Avraham Ben-Aroya (Tel Aviv University) Oded Regev (Tel Aviv University) Ronald de Wolf (CWI, Amsterdam) A Hypercontractive Inequality for Matrix-Valued Functions with Applications to Quantum Computing

Outline The new hypercontractive inequality The new hypercontractive inequality Main application: k-out-of-n random access codes Main application: k-out-of-n random access codes Another application: direct product theorem for one-way communication complexity Another application: direct product theorem for one-way communication complexity

The New Inequality

The Parallelogram Law For any two vectors a,b  R d, For any two vectors a,b  R d, Or equivalently, Or equivalently, a b a+b a-b

a b a+b a-b The Parallelogram Law This was for the  2 norm This was for the  2 norm What happens in the p norm, for 1  p<2? What happens in the p norm, for 1  p<2? The equality no longer holds, take, e.g., a=(1,0),b=(0,1) and p=1 The equality no longer holds, take, e.g., a=(1,0),b=(0,1) and p=1 But, we have the following powerful inequality for all a,b  R d and 1  p  2: But, we have the following powerful inequality for all a,b  R d and 1  p  2:

The Extended Parallelogram Law This inequality was proven by [Tomczak-Jaegermann74, BallCarlenLieb94] This inequality was proven by [Tomczak-Jaegermann74, BallCarlenLieb94] Originally used to prove the ‘sharp uniform convexity’ of p spaces Originally used to prove the ‘sharp uniform convexity’ of p spaces Implies the Bonami-Beckner hypercontractive inequality Implies the Bonami-Beckner hypercontractive inequality An extremely useful inequality in computer science (analysis of Boolean functions, hardness of approximation, learning theory, communication complexity, percolation, etc.) An extremely useful inequality in computer science (analysis of Boolean functions, hardness of approximation, learning theory, communication complexity, percolation, etc.) Recently used by [LeeNaor04] to prove a lower bound on the distortion of embeddings into 1 spaces Recently used by [LeeNaor04] to prove a lower bound on the distortion of embeddings into 1 spaces Amazingly, the same inequality also holds with a,b being matrices and norms being matrix p-norms (i.e., Schatten p- norms) Amazingly, the same inequality also holds with a,b being matrices and norms being matrix p-norms (i.e., Schatten p- norms)

Prelims: Fourier Transform Let f be a function from {0,1} n to R d (or ℂ d×d ) Let f be a function from {0,1} n to R d (or ℂ d×d ) Then we define its Fourier transform as Then we define its Fourier transform as So, e.g., So, e.g.,

The New Hypercontractive Ineq. Thm: For any vector- or matrix-valued f on {0,1} n and 1  p  2, Thm: For any vector- or matrix-valued f on {0,1} n and 1  p  2, Remark: This is the extension of the Bonami- Beckner inequality to vector/matrix-valued functions Remark: This is the extension of the Bonami- Beckner inequality to vector/matrix-valued functions

The New Hypercontractive Ineq. Thm: For any vector- or matrix-valued f on {0,1} n and 1  p  2, Thm: For any vector- or matrix-valued f on {0,1} n and 1  p  2, Proof: By induction on n. Proof: By induction on n. The case n=1 is exactly the [BCL94] inequality with a=f(0), b=f(1) The case n=1 is exactly the [BCL94] inequality with a=f(0), b=f(1) For simplicity, let’s see how to get the n=2 case. For simplicity, let’s see how to get the n=2 case. This involves four matrices, a=f(00), b=f(01), c=f(10), d=f(11) This involves four matrices, a=f(00), b=f(01), c=f(10), d=f(11)

The New Inequality (cont.) Using the case n=1 we get Using the case n=1 we get By averaging the two inequalities, we get By averaging the two inequalities, we get

The New Inequality (cont.) Using the case n=1 again, the left side is at least Using the case n=1 again, the left side is at least

Application 1: Random Access Codes Application 1: Random Access Codes

Compressing Information? Assume we are trying to store n (random) bits into n/8 bits or qubits Assume we are trying to store n (random) bits into n/8 bits or qubits Recovering all of the n original bits is ‘clearly’ impossible Recovering all of the n original bits is ‘clearly’ impossible The best success probability is obtained by storing, say, the first n/8 bits and is only 2 -  (n) The best success probability is obtained by storing, say, the first n/8 bits and is only 2 -  (n) Proving this is easy, both in the classical and quantum cases Proving this is easy, both in the classical and quantum cases 1?0????????????? nn/8

