Avraham Ben-Aroya

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A Hypercontractive Inequality for Matrix-Valued
Functions with Applications to Quantum Computing
Avraham Ben-Aroya (Tel Aviv University) Oded
Regev (Tel Aviv University) Ronald de
Wolf (CWI, Amsterdam)
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Outline
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The New Inequality
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The Parallelogram Law
ab
a-b
b
a

• For any two vectors a,b?Rd,
• Or equivalently,

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The Parallelogram Law
ab
a-b
b
a

• This was for the ??2 norm
• What happens in the ?p norm, for 1?plt2?
• The equality no longer holds, take, e.g.,
  a(1,0),b(0,1) and p1
• But, we have the following powerful inequality
  for all a,b?Rd and 1?p?2

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The Extended Parallelogram Law

• This inequality was proven by Tomczak-Jaegermann7
  4, BallCarlenLieb94
• Originally used to prove the sharp uniform
  convexity of ?p spaces
• 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.)
• 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)

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Prelims Fourier Transform

• Let f be a function from 0,1n to Rd (or Cdd)
• Then we define its Fourier transform as
• So, e.g.,

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The New Hypercontractive Ineq.

• Thm For any vector- or matrix-valued f on 0,1n
  and 1?p?2,
• Remark This is the extension of the
  Bonami-Beckner inequality to vector/matrix-valued
  functions

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The New Hypercontractive Ineq.

• Thm For any vector- or matrix-valued f on 0,1n
  and 1?p?2,
• Proof By induction on n.
• The case n1 is exactly the BCL94 inequality
  with af(0), bf(1)
• For simplicity, lets see how to get the n2
  case.
• This involves four matrices, af(00), bf(01),
  cf(10), df(11)

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The New Inequality (cont.)

• Using the case n1 we get

• By averaging the two inequalities, we get

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The New Inequality (cont.)

• Using the case n1 again, the left side is at
  least

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Application 1 Random Access Codes
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Compressing Information?

• 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
• 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

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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).
• Clearly impossible classically what happens
  quantumly?
• More formally
• A RAC is a function f0,1n?R2n/8 mapping each
  x?0,1n 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 dont exist AmbainisNayakTa-ShmaVazirani99,
  Nayak99
• Quantum entropy behaves a lot like classical
  entropy, so same proof applies also for quantum
  RAC

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k-out-of-n Random Access Codes

• Now assume we wish to recover some arbitrary k
  bits of x (think of klogn)
• One would expect the success probability to
  behave like 2-?(k)
• Entropy-based arguments no longer work!
• For instance, consider the encoding that given
  x?0,1n outputs x with probability 10 and 0000
  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 (2n/8)

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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, lets prove the classical k1
  case
• The quantum case is identical (using matrices
  instead of vectors)
• kgt1 case is similar
• Recall that the RAC is described by a function
• f0,1n?R2n/8
• Let us apply the inequality to f

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k-out-of-n Random Access Codes

• Since f(x) is a probability distribution, we have
• therefore the RHS is at most 1
• The LHS is at least

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k-out-of-n Random Access Codes

• By rearranging, we get
• Choosing p14/n yields
• in contradiction.

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Application 2 Communication Complexity
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Direct product theorem for one-way quantum
communication complexity
Alice
Bob

• Consider the Disjointness function
• 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)
• 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)

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Open Questions

• Find other applications of the inequality