Scott Aaronson Institut pour l' É tude Avançée

1. Le Principe de la Postselection Scott Aaronson Institut pour l'Étude Avançée

2. Could you ever learn enough about a person to predict his or her future behavior reliably?

3. Examples Good novels don’t just put their characters in random situations—they repeatedly subject the characters to “crucial tests” that reveal aspects of their personalities we didn’t already know (I guess)

5. The Karp-Lipton Theorem (1982) Suppose NP-complete problems were solvable in polynomial time, but only nonuniformly—that is, with polynomial-size circuits Then we could use those circuits to collapse the Polynomial-Time Hierarchy down to the seocnd level, NPNP This would be almost as shocking as if P=NP! “If pigs could whistle, then donkeys could fly”

6. We want to exploit a small circuit for solving NP-complete problems… but all we know is that it exists! Does there exist a circuit C of size nk, such that for all Boolean formulas  of size n, C correctly decides whether  is satisfiable, and C outputs “yes” on whatever problem we wanted to solve originally?

7. CIRCUIT LOWER BOUNDS But why should we care?

8. Proof: It’s not hard to show that doesn’t have circuits of size nk • So either • NP doesn’t have circuits of size nk, in which case NPNP doesn’t either, or • NP does have circuits of size nk, in which case • and we win again! Theorem (Kannan 1982): For every k, there exists a language in NPNP that does not have circuits of size nk

9. Corollary: ZPPNP does not have circuits of size nk Bshouty et al.’s Improvement (1994) If a function f:{0,1}n{0,1} has a polynomial-size circuit, then we can find the circuit in ZPPNP, provided we can somehow compute f (ZPP: Zero-Error Probabilistic Polynomial-Time) Idea: Iterative learning. Repeatedly find an input xt such that, among the circuits that correctly compute f on x1,…,xt-1, at least a 1/3 fraction get xt wrong This process can’t continue for long!

10. But what about quantum anthropic computing?

11. PostBQP I hereby define a newcomplexity class… (Postselected BQP) Class of languages decidable by a bounded-error polynomial-time quantum computer, if at any time you can measure a qubit that has a nonzero probability of being |1, and assume the outcome will be |1

12. Another Important Animal: PP Class of languages decidable by a nondeterministic poly-time Turing machine that accepts iff the majority of its paths do PSPACE P#P=PPP PP NP P

13. Theorem (A., 2004):PostBQP = PP Unexpectedly, this theorem turned out to have an implication for classical complexity: the simplest known proof of the “Beigel-Reingold-Spielman Theorem,” that PP is closed under intersection

14. But a key fact about quantum mechanics is that, given any orthogonal basis of n-qubit quantum states, we could just as well write In as the uniform distribution over that basis Detour The “maximally mixed state” In is just the uniform distribution over n-bit strings

15. Quantum Proofs QMA (defined by Kitaev and Watrous) is the quantum version of NP: “Does there exist a quantum state | accepted by such-and-such a circuit with high probability?” Unlike NP, QMA doesn’t seem to be “self-reducible”—we don’t know how to construct | given an oracle for QMA problems But we can construct | in PostBQP. (Why?)

16. Quantum Advice Mike & Ike:“We know that many systems in Nature ‘prefer’ to sit in highly entangled states of many systems; might it be possible to exploit this preference to obtain extra computational power?” BQP/qpoly: Class of languages decidable by polynomial-size, bounded-error quantum circuits, given a polynomial-size quantum advice state |n that depends only on the input length n

17. How powerful is quantum advice? Could it let us solve problems that are not even computable given classical advice of similar size?!

18. Closely related: for all (partial or total) Boolean functions f : {0,1}n {0,1}m {0,1}, Limitations of Quantum Advice NP  BQP/qpoly relative to an oracle(Uses direct product theorem for quantum search) BQP/qpoly  PostBQP/poly ( = PP/poly)

19. Given an input x, clearly lets Bob decide in PostBQP whether xL Alice’s Classical Advice Bob, suppose you used the maximally mixed state in place of your quantum advice. Then x1 is the lexicographically first input for which you’d output the right answer with probability less than ½. But suppose you succeeded on x1, and used the resulting reduced state as your advice. Then x2 is the lexicographically first input after x1 for which you’d output the right answer with probability less than ½... x1 x2

20. But how many inputs must Alice specify? We can boost a quantum advice state so that the error probability on any input is at most (say) 2-100n; then Bob can reuse the advice on as many inputs as he likes We can decompose the maximally mixed state on p(n) qubits as the boosted advice plus 2p(n)-1 orthogonal states Alice needs to specify at most p(n) inputs x1,x2,…, since each one cuts Bob’s total success probability by at least half, but the probability must be at least ~2-p(n) by the end

21. U: Picks a size-nk quantum circuit uniformly at random and runs it x0 x1 x2 x3 x4 x5 PPP Does Not Have Quantum Circuits of Size nk   Does U accept x0 w.p.  ½?If yes, set x0LIf no, set x0L Conditioned on deciding x0 correctly, does U accept x1 w.p.  ½?If yes, set x1LIf no, set x1L Conditioned on deciding x0 and x1 correctly, does U accept x2 w.p.  ½?If yes, set x2LIf no, set x2L

22. Why? Intuitively, each iteration cuts the number of potential circuits in half, but there were at most circuits to begin with Even works for quantum circuits with quantum advice! For any k, defines a language L that does not have quantum circuits of size nk On the other hand, clearly L  PPP

23. If PP  BQP/qpoly, then the counting hierarchy—consisting ofetc.—collapses to PP Quantum Karp-Lipton Theorem Also: PP does not have quantum circuits of size nk PEXP requires quantum circuits of size f(n), where f(f(n))2n

24. Concluding Thought: What Makes Science Possible? That which we can observe, we can understand That which we can observe, and then observe in a new situation where we can’t predict what it will do even given the earlier observation, and so on for a polynomial number of steps, we can understand (provided we can postselect a description consistent with our observations)

25. Yields in first qubit From we can easily prepare To Show PP  PostBQP… Given a Boolean function f:{0,1}n{0,1},let s=|{x : f(x)=1}|. Need to decide if s>2n-1 Goal: Decide if these amplitudes have the same or opposite signs Prepare |0|+|1H| for some ,.Then postselect on second qubit being |1

26. Suppose s and 2n-2s are both positive Then by trying / = 2i for all i{-n,…,n}, we’ll eventually get close to Yields in first qubit To Show PP  PostBQP… On the other hand, if 2n-2s is negative, then we won’t. QED

27. Beigel, Reingold, Spielman 1990: PP is “closed under intersection”Solved a problem that was open for 18 years… Observation: PostBQP is trivially closed under intersection  PP is too Given L1,L2PostBQP, to decide if xL1 and xL2, postselect on both computations succeeding, and accept iff they both accept Other classical results proved with quantum techniques: Kerenidis & de Wolf, A., Aharonov & Regev, …