Tsirelson's bound

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 ${\displaystyle A_{0}}$, ${\displaystyle A_{1}}$, ${\displaystyle B_{0}}$, ${\displaystyle B_{1}}$ (i.e., two observables for Alice and two for Bob) with outcomes ${\displaystyle +1,-1}$ such that ${\displaystyle [A_{i},B_{j}]=0}$ for all ${\displaystyle i,j}$, then

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

For comparison, in the classical case (or local realistic case) the upper bound is 2, whereas if any arbitrary assignment of ${\displaystyle +1,-1}$ 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.

Several proofs of this bound exist, but perhaps the most enlightening one is based on the Khalfin–Tsirelson–Landau identity. If we define an observable

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

and ${\displaystyle A_{i}^{2}=B_{j}^{2}=\mathbb {I} }$, i.e., if the observables' outcomes are ${\displaystyle +1,-1}$, then

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

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

Tsirelson also showed that for any bipartite full-correlation Bell inequality with m inputs for Alice and n inputs for Bob, the ratio between the Tsirelson bound and the local bound is at most ${\displaystyle K_{G}^{\mathbb {R} }(\lfloor r\rfloor ),}$ where ${\displaystyle r=\min \left\{m,n,-{\frac {1}{2}}+{\sqrt {{\frac {1}{4}}+2(m+n)}}\right\},}$ and ${\displaystyle K_{G}^{\mathbb {R} }(d)}$ is the Grothendieck constant of order d.[2] Note that since ${\displaystyle K_{G}^{\mathbb {R} }(2)={\sqrt {2}}}$, this bound implies the above result about the CHSH inequality.

In general, obtaining a Tsirelson bound for a given Bell inequality is a hard problem that has to be solved on a case-by-case basis. It is not even known to be decidable. The best known computational method for upperbounding it is a convergent hierarchy of semidefinite programs, the NPA hierarchy, that in general does not halt.[3][4] The exact values are known for a few more Bell inequalities:

For the Braunstein–Caves inequalities we have that

${\displaystyle \langle {\text{BC}}_{n}\rangle \leq n\cos \left({\frac {\pi }{n}}\right).}$

For the WWŻB inequalities the Tsirelson bound is

${\displaystyle \langle {\text{WWZB}}_{n}\rangle \leq 2^{(n-1)/2}.}$

For the ${\displaystyle I_{3322}}$ inequality[5] the Tsirelson bound is not known exactly, but concrete realisations give a lower bound of 0.250875384514,[6] and the NPA hierarchy gives an upper bound of 0.2508753845139766.[7] It is conjectured that only infinite-dimensional quantum states can reach the Tsirelson bound.

Significant research has been dedicated to finding a physical principle that explains why quantum correlations go only up to the Tsirelson bound and nothing more. Three such principles have been found: no-advantage for non-local computation,[8] information causality[9] and macroscopic locality.[10] That is to say, if one could achieve a CHSH correlation exceeding Tsirelson's bound, all such principles would be violated. Tsirelson's bound also follows if the Bell experiment admits a strongly positive quantal measure.[11]

There are two different ways of defining the Tsirelson bound of a Bell expression. One by demanding that the measurements are in a tensor product structure, and another by demanding only that they commute. Tsirelson's problem is the question of whether these two definitions are equivalent. More formally, let

${\displaystyle B=\sum _{abxy}\mu _{abxy}p(ab|xy)}$

be a Bell expression, where ${\displaystyle p(ab|xy)}$ is the probability of obtaining outcomes ${\displaystyle a,b}$ with the settings ${\displaystyle x,y}$. The tensor product Tsirelson bound is then the supremum of the value attained in this Bell expression by making measurements ${\displaystyle A_{x}^{a}:{\mathcal {H}}_{A}\to {\mathcal {H}}_{A}}$ and ${\displaystyle B_{y}^{b}:{\mathcal {H}}_{B}\to {\mathcal {H}}_{B}}$ on a quantum state ${\displaystyle |\psi \rangle \in {\mathcal {H}}_{A}\otimes {\mathcal {H}}_{B}}$:

${\displaystyle T_{t}=\sup _{|\psi \rangle ,A_{x}^{a},B_{y}^{b}}\sum _{abxy}\mu _{abxy}\langle \psi |A_{x}^{a}\otimes B_{y}^{b}|\psi \rangle .}$

The commuting Tsirelson bound is the supremum of the value attained in this Bell expression by making measurements ${\displaystyle A_{x}^{a}:{\mathcal {H}}\to {\mathcal {H}}}$ and ${\displaystyle B_{y}^{b}:{\mathcal {H}}\to {\mathcal {H}}}$ such that ${\displaystyle \forall a,b,x,y;[A_{x}^{a},B_{y}^{b}]=0}$ on a quantum state ${\displaystyle |\psi \rangle \in {\mathcal {H}}}$:

${\displaystyle T_{c}=\sup _{|\psi \rangle ,A_{x}^{a},B_{y}^{b}}\sum _{abxy}\mu _{abxy}\langle \psi |A_{x}^{a}B_{y}^{b}|\psi \rangle .}$

Since tensor product algebras in particular commute, ${\displaystyle T_{t}\leq T_{c}}$. In finite dimensions commuting algebras are always isomorphic to (direct sums of) tensor product algebras,[12] so only for infinite dimensions it is possible that ${\displaystyle T_{t}\neq T_{c}}$. Tsirelson's problem is the question of whether for all Bell expressions ${\displaystyle T_{t}=T_{c}}$.

This question was first considered by Boris Tsirelson in 1993, where he asserted without proof that ${\displaystyle T_{t}=T_{c}}$.[13] Upon being asked for a proof by Antonio Acín in 2006, he realized that the one he had in mind didn't work, and issued the question as an open problem.[14] Together with Miguel Navascués and Stefano Pironio, Antonio Acín had developed an hierarchy of semidefinite programs, the NPA hierarchy, that converged to the commuting Tsirelson bound ${\displaystyle T_{c}}$ from above,[4] and wanted to know whether it also converged to the tensor product Tsirelson bound ${\displaystyle T_{t}}$, the most physically relevant one.

Since one can produce a converging sequencing of approximations to ${\displaystyle T_{t}}$ from below by considering finite-dimensional states and observables, if ${\displaystyle T_{t}=T_{c}}$, then this procedure can be combined with the NPA hierarchy to produce a halting algorithm to compute the Tsirelson bound, making it a computable number (note that in isolation neither procedure halts in general). Conversely, if ${\displaystyle T_{t}}$ is not computable, then ${\displaystyle T_{t}\neq T_{c}}$. In January 2020, Ji, Natarajan, Vidick, Wright, and Yuen claimed to have proven that ${\displaystyle T_{t}}$ is not computable, thus solving Tsirelson's problem in the negative;[15] a finalized, but still unreviewed, version of the proof appeared in Communications of the ACM in November 2021.[16] Tsirelson's problem has been shown to be equivalent to Connes' embedding problem,[17] so the same proof also implies that the Connes embedding problem is false.[18]

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