Why does the CHSH game need complicated bases to show advantage?

Question

The CHSH game is the standard example of a game where two cooperating players Alice and Bob who cannot communicate, but who nevertheless can get an advantage by measuring an entangled quantum state in some basis which depends on its input.

Detailed information is available here: https://en.wikipedia.org/wiki/CHSH_inequality#CHSH_game

The optimal strategy in the game needs measurement in quite complicated bases, which are not just $\{ \vert 0 \rangle , \vert 1 \rangle \} $ combined with $\{ \vert + \rangle , \vert - \rangle \} $. I can verify the computation and it makes perfect sense to me, which this is optimal for the particular example of the CHSH game.

However, by measuring the quantum state $ \frac{ \vert 00 \rangle + \vert 11 \rangle}{\sqrt{2}}$ in the two bases mentioned I can also get some correlations that I cannot reproduce classically without communication. Then why doesn't there exist a standard example of a quantum game using these simpler bases?

Answer 1

It's not that hard to understand in terms of the Bloch sphere. If Alice measures along vector $\alpha$ on the Bloch sphere, and Bob measures along $\beta$, then the correlation between their answers is the dot product $\alpha \cdot \beta$. In other words, the probability that they give the same answer is $\tfrac{1+(\alpha\cdot \beta)}{2} =\tfrac{1+\cos \theta}{2} = \cos^2 \tfrac{\theta}{2}$ where $\theta$ is the angle between $\alpha$ and $\beta$.

Let Alice measure along $\alpha_0$ if she receives $0$ and along $\alpha_1$ if she receives $1$, and likewise for Bob along $\beta_0$ and $\beta_1$. Then we want the angles $\angle(\alpha_0, \beta_0)$, $\angle(\alpha_0, \beta_1)$ and $\angle(\alpha_1, \beta_0)$ to be as acute as possible, and we want $\angle(\alpha_1, \beta_1)$ to be as obtuse as possible. Fiddle around and you'll convince yourself that you want $(\alpha_1, \beta_0, \alpha_0, \beta_1)$ to lie on a great circle at $\pi/4$ apart (so the angle between $\beta_1$ and $-\alpha_1$ is also $\pi/4$).

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That said, I originally learned Bell's inequality from a book that used the following game: The referee sends Alice and Bob signals $a$ and $b$ from $\{ 1,2,3 \}$, and their goal is to always agree if $a=b$ and disagree as often as possible if $a \neq b$. Then the quantum strategy is to take $\alpha_1 = \beta_1$, $\alpha_2 = \beta_2$ and $\alpha_3=\beta_3$ at angles of $\tfrac{2 \pi}{3}$, which gives the probability of agreement $\cos^2 \tfrac{\pi}{3} = \tfrac{1}{4}$ when $a \neq b$.

The best classical strategy is for them both to choose the same surjection $f: \{ 1,2,3 \} \longrightarrow \{ 0,1 \}$ and return $f(a)$ and $f(b)$; this achieves $\tfrac{1}{3}$. (Or, more realistically, they pre-agree on the same sequence of surjections $f_1$, $f_2$, $f_3$, ... from $\{ 1,2,3 \}$ to $\{ 0,1 \}$, and both play function $f_k$ on turn $k$. The probability is still $\tfrac{1}{3}$, but the referee is less likely to detect the strategy.)

The book explained Bell's inequality as "if you have three socks, each black or white, and a referee demands you reveal two of them at random, the odds they will match are at least $\tfrac{1}{3}$, but quantumly it can be as low as $\tfrac{1}{4}$." I always found that variant more mnemonic than the CHSH version. (I wish I could remember which book this was!)