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# 物理代写|量子力学代写Quantum mechanics代考|Entanglement in the CHSH Game

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## 物理代写|量子力学代写Quantum mechanics代考|Entanglement in the CHSH Game

One of the simplest means for demonstrating the power of entanglement is with a two-player game known as the $\mathrm{CHSH}$ game (after Clauser, Horne, Shimony, and Holt), which is a particular variation of the original setup in Bell’s theorem. We first present the rules of the game, and then we find an upper bound on the probability that players operating according to a classical strategy can win. We finally leave it as an exercise to show that players sharing a maximally entangled Bell state $\left|\Phi^{+}\right\rangle$can have an approximately $10 \%$ higher chance of winning the game using a quantum strategy. This result, known as Bell’s theorem, represents one of the most striking separations between classical and quantum physics.
The players of the game are Alice and Bob, who are spatially separated from each other from the time that the game starts until it is over. The game begins with a referee selecting two bits $x$ and $y$ uniformly at random. The referee then sends $x$ to Alice and $y$ to Bob. Alice and Bob are not allowed to communicate with each other in any way at this point. Alice sends back to the referee a bit $a$, and Bob sends back a bit $b$. Since they are spatially separated, Alice’s response bit $a$ cannot depend on Bob’s input bit $y$, and similarly, Bob’s response bit $b$ cannot depend on Alice’s input bit $x$. After receiving the response bits $a$ and $b$, the referee determines if the AND of $x$ and $y$ is equal to the exclusive OR of $a$ and $b$. If so, then Alice and Bob win the game. That is, the winning condition is
$$x \wedge y=a \oplus b$$
Figure 3.9 depicts the $\mathrm{CHSH}$ game.
We need to figure out an expression for the winning probability of the $\mathrm{CHSH}$ game. Let $V(x, y, a, b)$ denote the following indicator function for whether they win in a particular instance of the game:
$$V(x, y, a, b)=\left{\begin{array}{cc} 1 & \text { if } x \wedge y=a \oplus b \ 0 & \text { else } \end{array}\right.$$

## 物理代写|量子力学代写Quantum mechanics代考|Classical Strategies

Let us suppose that they act according to a classical strategy. What is the most general form of such a strategy? Looking at the picture in Figure 3.10(i), there are a few aspects of it which are not consistent with our understanding of how the game works.

In a classical strategy, the random variable $\Lambda$ corresponds to classical correlations that Alice and Bob can share before the game begins. They could meet beforehand and select a value $\lambda$ of $\Lambda$ at random. According to the specification of the game, the input bits $x$ and $y$ for Alice and Bob are chosen independently at random, and so the random variable $\Lambda$ cannot depend on the bits $x$ and $y$. So the conditional distribution $p_{\Lambda \mid X Y}(\lambda \mid x, y)$ simplifies as follows:
$$p_{\Lambda \mid X Y}(\lambda \mid x, y)=p_{\Lambda}(\lambda)$$
and Figure 3.10 (ii) reflects this constraint.
Next, Alice and Bob are spatially separated and acting independently, so that the distribution $p_{A B \mid \Lambda X Y}(a, b \mid \lambda, x, y)$ factors as follows:
$$p_{A B \mid \Lambda X Y}(a, b \mid \lambda, x, y)=p_{A \mid \Lambda X Y}(a \mid \lambda, x, y) p_{B \mid \Lambda X Y}(b \mid \lambda, x, y)$$
But we also said that Alice’s strategy cannot depend on Bob’s input bit $y$ and neither can Bob’s strategy depend on Alice’s input $x$, because they are spatially separated. However, their strategies could depend on the random variable $\Lambda$, which they are allowed to share before the game begins. All of this implies that the conditional distribution describing their strategy should factor as follows:
$$p_{A B \mid \Lambda X Y}(a, b \mid \lambda, x, y)=p_{A \mid \Lambda X}(a \mid \lambda, x) p_{B \mid \Lambda Y}(b \mid \lambda, y)$$
and Figure 3.10(iii) reflects this change. Now Figure 3.10(iii) depicts the most general classical strategy that Alice and Bob could employ if $\Lambda$ corresponds to a random variable that Alice and Bob are both allowed to access before the game begins.

## 物理代写|量子力学代写Quantum mechanics代考|Entanglement in the CHSH Game

$$x \wedge y=a \oplus b$$

$$V(x, y, a, b)=\left{\begin{array}{cc} 1 & \text { if } x \wedge y=a \oplus b \ 0 & \text { else } \end{array}\right.$$

## 物理代写|量子力学代写Quantum mechanics代考|Classical Strategies

$$p_{\Lambda \mid X Y}(\lambda \mid x, y)=p_{\Lambda}(\lambda)$$

$$p_{A B \mid \Lambda X Y}(a, b \mid \lambda, x, y)=p_{A \mid \Lambda X Y}(a \mid \lambda, x, y) p_{B \mid \Lambda X Y}(b \mid \lambda, x, y)$$

$$p_{A B \mid \Lambda X Y}(a, b \mid \lambda, x, y)=p_{A \mid \Lambda X}(a \mid \lambda, x) p_{B \mid \Lambda Y}(b \mid \lambda, y)$$

## MATLAB代写

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