Avraham Ben-AroyaAvraham Ben-Aroya(Tel Aviv University)(Tel Aviv University)

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

Ronald de WolfRonald de Wolf(CWI, Amsterdam)(CWI, Amsterdam)

A Hypercontractive Inequality for A Hypercontractive Inequality for Matrix-Valued Functions Matrix-Valued Functions

with Applications to Quantum with Applications to Quantum ComputingComputing

OutlineOutline

•The new hypercontractive inequalityThe new hypercontractive inequality•Main application: k-out-of-n random Main application: k-out-of-n random

access codesaccess codes•Another application: direct product Another application: direct product

theorem for one-way communication theorem for one-way communication complexitycomplexity

The New InequalityThe New InequalityThe New InequalityThe New Inequality

The Parallelogram LawThe Parallelogram Law

•For any two vectors a,bFor any two vectors a,bRRdd,,

•Or equivalently,Or equivalently,

a

ba+ba-b

a

ba+ba-b

The Parallelogram LawThe Parallelogram Law

• This was for the This was for the 22 norm norm

• What happens in the What happens in the pp norm, for 1 norm, for 1p<2?p<2?

• The equality no longer holds, take, e.g., The equality no longer holds, take, e.g., a=(1,0),b=(0,1) and p=1a=(1,0),b=(0,1) and p=1

• But, we have the following powerful But, we have the following powerful inequality for all a,binequality for all a,bRRdd and and 11pp22::

The Extended Parallelogram The Extended Parallelogram LawLaw

• This inequality was proven by This inequality was proven by [Tomczak-[Tomczak-Jaegermann74, BallCarlenLieb94]Jaegermann74, BallCarlenLieb94]

• Originally used to prove the ‘sharp uniform Originally used to prove the ‘sharp uniform convexity’ of convexity’ of pp spaces spaces

• Implies the Bonami-Beckner hypercontractive Implies the Bonami-Beckner hypercontractive inequalityinequality

• An extremely useful inequality in computer An extremely useful inequality in computer science (analysis of Boolean functions, hardness science (analysis of Boolean functions, hardness of approximation, learning theory, of approximation, learning theory, communication complexity, percolation, etc.)communication complexity, percolation, etc.)

• Recently used by Recently used by [LeeNaor04][LeeNaor04] to prove a lower to prove a lower bound on the distortion of embeddings into bound on the distortion of embeddings into 11 spacesspaces

• Amazingly, the same inequality also holds with a,b Amazingly, the same inequality also holds with a,b being matrices and norms being matrix p-norms (i.e., being matrices and norms being matrix p-norms (i.e., Schatten p-norms)Schatten p-norms)

Prelims: Fourier TransformPrelims: Fourier Transform

• Let f be a function from {0,1}Let f be a function from {0,1}nn to to RRdd (or (or ℂℂdd××dd))• Then we define its Fourier transform asThen we define its Fourier transform as

• So, e.g.,So, e.g.,

The New Hypercontractive The New Hypercontractive Ineq. Ineq. • ThmThm: For any vector- or matrix-valued f on : For any vector- or matrix-valued f on

{0,1}{0,1}nn and 1 and 1pp2,2,

• Remark:Remark: This is the extension of the Bonami- This is the extension of the Bonami-Beckner inequality to vector/matrix-valued Beckner inequality to vector/matrix-valued functionsfunctions

The New Hypercontractive The New Hypercontractive Ineq. Ineq. • ThmThm: For any vector- or matrix-valued f on : For any vector- or matrix-valued f on

{0,1}{0,1}nn and 1 and 1pp2,2,

• Proof:Proof: By induction on n. By induction on n. • The case n=1 is exactly the The case n=1 is exactly the [BCL94][BCL94]

inequality with a=f(0), b=f(1)inequality with a=f(0), b=f(1)

• For simplicity, let’s see how to get the n=2 For simplicity, let’s see how to get the n=2 case.case.•This involves four matrices, a=f(00), This involves four matrices, a=f(00),

b=f(01), c=f(10), d=f(11)b=f(01), c=f(10), d=f(11)

The New Inequality (cont.)The New Inequality (cont.)• Using the case n=1 we getUsing the case n=1 we get

• By averaging the two inequalities, we getBy averaging the two inequalities, we get

The New Inequality (cont.)The New Inequality (cont.)

• Using the case n=1 again, the left side is at Using the case n=1 again, the left side is at leastleast

Application 1:Application 1:Random Access CodesRandom Access Codes

Application 1:Application 1:Random Access CodesRandom Access Codes

Compressing Information?Compressing Information?• Assume we are trying to store n (random) bits into n/8 Assume we are trying to store n (random) bits into n/8

bits or qubitsbits or qubits

• Recovering all of the n original bits is ‘clearly’ Recovering all of the n original bits is ‘clearly’ impossibleimpossible

• The best success probability is obtained by storing, The best success probability is obtained by storing, say, the first n/8 bits and is only 2say, the first n/8 bits and is only 2--(n)(n)

• Proving this is easy, both in the classical and Proving this is easy, both in the classical and quantum casesquantum cases

1 ?0 ? ? ? ? ? ? ? ? ? ? ? ? ?nn

n/8n/8

Random Access CodesRandom Access Codes• But assume we wish to recover only But assume we wish to recover only 1 bit1 bit of the original of the original

n bits. Such a primitive is called a random access code n bits. Such a primitive is called a random access code (RAC).(RAC).

