QMA/qpoly  PSPACE/poly: De-Merlinizing Quantum Protocols

1. QMA/qpoly  PSPACE/poly: De-Merlinizing Quantum Protocols Scott Aaronson University of Waterloo

2. The Story xi One-Way x{0,1}N i{1,…,N} Alice 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

3. One-Way Merlin The Story xi One-Way x{0,1}N i{1,…,N} Alice Bob 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…

4. The Story xi One-Way One-Way x{0,1}N i{1,…,N} Alice 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

5. 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 The Story xi One-Way One-Way x{0,1}N i{1,…,N} Alice Bob Merlin

6. Ran Raz’s curveball:QIP/qpoly = ALL Application to Quantum Advice BQP/qpoly: Class of problems solvable efficiently by a quantum computer with help from polynomial-size “quantum advice states” • , CCC 2004: BQP/qpoly  PostBQP/poly = PP/poly • Seemed to place a strong limit on quantum advice… Raz’s result actually has nothing to do with quantum mechanics, since IP/rpoly = ALL as well

7. Oded Regev: What about QMA/qpoly? Is that also equal to ALL? Or can you upper-bound it by (say) PP/poly? Where’s The Phase Transition?(the point in the complexity hierarchy where quantum advice starts acting like exponentially-long classical advice) QMA/qpoly: Class of languages L for which there exists a poly-time quantum verifier V, together with poly-size quantum advice states {|n}, such that for all x{0,1}n: (1) If xL then there exists a poly-size quantum witness | such that V accepts |x|n| w.p.  2/3 (2) If xL, then for all purported witnesses |, V rejects |x|n| w.p.  2/3 A few months later, I had my answer:QMA/qpoly  PSPACE/poly

8. The Quantum Advice Hypothesis: For any “natural” complexity class C, if C/qpoly=ALL, then C/rpoly=ALL as well “Sure, quantum advice is a weird resource, but so is classical randomized advice!” Four Confirming Instances So Far: 1. BQP/qpoly  PP/poly, BQP/rpoly = BQP/poly2. QIP/qpoly = QIP/rpoly = ALL3. PostBQP/qpoly = PostBQP/rpoly = ALL4. QMA/qpoly  PSPACE/poly, QMA/rpoly = QMA/poly

9. PSPACE/poly PostBQPSPACE/poly Plan of Attack Similar to Watrous’s result that BQPSPACE=PSPACE Similar to my result that BQP/qpolyPostBQP/poly BQPSPACE/qpoly Main difficulty of proof(Why doesn’t it follow trivially from QMAPSPACE??) QMA/qpoly

10. Warmup: The Classical Case xi x{0,1}N i{1,…,N} Claim: For all awN, 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.

11. 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).

12. Solution: Bob will detect | by looking for the “shadows” it casts on computational basis states | 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

13. But couldn’t the measurements destroy |? Sure. But that can only mean one of the measurements has already accepted with non-negligible probability Theorem:C| will accept at least one of the basisstates with probability at least Quantum OR Bound Let C| be a quantum verifier that takes | as advice Let |HN be a witness that C| accepts with probability at least . Suppose that, instead of feeding | to C|, we feed it TN/2 uniformly random basis states in sequence: |j1,…,|jTHN(reusing the same advice | throughout)

14. 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

15. 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

16. PP/rpoly = IP(2)/rpoly = ALL PSPACE/poly = PSPACE/rpoly PP/poly = PostBQP/poly QMA/qpoly QMA/poly = QMA/rpoly QCMA/qpoly QCMA/poly = QCMA/rpoly BQP/qpoly BQP/poly = BQP/rpoly

17. 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?