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)?
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