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Mar 26, 2015

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Limitations of Quantum Advice and One-Way Communication Scott Aaronson UC Berkeley IAS Useful? Slide 2 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) Slide 3 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? Slide 4 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 Slide 5 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! Slide 6 Slide 7 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 Slide 8 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 Slide 9 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 Slide 10 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. Slide 11 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? Slide 12 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)| . 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)| . y1y1 y2y2 Slide 13 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 Slide 14 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 Slide 15 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)) Slide 16 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) 0 1 012..... k N Slide 17 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) 0 1 012... k..... 2n2n |S| Theorem: Given the above, (Improved by Klauck et al.) Slide 18 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! Slide 19 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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