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

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

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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)

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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

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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?

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But why should we care? CIRCUIT LOWER BOUNDS

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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!

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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

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But what about quantum anthropic computing?

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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)

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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

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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

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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

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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?)

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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?

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

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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},

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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

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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 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

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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

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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!

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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

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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)

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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

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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

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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

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