• ‘‘Clearly’ impossible classically… what happens Clearly’ impossible classically… what happens quantumly?quantumly?

• More formally:More formally:

• A RAC is a function f:{0,1}A RAC is a function f:{0,1}nnRR22n/8n/8 mapping each mapping each

xx{0,1}{0,1}nn to a probability distribution on n/8 bits, to a probability distribution on n/8 bits,

with the property that for all iwith the property that for all i{1,…,n} {1,…,n}

• Using entropy-based arguments one can show that Using entropy-based arguments one can show that RACs don’t exist RACs don’t exist [[AmbainisNayakTa-ShmaVazirani99, AmbainisNayakTa-ShmaVazirani99, Nayak99]Nayak99]

• Quantum entropy behaves a lot like classical Quantum entropy behaves a lot like classical entropy, so same proof applies also for quantum entropy, so same proof applies also for quantum RACRAC

k-out-of-n Random Access k-out-of-n Random Access CodesCodes• Now assume we wish to recover some arbitrary Now assume we wish to recover some arbitrary

k bits of x (think of k=logn)k bits of x (think of k=logn)• One would expect the success probability to One would expect the success probability to

behave like 2behave like 2--(k)(k)

• Entropy-based arguments no longer work!Entropy-based arguments no longer work!• For instance, consider the encoding that For instance, consider the encoding that

given xgiven x{0,1}{0,1}nn outputs x with probability outputs x with probability 10% and 000…0 with probability 90%. Then 10% and 000…0 with probability 90%. Then it has low entropy (roughly 0.1n) yet we can it has low entropy (roughly 0.1n) yet we can recover all of x prefectly with probability recover all of x prefectly with probability 10%10%

• We therefore have to use the fact that the We therefore have to use the fact that the dimensiondimension of the encoding is low (2 of the encoding is low (2n/8n/8))

n/8n/8

1 ?0 ? ? ? ? ? ? ? ? ? ? ? ? ?nn

k-out-of-n Random Access k-out-of-n Random Access CodesCodes

ThmThm: For any k-out-of-n quantum random access : For any k-out-of-n quantum random access code on n/8 qubits, the success probability is 2code on n/8 qubits, the success probability is 2--

(k)(k)..

RemarkRemark: The classical case can be proven by ‘brute-force’: The classical case can be proven by ‘brute-force’

Proof:Proof:• For simplicity, let’s prove the classical k=1 For simplicity, let’s prove the classical k=1

case case • The quantum case is identical (using The quantum case is identical (using

matrices instead of vectors)matrices instead of vectors)• k>1 case is similark>1 case is similar

• Recall that the RAC is described by a functionRecall that the RAC is described by a function

f:{0,1}f:{0,1}nnRR22n/8n/8

• Let us apply the inequality to fLet us apply the inequality to f

k-out-of-n Random Access k-out-of-n Random Access CodesCodes

• Since f(x) is a probability distribution, we haveSince f(x) is a probability distribution, we have

therefore the RHS is at most 1therefore the RHS is at most 1• The LHS is at leastThe LHS is at least

k-out-of-n Random Access k-out-of-n Random Access CodesCodes

• By rearranging, we getBy rearranging, we get

• Choosing p=1+4/n yieldsChoosing p=1+4/n yields

in contradiction.in contradiction.

Application 2:Application 2:Communication Communication

ComplexityComplexity

Application 2:Application 2:Communication Communication

ComplexityComplexity

Direct product theorem for one-way Direct product theorem for one-way quantum communication quantum communication

complexitycomplexity

• Consider the Disjointness function:Consider the Disjointness function:• Alice and Bob are each given a subset of {1,Alice and Bob are each given a subset of {1,

…,n} and need to decide whether their …,n} and need to decide whether their subsets are disjointsubsets are disjoint

• A naïve one-way protocol requires n bits of A naïve one-way protocol requires n bits of one-way communication (Alice just sends her one-way communication (Alice just sends her subset)subset)• This is essentially optimal (even quantumly)This is essentially optimal (even quantumly)

• We show that if Alice and Bob try to solve k We show that if Alice and Bob try to solve k independent instances of the problem with independent instances of the problem with less than kn/2 (qu)bits of one-way less than kn/2 (qu)bits of one-way communication, then their success probability communication, then their success probability is 2is 2--(k)(k)

AliceAlice BobBob

Open QuestionsOpen Questions

• Find other applications of the inequalityFind other applications of the inequality