# Tag Info

9

If Alice and Bob have an entangled pair of qubits and Alice locally measures her qubit, it does not affect local state of the Bob's qubit in any way. Mathematically, if Alice measures but does not look at the measurement outcome, density matrix of the Bob's qubit does not change. The sole fact of Alice's measurement does not affect the Bob's qubit in any way....

8

A natural way to relate Toffoli gates and PR boxes is to see them both as representations of the AND function of two binary inputs, but in different ways. The connection with the AND function is evident and clearly acknowledged by the question, but I would express it in a slightly different way: The Toffoli gate is of course the natural way of representing ...

7

It is certainly true that, within the mathematical description of qubits, operations on one qubit can require the whole description to be updated. This therefore affects the description of every qubit. Those who take a 'epistemic' view of this mathematical description might say that we are just updating our knowledge about the other qubits, and that it ...

7

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

When people talk about a loophole free Bell test, what they really mean is that the two loopholes that most concern the majority of people are closed simultaneously: the measurement loophole and the locality loophole. Let us briefly review the protocol: A Bell state $(|00\rangle+|11\rangle)/\sqrt{2}$ is produced, and two parties, Alice and Bob, each take ...

5

This question was solved in 2014 by Vértesi and Brunner: they found a quantum state with positive partial transposition that violated a Bell inequality. The conjecture that all states with positive partial transposition do not violate any Bell inequality was known as the Peres conjecture, so they disproved it. As for your parenthetical question, whether all ...

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 ...

5

Assume this works. Then, nothing prevents Alice from applying the same protocol to a quantum state that is known to her, such as $|0\rangle$ or $|1\rangle$. This way, she could send information to Bob instantaneously. Thus, it violates faster-than-light communication and thus is impossible.

5

The difference is as follows: the original Bell inequality requires that outcomes from the same setting are always perfectly anti-correlated. It says nothing about the case where they are even marginally different. by contrast, in CHSH, the ideal (giving maximum violation) is that outcomes from the same setting would be anti-correlated, but it is not ...

5

I want to emphasize that you are wasting your time with this historical stuff. But since you insist I'll answer anyway. First of all, note that the inequality \left|\int d\lambda p(\lambda)(1-A(\vec a', \lambda)B(\vec b',\lambda))\right|+ \left|\int d\lambda p(\lambda)(1-A(\vec a', \lambda)B(\vec b,\lambda))\right| \\ \le 2-| E(\vec a',\vec ...

5

It's easy to generate such a model for specific cases. For example, take the maximally entangled state $|\phi^+\rangle = \frac1{\sqrt2} (|00\rangle + |11\rangle)$, and let the observables in the CHSH inequality be $A_0=B_1=Z$, and $A_1=B_0=X$. Now if Alice and Bob measure in the same basis, they'll get results that are random but equal with probability 1, ...