Random Access Codes But assume we wish to recover only 1 bit of the original n bits. Such a primitive is called a random access code (RAC). But assume we wish to recover only 1 bit of the original n bits. Such a primitive is called a random access code (RAC). ‘Clearly’ impossible classically… what happens quantumly? ‘Clearly’ impossible classically… what happens quantumly? More formally: More formally: A RAC is a function f:{0,1} n  R 2 n/8 mapping each x  {0,1} n to a probability distribution on n/8 bits, A RAC is a function f:{0,1} n  R 2 n/8 mapping each x  {0,1} n to a probability distribution on n/8 bits, with the property that for all i  {1,…,n} Using entropy-based arguments one can show that RACs don’t exist [AmbainisNayakTa-ShmaVazirani99, Nayak99] Using entropy-based arguments one can show that RACs don’t exist [AmbainisNayakTa-ShmaVazirani99, Nayak99] Quantum entropy behaves a lot like classical entropy, so same proof applies also for quantum RAC Quantum entropy behaves a lot like classical entropy, so same proof applies also for quantum RAC

k-out-of-n Random Access Codes Now assume we wish to recover some arbitrary k bits of x (think of k=logn) Now assume we wish to recover some arbitrary k bits of x (think of k=logn) One would expect the success probability to behave like 2 -  (k) One would expect the success probability to behave like 2 -  (k) Entropy-based arguments no longer work! Entropy-based arguments no longer work! For instance, consider the encoding that given x  {0,1} n outputs x with probability 10% and 000…0 with probability 90%. Then it has low entropy (roughly 0.1n) yet we can recover all of x prefectly with probability 10% For instance, consider the encoding that given x  {0,1} n outputs x with probability 10% and 000…0 with probability 90%. Then it has low entropy (roughly 0.1n) yet we can recover all of x prefectly with probability 10% We therefore have to use the fact that the dimension of the encoding is low (2 n/8 ) We therefore have to use the fact that the dimension of the encoding is low (2 n/8 )n/8 1?0????????????? n

k-out-of-n Random Access Codes Thm: For any k-out-of-n quantum random access code on n/8 qubits, the success probability is 2 -  (k). Remark: The classical case can be proven by ‘brute-force’ Proof: For simplicity, let’s prove the classical k=1 case For simplicity, let’s prove the classical k=1 case The quantum case is identical (using matrices instead of vectors) The quantum case is identical (using matrices instead of vectors) k>1 case is similar k>1 case is similar Recall that the RAC is described by a function Recall that the RAC is described by a function f:{0,1} n  R 2 n/8 Let us apply the inequality to f Let us apply the inequality to f

k-out-of-n Random Access Codes Since f(x) is a probability distribution, we have Since f(x) is a probability distribution, we have therefore the RHS is at most 1 The LHS is at least The LHS is at least

k-out-of-n Random Access Codes By rearranging, we get By rearranging, we get Choosing p=1+4/n yields Choosing p=1+4/n yields in contradiction.

Application 2: Communication Complexity Application 2: Communication Complexity

Direct product theorem for one-way quantum communication complexity Consider the Disjointness function: Consider the Disjointness function: Alice and Bob are each given a subset of {1,…,n} and need to decide whether their subsets are disjoint Alice and Bob are each given a subset of {1,…,n} and need to decide whether their subsets are disjoint A naïve one-way protocol requires n bits of one- way communication (Alice just sends her subset) A naïve one-way protocol requires n bits of one- way communication (Alice just sends her subset) This is essentially optimal (even quantumly) This is essentially optimal (even quantumly) We show that if Alice and Bob try to solve k independent instances of the problem with less than kn/2 (qu)bits of one-way communication, then their success probability is 2 -  (k) We show that if Alice and Bob try to solve k independent instances of the problem with less than kn/2 (qu)bits of one-way communication, then their success probability is 2 -  (k) AliceBob

Open Questions Find other applications of the inequality Find other applications of the inequality