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

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