7
$\begingroup$

I'm familiar with CHSH game and the strategy that allows Alice and Bob to succeed with a probability of $$\frac{1+\tfrac{1}{\sqrt 2}}{2}\approx 85\%$$ if they share a maximally entangled state such as $\lvert\Phi_+\rangle$ and use the measurements described by the observables $\sigma_x,\sigma_y$ on their respective qubits to determine their choice of move, depending on the bit they receive from Charlie. I've also seen the proof of the Tsirelson bound which effectively shows that $$\big\lVert A_1 \otimes B_1 + A_1\otimes B_2 + A_2\otimes B_1 - A_2\otimes B_2\big\rVert_2 \leq 2\sqrt{2}$$ if $A_1,A_2$ are Hermitian operators on a space $\mathcal H_A$ with spectrum $\pm 1$ and $B_1, B_2$ are Hermitian operators on $\mathcal H_B$ with spectrum $\pm 1$, and it happens that this maximum is attained when $A_1=B_1=\sigma_x$ and $A_2=B_2=\sigma_y$ with $\mathcal H_A=\mathcal H_B = \mathbb C^2$. (And the singular vector that gives the matrix norm its value happens to be maximally entangled.)

My question: Suppose we are only allowing Alice and Bob to start with some given quantum state that is entangled but not maximally entangled. For instance, say we give them the state $$\lvert\Psi\rangle = \tfrac{3}{5}\lvert 00\rangle + \tfrac{4}{5}\lvert 11\rangle$$ They can still outperform the classical maximum winning probability of $75\%$, and in fact by performing a $e^{\pi i/4}$ phase shift on the second component of their shared state and using the Pauli matrices just as is done in the optimal solution to the CHSH game, they can attain a winning probability of $$\frac{1+\tfrac{1}{\sqrt 2}\cdot\tfrac{24}{25}}{2}\approx 84\%$$ but I'd like to know if this solution is optimal for the given entangled state, or if some other slightly different choice of observables could give a better performance. More generally, does being given a different pure entangled state change their optimal strategy for the CHSH game, and how can we prove the answer?

$\endgroup$
1
  • $\begingroup$ Why the downvote? Can I improve this question somehow? $\endgroup$ Mar 18, 2023 at 8:35

1 Answer 1

9
$\begingroup$

What you're after is Gisin's theorem. He proved that the maximal violation of the CHSH inequality with a fixed state $|\psi\rangle = c_0 |00\rangle + c_1|11\rangle$ is given by $2\sqrt{1+4 |c_0 c_1|^2}$. He also gives the optimal strategy.

This implies that the violation you found for the Pythagorean state is not optimal, one can achieve 0.846554, whereas your strategy gives 0.839411.

$\endgroup$
3
  • $\begingroup$ This is exactly what I was looking for, thanks! $\endgroup$ Mar 18, 2023 at 0:52
  • $\begingroup$ Am I being silly, or does this paper have a small error? I'm trying to follow his algebra, but I'm finding that the expression $|\cos\beta-\cos\beta'| + 2c_1c_2[\sin\beta + \sin\beta']$ is maximized at $x = (1+4|c_1c_2|^2)^{-1/2}$, as you write in your question, not $x = (1+4|c_1c_2|)^{-1/2}$. Maybe there is some strange notational issue I'm missing in the paper? $\endgroup$ Mar 18, 2023 at 17:20
  • $\begingroup$ It's just a typo. $\endgroup$ Mar 19, 2023 at 6:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.