# Tag Info

9

Quantum xor games are a method of greatly simplifying the ideas behind Bell's theorem, which states that no physical theory of local hidden variables can ever reproduce all of the predictions of quantum mechanics. Basically, when two qbits are entangled, measurements on them appear correlated even if they are vastly far apart. The question then is whether ...

8

We simply translate the binary result of a qubit measurement to our guess whether it's the first state or the second, calculate the probability of success for every possible measurement of the qubit, and then more find the maximum of a function of two variables (on the two-sphere). First, something that we won't really need, the precise description of the ...

6

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 ...

5

I would argue that this is the critical issue to understand for Bell inequalities. Finding a violation of a Bell inequality tells you that the system is not classical (note: it does not prove that it is quantum), so you need to understand what the classical thing is that the world is not. Let's state the CHSH random variable that we're interested in: $$S=... 5 We translate this game into conventional QC terminology as follows: The coin is a single qbit state |\psi\rangle=\begin{bmatrix}\alpha \\ \beta\end{bmatrix} where \alpha, \beta \in \mathbb{C} and |\alpha|^2 + |\beta|^2 = 1 "Flipping" the coin is application of the bit-flip operator X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} The "heads" ... 5 As far as I have been researching from the intrenet about the CHSH game, the first experimental realization seems to be the one published by Aspect in Experimental Realization of Einstein-Podolsky-Rosen-Bohm Gedankenexperiment: A New Violation of Bell's Inequalities. The experiment is based on photon polarization and is the proof of the universe's ... 3 One way to go about proving this is to characterise the set of all possible strategies that Alice & Bob can adopt. By "strategy" here I mean a possible relation between inputs and outputs, encoded in the set of four binary numbers A_0,A_1,B_0,B_1. It is worth noting that it doesn't matter whether we are considering deterministic or probabilistic ... 3 Nonlocal games such as the CHSH game are not impartial games in the sense of Sprague-Grundy. Alice and Bob are thought to be cooperating rather than competing, and randomness is central to the study of nonlocal games. Impartial games are competitive, deterministic, and perfect information. A good sanity check when asking about quantizing some classical ... 3 Imagine you're playing a CHSH game with someone, although you don't know what quantum system it is that you're playing with, or even what measurements it is that you're doing on the system. You just know that you're getting the average value$$ \langle A_1(B_1+B_2)+A_2(B_1-B_2)\rangle=2\sqrt{2}  (where measurement results of $\pm 1$ are recorded in $A_1$ ...

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

2

If you look at the literature for blind quantum computation, there is the concept of a "trap state". Basically, something that isn't part of the main computation that is supposed to give specific results so that you can easily verify that the computer is behaving as expected. I believe some of these trap states are Bell pairs, and the measurements performed ...

2

In the CHSH game we have 2 players Alice and Bob. Can we proof in the form of a noncommunication pair of TMs which takes as input independant random bits x and y plus an arbitrary shared bitstring, that ALice and Bob win the CHSH game with probability greater than 75%. We ask Alice and Bob questions x and y with probability p(xy) they give answers a and b. ...

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. ...

1

The key is in the optimal strategy for distinguishing two non-orthogonal states. This is something called the Helstrom measurement, which I described here.

Only top voted, non community-wiki answers of a minimum length are eligible