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

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

Scott Aaronson

University of Waterloo

The Story

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

One-Way

xi

The Story

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…

One-Way

Merlin

One-Way

xi

The Story

x{0,1}N i{1,…,N}

Alice Bob

On the plus side: Merlin knows both x1…xn and i

Merlin

xi

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?

One-Way One-Way

On the minus side: He’s a lying weasel

Not hard to prove that this is classically impossible

The Story

x{0,1}N i{1,…,N}

Alice Bob Merlin

xi

One-Way One-Way

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

Nwa

2log1

Application to Quantum Advice

A., 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

BQP/qpoly: Class of problems solvable efficiently by a quantum computer with help from polynomial-size “quantum advice states”

Ran Raz’s curveball: QIP/qpoly = ALL

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 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

Oded Regev: What about QMA/qpoly? Is that also equal to ALL? Or can you upper-bound it by (say) PP/poly?

A few months later, I had my answer:QMA/qpoly PSPACE/poly

A few months later, I had my answer:QMA/qpoly PSPACE/poly

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

Plan of Attack

QMA/qpoly

BQPSPACE/qpoly

PostBQPSPACE/poly

PSPACE/poly

Main difficulty of proof(Why doesn’t it follow trivially from QMAPSPACE??)

Similar to my result that BQP/qpolyPostBQP/poly

Similar to Watrous’s result that BQPSPACE=PSPACE

Warmup: The Classical Case

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.

xi

Warmup: The Classical Case

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).

xi

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!

12

z

z

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

|

And Alice can’t afford to amplify her message exponentially many times

Quantum OR BoundLet 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 TN/2 uniformly random basis states in sequence: |j1,…,|jTHN

(reusing the same advice | throughout)

.

2

T

N

Theorem: C| will accept at least one of the basisstates with probability at least

But couldn’t the measurements

destroy |?

Sure. But that can only mean one of the

measurements has already accepted with non-

negligible probability

QMA/qpoly BQPSPACE/qpolySimulation 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?

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

Yes, if either the advice or the witness is classical

BQP/poly = BQP/rpoly

QCMA/poly = QCMA/rpoly

QMA/poly = QMA/rpoly

PP/poly = PostBQP/poly

BQP/qpoly

QCMA/qpoly

QMA/qpoly

PSPACE/poly = PSPACE/rpoly

PP/rpoly = IP(2)/rpoly = ALL

Open Problems

Is the Quantum Advice Hypothesis true? What about for QMA(2) (QMA with two unentangled yes-provers)?

Is QMA/qpoly PP/poly?

Can we tighten the de-Merlinization result from a(w+1)=(N/log2N) to a(w+1)=(N)?