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33
Bell inequalities and operator algebras
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Bell inequalities and operator algebras |
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| contact: | B. S. Tsirelson | date: | 6 July 2006 | last progress: | - | solved by: | - |
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| Problem |
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?
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| Background |
Quantum Bell-type inequalities are discussed in Problem 26 "Bell inequalities holding for all quantum states'' and Partial Solutions to Problem 1 "All the Bell inequalities'' (and many other texts, including my own [Ts93]) without a clear indication of the approach used.
Some relevant formal definitions (for two subsystems) may be found in [Ts93], Sect. 2 (and Sect. 1). Formulas (2.1-2.3) of [Ts93] formalize the less restrictive approach (via commuting operators), while formulas (2.4-2.6) formalize a more restrictive approach, requiring a tensor product Hilbert space (which could be essential) and some additional restrictions (these are not essential). Unfortunately, it is claimed (without proof) in [Ts93] that the two approaches are equivalent! This is my error. I regret deeply for my irresponsible claim. A failure of my old would-be-proof came to light in May 2006. We must thank Antonio Acin who requested from me a proof.
Three convex sets are considered in [Ts93] and Problem 26, denoted XHDB \subset XQB \subset XB 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' != 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' != Q'' : which properties of operator algebras are discerned by Q ? One may introduce QI, QII, QIII for type I, II, III factors, then QI = Q' , but what about QII, QIII ? Are they equal?
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| Partial Results |
If the given (global) Hilbert space H is finite-dimensional and only two subsystems are dealt with, then the two approaches are equivalent. Proof (sketch). The given operators of the first subsystem generate an operator algebra \mathcal A1 . Its center decomposes H into the direct sum H1 \oplus \dots \oplus Hn of subspaces (sectors) Hk . On each sector, \mathcal A1 boils down to a factor. Thus, Hk = 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'' whereH' = H'1 \oplus \dots \oplus H'n , \quad H'' = H''1 \oplus \dots \oplus H''n .
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| Remarks |
Three decades ago some mathematicians told me that quantum Bell-type inequalities belong to physics, not mathematics, because they are not related to the advanced mathematical theory of operator algebras (classification etc). What now?
If you wonder, where to find commuting operators that do not fit into the tensor product paradigm, here is a candidature. Consider the free group G with (say) two generators a,b . (Elements of G are like this: b3 a-2 b a b-2 a .) The countable group G acts on the Hilbert space H = l2(G) (of all square summable functions on G ) by left shifts as well as right shifts: Ux \psi(z) = \psi(xz) , Vx \psi(z) = \psi(zx-1) for x,z in G , \psi in H . Every Ux commutes with every Vy . However, it does not decompose H into H' \otimes H'' .
Let us choose the simplest state vector \psi , namely, \psi(x) = 1 if x is the unit of G , otherwise \psi(x) = 0 . Strong correlations appear: < Ux Vx >\psi = 1 . Do they lead to a point of Q'' not approximable by Q' ? I do not know.
We (the site maintainers) apologize for posting the problem only now (20 Nov 2007). The question was originaly posed by B. S. Tsirelson in July 2006. In this year, the former site administrator left the institute. By now, the dust has settled and we try to implement again a reliable update schedule for the problem site.
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| Literature |
| [Ts93] | B.S. Tsirelson, »Some results and problems on quantum Bell-type inequalities,« Hadronic Journal Supplement 8, 329-345 (1993). Available also on my site. |
| Questions and comments | Last modified: 20 Nov 2007 |
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