Alice and Bob are each in possession of one half of a maximally entangled pair of particles. Alice can make either of two observations, $A_1$ or $A_2$. Bob can make either of two observations, $B_1$ or $B_2$. (Observations have values $1$ or $-1$). Write $E(x,y)$ for the expected value of the product $xy$. Then Tsirelson's Bound says that $|E(A1,B2)-E(A1,B1)-E(A2,B1)-E(A2,B2)|$ is bounded above by $2 \sqrt{2}$.

Question: Is the converse true? That is, suppose I have four numbers $x,y,z,w$. Suppose they are all bounded by one in absolute value and that they satisfy $|x-y-z-w| < 2 \sqrt{2}$. Does it follow that there are observables $A_1, A_2, B_1, B_2$ such that $x=E(A_1,B_2)$, $y=E(A_1,B_1)$, etc.?

If not, what other conditions do I need on $x, y, z, w$ in order for such observables to exist?


2 Answers 2


No. The necessary and sufficient condition is well-known; see for instance Tsirelson 1993, or L. Landau.

  • $\begingroup$ The Tsirelson condition is necessary but not sufficient. If the Tsirelson condition had been sufficient, the set of quantum correlations would have been a polytope in a 8-dimensional space, but it is not so. In fact, the quantum boundary has curved regions. See arxiv.org/pdf/1710.05892.pdf $\endgroup$
    – abir
    Jul 5, 2020 at 12:07

No, the converse is not true. Even if a correlation satisfies the Tsirelson's bound, even then it might not be reproducible by measurements on quantum states. One of the most prominent examples is the Hardy correlation. Imposing the Tsirelson's bound, it seems that the maximum value of q is 0.207. However, it turns out that with quantum states, we cannot obtain a q value greater than 0.09.


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