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Oded Regev (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Simulating Quantum Correlations with Finite Communication. Oded Regev (Tel Aviv University) Ben Toner (CWI, Amsterdam). The problem Getting strong enough correlations Getting the right correlations. Outline. The Problem. b 0. a 1. a 0. b 1. The CHSH game.

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Oded Regev (Tel Aviv University) Ben Toner (CWI, Amsterdam)

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  1. Simulating Quantum Correlations with Finite Communication Oded Regev (Tel Aviv University) Ben Toner (CWI, Amsterdam)

  2. The problem • Getting strong enough correlations • Getting the right correlations Outline

  3. The Problem

  4. b0 a1 a0 b1 The CHSH game • Alice gets a bit a and outputs a bit  • Bob gets a bit b and outputs a bit  • Goal: =ab (i.e., output bits should be equal unless a=b=1) • No communication is allowed • Best strategy is to always output 0: they get 3 out of the 4 possible questions right • Moreover, even if they share a random string, their average success probability is at most 75% • However, if they share an EPR state, they can get success probability ~85% for each of the 4 questions

  5. Simulating Quantum Correlations • Fix some bipartite quantum state  • Alice gets a matrix A with 1 eigenvalues; outputs a bit {-1,1} • Bob gets a matrix B with 1 eigenvalues;outputs a bit {-1,1} • Goal: the correlation E[] should satisfy • E[] = Tr(AB ) • If the parties share , this is easy • Without shared entanglement, impossible • However, what happens if we allow classical communication between Alice and Bob? How many bits do they need to exchange to simulate quantum correlations?

  6. Simulating Quantum Correlations(classical reformulation [Tsirelson87]) • Alice gets a unit vector aRn and • outputs a bit {-1,1} • Bob gets a unit vector bRn and • outputs a bit {-1,1} • Goal: the correlation E[] should satisfy • E[] = a,b a,b=1 a,b=0 a,b=-1 a a a b b b

  7. Previous Work • Problem introduced by several authors [Maudlin92,Steiner00,BrassardCleveTapp99] • In the naïve protocol, Alice simply sends her vector to Bob; this requires infinite communication • For the case n=3 (EPR state), several protocols were developed [BrassardCleveTapp99, Csirek00, CerfGisinMassar00] with the best one requiring only one bit of communication [TonerBacon03] • For the general problem, best known protocol requires n/2 bits [TonerBacon06] • Another protocol achieves only logn/2 bits, but only on average (worst case communication is unbounded) [DegorreLaplanteRoland07]

  8. New Result: The problem can be solved with only 2 bits of communication

  9. Getting strong enough correlations

  10. Alice and Bob share a random unit vector Rn • Alice outputs sign(,a) • Bob outputs sign(,b) A Naïve Protocol with No Communication +1 -1

  11. Alice and Bob share a random unit vector Rn • Alice outputs sign(,a) • Bob outputs sign(,b) • Analysis: if =a,b then • therefore A Naïve Protocol with No Communication -1 +1 +1 a -1 b

  12. Resulting Correlation Function desired result

  13. The ‘Orthant’ Protocol • Alice and Bob project their vectors on a random k-dimensional subspace • Alice tells Bob which of the 2k orthants her vector lies in, and outputs +1 • Bob outputs +1 or -1 depending on whether his vector lies in the half-space determined by the orthant • This uses k bits of • communication • (easy to improve to k-1) +1 -1 a

  14. Analysis of the ‘Orthant’ Protocol • By using Gaussian random variables, we find out that the correlation function is given by certain areas on the sphere in k+1 dimensions • For k=1 we get arcs on • the circle; area = angle k=1 k=2 • For k=2 we get spherical • triangles: • area = 1+2+3- • For k=3, we get spherical • tetrahedra…

  15. Resulting Correlation Function k=3 k=2 k=1 Strong enough! Requires only 2 bits of communication!!

  16. Getting the right correlations

  17. Our goal is to have a protocol with correlations h()= • However, all protocols we tried were either too weak or too strong • We now show how to take any protocol with ‘strong enough’ correlations, and transform it into a protocol with the right correlation function h()= Getting the Right Correlations

  18. We define a transformation C from Rn to some other Hilbert space with the property that for all a,bRn, • C(a),C(b)=f(a,b) • where f:[-1,1][-1,1] is some function with f(1)=1. • Alice and Bob now run the original protocol on the vectors C(a) and C(b) • The resulting correlation function is • h(f()) • where h is the original correlation function. • If we take f=h-1, we obtain the right correlation function! The Idea

  19. Idea - Continued • Our goal is, therefore, to find a transformation C on vectors such that for all a,bRn, • C(a),C(b)=h-1(a,b) • Assume, for example, that h-1(x)=x3 • Then we can choose C to be the mapping • v  vvv • and then for any vectors a,b, • C(a),C(b)=aaa,bbb=a,b3=h-1(a,b) • as required.

  20. Extending this Idea • Now assume that h-1(x)=(x3+x)/2 • We can choose C to be the mapping • v  (vvv v)/2 • and this gives • C(a),C(b) = ½aaa  a , bbb b • = ½a,b3 + ½a,b • = h-1(a,b) • as required.

  21. Extending this Idea • In general, we can find a mapping C as long as the power series expansion of h-1 has only nonnegative coefficients • In order to apply this idea to the 2-bit ‘orthant’ protocol, we ‘simply’ have to analyze the power series of the inverse of • We omit the details…

  22. Open Questions • Is there any 1-bit protocol? • We conjecture that there isn’t any… • Extend to the more general problem of simulating local measurements on quantum states

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