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QMA/qpoly PSPACE/poly: De-Merlinizing Quantum Protocols Scott Aaronson University of Waterloo The Story xi One-Way x{0,1}N Alice i{1,…,N} Bob Bob, a grad student, has a thesis problem i{1,…,N} Alice, Bob’s omniscient advisor, knows the binary answer xi to every thesis problem i But she’s too busy to find out which specific problems her students are working on So instead, she just doles out the same generic advice ax to all of them The Story xi One-Way One-Way x{0,1}N Alice i{1,…,N} Bob Merlin Clearly ax needs to be (N) bits long, for Bob to be able to learn xi with probability 2/3 for any i Ambainis et al., Nayak: Indeed, this is true even if Alice can send a quantum message |x So in desperation, Bob turns for help to Merlin, the star student in his department… The Story xi One-Way One-Way x{0,1}N Alice i{1,…,N} Bob Merlin On the plus side: Merlin knows both x1…xn and i On the minus side: He’s a lying weasel Can Bob play Alice’s vague but reliable advice against Merlin’s specific but unreliable witness, to learn xi using polylog(N) bits from both? Not hard to prove that this is classically impossible The Story xi One-Way One-Way x{0,1}N Alice i{1,…,N} Bob Merlin Main Result: Even in the quantum case, if Alice sends a qubits and Merlin sends w qubits, for Bob to learn xi w.h.p. we need N aw 1 2 log N Application to Quantum Advice BQP/qpoly: Class of problems solvable efficiently by a quantum computer with help from polynomial-size “quantum advice states” A., CCC 2004: BQP/qpoly PostBQP/poly = PP/poly Seemed to place a strong limit on quantum advice… Ran Raz’s curveball: QIP/qpoly = ALL Raz’s result actually has nothing to do with quantum mechanics, since IP/rpoly = ALL as well Where’s The Phase Transition? (the point in the complexity hierarchy where quantum advice starts acting like exponentially-long classical advice) Oded Regev: What about QMA/qpoly? Is that also equal to ALL? Or can you upper-bound it by (say) PP/poly? QMA/qpoly: Class of languages L for which there exists a poly-time quantum verifier V, together with poly-size n: A few months later, I had my answer: quantum advice states {| }, such that for all x{0,1} n QMA/qpoly PSPACE/poly (1) If xL then there exists a poly-size quantum witness | such that V accepts |x|n| w.p. 2/3 (2) If xL, then for all purported witnesses |, V rejects |x|n| w.p. 2/3 “Sure, quantum advice is a weird resource, but so is classical randomized advice!” The Quantum Advice For any “natural” complexity class C, if Hypothesis: C/qpoly=ALL, then C/rpoly=ALL as well Four Confirming Instances So Far: 1. BQP/qpoly PP/poly, BQP/rpoly = BQP/poly 2. QIP/qpoly = QIP/rpoly = ALL 3. PostBQP/qpoly = PostBQP/rpoly = ALL 4. QMA/qpoly PSPACE/poly, QMA/rpoly = QMA/poly Plan of Attack PSPACE/poly Similar to Watrous’s result that BQPSPACE=PSPACE PostBQPSPACE/poly Similar to my result that BQP/qpolyPostBQP/poly BQPSPACE/qpoly Main difficulty of proof QMA/qpoly (Why doesn’t it follow trivially from QMAPSPACE??) Warmup: The Classical Case xi x{0,1}N i{1,…,N} Claim: For all awN, there’s a randomized protocol where Alice sends a+O(log N) bits and Merlin sends w bits Proof: Alice divides x into w-bit substrings. She then encodes each one with an error-correcting code, and sends Bob a random k along with the kth bit of each codeword. Merlin sends the substring containing xi. Warmup: The Classical Case xi x{0,1}N i{1,…,N} Claim: The previous protocol is optimal. Proof: Suppose Alice amplifies her a-bit randomized advice O(w+1) times. Then Bob’s error probability becomes 2-w. So Bob no longer needs Merlin—he can just loop over all possible w-bit witnesses. Hence a(w+1)=(N). Trouble in QuantumLand If Bob wants to eliminate Merlin’s w-qubit quantum witness, the number of states he needs to loop through is doubly exponential in w! And Alice can’t afford to amplify her message exponentially many times Solution: Bob will detect | by looking for the “shadows” it casts on computational basis states z z 2 1 | Quantum OR Bound Let C| be a quantum verifier that takes | as advice But couldn’t the Let |HN be a witness that C| accepts with measurements probability at least . destroy |? Suppose that, instead of feeding | to C|, we feed it TN/2 uniformly random basis states in sequence: |j1,…,|jTHN Sure. But that can only (reusing the same advice | throughout) mean one of the Theorem: C| measurements will accept at least has one ofalready the basis 2 accepted states with probability at least with non N . negligible probability T QMA/qpoly BQPSPACE/qpoly Simulation algorithm: Repeatedly choose a random basis state |j, then simulate the QMA machine with | as advice and |j as witness By the quantum OR bound, if there’s a valid witness |, then w.h.p. some iteration will accept And what if there’s no valid witness? To control soundness error, we use an unusual amplification procedure—one that involves amplifying Alice’s message poly(n) times and Merlin’s message only log(n) times Can we get below PSPACE/poly? Yes, if either the advice or the witness is classical Theorem: QMA/rpoly = QMA/poly Idea: First amplify, then find a single random string r that works for all inputs of size n and all quantum witnesses (doubly-exponentially many, but OK) Chicken & egg problem: The more we amplify the witness, the more we need to amplify Solution: In-place amplification [Marriott & Watrous] Theorem: QCMA/qpoly PP/poly PP/rpoly = IP(2)/rpoly = ALL PSPACE/poly = PSPACE/rpoly PP/poly = PostBQP/poly QMA/poly = QMA/rpoly QCMA/poly = QCMA/rpoly BQP/poly = BQP/rpoly QMA/qpoly QCMA/qpoly BQP/qpoly Open Problems Is the Quantum Advice Hypothesis true? What about for QMA(2) (QMA with two unentangled yes-provers)? Can we tighten the de-Merlinization result from a(w+1)=(N/log2N) to a(w+1)=(N)? Is QMA/qpoly PP/poly?