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The Power of Quantum Advice Scott Aaronson Andrew Drucker.

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1 The Power of Quantum Advice Scott Aaronson Andrew Drucker

2 Freeze-Dried Computation Motivating Question: How much useful computational work can one store in a quantum state, for later retrieval? If quantum states are exponentially large objects, then possibly a huge amount! Yet we also know that quantum states have no more general- purpose storage capacity than classical strings of the same size

3 Cast of Characters BQP/qpoly is the class of problems solvable in quantum polynomial time, with the help of polynomial-size quantum advice states Formally: a language L is in BQP/qpoly if there exists a polynomial time quantum algorithm A, as well as quantum advice states {| n } n on poly(n) qubits, such that for every input x of size n, A(x,| n ) decides whether or not x L with error probability at most 1/3 YQP (Yoda Quantum Polynomial-Time) is the same, except we also require that for every alleged advice state, A(x, ) outputs either the right answer or FAIL with probability at least 2/3 BQP YQP QMA BQP/qpoly

4 Watrous 2000: For any fixed, finite black-box group G n and subgroup H n G n, deciding membership in H n is in BQP/qpoly The quantum advice state is just an equal superposition |H n over the elements of H n We dont know how to solve the same problem in BQP/poly A. 2004: BQP/qpoly PP/poly = PostBQP/poly Quantum advice can be simulated by classical advice, combined with postselection on unlikely measurement outcomes A. 2006: HeurBQP/qpoly = HeurYQP/poly Trusted quantum advice can be simulated on most inputs by trusted classical advice combined with untrusted quantum advice A.-Kuperberg 2007: There exists a quantum oracle separating BQP/qpoly from BQP/poly Q UANTUM ADVICE IS POWERFUL N O I T I SN T

5 New Result: BQP/qpoly = YQP/poly Trusted quantum advice is equivalent in power to trusted classical advice combined with untrusted quantum advice. (Quantum states never need to be trusted) Given an n-qubit state and parameters m,, there exists a local Hamiltonian H on poly(n,m,1/ ) qubits (e.g., a sum of 2- qubit interactions) for which the following holds: For any ground state | of H, and any binary measurement E on performed by a circuit with m gates, theres an efficient measurement f(E) that we can perform on | such that F OR THE P HYSICISTS

6 What Does It Mean? Preparing quantum advice states is no harder than preparing ground states of local Hamiltonians This explains a once-mysterious relationship between quantum proofs and quantum advice: efficient preparability of ground states would imply both QMA=QCMA and BQP/qpoly=BQP/poly Quantum Karp-Lipton Theorem: NP-complete problems are not efficiently solvable using quantum advice, unless some uniform complexity classes collapse unexpectedly QCMA/qpoly QMA/poly: classical proofs and quantum advice can be simulated with quantum proofs and classical advice

7 BQP YQPQCMABQP/poly BQP/qpoly =YQP/poly QCMA/polyQMA QCMA/qpoly QMA/polyPP PP/polyQMA/qpoly PSPACE/poly A.06 This work

8 Majority- Certificates Lemma Real Majority- Certificates Lemma Circuit Learning (Bshouty et al.) Minimax Theorem Safe Winnowing Lemma Holevos Theorem Random Access Code Lower Bound (Ambainis et al.) BQP/qpoly=YQP/poly HeurBQP/qpoly=HeurYQP/poly (A.06) Quantum advice no harder than ground state preparation Fat-Shattering Bound (A.06) Covering Lemma (Alon et al.) Learning of p- Concept Classes (Bartlett & Long) L OCAL H AMILTONIANS is QMA-complete (Kitaev) Cook-Levin Theorem QMA=QMA+ (Aharonov & Regev) Used as lemma Generalizes

9 that computes some Boolean function f:{0,1} n {0,1} belonging to a small set S (meaning, of size 2 poly(n) ). Someone wants to prove to us that f equals (say) the all-0 function, by having us check a polynomial number of outputs f(x 1 ),…,f(x m ). Intuition: Were given a black box (think: quantum state) f xf(x) This is trivially impossible! f0f0 f1f1 f2f2 f3f3 f4f4 f5f5 x1x1 010000 x2x2 001000 x3x3 000100 x4x4 000010 x5x5 000001 But … what if we get 3 black boxes, and are allowed to simulate f=f 0 by taking the point-wise MAJORITY of their outputs?

10 Majority-Certificates Lemma Lemma: Let S be a set of Boolean functions f:{0,1} n {0,1}, and let f * S. Then there exist m=O(n) certificates C 1,…,C m, each of size k=O(log|S|), such that (i)Some f i S is consistent with each C i, and (ii)If f i S is consistent with C i for all i, then MAJ(f 1 (x),…,f m (x))=f * (x) for all x {0,1} n. Definitions: A certificate is a partial Boolean function C:{0,1} n {0,1,*}. A Boolean function f:{0,1} n {0,1} is consistent with C, if f(x)=C(x) whenever C(x) {0,1}. The size of C is the number of inputs x such that C(x) {0,1}.

