RMT and boson computers

Transcription

1 RMT and boson computers [Aaronson-Arkhipov 2011] John Napp May 11, 2016

2 Introduction Simple type of quantum computer proposed in 2011 by Aaronson and Arkhipov based on the statistics of noninteracting bosons The computer A outputs a sample from a probability distribution D A They prove that if there exists a classical algorithm that can efficiently output a sample from a distribution close to D A, then P #P = BPP NP : a drastic consequence for complexity theory! Among the strongest evidence to date that quantum computers have capabilities beyond classical computers. Proof relies on random matrix techniques, and requires two unproven RMT conjectures

3 Complexity Preliminaries Definition (#P) A function f : {0, 1} N is in #P if there exists a polynomial p : N N and a polynomial-time TM M such that for every x {0, 1} : { f (x) = y {0, 1} p( x ) : M(x, y) = 1} Theorem (Valiant) The following problem is #P-complete: given a matrix X {0, 1} n n, compute Per(X). Theorem (Aaronson-Arkhipov) The following problem is #P-hard, for any g [1, poly(n)]: given a matrix X R n n, approximate Per(X) 2 to within a multiplicative factor of g.

4 Complexity Preliminaries (2) BPP: class of languages efficiently decided with high probability by a probabilistic TM BPP NP machine: a BPP machine that can also solve NP-complete problems in a single step P #P machine: a P machine that can compute #P-complete functions in a single step Stockmeyer: a BPP NP machine can efficiently estimate the acceptance probability of a BPP machine Widely believed that BPP NP P #P (can only prove BPP NP P #P )

5 Preview: Gaussian Permanent Estimation problems GPE 2 ±: Given as input a matrix X N (0, 1) n n C of iid Gaussians, together with error bounds ε, δ > 0, estimate Per(X) 2 to within additive error ±ε n!, with probability at least 1 δ over X, in poly(n, 1/ε, 1/δ) time. GPE : Given as input a matrix X N (0, 1) n n C of iid Gaussians, together with error bounds ε, δ > 0, estimate Per(X) to within ±ε Per(X), with probability at least 1 δ over X, in poly(n, 1/ε, 1/δ) time.

6 Preview: RMT conjectures Permanent-of-Gaussians Conjecture (PGC): GPE is #P-hard. Permanent Anti-Concentration Conjecture (PACC): There exists a polynomial p such that for all n and δ > 0, [ ] n! Per(X) < < δ p(n, 1/δ) Pr X N (0,1) n n C Theorem: if PACC is true, then GPE and GPE 2 ± are equivalent

7 BosonSampling n photons sent through linear optical network. Can end up in m possible photodetectors, for m n. Description of network encoded by m n column-orthonormal complex matrix A U m,n Output of computer: measurement of how many photons end up in each photodetector. S Φ m,n, where Φ m,n is the set of tuples (s 1,..., s m ) s.t. s i 0 and s i = n By quantum mechanics, output distribution of computer is Pr[S] = Per(A S) 2 D A s 1! s m! where A S is n n matrix constructed by keeping s i copies of row i of A

8 Exact BosonSampling = P #P = BPP NP Assume a classical algorithm O(A, r) for A U m,n and r a string s.t. the distribution of O over r is D A. Then with BPP NP machine, we can compute the squared permanent of an arbitrary real matrix X: a #P-hard problem! Embed scaled X as a submatrix of A U m,n (can prove this is possible) Now a certain output probability is proportional to the squared permanent of X (exponentially small due to scaling during the embedding) Use Stockmeyer result to compute this probability with a BPP NP machine P #P = BPP NP

9 The problem with the above result The classical algorithm O used above was assumed to sample exactly from D A. Not physically reasonable - due to noise, even a boson computer can t sample exactly from D A! To be reasonable, let O sample from some distribution D A s.t. D A D A < ε in variation distance. But this ruins the above result! If O were adversarial and knew where we embedded X in A, it could concentrate its error on the probability corresponding to the permanent of X, and so a BPP NP machine would no longer be able to use O to estimate the squared permanent. Solution: smuggle X into A with the help of RMT, so O has no way of detecting where the embedded X is in A.

10 Haar-Unitary Hiding Theorem H m,n : Haar measure over m n column-orthonormal matrices S m,n : Distribution obtained by drawing U H m,m, and outputting mu n,n G n n : Distribution of complex n n matrices with iid standard complex Gaussian entries Theorem Let m n5 δ log2 n δ, for any δ > 0. Then S m,n G n n = O(δ). m 1/6 m 1/6 truncations of m m unitaries look like iid Gaussians

11 Hiding Lemma Lemma Let m n5 δ log2 n δ for some δ > 0. Then there exists a BPPNP algorithm A that takes as input a matrix X G n n, that succeeds with probability 1 O(δ) over X, and that, conditioned on succeeding, samples a matrix A U m,n from a probability distribution D X, such that the following properties hold: i) X/ m occurs as a uniformly-random n n submatrix of A D X, for every X such that Pr[A(X) succeeds] > 0. ii) The distribution over A C m n induced by drawing X G n n, running A(X), and conditioning on A(X) succeeding is simply H m,n.

12 Hiding Lemma - proof strategy Sample X G n n Using rejection sampling and the previous theorem, with high probability turn X into a sample from S m,n Define D X as sampling from H m,n, conditioned on X appearing as a submatrix (up to scaling). Complexity theory: BPP NP machine can produce sample from D X By symmetry, distribution of outputs is H m,n

13 Approximate BosonSampling Assuming approximate classical sampler O, want to prove GPE 2 ± BPP NP Generate sample X G n n With high probability, use the Hiding Lemma to smuggle X into a matrix A U m,n With X smuggled into A, we can compute the squared permanent of X with BPP NP machine as before, up to ± error Since A H m,n and X S m,n, adversarial O can t corrupt: no way of knowing where the smuggled X is! Assuming RMT conjectures, GPE 2 ± is #P-hard problem, hence P #P = BPP NP as desired.

14 Permanent of Gaussians Conjecture PGC: GPE is #P-hard. Plausible for complexity-theoretic reasons Analogous result is true for finite fields. Just need to generalize to C

15 Permanent Anti-Concentration Conjecture PACC: There exists a polynomial p such that for all n and δ > 0, [ ] n! Per(X) < < δ p(n, 1/δ) Pr X N (0,1) n n C Standard deviation of Per(X) is n!. PACC: probability mass of Per(X) does not concentrate at very small values relative to σ

16 Some PACC Evidence Tao-Vu (2009): For all ε > 0 and sufficiently large n, [ ] n! Pr Per(X) < X { 1,1} n n n εn < 1 n 0.1 Aaronson-Arkhipov (Weak PACC): For all α < 1, [ ] Pr Per(X) 2 (1 α)2 α n! > X G n n n + 1. Conjecture is proven for determinant instead of permanent Supported by numerics

17 Figure 1: Distribution of Per(X) for X G ,000 samples were taken. Note that 7! 71.