7

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....


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

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 ...


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 ...


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

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 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 $\mathsf{MIP^*}$ ...


3

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. ...


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

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 ...


3

This question was solved in 2014 by VĂ©rtesi 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 ...


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 (...


2

If you look at these states in the $X$ basis, they are $$ |++\rangle+|--\rangle,\qquad |+-\rangle+|-+\rangle. $$ Thus, by both measuring in the $X$ basis and computing the parity of the answers, you can tell $C$'s bit value.


2

How then, would one be able to simulate say CHSH, which produces fundamentally quantum probabilities that cannot be explained locally/classically? Am I misinterpreting the meaning of simulate? Quantum phenomena cannot be "explained classically" only when locality is taken into consideration. In other words, classical phenomena cannot reproduce (some types ...


2

There are two definitions of simulation that are commonly used in this context. We consider a quantum computation to be: 1. loading an input 2. performing some processing 3. doing a measurement This defines a distribution on possible measurement outcomes for each input. Weak Simulation would be a classical randomised algorithm that could sample from these ...


2

First, is this value correct? Yes, it is. If you expand out the calculation you're doing, this is the same as $$ \sqrt{2}\langle\psi|X\otimes X+Z\otimes Z|\psi\rangle $$ for any two-qubit state $|\psi\rangle$. In the particular case $|\psi\rangle=|00\rangle$, it's easy to extract that this is $\sqrt{2}$. Indeed, for any separable state of the form $$ |\psi\...


2

Imagine you had a general formula $$ C=a_1QS+a_2RS+a_3RT+a_4QT. $$ Algebraically, we know that if $Q$, $S$, $R$ and $T$ are random variables with values $\pm 1$, then each term such as $QS\in\{\pm 1\}$. Hence, there is a trivial bound $$ C\leq |a_1|+|a_2|+|a_3|+|a_4| =C_\max. $$ This can never be beaten by any model, be it local hidden variable, quantum, ...


2

Consider these two states $$ \sigma_0 = \frac{1}{2}(|11\rangle\langle 11| + |++\rangle\langle ++|) $$ $$ \sigma_1 = \frac{1}{2}(|0-\rangle\langle 0-| + |-0\rangle\langle -0|) $$ I believe they are indistinguishable (with certainty), though to be sure it's better to find the exact proof. Also check this paper https://arxiv.org/abs/quant-ph/9804053. It's ...


1

We can distinguish these states using SingleQubit Unitaries and Measurements. Lets the 2 states be $|\psi^{00}\rangle = \frac{1}{\sqrt2}(|00\rangle + |11\rangle)$ and $|\psi^{01}\rangle = \frac{1}{\sqrt2}(|00\rangle - |11\rangle)$. Let Alice be in possession of the 1st qubit and Bob be in possession of the 2nd qubit. Bob applies a $H$ gate on his qubit. This ...


1

Original paper: Going beyond Bell's theorem Papers by Mermin: Quantum mysteries revisited Hidden variables and the two theorems of John Bell We can give a non-probabilistic proof that local hidden variables theories are incompatible with Quantum Mechanics. The proof does not explain how experimentally decide which theory is correct; it just shows more ...


1

It is equivalent to saying that Alice and Bob share a state of the form $$\rho^{AB} = \int \mathrm{d}\lambda \mu(\lambda) |\lambda\rangle\langle \lambda| \otimes \rho_\lambda, $$ without any restriction whatsoever on Alice's measurements (you forgot the normalisation, without which it is not a valid quantum state). The expression "classical message"...


1

I figured out the answer while writing the question, but figured I'd still post it for future reference. The problem with the calculation is that it was not taking into account the locality constraint. Namely, the fact that local deterministic behaviours also have to satisfy $$p(ab|xy)=p(a|x)p(b|y).$$ Taking this into account, we notice that there are $\...


1

Not quite. Consider the following no-signalling distribution $PR_1$ which I will write in the form $$ \begin{pmatrix} p(00|00) & p(01|00) & p(00|01) & p(01|01) \\ p(10|00) & p(11|00) & p(10|01) & p(11|01) \\ p(00|10) & p(01|10) & p(00|11) & p(01|11) \\ p(10|10) & p(11|10) & p(10|11) & p(11|11) \\ \end{pmatrix}...


1

Yes. As you've effectively said, all cases satisfying (2) are in a polytope and therefore convex. All the vertices of that polytope are deterministic strategies, and so every point inside the polytope can be described as a convex combination of these, and that gives you (at least) one such local realistic explanation.


1

The Quantum Reality group at the Centre for Quantum Technologies (National University of Singapore) https://qreality.quantumlah.org/


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