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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 qubits are entangled, measurements on them appear correlated even if they are vastly far apart. The question then is whether ...
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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 ...
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This is definitely doable, but the tasks seem quite simple and they only introduce single-qubit measurement and the X gate, while quantum state preparation usually involves some superposition and entanglement generation. One can get rid of the input superposition by measuring each qubit and then use a bunch of X gates to set each qubit to the right state, ...
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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 ...
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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 ...
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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" ...
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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:
$$
S=...
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Nonlocal games such as the CHSH game are not impartial games in the sense of Sprague-Grundy. Alice and Bob are thought to be cooperating rather than competing, and randomness is central to the study of nonlocal games. Impartial games are competitive, deterministic, and perfect information.
A good sanity check when asking about quantizing some classical ...
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There are a number of equivalent formulations to CEP. One of them is Tsirelson's problem from quantum information theory, which asks whether the infinite dimensional commuting operator correlations can be approximated arbitrarily well by finite dimensional quantum correlations. This was shown in a combination of papers by Junge, et al (https://arxiv.org/pdf/...
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One way to go about proving this is to characterise the set of all possible strategies that Alice & Bob can adopt. By "strategy" here I mean a possible relation between inputs and outputs, encoded in the set of four binary numbers $A_0,A_1,B_0,B_1$.
It is worth noting that it doesn't matter whether we are considering deterministic or probabilistic ...
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It seems that using more rounds will not be such helpful for us to get something more powerful from complexity perspectives. There are a few comments about the number of rounds and the number of players for $\mathsf{MIP}^*$ in Thomas Vidick's lecture note regarding quantum mutli-prover interactive proofs. Note that the non-local games are $\mathsf{MIP^*}$ ...
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Imagine you're playing a CHSH game with someone, although you don't know what quantum system it is that you're playing with, or even what measurements it is that you're doing on the system. You just know that you're getting the average value
$$
\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$ ...
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They certainly seem doable. I'd suggest the first one, as it is a little more complex. For added complexity you could also include constraints on the allowed gates, such as not allowing X or Y on certain qubits, but instead supplying CNOTs (so a $|1\rangle$ can be copied from other qubits) or partial rotations around the X and Y axis, such that the X or Y ...
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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 ...
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Consider the task on one qubit. You are given a state that is either $H|0\rangle$ or $H|1\rangle$ and your task before the measurement is to get it to be either more probably $|0\rangle$ or $|1\rangle$ as instructed. Say you were instructed to get 0, then you are trying for a unitary that takes both $H \mid 0 \rangle$ and $H | 1 \rangle$ to states close to $|...
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It is relevant to consider the following research paper from Mashiko Fukiyama on Nim Game on Graph to come up with an algorithmic approach . At first, to set a starting position of the game, we fix some finite undirected graph and assign to each edge a non-negative integer. Further we take one piece and put it at a vertex of the graph. From this given ...
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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. ...
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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. ...
2
import sys
sys.exec_prefix
will show the path of python used by blender. Rename this python folder to something else. Now blender will automatically use the python version in windows path. Make sure that your Python runs in command prompt when you type python. If it doesn't, type environment variables in search and add python.exe to the path variable. ...
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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 ...
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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 ...
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Not an answer yet, but to long for comment:
Alice and Bob share two entangled pairs. Now look only at Alice's.
The last line gives the state that only one of'em has:
$$\frac1{\sqrt 2}(|00〉+|11〉) =\frac1{\sqrt 2}(|\phi^x_0\phi^x_0〉+|\phi^x_1\phi^x_1〉) =\frac1{\sqrt 2}(|\phi^y_0\phi^y_1〉+|\phi^y_1\phi^y_0〉).$$
You can see that when you trace out Bob.
Now all ...
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Even though Alice and Bob use the "same" observable, it acts on different systems. To avoid confusion it's better to denote observables as $D_{ij}^A, D_{ij}^B$ and $D_{ij} = D_{ij}^A \otimes D_{ij}^B$ for the joint observable that acts on 4 qubit space.
$\text{Tr}(\rho^A D_{ij}^A)$ and $\text{Tr}(\rho^B D_{ij}^B)$ are average values (expectations) ...
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General setup
Suppose you have a bipartite state $\rho$, and a collection of measurements $\newcommand{\bs}[1]{\boldsymbol{#1}}\{\bs\mu^{(j)}\}_j$. Each $\bs\mu^{(j)}$ represents a way to measure the two sides of the state - meaning $\bs\mu^{(j)}$ represent ways to measure the partial states $\rho^A$ and $\rho^B$. Denote with $\mu^{(j)}_a$ the operators ...
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What a Franson interferometer does is to superpose two-photon wavepackets generated at different times (within the coherence time of the pump, supposing we are generating a pair of entangled photons by e.g. SPDC). This superposition is realized by an unbalanced MZI where the time delay between short and long arm should be in principle orders of magnitude ...
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Von Neumann measurement is a measurement that corresponds to an orthonormal basis, not necessary standard. Though it's related to the standard by a unitary transformation.
As for the solution, at first you need to construct a POVM from the vectors orthogonal to those $|A\rangle, |B\rangle, |C\rangle$. For example, you can take
$$
|E_0\rangle = \frac{1}{\...
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Are you using Qiskit from within a conda virtual environment? If Blender isn't launched from within the same venv they may not be able to communicate (or share libraries).
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The key is in the optimal strategy for distinguishing two non-orthogonal states. This is something called the Helstrom measurement, which I described here.
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