Limitations of Quantum Advice and One-Way Communication

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

2. 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 What Are Quantum States?   We give complexity-theoretic evidence that quantum states lie to the left end of this spectrumSupplements information-theoretic evidence (e.g. Holevo)

3. 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

4. Not known to be in BQP/poly Solvable in BQP/qpoly using the advice state Idea: Check whether Hn|xHn is 1 or 0 Example (Watrous) For each n, fix a group Gn and subgroup HnGn(|Gn|2n, but group operations are polytime) Given an element xGn as input, is xHn?

5. 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!

7. Corollary: Can’t show BQP/poly  BQP/qpoly without also showing PP  P/poly 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),D1(f) = O(m Q1(f) logQ1(f))D1(f) = deterministic 1-way communication complexity of fQ1(f) = bounded-error quantum 1-way complexity

8. 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 #2 NPA BQPA/qpolyfor some oracle A (actually, a random oracle)

9. Result #3(Won’t say any more about this one) Ambainis: Suppose Alice has x,yFp and Bob has a,bFp. They want to know whether yax+b.1-way quantum communication complexity? Theorem: Alice must send (log p) qubits to Bob Invented new “trace distance method” to show this Previously, even randomized complexity was unknown Alice’s point Bob’s line

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. 

11. f(x,y1)f(x,y2) x = maximally mixed state? D1(f) = O(m Q1(f) logQ1(f)) for allf : {0,1}n{0,1}m {0,1} x y1,y2,… f(x,y) Alice 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

12. Alice’s Classical Message 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. y1 y2

13. 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

14. Improves earlier result:BQP/qpoly  EXP/poly BQP/qpoly  PP/poly Bob is the PP algorithm Alice is the “advisor” 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 advice (as before) Adleman, DeMarrais, Huang: In PP, we can decide which of two sequences of measurement outcomes has greater probability

15. NPA BQPA/qpoly Oracle: A(x)=1 iff xS, where S  {0,1}n is chosen uniformly at random subject to |S|=2n/10 Language: (y,z)LA iff there exists an xS between y and z lexicographically (clearly LANPA) Claim: If LABQPA/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))

16. 1 p(k) 0 . . . . . . . . . . 0 1 2 N k 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 INTUITIVELY PLAUSIBLE 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

17. 1 p(k) 0 . . . . . . . . 0 1 2 |S| 2n k Theorem: Given the above, 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} (Improved by Klauck et al.)

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2n. On the other hand, one can show by induction on m that r(m)  2-O(poly(n))/m!

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) ?