Bell inequalities and operator algebras

Cite this problem as Problem 33.


Quantum Bell-type inequalities are defined in terms of two (or more) subsystems of a quantum system. The subsystems may be treated either via (local) Hilbert spaces, — tensor factors of the given (global) Hilbert space, or via commuting (local) operator algebras. The latter approach is less restrictive, it just requires that the given operators commute whenever they belong to different subsystems.

Are these two approaches equivalent?


Three convex sets are considered in [1] and Problem 26, denoted X_{\text{HDB}} \subset X_{\text{QB}} \subset X_{\text{B}} and C \subset Q \subset P respectively. Their elements are called `behaviors’ and `correlations’, respectively. However, in both cases one should consider four sets, say,  C \subset Q' \subset Q'' \subset P, where Q', Q'' correspond to the two approaches (more restrictive and less restrictive, respectively).

The question is, whether Q' = Q'', or not.

If Q' \ne Q'' then another (even more important) question arises naturally: is Q' dense in Q'', or not?

If Q' is essentially different from Q'' (that is, not dense) then we should decide, which one is more relevant to physics. I believe that Q' is. Here is why. In physics we deal not just with commuting operators, but with operators belonging to local algebras corresponding to disjoint domains (on a positive distance). True, it is usually believed that local algebras are type III factors.  But it is also usually believed that they are separated by type I factors, provided that the domains are disjoint.  (This is called “F property” or “funnel property” in the local quantum field theory.)  And type I factors mean a tensor product Hilbert space.

Still another question appears if Q' \ne Q'': which properties of operator algebras are discerned by Q ? One may introduce Q_{\text{I}}, Q_{\text{II}},Q_{\text{III}} for type I, II, III factors, then  Q_{\text{I}} = Q', but what about Q_{\text{II}}, Q_{\text{III}}? Are they equal?

Partial Results

Finite dimensions

If the given (global) Hilbert space H is finite-dimensional and only two subsystems are dealt with, then the two approaches are equivalent. The given operators of the first subsystem generate an

operator algebra \mathcal A_1. Its center decomposes H into the direct sum H_1 \oplus \dots \oplus H_n of subspaces (sectors) H_k. On each sector,  \mathcal A_1 boils down to a factor. Thus, H_k = H'_k \otimes H''_k and

H = H'_1 \otimes H''_1 \oplus \dots \oplus H'_n \otimes H''_n \, .

It remains to embed H into H' \otimes H'' where

H' = H'_1 \oplus \dots \oplus H'_n \, , \quad H'' = H''_1 \oplus \dots \oplus H''_n \, .

Infinite Dimensions

This problem is connected to questions of finite approximability of C*-algebras [2]. In [3], [4], the problem is shown to be related to one major open question in operator algebra theory, the embedding problem of Connes’, asking whether any type II_1 factor can be embedded into an ultrapower of the hyperfinite type II_1 factor. More specifically, if Connes’ embedding conjecture is true, then the closure of Q' must equal Q''. The converse relation, namely, that \mbox{closure}(Q')=Q'' implies that Connes’ embedding conjecture is true, is proven in [5].

In [6], Slofstra solved part of the problem, by demonstrating that Q'\not= Q''. The question of whether Q' is dense in Q'' remains open.


[1] B.S. Tsirelson, Hadronic Journal Supplement 8, pp 329, (1993)

[2] V.B. Scholz, R.F. Werner, arXiv:0812.4305

[3] M. Junge, M. Navascues, C. Palazuelos, D. Perez-Garcia, V. B. Scholz, R. F. Werner, J. Math. Phys. 52, 012102 (2011).

[4] T. Fritz, Rev. Math. Phys. 24(5), 1250012 (2012).

[5] N. Ozawa, Jpn. J. Math. 8, no. 1, 147–183 (2013).

[6] W. Slofstra, arXiv:1606.03140.