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

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

Examples Good novels dont just put their characters in random situationsthey repeatedly subject the characters to crucial tests that reveal aspects of their personalities we didnt already know (I guess)

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, NP NP This would be almost as shocking as if P=NP! If pigs could whistle, then donkeys could fly

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 n k, 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?

But why should we care? CIRCUIT LOWER BOUNDS

Theorem (Kannan 1982): For every k, there exists a language in NP NP that does not have circuits of size n k Proof: Its not hard to show that doesnt have circuits of size n k So either NP doesnt have circuits of size n k, in which case NP NP doesnt either, or NP does have circuits of size n k, in which case and we win again!

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 ZPP NP, provided we can somehow compute f (ZPP: Zero-Error Probabilistic Polynomial-Time) Idea: Iterative learning. Repeatedly find an input x t such that, among the circuits that correctly compute f on x 1,…,x t-1, at least a 1/3 fraction get x t wrong This process cant continue for long! Corollary: ZPP NP does not have circuits of size n k

But what about quantum anthropic computing?

PostBQP 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 I hereby define a new complexity class… (Postselected BQP)

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

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

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

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 doesnt seem to be self- reduciblewe dont know how to construct | given an oracle for QMA problems But we can construct | in PostBQP. (Why?)

Quantum Advice 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 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?

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

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

Alices Classical Advice Bob, suppose you used the maximally mixed state in place of your quantum advice. Then x 1 is the lexicographically first input for which youd output the right answer with probability less than ½. But suppose you succeeded on x 1, and used the resulting reduced state as your advice. Then x 2 is the lexicographically first input after x 1 for which youd output the right answer with probability less than ½... x1x1 x2x2 Given an input x, clearly lets Bob decide in PostBQP whether x L

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) n ; 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 2 p(n) -1 orthogonal states Alice needs to specify at most p(n) inputs x 1,x 2,…, since each one cuts Bobs total success probability by at least half, but the probability must be at least ~2 -p(n) by the end

P PP Does Not Have Quantum Circuits of Size n k Does U accept x 0 w.p. ½? If yes, set x 0 L If no, set x 0 L U : Picks a size-n k quantum circuit uniformly at random and runs it x0x0 x1x1 x2x2 x3x3 x4x4 x5x5 Conditioned on deciding x 0 correctly, does U accept x 1 w.p. ½? If yes, set x 1 L If no, set x 1 L Conditioned on deciding x 0 and x 1 correctly, does U accept x 2 w.p. ½? If yes, set x 2 L If no, set x 2 L

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

Quantum Karp-Lipton Theorem If PP BQP/qpoly, then the counting hierarchyconsisting of etc.collapses to PP Also: PP does not have quantum circuits of size n k PEXP requires quantum circuits of size f(n), where f(f(n)) 2 n

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 cant 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)

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>2 n-1 Fromwe can easily prepare Goal: Decide if these amplitudes have the same or opposite signs Prepare |0 | + |1 H| for some,. Then postselect on second qubit being |1 Yieldsin first qubit

To Show PP PostBQP… Yieldsin first qubit Suppose s and 2 n -2s are both positive Then by trying / = 2 i for all i {-n,…,n}, well eventually get close to On the other hand, if 2 n -2s is negative, then we wont. QED

Beigel, Reingold, Spielman 1990: PP is closed under intersection Solved a problem that was open for 18 years… Other classical results proved with quantum techniques: Kerenidis & de Wolf, A., Aharonov & Regev, … Observation: PostBQP is trivially closed under intersection PP is too Given L 1,L 2 PostBQP, to decide if x L 1 and x L 2, postselect on both computations succeeding, and accept iff they both accept