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Limitations of Quantum Advice and One-Way Communication Scott Aaronson UC Berkeley IAS Useful?

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What Are Quantum States? To many quantum computing skeptics, theyre exponentially long vectorsand therefore a bad description of Nature Yet a classical probability distribution over {0,1} n also takes 2 (n) bits to specify! Sure, but each sample is only n bits… Distributions over n-bit strings 2 n -bit strings We give complexity-theoretic evidence that quantum states lie to the left end of this spectrum Supplements information-theoretic evidence (e.g. Holevo)

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Quantum Advice BQP/qpoly: Class of languages decidable by polynomial-size, bounded-error quantum circuits, given a polynomial-size quantum advice state | n that depends only on the input length n Nielsen & Chuang: We know that many systems in Nature prefer to sit in highly entangled states of many systems; might it be possible to exploit this preference to obtain extra computational power?

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Example (Watrous) For each n, fix a group G n and subgroup H n G n (|G n | 2 n, but group operations are polytime) Given an element x G n as input, is x H n ? Solvable in BQP/qpoly using the advice state Idea: Check whether H n |xH n is 1 or 0 Not known to be in BQP/poly

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Maybe BQP/qpoly even contains NP! Obvious Challenge: Prove an oracle separation between BQP/poly and BQP/qpoly Buhrman: Hey Scottwhy not try for an unrelativized separation? After all, if quantum states are like 2 n -bit classical strings, then maybe BQP/qpoly NEEEEE/poly!

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Result #1 BQP/qpoly PP/poly Proof based on new communication result: Given f:{0,1} n {0,1} m {0,1} (partial or total), D 1 (f) = O(m Q 1 (f) logQ 1 (f)) D 1 (f) = deterministic 1-way communication complexity of f Q 1 (f) = bounded-error quantum 1-way complexity Corollary: Cant show BQP/poly BQP/qpoly without also showing PP P/poly

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Result #2 NP A BQP A /qpoly for some oracle A (actually, a random oracle) Proof based on new Direct Product Theorem for quantum search: Theorem: With few ( N) quantum queries, the probability of finding all K marked items is 2 - (K) Fixes a wrong result of Klauck N items, K of them marked

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Result #3 (Wont say any more about this one) Ambainis: Suppose Alice has x,y F p and Bob has a,b F p. They want to know whether y ax+b. 1-way quantum communication complexity? Alices point Bobs line Theorem: Alice must send (log p) qubits to Bob Invented new trace distance method to show this Previously, even randomized complexity was unknown

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Then after the measurement, we can recover a such that The Almost As Good As New Lemma Suppose a 2-outcome measurement of a mixed state yields 0 w.p. 1- and 1 w.p.

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D 1 (f) = O(m Q 1 (f) logQ 1 (f)) for all f : {0,1} n {0,1} m {0,1} xy 1,y 2,… f(x,y) BobAlice Alice can decrease the error probability to 1/Q 1 (f) 10, by sending K=O(Q 1 (f)logQ 1 (f)) qubits Bob can then compute f(x,y) for Q 1 (f) 2 values of y simultaneously, with probability 0.9 With no communication, he can still do that with probability 0.9/2 K, by guessing x =I f(x,y 1 ) f(x,y 2 ) x = maximally mixed state?

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Alices Classical Message Bob, let p 0 (y) be the probability youd guess f(x,y)=1 using I in place of x. Then y 1 is the lexicographically first y for which |p 0 (y)-f(x,y)| ½. Now let I 1 be the reduced state assuming you guessed f(x,y 1 ) correctly. Let p 1 (y) be the probability youd guess f(x,y)=1 using I 1 in place of x. Then y 2 is the first y after y 1 for which |p 1 (y)-f(x,y)| ½. y1y1 y2y2

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Clearly Alices message lets Bob compute f(x,y) for any y in his range Claim: Alice never has to send more than K y i sso her total message length is O(mK) Suppose not. Then Bob would succeed on y 1,…,y K+1 simultaneously with probability 1/2 K+1 But we already know he succeeds with probability 0.9/2 K, contradiction

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BQP/qpoly PP/poly Alice is the advisor Bob is the PP algorithm Suppose quantum advice has p(n) qubits. Then classical advice consists of K = O(p(n) log p(n)) inputs x 1,…,x K {0,1} n, on which algorithm would make the wrong guess using maximally mixed state in place of advice (as before) Adleman, DeMarrais, Huang: In PP, we can decide which of two sequences of measurement outcomes has greater probability Improves earlier result: BQP/qpoly EXP/poly

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NP A BQP A /qpoly Claim: If L A BQP A /qpoly, then using boosted advice, we can find all 2 n/10 elements of S w.h.p. using 2 n/10 poly(n) quantum queries Oracle: A(x)=1 iff x S, where S {0,1} n is chosen uniformly at random subject to |S|=2 n/10 Language: (y,z) L A iff there exists an x S between y and z lexicographically (clearly L A NP A ) Now replace advice by maximally mixed state. Success probability becomes 2 -O(poly(n))

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Direct Product Theorem Goal: Show that with o(2 n/2 ) quantum queries, the probability of finding all 2 n/10 marked items must be doubly exponentially small in n Beals et al: If a quantum algorithm makes T queries to X {0,1} N, then the probability it accepts a random X with |X|=k is a univariate polynomial p(k) of degree 2T INTUITIVELY PLAUSIBLE p(k) k N

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Have the algorithm accept iff it finds |S|=2 n/10 marked items. Then (1) p(k)=0 for all k {0,…,|S|–1} (2) p(|S|) = 2 -O(poly(n)) (3) p(k) [0,1] for all k {0,…,2 n } p(k) k n2n |S| Theorem: Given the above, (Improved by Klauck et al.)

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Idea: Let Then V.A. Markov (younger brother of A.A. Markov) showed in 1892 that provided -1 p(x) 2 for all 0 x 2 n. On the other hand, one can show by induction on m that r (m) 2 -O(poly(n)) /m!

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Open Questions Can we show BQP/poly BQP/qpoly relative to an oracle? What about SZK BQP/qpoly? Are randomized and quantum 1-way communication complexities polynomially related for all total Boolean functions? (No asymptotic gap is known) ? ? ? ?

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