11 Proof Idea By symmetry, we can assume f * is the all-0 function. Consider a two-player, zero-sum matrix game: Alice picks a certificate C of size k consistent with some f S Bob picks an input x {0,1} n Alice wins this game if f(x)=0 for all f S consistent with C. Crucial Claim: Alice has a mixed strategy that lets her win >90% of the time. The lemma follows from this claim! Just choose certificates C 1,…,C m independently from Alices winning distribution. Then by a Chernoff bound, almost certainly MAJ(f 1 (x),…,f m (x))=0 for all f 1,…,f m consistent with C 1,…,C m respectively and all inputs x {0,1} n. So clearly there exist C 1,…,C m with this property.

12 Proof of Claim Use the Minimax Theorem! Given a distribution D over x, its enough to create a fixed certificate C such that Stage I: Choose x 1,…,x t independently from D, for some t=O(log|S|). Then with high probability, requiring f(x 1 )=…=f(x t )=0 kills off every f S such that Stage II: Repeatedly add a constraint f(x i )=b i that kills at least half the remaining functions. After log 2 |S| iterations, well have winnowed S down to just a single function f S.

13 Lifting the Lemma to Quantumland Boolean Majority-CertificatesBQP/qpoly=YQP/poly Proof Set S of Boolean functionsSet S of p(n)-qubit mixed states True function f * STrue advice state | n Other functions f 1,…,f m Other states 1,…, m Certificate C i to isolate f i Measurement E i to isolate I New DifficultySolution The class of p(n)-qubit quantum states is infinitely large! And even if we discretize it, its still doubly-exponentially large Result of A.06 on learnability of quantum states (building on Ambainis et al. 1999) Instead of Boolean functions f:{0,1} n {0,1}, now we have real functions f :{0,1} n [0,1] representing the expectation values Learning theory has tools to deal with this: fat-shattering dimension, -covers… (Alon et al. 1997) How do we verify a quantum witness without destroying it? QMA=QMA+ (Aharonov & Regev 2003) What if a certificate asks us to verify Tr(E )a, but Tr(E ) is right at the knife-edge? Safe Winnowing Lemma

14 Theorem: BQP/qpoly = YQP/poly. Proof Sketch: YQP/poly BQP/qpoly is immediate. For the other direction, let L BQP/qpoly. Let M be a quantum algorithm that decides L using advice state | n. Define Let S = {f : }. Then S has fat-shattering dimension at most poly(n), by A.06. So we can apply a real analogue of the Majority-Certificates Lemma to S. This yields certificates C 1,…,C m (for some m=poly(n)), such that any states 1,…, m consistent with C 1,…,C m respectively satisfy for all x {0,1} n (regardless of entanglement). To check the C i s, we use the QMA+ super-verifier of Aharonov & Regev.

15 Promised Application to Physics Furthermore, in their reduction, the witness is a history state So given any language L BQP/qpoly=YQP/poly, we can use the Kitaev et al. reduction to get a local Hamiltonian H whose unique ground state is |. We can then use | to recover the YQP witness |, and thereby decide L By Kitaev et al., we know L OCAL H AMILTONIANS is QMA-complete. Measuring this state yields the original QMA witness | 1 with (1/poly(n)) probability. Hence | 1 can be recovered from

16 Quantum Karp-Lipton Theorem Our quantum analogue: If NP BQP/qpoly, then coNP NP QMA PromiseQMA. Karp-Lipton 1982: If NP P/poly, then coNP NP = NP NP. Proof Idea: A coNP NP statement has the form x y R(x,y). By the hypothesis and BQP/qpoly = YQP/poly, there exists an advice string s, such that any quantum state consistent with s lets us solve NP problems (and some such is consistent). In QMA PromiseQMA, first guess an s thats consistent with some state. Then use the oracle to search for an x and such that, if is consistent with s, then R(x,Q(x, )) holds, where Q is a quantum algorithm that searches for a y such that R(x,y).

17 A Theory of Isolatability Which classes of functions C are isolatablein the sense that for any f C, one can give a small number of conditions such that any f 1,…,f m C satisfying the conditions can be used to compute f efficiently on all inputs? We can generalize the majority-certificates idea well beyond what we have any application for Another application of the Majority-Certificates Lemma: it substantially simplifies the proof that BQPSPACE/coin = PSPACE/poly We study the following abstract question, inspired by computational learning theory:

18 Open Problems Improve QMA/qpoly PSPACE/poly to QMA/qpoly P #P /poly Find other applications of the majority-certificates technique Circuit complexity? Communication complexity? Learning theory? Quantum information? Is the dependence on n, log|S|, and 1/ optimal? Prove a classical oracle separation between BQP/poly and BQP/qpoly=YQP/poly Although this work closes off a chapter in the quantum advice story, there are still

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