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

Outline The problem The problem Getting strong enough correlations Getting strong enough correlations Getting the right correlations Getting the right correlations

The Problem

The CHSH game Alice gets a bit a and outputs a bit  Alice gets a bit a and outputs a bit  Bob gets a bit b 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 ) Goal:  =a  b (i.e., output bits should be equal unless a  b=1 ) No communication is allowed No communication is allowed Best strategy is to always output 0: they get 3 out of the 4 possible questions right 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% 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 However, if they share an EPR state, they can get success probability ~85% for each of the 4 questions a1a1a1a1 a0a0a0a0 b0b0b0b0 b1b1b1b1

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

Simulating Quantum Correlations (classical reformulation [Tsirelson87] ) Alice gets a unit vector a  R n and Alice gets a unit vector a  R n and outputs a bit  Bob gets a unit vector b  R n and Bob gets a unit vector b  R n and outputs a bit  Goal: the correlation E [  ] should satisfy Goal: the correlation E [  ] should satisfy E [  ] =  a,b   a,b  =1  a,b  =0  a,b  =-1 a b a b a b

Example – CHSH Consider the special case in which Alice gets either Consider the special case in which Alice gets either a 0 =(1,0) or a 1 =(0,1) and Bob gets either b 0 =(1,1)/  2 orb 1 =(1,-1)/  2. b 0 =(1,1)/  2 or b 1 =(1,-1)/  2. Then  a i,b j  = -½ if i=j=1 and ½ otherwise Then  a i,b j  = -½ if i=j=1 and ½ otherwise Hence their goal is to output Hence their goal is to output bits ,  such that  with probability 85% if i=j=1, and  with probability 85% otherwise a1a1a1a1 a0a0a0a0 b0b0b0b0 b1b1b1b1

Previous Work Problem introduced by several authors [ Maudlin92,Steiner00,BrassardCleveTapp99 ] Problem introduced by several authors [ Maudlin92,Steiner00,BrassardCleveTapp99 ] In the naïve protocol, Alice simply sends her vector to Bob; this requires infinite communication 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 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 ] 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 ] Another protocol achieves only logn/2 bits, but only on average (worst case communication is unbounded) [ DegorreLaplanteRoland07 ]

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

Getting strong enough correlations

A Na ï ve Protocol with No Communication Alice and Bob share a random unit vector  R n Alice and Bob share a random unit vector  R n Alice outputs sign( ,a  ) Alice outputs sign( ,a  ) Bob outputs sign( ,b  ) Bob outputs sign( ,b  ) +1

A Na ï ve Protocol with No Communication Alice and Bob share a random unit vector  R n Alice and Bob share a random unit vector  R n Alice outputs sign( ,a  ) Alice outputs sign( ,a  ) Bob outputs sign( ,b  ) Bob outputs sign( ,b  ) Analysis: if  =  a,b  then Analysis: if  =  a,b  thenthereforea +1 b +1

Resulting Correlation Function desired result

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

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 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 For k=1 we get arcs on the circle; area = angle k=1 k=2 For k=2 we get spherical For k=2 we get sphericaltriangles: area =  1 +  2 +  3 -  For k=3, we get spherical For k=3, we get sphericaltetrahedra…

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

Getting the right correlations

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

The Idea We define a transformation C from R n to some other Hilbert space with the property that for all a,b  R n, We define a transformation C from R n to some other Hilbert space with the property that for all a,b  R n,  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) Alice and Bob now run the original protocol on the vectors C(a) and C(b) The resulting correlation function is 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! If we take f=h -1, we obtain the right correlation function!

Idea - Continued Our goal is, therefore, to find a transformation C on vectors such that for all a,b  R n, Our goal is, therefore, to find a transformation C on vectors such that for all a,b  R n,  C(a),C(b)  =h -1 (  a,b  ) Assume, for example, that h -1 (x)=x 3 Assume, for example, that h -1 (x)=x 3 Then we can choose C to be the mapping 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.

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

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 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 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… We omit the details…

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