Ben Toner (CWI, Amsterdam) Oded Regev (Tel Aviv University) Proceedings of FOCS 2007,

1. Simulating Quantum Correlations with Finite Communication Ben Toner (CWI, Amsterdam) Oded Regev (Tel Aviv University) Proceedings of FOCS 2007, quant-ph/0708.0827 QIP, 21 December, 2007 * Original slides prepared by Oded.

2. Warm up problem: simulating EPR correlations 01001000011000 a 2 R3 b 2 R3 2 {0,1} 2 {0,1} Bob Alice • Alice gets a unit vector aR3and must outputs a bit . • Bob gets a unit vector bR3and must outputs a bit . • Goal: the correlation E[] should satisfy • Easy if Alice and Bob share |-i. • Impossible with a shared random string and no communication (Bell’s theorem)

3. Bell’s theorem 01001000011000 A B 2 {0,1} 2 {0,1} Bob Alice • Bell’s theorem: It is impossible to simulate these correlations with a shared random string and no communication • Werner & Wolf, quant-ph/0107093 (physics version); • Arora & Barak, Complexity Theory: A Modern Approach (interactive proofs); • But can we quantify Bell’s theorem? • Yes! For example, allow some communication • between Alice and Bob after they receive questions. • How much communication is required?

4. More general problem A B 2 {0,1} 2 {0,1} Bob Alice • Fix some bipartite quantum state  • Alice gets a matrix A with 1 eigenvalues; outputs bit {-1,1} • Bob gets a matrix B with 1 eigenvalues;outputs bit {-1,1} • Goal: the correlation should be • Easy if Alice and Bob share . • Impossible with a shared random string and no communication • How much communication is required?

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

6. 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 [BaconToner0?]

7. Our Result: The problem can be solved with only 2 bits of communication

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

9. Getting strong enough correlations

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

12. Resulting Correlation Function desired result no correlation

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).

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. Analysis of the ‘Orthant’ Protocol • Problem:No closed formula is known for the volume of a spherical tetrahedra. • Solution:There is an expression for the derivative. [Schlaefli1858]

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

17. Getting the right correlations

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

19. 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(r)) • where h is the original correlation function. • If we take f=h-1, we obtain the right correlation function! The Idea [Krivine79]

20. 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.

21. 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.

22. 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…

23. Open Questions • Is there any 1-bit protocol? • We conjecture that there is not. • Extend to the more general problem of simulating local measurements on quantum states. • Are there states for which there is no exact finite bit simulation protocol?

24. 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. True even with randomness. • If, however, they share an EPR state, they can get success probability ~85% for each of the 4 questions.