A Theory of Isolatability

1. A Theory of Isolatability Scott Aaronson Andrew Drucker MIT

2. Freeze-Dried Computation Motivating Question: How much useful computational work can one “store” in (say) an n-qubit quantum state, or a coin whose bias is an arbitrary real number? Potentially a huge amount! We give a new tool—called “isolatability”—for ruling out the possibility of such extravagant encodings.

3. Idea: Take some advice resource (such as a coin or a quantum state), and simulate it using a short classical string, together with a polynomial number of untrusted advice resources. In other words, all the relevant information in the advice resource gets packed into an ordinary string (which we say “isolates” the resource), and we're left with just a computational search problem—of finding coins, quantum states, etc. that are consistent with the string.

4. Part I: The Majority-Certificates Lemma Our basic tool Part II: Application to Quantum Advice BQP/qpolyQMA/poly Part III: Application to Advice Coins PSPACE/coinPSPACE/poly

5. Part I: The Majority-Certificates Lemma

6. Intuition: We’re given a black box 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?

7. 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 O(log|S|), such that • Some fiS is consistent with each Ci, and • If f1S is consistent with C1, f2S is consistent with C2, and so on, then MAJ(f1,…,fm)=f*, where MAJ denotes pointwise majority.

8. 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.

9. 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.

10. Part II: Application to Quantum Advice

11. 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

12. 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

13. 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”) “Physics” Implication Let ρ be any quantum state on n qubits. Then for all m,ε, there exists a 2-local Hamiltonian H=H1+⋯+HL on poly(n,m,1/ε) qubits, such that any ground state |φ of H can be used to simulate ρ (with error ε) on all quantum circuits of size at most m. In other words, there exists an efficient mapping C→C′ such that for all circuits C of size m,

14. 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

15. 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

16. 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

17. “Lifting” the Majority-Certificates Lemma

18. Majority-Certificates Lemma, Real Case Lemma: Let S be a set of functions f:{0,1}ⁿ→[0,1], let f∗∈S, and let ε>0. Then we can find m=O(n/ε²) functions f1,…,fm∈S, sets X1,…,Xm⊆{0,1}ⁿ each of size and for which the following holds. All functions g1,…,gm∈S that satisfy for all i[m] also satisfy where

19. 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 the real version 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.

20. 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).

21. Part III: Application to Advice Coins

22. Erik Demaine(motivated by a computational genetics problem): “Suppose a PSPACE machine can flip a coin with Bernoulli probability p an unlimited number of times. Can it extract an exponential amount of information (or even more) about p?” Me: “I’m sure whatever the answer is, it’s obvious...” Didn’t seem too likely there could be superpowerful “Advice Coins”

23. Indeed, Hellman & Cover proved the following in 1970... Suppose a finite automaton M is trying to decide whether a coin has p=½ or p=½+. Then even if it can flip the coin an unlimited number of times, M needs (1/) states to succeed with probability (say) 2/3. This result seems to imply PSPACE/coinPSPACE/poly, since we could take the first poly(n) bits of p as the advice. But it breaks down if p is close to 0 or 1!

24. Furthermore, quantum mechanics nullifies the Hellman-Cover Theorem! Halt with probability ~2/100 at each time step Expected difference in final angle after halting, in p=½ vs. p=½+ cases: 1 radian Standard deviation in angle: Keep flipping the coin. Whenever the coin lands heads, rotate /100 radians counterclockwise. Whenever it lands tails, rotate /100 radians clockwise. Theorem:For any >0, it’s possible to distinguish a coin with p=½ from a coin with p=½+ using a single qubit of memory, with error probability independent of .

25. Proof:Suffices to show BQPSPACE/coin  PSPACE/poly. Let 0 = quantum operation applied to our memory qubits whenever coin lands heads, 1 = operation applied when it lands tails Then induced operation at each time step: Theorem:Despite these obstacles, BQPSPACE/coin = PSPACE/coin = PSPACE/poly. We’re interested in a fixed-point of p: a mixed state p such that

26. Fixed-Points of Superoperators Already studied by[A.-Watrous 2008], in the context of quantum computing with closed timelike curves Our result there:BQPCTC = PCTC = PSPACE Quantum computers with CTCs have exactly the same power as classical computers with CTCs, namely PSPACE (or: “CTCs make time and space equivalent as computational resources”) 0.000000000000000110101111101

27. Key Lemma:Let a(p) be the probability that an S-state quantum finite automaton accepts, if each input bit is 1 with independent probability p. Then a(p) is a degree-S2rational function of p. Proof Idea: We can write a(p) as Furthermore, each entry of p can be written as a degree-S2 rational function of p, by using Cramer’s Rule on S2S2 matrices.

28. Hence a(p) is itself a degree-S2 rational function of p, except possibly at a finite number of singularities: ax(p) p A further continuity argument rules out the singularities, except possibly at p=0 and p=1. Now, by calculus, a degree-S2rational function can cross the line y=½ at most 2S2 times…

29. Given a BQPSPACE/coin machine M, let ax(p) be its acceptance probability on input x{0,1}n and a coin with Bernoulli probability p. Challenge: How can we describe p well enough to compute ax(p) for every x, using only poly(n) bits? ax(p) p

30. Finishing the Argument (Sketch) We’ve upper-bounded how many distinct Boolean functions f:{0,1}n{0,1} can be expressed, as we vary p from 0 to 1. So by the Majority-Certificates Lemma, we can simulate PSPACE/coin using a PSPACE/poly machine, combined with poly(n) untrusted advice coins. We then get a computational search problem: finding coin biases p1,…,pm that are consistent with the /poly advice string. We can solve that problem in PSPACE, using NC algorithms for root-finding developed by Neff and Pan

31. When Can An O(1)-State Finite Automaton Detect an  Change to the Bias of a Coin?

32. Open Problems Find other applications of isolatability Circuit complexity? Communication complexity? Learning theory? Quantum information? Optimality of the Majority-Certificates Lemma? Quantum finite automata: are their limiting acceptance probabilities continuous functions of p, for p(0,1)? Prove a classical oracle separation between BQP/poly and BQP/qpoly=YQP/poly

33. 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