Definition of locality in Bell experiments

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 ...
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4 votes

Determining whether $P(ab|xy)$ factorizes in Bell experiments

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 ...
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4 votes

In Bell nonlocality, why does $P(ab|xy)\neq P(a|x)P(b|y)$ mean the variables are not statistically independent?

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 ...
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3 votes
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Example of a two-qudit state whose measurement outcomes are independent in one basis but dependent in another

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 ...
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How are EPR Pairs used in quantum computing?

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 ...
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3 votes

In Bell nonlocality, why does $P(ab|xy)\neq P(a|x)P(b|y)$ mean the variables are not statistically independent?

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 $b$ ...
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Why is correlation in the $X$ basis represented as $X\otimes X = 1$?

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 ...
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How is Bell’s Inequality converted to the CHSH inequality?

... 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 ...
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Do entangled states always result in some form of correlation?

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 ...
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$\rho_{SE}(0)=\rho_S(0)\otimes\rho_E(0)$: No coupling or no entanglement?

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 ...
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2 votes

Example of a two-qudit state whose measurement outcomes are independent in one basis but dependent in another

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}[1]{\lvert #1\rangle}...
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Why is $P(1,2)_{\text{same}} = \frac{1}{4}$ and not $\frac{1}{2}$ in Preskill's Bell experiment?

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 ...
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2 votes
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Can pairwise entanglement be converted into tripartite correlations?

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'...
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General Master Equation with Decoherence Query

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\...
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What is the no-signaling set and how can it be related to other types of correlations?

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 ...
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2 votes
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Classical versus quantum correlations and partial traces

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 ...
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1 vote
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Why does measurement in computational basis result in classic probability?

The ideal probability of obtaining some measurement result $i$ (with associated measurement operator $M_i$) from some state $\rho$ is $Tr(M_i \rho) = p_i$. For the computational basis, $M_0 = |0\...
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1 vote
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Quantum discord of a tripartite system A:BC

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,...
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