# Lecture 015 - CHSH Game

Here is a good video about Bell's Test that rule out the world is "local and has hidden property".

## CHSH Game

Game: a co-op game

1. Let Alice and Bob far apart.
2. Alice flip a coin with side RED, YELLOW, and based on the result, pick a response a
3. Bob flip a coin with side ORANGE, GREEN, and based on the result, pick a response b
4. If the coin was RED and GREEN, they win iff $a \neq b$
5. If the coin was not RED and GREEN, they win iff $a = b$

Notice Alice(R/Y), Bob(O/G) can be viewed as a deterministic function with fixed global variables. Classically, the best chance of winning is $75\%$.

Strategy:

• If Alice see RED, rotate $0^\circ$

• If Alice see YELLOW, rotate $45^\circ$

• If Bob see ORANGE, rotate $22.5^\circ$ (equivalent to Alice rotate $-22.5^\circ$)

• If Bob see GREEN, rotate $67.5^\circ$ (equivalent to Alice rotate $-67.5^\circ$)

• Alice set a equal to the measurement of qubit A

• Bob set b equal to the measurement of qubit B

Observe when we don't get RED and GREEN, the overall rotation is like Alice rotate $\pm 22.5^\circ$. When we get RED and GREEN, the overall rotation is like Bob rotate $67.5^\circ$.

Therefore, when we get RED and GREEN, the outcome of qubit $A$ and $B$ are less correlated. We win with overall probability $85\%$. (which is proven to be optimal by Boris Tsirelson)

### Rotation of Quantum Operation

Let $U$ be a unitary operation on $\mathbb{R}^N$, then there exists

• orthogonal 2D subspace $P_1, P_2, ...$ such that $U$ will rotate all vectors in the subspace by $\theta_1, \theta_2, ...$

• orthogonal 1D subspace (that is the rotational axis) such that $U$ will do nothing to all vectors in the subspace

• remaining 1D subspace such that $U$ will negate ($180^\circ$ rotation) all vectors in the subspace

It is fine to miss some categories

Let $U$ be a unitary operation on $|v\rangle \in \mathbb{R}^N$, then

• $U$ is a rotation $\iff (\exists k \neq 0)(U^k |v\rangle = |v\rangle)$

• $U$ rotate $|v\rangle$ in 2D subspace $\iff \text{avg}\{UU|v\rangle, |v\rangle\} = U|v\rangle$

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