oded regev (tel aviv university) ben toner (cwi, amsterdam)

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

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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. - PowerPoint PPT Presentation

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Page 1: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Oded Regev Oded Regev (Tel Aviv University)(Tel Aviv University)

Ben TonerBen Toner(CWI, Amsterdam)(CWI, Amsterdam)

Simulating Quantum Simulating Quantum Correlations Correlations with Finite with Finite CommunicationCommunication

Page 2: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

OutlineOutline

•The problemThe problem•Getting strong enough correlationsGetting strong enough correlations•Getting the right correlationsGetting the right correlations

Page 3: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

The ProblemThe ProblemThe ProblemThe Problem

Page 4: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

The CHSH gameThe CHSH game• AliceAlice gets a bit gets a bit a a and outputs a bit and outputs a bit • BobBob gets a bit gets a bit b b and outputs a bit and outputs a bit • Goal: Goal: =a=ab b (i.e., output bits should be equal (i.e., output bits should be equal

unless unless aab=1b=1))• No communication is allowedNo communication is allowed

• Best strategy is to always output 0: they get 3 Best strategy is to always output 0: they get 3 out of the 4 possible questions rightout of the 4 possible questions right

• Moreover, even if they share a random string, Moreover, even if they share a random string, their average success probability is at most their average success probability is at most 75%75%

• However, if they share an EPR state, they can However, if they share an EPR state, they can get success probability ~85% for each of the 4 get success probability ~85% for each of the 4 questionsquestions

aa11

aa00

bb00

bb11

Page 5: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Simulating Quantum Simulating Quantum CorrelationsCorrelations• Fix some bipartite quantum state Fix some bipartite quantum state

• AliceAlice gets a matrix A with gets a matrix A with 1 eigenvalues; 1 eigenvalues; outputs a bit outputs a bit

• BobBob gets a matrix B with gets a matrix B with 1 eigenvalues;1 eigenvalues; outputs a bit outputs a bit

• GoalGoal: the correlation : the correlation EE[[] should satisfy ] should satisfy EE[[] = Tr(A] = Tr(AB B ) )

• If the parties share If the parties share , this is easy, this is easy• Without shared entanglement, impossibleWithout shared entanglement, impossible• However, what happens if we allow classical However, what happens if we allow classical

communication between Alice and Bob? How communication between Alice and Bob? How many bits do they need to exchange to many bits do they need to exchange to simulate quantum correlations?simulate quantum correlations?

Page 6: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Simulating Quantum CorrelationsSimulating Quantum Correlations(classical reformulation (classical reformulation

[Tsirelson87][Tsirelson87]))•AliceAlice gets a unit vector a gets a unit vector aRRn n and and

outputs a bit outputs a bit •BobBob gets a unit vector b gets a unit vector bRRn n and and

outputs a bit outputs a bit

•GoalGoal: the correlation : the correlation EE[[] should satisfy ] should satisfy EE[[] = ] = a,ba,b

a,ba,b=1=1 a,ba,b=0=0 a,ba,b=-1=-1aa

bbaa

bb

aa

bb

Page 7: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Previous WorkPrevious Work• Problem introduced by several authors Problem introduced by several authors

[[Maudlin92,Steiner00,BrassardCleveTapp99Maudlin92,Steiner00,BrassardCleveTapp99]]• In the naïve protocol, Alice simply sends her In the naïve protocol, Alice simply sends her

vector to Bob; this requires infinite vector to Bob; this requires infinite communicationcommunication

• For the case n=3 (EPR state), several protocols For the case n=3 (EPR state), several protocols were developed were developed [BrassardCleveTapp99, Csirek00, [BrassardCleveTapp99, Csirek00, CerfGisinMassar00] CerfGisinMassar00] with the best one requiring with the best one requiring only one bit of communicationonly one bit of communication [TonerBacon03] [TonerBacon03]

• For the general problem, best known protocol For the general problem, best known protocol requires requires n/2n/2 bits bits [[TonerBacon06TonerBacon06]]

• Another protocol achieves only logn/2 bits, but Another protocol achieves only logn/2 bits, but only on average (worst case communication is only on average (worst case communication is unbounded) unbounded) [[DegorreLaplanteRoland07DegorreLaplanteRoland07]]

Page 8: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

New Result:New Result:The problem can be solved The problem can be solved

with only 2 bits of with only 2 bits of communication communication

Page 9: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Getting strong Getting strong enough enough

correlationscorrelations

Getting strong Getting strong enough enough

correlationscorrelations

Page 10: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

A Naïve Protocol with No A Naïve Protocol with No CommunicationCommunication

•AliceAlice and and BobBob share a random unit vector share a random unit vector RRnn

•AliceAlice outputs sign( outputs sign(,a,a))•BobBob outputs outputs sign(sign(,b,b))

+1

-1

Page 11: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

A Naïve Protocol with No A Naïve Protocol with No CommunicationCommunication

•Alice and Bob share a random unit vector Alice and Bob share a random unit vector RRnn

•Alice outputs sign(Alice outputs sign(,a,a))•Bob outputs Bob outputs sign(sign(,b,b))

