1 / 18

The Power of Quantum Advice

The Power of Quantum Advice. Scott Aaronson Andrew Drucker. 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!.

egil
Download Presentation

The Power of Quantum Advice

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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. Quantum advice is powerful Watrous 2000: For any fixed, finite black-box group Gn and subgroup Hn≤Gn, deciding membership in Hn is in BQP/qpolyThe quantum advice state is just an equal superposition |Hn over the elements of Hn We don’t know how to solve the same problem in BQP/poly A.-Kuperberg 2007: There exists a “quantum oracle” separating BQP/qpolyfromBQP/poly No It Isn’t 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/polyTrusted quantum advice can be simulated on most inputs by trusted classical advice combined with untrusted quantum advice

  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”) For the Physicists 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, there’s an efficient measurement f(E) that we can perform on | such that

  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. PSPACE/poly A.’06 QMA/qpoly PP/poly QMA/poly PP This work QCMA/qpoly BQP/qpoly=YQP/poly QCMA/poly QMA BQP/poly YQP QCMA BQP

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

  9. Intuition: We’re given a black box (think: quantum state) f x f(x) that computes some Boolean function f:{0,1}n{0,1} belonging to a “small” set S (meaning, of size 2poly(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(x1),…,f(xm). This is trivially impossible! But … what if we get 3 black boxes, and are allowed to simulate f=f0 by taking the point-wise MAJORITY of their outputs?

  10. Majority-Certificates Lemma 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}. • 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 C1,…,Cm, each of size k=O(log|S|), such that • Some fiS is consistent with each Ci, and • If fiS is consistent with Ci for all i, then MAJ(f1(x),…,fm(x))=f*(x) for all x{0,1}n.

  11. Proof Idea By symmetry, we can assume f* is the all-0 function. Consider a two-player, zero-sum matrix game: Bob picks an input x{0,1}n The lemma follows from this claim! Just choose certificates C1,…,Cm independently from Alice’s winning distribution. Then by a Chernoff bound, almost certainly MAJ(f1(x),…,fm(x))=0 for all f1,…,fm consistent with C1,…,Cm respectively and all inputs x{0,1}n. So clearly there exist C1,…,Cm with this property. Alice picks a certificate C of size k consistent with some fS 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.

  12. Proof of Claim Use the Minimax Theorem! Given a distribution D over x, it’s enough to create a fixed certificate C such that Stage I: Choose x1,…,xt independently from D, for some t=O(log|S|). Then with high probability, requiring f(x1)=…=f(xt)=0 kills off every fS such that Stage II: Repeatedly add a constraint f(xi)=bi that kills at least half the remaining functions. After ≤ log2|S| iterations, we’ll have winnowed S down to just a single function fS.

  13. “Lifting” the Lemma to Quantumland

  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 C1,…,Cm (for some m=poly(n)), such that any states 1,…,m consistent with C1,…,Cm respectively satisfy for all x{0,1}n (regardless of entanglement). To check the Ci’s, we use the “QMA+ super-verifier” of Aharonov & Regev.

  15. Promised Application to “Physics” By Kitaev et al., we know Local Hamiltonians is QMA-complete. Furthermore, in their reduction, the witness is a “history state” Measuring this state yields the original QMA witness |1 with (1/poly(n)) probability. Hence |1 can be recovered from 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

  16. Quantum Karp-Lipton Theorem Karp-Lipton 1982: If NPP/poly, then coNPNP = NPNP. Our quantum analogue: If NPBQP/qpoly, then coNPNPQMAPromiseQMA. Proof Idea:A coNPNP 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 QMAPromiseQMA, first guess an s that’s 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 We can generalize the majority-certificates idea well beyond what we have any application for We study the following abstract question, inspired by computational learning theory: Which classes of functions C are “isolatable”—in the sense that for any fC, one can give a small number of conditions such that any f1,…,fmC satisfying the conditions can be used to compute f efficiently on all inputs? Another application of the Majority-Certificates Lemma: it substantially simplifies the proof thatBQPSPACE/coin = PSPACE/poly

  18. Although this work closes off a chapter in the quantum advice story, there are still Open Problems 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? Improve QMA/qpolyPSPACE/poly to QMA/qpolyP#P/poly Prove a classical oracle separation between BQP/poly and BQP/qpoly=YQP/poly

More Related