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Quantum Coloring and related problems K. Svozil (ITP-TUW May 22nd, 2001)

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Presentation on theme: "Quantum Coloring and related problems K. Svozil (ITP-TUW May 22nd, 2001)"— Presentation transcript:

1 Quantum Coloring and related problems K. Svozil (ITP-TUW May 22nd, 2001)

2 Consistent description of observables Three-coloring of the rational quantum sphere Hans Havlicek, Guenther Krenn, Johann Summhammer and KS, ``Coloring the rational quantum sphere and the Kochen-Specker theorem,'' J. Phys. A:Math. Gen. 34(14), 3071-3077 (13 April 2001) Boole-Bell inequalities via Pitowsky polytopes Itamar Pitowsky and KS,``New optimal tests of quantum nonlocality,'' Phys. Rev. A, in print; Stefan Filipp and KS, quant-ph/0105083 Haystack problem Niko Donath & KS http://arxiv.org/abs/quant-ph/0105046

3 The quantum unit ball

4 Every point of the ball is a yes/no proposition (von Neumann, "Mathematische Grundlagen der Quantenmechanik,“ 1932) ``  To each property E we can assign a quantity which we define as follows: each measurement which distinguishes between the presence or absence of E is considered as a measurement of this quantity, such that its value is 1 if E is verified, and zero in the opposite case. This quantity which corresponds to E will also be denoted by E. Such quantities take only the values of 0 and 1, and conversely, each quantity R which is capable of these two values only, corresponds to a property E which is evidently this: ``the value of R is  0.'' The quantities E that correspond to the properties are therefore characterized by this behavior. That E takes on only the values 0,1 can also be formulated as follows: Substituting E into the polynomial F( ) =  2 makes it vanish identically. If E has the operator E, then F(E) has the operator F(E) = E  E 2, i.e., the condition is that E  E 2 = 0 or E = E 2. In other words: the operator E of E is a projection. The projections E therefore correspond to the properties E (through the agency of the corresponding quantities E which we just defined). If we introduce, along with the projections E, the closed linear manifold M, belonging to them (E = P M ), then the closed linear manifolds correspond equally to the properties of E.''

5 Kochen-Specker theorem Under certain „reasonable“ assumptions... there is no 1-0-0 coloring of systems of orthogonal tripods in Hilbert spaces of Dimensions > 2. The existence of elements of physical reality independent of the „context“ is inconsistent. Scholastic futurabilities: does God know also things which could have happened but did not happen ?

6 2   2, 1  _

7 SURPRISE: Three coloring of the rational quantum sphere Consider rays from the origin which meet the rational unit sphere S 2  Q 3. The following statements on a triple (x,y,z)  Z 3 \{(0,0,0)} are equivalent: (i) The ray Sp(x,y,z) intersects the unit sphere at two rational points; i.e., it contains the rational points  ( x,y,z)/  [(x 2 +y 2 +z 2 )]  S 2  Q 3. (ii) The Pythagorean property holds, i.e., x 2 +y 2 +z 2 = n 2, n  N. This equivalence can be demonstrated as follows. All points on the rational unit sphere can be written as r = ([a/(a)],[b/(b)],[c/(c)] ) with a,b,c  Z, a,b,c  Z\{0}, and ([a/(a)]) 2 +([b/(b)]) 2 +([c/(c)]) 2 = 1. Multiplication of r with a 2 b 2 c 2 results in a vector of Z 3 satisfying (ii). Conversely, from x 2 +y 2 +z 2 =n 2, n  N, we obtain the rational unit vector ( x / n, y / n, z / n )  S 2  Q 3.

8 How-To? If x,y,z are chosen coprime then a necessary condition for x 2 +y 2 +z 2 being a non- zero square is that precisely one of x, y, and z is odd. This is a direct consequence of the observation that any square is congruent to 0 or 1, modulo 4, and from the fact that at least one of x, y, and z is odd. Hence our coloring of the rational rays induces the following coloring of the rational unit sphere with those three colors that are represented by the standard basis of Z 2 3 : color #1 if x is odd, y and z are even, color #2 if y is odd, z and x are even, color #3 if z is odd, x and y are even. All three colors occur, since the vectors of the standard basis of R 3 are colored differently. They are also dense on the unit sphere.

9 Problems Propositional structures not closed under NORs:  S 2  Q 3 but the cross product thereof is (12/25, -9/25,12/25)  S 2  Q 3. Arbitraryness of approximation

10 Boole-Bell type inequalities and Pitowsky correlation polytopes Truth table of events and their joint events Rows represent vectors All row vectors define a convex correlation polytope By the Minkowski-Weyl representation theorem, an equivalent representation is via the inequalities corresponding to the faces of the polytope (the „hull problem“).

11 Hull problem for 3-3 and 2-2-2 2-2-2: GHZ case: three particles at two angles: 53856 inequalities, e.g., 0  - P(A 1 B 1 C 1 ) - 2P(A 2 B 1 C 1 ) - 3P(A 1 B 2 C 1 ) - 3P(A 1 B 1 C 2 ) - P(A 2 B 2 C 1 ) - P(A 2 B 1 C 2 ) - P(A 1 B 2 C 2 ) + 4P(A 2 B 2 C 2 ), 3-3: two particles at three angles: 684 inequalies, e.g., 3  2P(A 1 ) + P(A 2 ) +P(B 2 ) +2P(B 3 ) - P(A 1 B 1 ) - P(A 1 B 2 ) - P(A 1 B 3 ) + P(A 2 B 1 ) - P(A 2 B 2 ) - P(A 2 B 3 ) + P(A 3 B 2 ) - P(A 3 B 3 )

12 Parameter studies 2-2-2

13 Parameter studies 3-3

14 Haystack problem... to single out a particular state among 2 n orthogonal pure states. As it turns out, in general the optimal strategy is not to measure the particles separately, but to consider joint properties of the n-particle system. The required number of propositions is n. There exist 2 n ! equivalent operational procedures to do so. We enumerate some configurations for three particles, in particular the Greenberger-Horne- Zeilinger (GHZ)- and W-states, which are specific cases of a unitary transformation...

15 Instead of a summary: the generalized urn puzzle


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