The Power of Quantum Advice

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