Download presentation

Presentation is loading. Please wait.

Published byAlexa White Modified over 4 years ago

1
Quantum Versus Classical Proofs and Advice Scott Aaronson Waterloo MIT Greg Kuperberg UC Davis | x {0,1} n ?

2
Can quantum proofs let us verify certain theorems exponentially faster than classical proofs? Yes (we think!) But to argue for the power of quantum proofs, well have to introduce a new kind of evidence: Quantum Oracle Separations (Its not just that we failed to find the old kind of evidencewe can tell you exactly why we failed)

3
QMA: Quantum Merlin-Arthur Class of problems for which a yes answer can be verified in quantum polynomial-time, with help from a polynomial-size quantum witness state Schrödingers Zoo QCMA: Quantum Classical Merlin-Arthur Same, except now the witness has to be classical BQP/qpoly: Class of problems solvable in quantum polynomial time, with help from a quantum advice state | n that depends only on the input length n BQP/poly: Same, except now advice has to be classical Closely related to quantum proofs is quantum advice…

4
BQP QCMA QMA BQP/poly BQP/qpoly PP PP/poly MAP/poly Surely it should at least be easy to separate these classes by oracles… Dream on!

5
Theorem: There exist quantum oracles U and V such that QMA U QCMA U and BQP V /qpoly BQP V /poly Quantum oracle: A sequence of unitary transformations {U n } that a quantum algorithm can apply in a black-box fashion Models subroutines that take quantum input and produce quantum output A new kind of evidence that two complexity classes are different Idea has already found other applications in quantum computing [A07] [MS07] This Talk: Quantum Oracle Separations

6
Choose an n-qubit state | uniformly at random Let U be the unitary that maps | |0 to | |1, and | |0 to | |0 whenever | =0 Problem: Given oracle access to U, decide whether (YES) U=U for some, or (NO) U=I is the identity transformation Clearly this problem is in QMA U (The witness: | itself) Claim: The problem is not in QCMA U The Oracle Problem Well Use

7
Intuition: Not much! Underlying Question: How much does an n k -bit classical hint help in searching for an unknown 2 n -dimensional unit vector? advice regions 2 n -dimensional unit sphere

8
Let be a probability measure over N-dimensional unit vectors Call p-uniform if it can be obtained by starting from the uniform measure, and then conditioning on an event that occurs with probability p Lemma: If is p-uniform, then for every fixed quantum state |, To prove the intuition, we need a geometric lemma…

9
Intuition: Best you can do is let be the uniform measure over the fraction p of states that are closest to | |

10
Theorem: Suppose were given oracle access to an n-qubit unitary U, and want to decide whether (i) U=I is the identity operator, or (ii) U=U for some secret marked state |. Then even if were given an m-bit classical witness in support of case (ii), we still need Lower Bound queries to U to verify the witness. Proof uses BBBV hybrid argument Quantum oracles relative to which QMA U QCMA U and BQP U /qpoly BQP U /poly now follow by standard arguments

11
Theorem: We can find an n-qubit marked state | using an m-bit classical hint, together with Almost-Matching Upper Bound queries to the quantum oracle U. (Provided m 2n) Idea: A mesh of 2 m states. Merlin tells Arthur the state closest to |, which Arthur then uses as a starting point for Grovers algorithm

12
But What About A Classical Oracle Separation Between QMA and QCMA? Weve had essentially one candidate problem for this: Group Non-Membership (Babai) Problem: Given a group G, a subgroup H G, and an element x G, is x H? Here G and H are specified as black-box groups I.e. every x G is labeled by a meaningless string s(x), and were given an oracle that maps s(x) and s(y) to s(xy) and s(x -1 )

13
Watrous (2000) showed how to solve GNM in QMA, using the state as a witness Group Non-Membership (as an oracle problem) is known to be in AM but outside MA Our result: Arthur can verify x H using (1) a polynomial-size classical witness from Merlin, and (2) polynomially many quantum queries to the group oracle (but possibly an exponential amount of computation) Conclusion: The Group Non-Membership problem cannot, alas, lead to an oracle separation between QMA and QCMA.

14
Idea: Pull the group out of the black box Explicit group Black-box group Isomorphism claimed by Merlin

15
Merlin gives Arthur an explicit group, together with a claimed isomorphism f: G (defined by its action on generators) Arthur checks that f is a homomorphism using the BCLR tester He checks that f is one-to-one by solving an instance of the Hidden Subgroup Problem (f is one-to-one kernel of f is trivial) Ettinger-Høyer-Knill: Hidden Subgroup Problem has polynomial quantum query complexity Once weve replaced G by an explicit group, no more queries to the group oracle are needed

16
Can we prove a classical oracle separation between QMA and QCMA? Bigger question: Whenever we prove a quantum oracle separation, can we also prove a classical one? Is Group Non-Membership in QCMA? (I.e. is the computational complexity polynomial, in addition to the query complexity?) Other quantum oracle separations? QMA vs. QMA(2) Open Problems

Similar presentations

OK

Oracles Are Subtle But Not Malicious Scott Aaronson University of Waterloo.

Oracles Are Subtle But Not Malicious Scott Aaronson University of Waterloo.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google