The Learnability of Quantum States

1. The Learnability of Quantum States  Scott Aaronson University of Waterloo

2. Quantum State Tomography Suppose we have a physical process that produces a quantum mixed state  By applying the process repeatedly, we can prepare as many copies of  as we want To each copy, we then apply a binary measurement E, obtaining ‘1’ with probability Tr(E) and ‘0’ otherwise Our goal is to learn an approximate description of  EXPERIMENTALISTS ACTUALLY DO THISTo learn about chemical reactions (Skovsen et al. 2003), test equipment (D’Ariano et al. 2002), study decoherence mechanisms (Resch et al. 2005), …

3. But there’s a problem… Fear not! Why would he be raising this “problem” if he wasn’t gonna demolish it? To do tomography on an entangled state of n qubits, we need (4n) measurements The current record: 8 qubits (Häffner et al. 2005), requiring 656,100 experiments (!) Does this mean that a generic 10,000-particle state can never be “learned” within the lifetime of the universe? If so, would call into question the operational status of quantum states themselves (and make quantum computing skeptics extremely happy)…

4. The Quantum Occam’s Razor Theorem Let  be an n-qubit mixed state. Let D be a distribution over two-outcome measurements. Suppose we draw m measurements E1,…,Em independently from D, and then output a “hypothesis state”  such that|Tr(Ei)-Tr(Ei)|≤ for all i. Then provided /10 and we’ll have with probability at least 1- over E1,…,Em

5. Can make dependence and  and  more reasonable, at the cost of a log2n factor: The above bound is nearly tight: Remarks Implies that we can do “pretty good tomography,” using a number of measurements that grows only linearly (!) with the number of qubits n Result says nothing about the computational complexity of preparing a hypothesis state that agrees with measurement results

6. To prove the theorem, we need a notion introduced by Kearns and Schapire called Fat-Shattering Dimension Let C be a class of functions from S to [0,1]. We say a set {x1,…,xk}S is -shattered by C if there exist reals a1,…,ak such that, for all 2k possible statements of the form f(x1)a1-  f(x2)a2+  …  f(xk)ak-, there’s some fC that satisfies the statement. Then fatC(), the -fat-shattering dimension of C, is the size of the largest set -shattered by C.

7. Small Fat-Shattering Dimension Implies Small Sample Complexity Let C be a class of functions from S to [0,1], and let fC. Suppose we draw m elements x1,…,xm independently from some distribution D, and then output a hypothesis hC such that |h(xi)-f(xi)| for all i. Then provided /7 and we’ll have with probability at least 1- over x1,…,xm. Proof uses a 1996 result of Bartlett and Long—building on Alon et al., building on Blumer et al., building on Valiant

8. No need to thank me! Upper-Bounding the Fat-Shattering Dimension of Quantum States Nayak 1999: If we want to “encode” k classical bits into n qubits, in such a way that any bit can be recovered with probability 1-p, then we need n(1-H(p))k Corollary (“turning Nayak’s result on its head”):Let Cn be the set of functions that map an n-qubit measurement E to Tr(E), for some . Then Quantum Occam’s Razor Theorem follows easily…

9. GBUSTERS L Simple Application of Quantum Occam’s Razor Theorem to Communication Complexity x y f(x,y) Alice Bob • f: Boolean function mapping Alice’s N-bit string x and Bob’s M-bit string y to a binary output • D1(f), R1(f), Q1(f): Deterministic, randomized, and quantum one-way communication complexities of f • How much can quantum communication save? • It’s known that D1(f)=O(M Q1(f)) for all total f • In 2004 I showed that for all f,D1(f)=O(M Q1(f)logQ1(f))

10. Theorem: R1(f)=O(M Q1(f)) for all f, partial or total Proof: Fix Alice’s input x By Yao’s minimax principle, Alice can consider a worst-casedistribution D over Bob’s input y Alice’s classical message will consist of y1,…,yT drawn from D, together with f(x,y1),…,f(x,yT), where T=(Q1(f)) Bob searches for a quantum message  that yields the right answers on y1,…,yT By the Quantum Occam’s Razor Theorem, with high probability such a  yields the right answers on most y drawn from D

11. What about computational complexity? BQP/qpoly: Class of problems solvable in quantum polynomial time, with help from poly-size “quantum advice state” that depends only on input length n A. 2004:BQP/qpolyPP/poly“Classical advice can always simulate quantum advice, provided we use exponentially more computation” Can this result be improved to BQP/qpolyQMA/poly?(QMA: Quantum Merlin-Arthur) Theorem:HeurBQP/qpolyHeurQMA/polyOr in English: We can use trusted classical advice to verify that untrusted quantum advice will work on most inputs

12. Stronger Result: HeurBQP/qpoly = HeurYQP/polyHere YQP (“Yoda Quantum Polynomial-Time”) is like QMAcoQMA, except that a single witness must work for all inputs of length n Proof Idea:The classical advice to the HeurQMA/poly verifier will consist of “training inputs” x1,…,xm where m=poly(n), as well as whether xiL for all i Given a purported quantum advice state |, the verifier first checks that | yields the right answers on the training inputs, and only then uses it on its real input x By the Quantum Occam’s Razor Theorem, if | passes the initial test, then w.h.p. it works on most inputs Technical part is to do the verification without destroying |

13. r r DUNCE DUNCE Open Problems Computationally-efficient learning algorithms Experimental implementation! Tighter bounds on number of measurements Does BQP/qpoly = YQP/poly? Is D1(f) = O(M Q1(f))?