•Analysis: if Analysis: if ==a,ba,b then then

thereforetherefore

aa++11

--11

bb

++11--11

Page 12: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Resulting Correlation FunctionResulting Correlation Function

desired

desired

resultresult

Page 13: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

The ‘Orthant’ ProtocolThe ‘Orthant’ Protocol•AliceAlice and and BobBob project their vectors on a project their vectors on a

random k-dimensional subspace random k-dimensional subspace •AliceAlice tells tells BobBob which of the 2 which of the 2kk orthants orthants

her vector lies in, and outputs +1her vector lies in, and outputs +1•BobBob outputs +1 or -1 depending on outputs +1 or -1 depending on

whether his vector lies in the half-space whether his vector lies in the half-space determined by the orthantdetermined by the orthant

•This uses k bits of This uses k bits of communication communication (easy to improve to k-1)(easy to improve to k-1)

aa

++11

--11

Page 14: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Analysis of the ‘Orthant’ ProtocolAnalysis of the ‘Orthant’ Protocol•By using Gaussian random variables, we By using Gaussian random variables, we

find out that the correlation function is find out that the correlation function is given by certain areas on the sphere in given by certain areas on the sphere in k+1 dimensionsk+1 dimensions

•For k=1 we get arcs on For k=1 we get arcs on the circle; area = anglethe circle; area = angle

k=1k=1

k=2k=2•For k=2 we get spherical For k=2 we get spherical triangles:triangles:

area = area = 11++22++33--•For k=3, we get sphericalFor k=3, we get spherical

tetrahedra…tetrahedra…

Page 15: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Resulting Correlation FunctionResulting Correlation Function

k=3k=3

k=1k=1k=2k=2

Strong enough!Strong enough!Requires only 2 Requires only 2

bits of bits of communication!!communication!!

Page 16: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Getting the right Getting the right correlationscorrelations

Getting the right Getting the right correlationscorrelations

Page 17: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Getting the Right CorrelationsGetting the Right Correlations•Our goal is to have a protocol with Our goal is to have a protocol with

correlations h(correlations h()=)=•However, all protocols we tried were However, all protocols we tried were

either too weak or too strongeither too weak or too strong

•We now show how to take any protocol We now show how to take any protocol with ‘strong enough’ correlations, and with ‘strong enough’ correlations, and transform it into a protocol with the right transform it into a protocol with the right correlation function correlation function h(h()=)=

Page 18: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

The IdeaThe Idea• We define a transformation C from We define a transformation C from RRnn to some to some

other Hilbert space with the property that for other Hilbert space with the property that for all a,ball a,bRRnn,,

C(a),C(b)C(a),C(b)=f(=f(a,ba,b))where f:[-1,1]where f:[-1,1][-1,1] is some function with [-1,1] is some function with f(1)=1.f(1)=1.

• Alice and Bob now run the original protocol on Alice and Bob now run the original protocol on the vectors C(a) and C(b)the vectors C(a) and C(b)

• The resulting correlation function is The resulting correlation function is h(f(h(f())))

where h is the original correlation function.where h is the original correlation function.• If we take f=hIf we take f=h-1-1, we obtain the right correlation , we obtain the right correlation

function!function!

Page 19: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Idea - ContinuedIdea - Continued•Our goal is, therefore, to find a Our goal is, therefore, to find a

transformation C on vectors such that for transformation C on vectors such that for all a,ball a,bRRnn,,

C(a),C(b)C(a),C(b)=h=h-1-1((a,ba,b))

•Assume, for example, that hAssume, for example, that h-1-1(x)=x(x)=x33

•Then we can choose C to be the mapping Then we can choose C to be the mapping v v v vvvvv

and then for any vectors a,b,and then for any vectors a,b,C(a),C(b)C(a),C(b)==aaaaa,ba,bbbbb==a,ba,b33=h=h--

11((a,ba,b))as required.as required.

Page 20: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Extending this IdeaExtending this Idea

•Now assume that hNow assume that h-1-1(x)=(x(x)=(x33+x)/2+x)/2•We We can choose C to be the mapping can choose C to be the mapping

v v (v (vvvv v v)/v)/22and this gives and this gives C(a),C(b)C(a),C(b) = ½ = ½aaaaa a a , b a , bbbbb bb

= ½= ½a,ba,b33 + ½ + ½a,ba,b = h= h-1-1((a,ba,b))

as required.as required.

Page 21: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Extending this IdeaExtending this Idea

• In general, we can find a mapping C as In general, we can find a mapping C as long as the power series expansion of hlong as the power series expansion of h-1-1 has only nonnegative coefficientshas only nonnegative coefficients

• In order to apply this idea to the 2-bit In order to apply this idea to the 2-bit ‘orthant’ protocol, we ‘simply’ have to ‘orthant’ protocol, we ‘simply’ have to analyze the power series of the inverse analyze the power series of the inverse ofof

•We omit the details…We omit the details…

Page 22: Oded Regev  (Tel Aviv University) Ben Toner (CWI, Amsterdam)

Open QuestionsOpen 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 Extend to the more general problem of simulating local measurements on simulating local measurements on quantum statesquantum states