5

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 $... 4 I think you're doing things a little bit backwards. You probably shouldn't be calculating$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$? So, how do you evaluate the probability of getting ... 4 Let$S_x$be a permutation between qubits 2,4 and 6, and acts as identity on the other qubits. Use$x$to index all possible permutations of that form, of which there are 6. Then you do indeed have a mixed state overall, and it's of the form $$\rho=\frac{1}{6}\sum_{x=1}^6S|\Psi\rangle\langle\Psi|S^T$$ (Note that the$S$matrix will be real so I can use the ... 4 It seems that using more rounds will not be such helpful for us to get something more powerful from complexity perspectives. There are a few comments about the number of rounds and the number of players for$\mathsf{MIP}^*$in Thomas Vidick's lecture note regarding quantum mutli-prover interactive proofs. Note that the non-local games are$\mathsf{MIP^*}$... 4 I give this second answer to address a misconception that might be lurking in the question and to look at the local Hamiltonian problem: I am not entirely convinced by my answer and happy to hear what others have to say about it. A Hamiltonian$H$, is first and foremost a Hermitian operator on the Hilbert space$\mathcal{H}=(\mathbb{C}^2)^{\otimes n}$, i.e. ... 4 The computational basis is "natural" in the sense that it provides a practical representation of measurement outcomes. Other bases are also "natural" for other tasks, and bases cannot be interchanged aribtrarily if such a change impacts the underlying tensor structure of the system. For example the Bell states form a basis in four dimensional Hilbert space ... 4 This is a partial answer addressing only what I know: Stoquastic Hamiltonians in the Monte Carlo sign problem. The TLDR is, yes, complexity may depend on the choice of basis. A stoquastic Hamiltonian is one that has non-positive off-diagonal matrix elements in the standard basis. This class of Hamiltonians was quite famously studied first here. In the ... 4 No signaling means can't be used for communication. Can't be used to move messages from a sender to a receiver. Local means doesn't require communication to implement. Some processes, such as phenomena violating bell inequalities, don't enable communication but also can't be emulated classically without communication. They take without giving back. They are ... 4 I think the best way to understand this is by showing that one can violate (ontological) locality while respecting (operational) no-signalling. Take the case of Bohmian mechanics. In it the result of Alice's measurement is deterministic, and it will depend Bob's choice of measurement, so it is clearly nonlocal. Nevertheless, the "quantum equilibrium&... 4 This simultaneous block diagonalization is known as Jordan's lemma (not the complex analysis one though). It's a very common technique in the analysis of device-independent protocols as it allows you (under certain conditions) to reduce the analysis from an arbitrary dimension system to qubit systems. You should take a look at Section 4.3.1 in Scarani15 for ... 3 Yes, there are. I just wrote a paper about it, actually. You need to define carefully what you mean by obtaining a Bell violation or ruling out local hidden variables. You can't demand to have a result which is impossible to explain with local hidden variables: the local bound of a Bell inequality is inherently probabilistic, so it is possible to obtain any ... 3 I believe you're thinking of the all-versus-nothing proofs based on GHZ states. You start with a state such as $$|\Psi\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle)$$ and you select at random one of the four measurements to implement:$X\otimes X\otimes X$,$X\otimes Y\otimes Y$,$Y\otimes X\otimes Y$or$Y\otimes Y\otimes X$. Assuming every one of ... 3 Are there scenarios in which Bell nonlocality can be observed without such averages, that is, in a single-shot scenario? No, you have to collect statistics. Any single result you see could have been due to classical players picking completely at random and getting lucky. Making the chance of classical luck arbitrarily close to zero requires repetition (or ... 3 First of all, you inverted$a,b$with$x,y$when trying to draw the analogy. In Bell's original paper,$\vec a,\vec b$are used to denote the measurement directions, so the underlying probability distribution should be written as $$p(x,y|\vec a,\vec b,\lambda)=p(x|\vec a,\lambda)p(y|\vec b,\lambda).$$ The expectation values$A(\vec a,\lambda)$and$B(\vec b,...

3

I don't think I can answer this question precisely, but I would like to say some things. Since this question was asked, much has changed in the field. We have the following known sources that may interest any reader passing by (since probably by now the OP is an expert in the subject, I hope this answer can be useful): Two very broad reviews on the notion ...

3

In schemes like E91, the idea behind using an entangled state is that: in a particular measurement basis (for both parties), the measurement outcomes are perfectly correlated but completely random (50:50 outcomes). you can perform a Bell test on the state to verify its nature. Using a maximally entangled state gives you the property of the 50:50 outcomes (...

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

Answering your precise question: mixing the four scenarios is not particular to Hardy's argument, it is done in all nonlocality proofs. The fundamental assumption is that the distribution of the hidden variables doesn't depend on the measurement setting, i.e., whether BS2$^+$ or BS2$^-$ are there, so that we can actually use the different measurements to ...

3

This, in a nutshell, is the whole issue of quantum. You say "of course an electron will have either spin up or spin down". It might seem like it should, but this is founded purely on your intuition about what the world around you is like. You have no a priori law of the universe that rules out other options. And hence, when an experiment shows you ...

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