1024 (number)

A Tsirelson bound is an upper limit to quantum mechanical correlations between distant events. Given that quantum mechanics is non-local, i.e., that quantum mechanical correlations violates Bell inequalities, a natural question to ask is "how non-local can quantum mechanics be?", or, more precisely, by how much can the Bell inequality be violated. The answer is precisely the Tsirelson bound for the particular Bell inequality in question. In general this bound is lower than what would be algebraically possible, and much research has been dedicated to the question of why this is the case.

The Tsirelson bounds are named after B. S. Tsirelson, (or Boris Cirel'son, in a different transliteration) the author of the paper in which the first one was derived.

Tsirelson bound for the CHSH inequality

The first Tsirelson bound was derived as an upper bound on the correlations measured in the CHSH inequality. It states that if we have four (Hermitian) dichotomic observables $A_{0}$, $A_{1}$, $B_{0}$, $B_{1}$ (i.e., two observables for Alice and two for Bob) with outcomes $+1,-1$ such that $[A_{i},B_{j}]=0$ for all $i,j$, then

$\langle A_{0}B_{0}\rangle +\langle A_{0}B_{1}\rangle +\langle A_{1}B_{0}\rangle -\langle A_{1}B_{1}\rangle \leq 2{\sqrt {2}}$

For comparison, in the classical (or local realistic case) the upper bound is 2, whereas if any arbitrary assignment of $+1,-1$ is allowed it is 4. The Tsirelson bound is attained already if Alice and Bob each makes measurements on a qubit, the simplest non-trivial quantum system.

Lots of proofs have been developed for this bound, but perhaps the most enlightening one is based on the Khalfin-Tsirelson-Landau identity. If we define an observable

${\mathcal {B}}=A_{0}B_{0}+A_{0}B_{1}+A_{1}B_{0}-A_{1}B_{1}$

and $A_{i}^{2}=B_{j}^{2}=\mathbb {I}$, i.e., if the outcomes of the observables are associated to projective measurements, then

${\mathcal {B}}^{2}=4\mathbb {I} -[A_{0},A_{1}][B_{0},B_{1}]$

If $[A_{0},A_{1}]=0$ or $[B_{0},B_{1}]=0$, which can be regarded as the classical case, it already follows that $\langle {\mathcal {B}}\rangle \leq 2$. In the quantum case, we need only notice that $\|[A_{0},A_{1}]\|\leq 2\|A_{0}\|\|A_{1}\|\leq 2$ and the Tsirelson bound $\langle {\mathcal {B}}\rangle \leq 2{\sqrt {2}}$ follows.

Tsirelson bounds for other Bell inequalities

Obtaining a Tsirelson bound for a given Bell inequality is in general a hard problem that has to be solved in a case-by-case basis, although we do have numerical algorithms than can upperbound it. The exact values are known for a few more Bell inequalities:

For the Braunstein-Caves inequalities we have that

$\langle BC_{n}\rangle \leq n\cos(\pi /n)$

For the WWŻB inequalities the Tsirelson bound is

$\langle WWZB_{n}\rangle \leq 2^{\frac {n-1}{2}}$

Finding the Tsirelson problem for the $I_{3322}$ inequality is a notorious open problem in quantum information theory.

Tsirelson bounds from physical principles

Lots of research have been dedicated to find a physical principle that explains why quantum correlations go only up to the Tsirelson bound and nothing more. Two such principles have been found, the non-triviality of communication complexity and information causality. Another relevant principle is the absence of 3rd-order interference.

See also

• Bell's theorem
• EPR paradox
• CHSH inequality
• Quantum pseudo-telepathy