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 ...


6

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

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

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 ...


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

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

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. ...


2

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 ...


2

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 ...


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

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


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