# CHSH game There are two cooperative players that can't communicate: Alice and Bob. A referee sample $x,y\in\{0,1\}$ uniformly independently and sends $x$ to Alice and $y$ to Bob. Alice and Bob reply $a,b\in\{0,1\}$. They win iff $x\land y=a\oplus b$ . | x | y | $a\oplus b$ | | --- | --- | ----------- | | 0 | 0 | 0 | | 0 | 1 | 0 | | 1 | 0 | 0 | | 1 | 1 | 1 | ## Classic Strategy The strategy of replying $a=b=0$ will win with probability 0.75. This is the optimal strategy for winning classically (needs proof). ## Quantum strategy We will show that if we provide Alice and Bob with [[Bell state]] qubits - they can win with probability $\cos^2(\frac{\pi}{8})=\frac{2+\sqrt 2}{4}\ge 0.85$. Alice and Bob will translate their bit to measure their half qubit from the Bell state w.r.t to some basis. More specific, Alice will measure w.r.t the computational basis $\{|0\rangle,|1\rangle\}$ for $x=0$ and phase basis $\{|+\rangle,|-\rangle\}$ for $x=1$. Alice will set $a$ to be the result of the measurement. ![[Pasted image 20260526112349.png|331]] Bob will set b to be the measurement results according to the following basis: ![[Pasted image 20260526112444.png|332]] The case $x=y=0$ Alice measure her qubit. if it is $0$ then Bob state is also collapse to 0 (due to Bell state) before he measure. According to [[Born rule]] he has probability of $\cos^2(\frac{\pi}{8})$ to measure and report $b=0$. Same if Alice measured "1" in her basis. ![[Pasted image 20260526121909.png|317]] The case $x=0, y=1$: same as above. see diagram. ![[Pasted image 20260526122556.png|256]] The case $x=1,y=0$ As $\frac{1}{\sqrt{2}}\big(|00\rangle+|11\rangle\big) =\frac{1}{\sqrt{2}}\big(|++\rangle+|--\rangle\big)$ , If Alice is measuring in phase basis - Bob's qubit will collapse to the same phase. As in previous cases, the following diagram shows that the probability of Bob to reply the same answer as Alice is $\cos^2(\frac{\pi}{8})$ . ![[Pasted image 20260526123620.png|286]] The case $x=1,y=1$ If Alice measure $|+\rangle$ , she replies $a=0$. Bob measure $|+\rangle$ in the direction of $b=1$ with probability $\cos^2(\frac{\pi}{8})$ . If Alice measure $|-\rangle$ she replies $a=1$. Bob measure $b=0$ with probability $\cos^2(\frac{\pi}{8})$ in the direction of $b=0$ . note that the basis for $a=1$ continues to the opposite direction as well (see dashed line below) ![[Pasted image 20260526124125.png|306]] Reference: [Gil's course](https://www.youtube.com/watch?v=9yUmmKCAicY&list=PL-FpbJb6Ix_MTBEyIujyaUtcj-J4xqWnt&index=28) ## Created 2026-05-26 10:46