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The Cryptographic Hardness of Decoding Hawking Radiation Scott Aaronson (MIT)

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Black Holes and Computational Complexity?? YES! Amazing connection made last year by Harlow & Hayden But first, let’s review 40 years of black hole history SZK QSZK BPP BQP AM QAM

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Bekenstein, Hawking 1970s: Black holes have entropy and temperature! They emit radiation The Information Loss Problem: Calculations suggest that Hawking radiation is thermal—uncorrelated with whatever fell in. So, is infalling information lost forever? Would violate the unitarity / reversibility of QM OK then, assume the information somehow gets out! The Xeroxing Problem: How could the same qubit | fall inexorably toward the singularity, and emerge in Hawking radiation? Would violate the No-Cloning Theorem Black Hole Complementarity (Susskind, ‘t Hooft): An external observer can describe everything unitarily without including the interior at all! Interior should be seen as “just a scrambled re-encoding” of the exterior degrees of freedom

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Violates monogamy of entanglement! The same qubit can’t be maximally entangled with 2 things The Firewall Paradox (AMPS 2012) B = Interior of “Old” Black Hole R = Faraway Hawking Radiation H = Just-Emitted Hawking Radiation Near-maximal entanglement Also near-maximal entanglement

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Harlow-Hayden 2013 (arXiv: ): Striking argument that Alice’s first task, decoding the entanglement between R and H, would require exponential time Complexity theory to the rescue of quantum field theory?? Two obvious questions: (1) Who cares if this is true? (2) Is it true? Does the decoding task require exponential time?

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Caveats of Complexity Arguments 1.Asymptotic E.g., 8 8 chess takes O(1) time! Only for n n chess can we give evidence of hardness. But for black holes, n … 2.(Usually) Conjectural Right now, we can’t even prove P≠NP! To get where we want, we almost always need to make assumptions. Question is, which assumptions? 3.Worst-Case We can argue that a natural formalization of Alice’s decoding task is “generically” hard. We can’t rule out that a future quantum gravity theory would make her task easy, for deep reasons not captured by our formalization.

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Quantum Circuits

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Given a description of a quantum circuit C, such that Promised that, by acting only on R (the “Hawking radiation part”), it’s possible to distill an EPR pair between R and H Problem: Distill such an EPR pair, by applying a unitary transformation U R to the qubits in R The HH Decoding Problem

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Problem: That would require waiting until the black hole was fully evaporated ( no more firewall problem) When the BH is “merely” >50% evaporated, we know from Page’s argument that “generically,” there will exist a U R that distills an EPR pair between R and B But interestingly, Page’s argument doesn’t suggest any efficient procedure to find U R or apply it! Isn’t the Decoding Task Trivial? Just invert C!

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Set Equality: Given two efficiently-computable injective functions f,g:{0,1} n {0,1} p(n). Promised that Range(f) and Range(g) are either equal or disjoint. Decide which. In the “black-box” setting, this problem requires at least ~2 n/3 steps, even with a QC (A Zhandry 2013). For explicit f,g, we can’t prove unconditional hardness, but solving it would give Graph Isomorphism, SZK… Theorem (Harlow-Hayden): Suppose there’s a polynomial-time quantum algorithm for HH decoding. Then there’s also a polynomial-time quantum algorithm for Set Equality (and indeed, QSZK=BQP) The HH Hardness Result

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Intuition: If Range(f) and Range(g) are disjoint, then the B register decoheres all entanglement between R and H, leaving only classical correlation If, on the other hand, Range(f)=Range(g), then there’s some permutation of the |x,1 R states that puts the last qubit of R into an EPR pair with H Thus, if we had a reliable way to distill EPR pairs whenever possible, then we could also decide Set Equality The HH Construction (easy to prepare in poly(n) time given f,g)

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My Alternative Hardness Result Theorem (A. 2013): Suppose there’s a poly-time quantum algorithm for the HH decoding problem. Then there’s also a poly-time quantum algorithm to invert any injective OWF One-Way Function (OWF): A collection of functions f:{0,1} n {0,1} p(n) (one for each n) such that: 1.f(x) is computable in poly(n) time 2.For all polynomial-time adversaries A, “Standard workhorses” of modern cryptography. Widely believed that there exist OWFs secure even against QCs

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Suppose applying U R to R decodes an EPR pair between R and H. Then for some states {| x } x, we must have So from U R, we can get unitaries V,W such that My Construction (again, easy to prepare in poly(n) time given f)

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Generalization to Arbitrary OWFs Known Classical Fact: Given any OWF f, it’s possible to produce another OWF g that, with probability at least ~1/n, is injective on at least a ~1/n fraction of its range Now H is many qubits, not just one But a more complicated argument shows that, if we can distill even 1 EPR pair between R and H, we must be able to invert g, and hence f

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Comparison of Arguments Advantages of my argument: Existence of OWFs is a “safer” assumption than hardness of finding collisions Works even against “nonuniform” algorithms (which spend exponential preparation time before the black hole is formed) Advantage of HH argument: Works even if Alice gets access to H as well as R

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Concluding Remarks There’s good evidence that decoding Hawking radiation requires an exponentially-long quantum computation Admittedly, “real” black holes won’t produce states that look anything like But intuitively, greater genericity seems like it should only make Alice’s decoding task harder! Would be great to formalize this (connections to quantum money?) Would also be great if computational considerations could give any clues about the black hole interior…

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A Curious Open Problem We’ve seen what computational powers are necessary for Alice to solve the HH decoding problem (inverting one-way functions, solving Set Equality) What computational powers would suffice to let her solve it in quantum polynomial time? (Not obvious how to do it even if given, say, an oracle for the halting problem…)

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