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QMA/qpoly  PSPACE/poly: De-Merlinizing Quantum Protocols. Scott Aaronson University of Waterloo. The Story. x i. One-Way. x{0,1} N. i{1,…,N}. Alice. Bob. Bob, a grad student, has a thesis problem i {1,…,N}

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slide2

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

slide3

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…

slide4

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

slide5

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

slide6

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

slide7

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

slide8

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

slide9

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

slide10

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.

slide11

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).

slide12

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

slide13

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)

slide14

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

slide15

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

slide16

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

slide17

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?