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Limitations of Quantum Advice and One-Way Communication Useful? Scott Aaronson UC Berkeley IAS What Are Quantum States? To many quantum computing skeptics, they’re exponentially long vectors—and 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 2n-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) Quantum Advice 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?” 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 Example (Watrous) For each n, fix a group Gn and subgroup HnGn (|Gn|2n, but group operations are polytime) Given an element xGn as input, is xHn? Solvable in BQP/qpoly using the advice state 1 to h H nNot known H n hH n be in BQP/poly Idea: Check whether Hn|xHn is 1 or 0 Obvious Challenge: Prove an oracle separation between BQP/poly and BQP/qpoly Buhrman: Hey Scott—why not try for an unrelativized separation? After all, if quantum states are like 2n-bit classical strings, then maybe BQP/qpoly NEEEEE/poly! Maybe BQP/qpoly even contains NP! Result #1 BQP/qpoly PP/poly Corollary: Can’t show BQP/poly BQP/qpoly without also showing PP P/poly Proof based on new communication result: Given f:{0,1}n{0,1}m{0,1} (partial or total), D1(f) = O(m Q1(f) logQ1(f)) D1(f) = deterministic 1-way communication complexity of f Q1(f) = bounded-error quantum 1-way complexity Result #2 A NP A BQP /qpoly for some oracle A (actually, a random oracle) Proof based on new Direct Product Theorem for quantum search: N items, K of them marked Theorem: With few (N) quantum queries, the probability of finding all K marked items is 2-(K) Fixes a wrong result of Klauck Result #3 (Won’t say any more about this one) Ambainis: Suppose Alice has x,yFp and Bob has a,bFp. They want to know whether yax+b. 1-way quantum communication complexity? Theorem: Alice must send (log p) qubits to Bob Invented new “trace distance method” to show this Alice’s point Bob’s line Previously, even randomized complexity was unknown 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. Then after the measurement, we can recover a ’ such that ' tr D1(f) = O(m Q1(f) logQ1(f)) for all f : {0,1}n{0,1}m {0,1} x Alice logQ f x 1 y1,y2,… f(x,y1) f(x,y) f(x,y2) Bob Alice can decrease the error probability to 1/Q1(f)10, by sending K=O(Q1(f)logQ1(f)) qubits Bob can then compute f(x,y) for Q1(f)2 values of y simultaneously, with probability 0.9 With no communication, he can still do that with probability 0.9/2K, by guessing x=I Alice’s Classical Message y1 y2 Bob, let p0(y) be the probability you’d guess f(x,y)=1 using I in place of x. Then y1 is the lexicographically first y for which |p0(y)-f(x,y)|½. Now let I1 be the reduced state assuming you guessed f(x,y1) correctly. Let p1(y) be the probability you’d guess f(x,y)=1 using I1 in place of x. Then y2 is the first y after y1 for which |p1(y)-f(x,y)|½. Clearly Alice’s message lets Bob compute f(x,y) for any y in his range Claim: Alice never has to send more than K yi’s—so her total message length is O(mK) Suppose not. Then Bob would succeed on y1,…,yK+1 simultaneously with probability 1/2K+1 But we already know he succeeds with probability 0.9/2K, contradiction 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 x1,…,xK{0,1}n, on which algorithm would make the wrong guess using maximally mixed state in place of adviceearlier (as before) Improves result: BQP/qpolyHuang: EXP/poly Adleman, DeMarrais, In PP, we can decide which of two sequences of measurement outcomes has greater probability NPA BQPA/qpoly Oracle: A(x)=1 iff xS, where S {0,1}n is chosen uniformly at random subject to |S|=2n/10 Language: (y,z)LA iff there exists an xS between y and z lexicographically (clearly LANPA) Claim: If LABQPA/qpoly, then using boosted advice, we can find all 2n/10 elements of S w.h.p. using 2n/10poly(n) quantum queries Now replace advice by maximally mixed state. Success probability becomes 2-O(poly(n)) Direct Product Theorem Goal: Show that with o(2n/2) quantum queries, the probability of finding all 2n/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 1 p(k) 0 0 1 2 . . . . . k . . . . . N Have the algorithm accept iff it finds |S|=2n/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,…,2n} 1 p(k) 0 0 1 2 . . . |S| k . . . . . 2n 2n S Theorem: Given the above, deg p poly n (Improved by Klauck et al.) Idea: Let r m dmp maxn . m 0 x 2 dx Then V.A. Markov (younger brother of A.A. Markov) showed in 1892 that r m 3deg p n 2 2 m 2m1 m 1! 2m 1 ! provided -1p(x)2 for all 0x2n. On the other hand, one can show by induction on m that r(m) 2-O(poly(n))/m! 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)