# Are Quantum States Exponentially Long Vectors? Scott Aaronson (who did and will have an affiliation) (did: IASwill: Waterloo) Distributions over n-bit.

## Presentation on theme: "Are Quantum States Exponentially Long Vectors? Scott Aaronson (who did and will have an affiliation) (did: IASwill: Waterloo) Distributions over n-bit."— Presentation transcript:

Are Quantum States Exponentially Long Vectors? Scott Aaronson (who did and will have an affiliation) (did: IASwill: Waterloo) Distributions over n-bit strings 2 n -bit strings

The Computer Science Picture of Reality Quantum computing challenges this picture Thats why everyone should care about it, whether or not quantum factoring machines are ever built + details

As far as I am concern[ed], the QC model consists of exponentially-long vectors (possible configurations) and some uniform (or simple) operations (computation steps) on such vectors … The key point is that the associated complexity measure postulates that each such operation can be effected at unit cost (or unit time). My main concern is with this postulate. My own intuition is that the cost of such an operation or of maintaining such vectors should be linearly related to the amount of non-degeneracy of these vectors, where the non-degeneracy may vary from a constant to linear in the length of the vector (depending on the vector). Needless to say, I am not suggesting a concrete definition of non-degeneracy, I am merely conjecturing that such exists and that it capture[s] the inherent cost of the computation. Oded Goldreich

(1) Its not easy to explain current experiments (let alone future ones!), if you dont think that quantum states are exponentially long vectors [A. 2004, Multilinear Formulas and Skepticism of Quantum Computing] (2) But its not that bad [A. 2004, Limitations of Quantum Advice and One-Way Communication] My Two-Pronged Response

Prong (1) Quantum states are exponentially long vectors

How Good Is The Evidence for QM? (1)Interference: Stability of e - orbits, double-slit, etc. (2)Entanglement: Bell inequality, GHZ experiments (3)Schrödinger cats: C 60 double-slit experiment, superconductivity, quantum Hall effect, etc. C 60 Arndt et al., Nature 401:680-682 (1999)

Exactly what property separates the Sure States we know we can create, from the Shor States that suffice for factoring? DIVIDING LINE

Not precision in amplitudes:Not entanglement across hundreds of particles:Not a combination of the two:

Intuition: Once we accept | and | into our set of possible states, were almost forced to accept | | and | + | as well But we might restrict ourselves to tree states: n-qubit states obtainable from |0 and |1 by a polynomial number of linear combinations and tensor products + |0 1 |1 2 ++ |0 1 |1 1 |0 2 |1 2

Definition: The tree size of a pure state | is the minimum number of linear combinations and tensor products in any tree representing | Observation: If then the tree size of | equals the multilinear formula size of f up to a constant factor Question: Can we show that quantum states arising in real (or at least doable) experiments have superpolynomial tree size?

Main Result Even to approximate |C takes tree size n (log n) Can derandomize using Reed-Solomon codes Yields the first function f:{0,1} n with a superpolynomial gap between formula size and multilinear formula size If over the codewords of a binary linear code, then |C has tree size n (log n), with high probability if the generator matrix is chosen uniformly at random from is a uniform superposition

Conjectures States arising in Shors algorithm also have superpolynomial tree size Can show this assuming a number-theoretic conjecture: n (log n) tree size lower bounds can be improved to exponential Can show this under the restriction that linear combinations are manifestly orthogonal

Conjectures (cont) 2D and 3D cluster states (lattices of spins with pairwise interactions) have large (2 (n) ?) tree size Would mean that states with enormous tree sizes have already been seen in the lab, if we believe the condensed- matter physicists (e.g. Ghosh et al., Entangled quantum state of magnetic dipoles, Nature 425:48-51, 2003) Can show that a 1D cluster state of n spins has tree size O(n 4 ) If a quantum computer is in a tree state at every time step, then that computer has a nontrivial classical simulation Can show its simulable in the third level of PH

Prong (2) Its not that bad

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 Nielsen & Chuang: 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?

Maybe BQP/qpoly even contains NP! Obvious Challenge: Prove an oracle separation between BQP/poly and BQP/qpoly Harry Buhrman: Hey Scottwhy not try for an unrelativized separation? After all, if quantum states are like 2 n -bit classical strings, then maybe BQP/qpoly NEEEEE/poly!

Result: BQP/qpoly PP/poly Proof based on new communication result: Given f:{0,1} n {0,1} m {0,1} (partial or total), D 1 (f) = O(m Q 1 (f) logQ 1 (f)) D 1 (f) = deterministic 1-way communication complexity of f Q 1 (f) = bounded-error quantum 1-way complexity Corollary: Cant show BQP/poly BQP/qpoly without also showing PP P/poly Like BPP but with no gap

Alices Classical Message Bob, if you use the maximally mixed state in place of my quantum message, then y 1 is the lexicographically first input for which youll output the wrong answer with probability at least 1/3. But if you condition on succeeding on y 1, then y 2 is the next input for which youll output the wrong answer with probability at least 1/3. But if you condition on succeeding on y 1 and y 2, then y 3 is the … y1y1 y2y2

Technicality: We assume Alices quantum message was boosted, so that the error probability is negligible Claim: Alice only needs to send T=O(Q) inputs y 1,…,y T, where Q = size of her quantum message Proof Sketch: Bob succeeds on y 1,…,y T simultaneously with probability at most (2/3) T. But we can decompose the Q-qubit maximally mixed state as where | 1 is Alices true quantum message. Therefore Bob also succeeds with probability (1/2 Q ).

BQP/qpoly PP/poly Alice is the advisor Bob is the PP algorithm But why can Bobs procedure be implemented in PP? A. 2005: PostBQP = PP, where PostBQP = BQP with postselected measurement outcomes (The PostBQP PP direction is an easy modification of Adleman et al.s proof that BQP PP)

Remarks Contrast with PQP/qpoly = Everything, and Razs recent result that QIP/qpoly = Everything Using similar techniques, I get that for every k, there exists a language in PP that does not have quantum circuits of n k, not even with quantum advice A second limitation of quantum advice (A. 2004): there exists an oracle relative to which NP BQP/qpoly

Conjectures The question of whether classical advice can always replace quantum advice (i.e. BQP/poly = BQP/qpoly) is not independent of ZF set theory Similarly for whether classical proofs can replace quantum proofs (i.e. QCMA = QMA) QMA/qpoly PP/poly Randomized and quantum one-way communication complexities are asymptotically equal for all total Boolean functions f

Summary The state of n particles is an exponentially long vector. Welcome to Quantum World! But for most purposes, its no worse than a probability distribution being an exponentially long vector If youre still not happy, suggest a Sure/Shor separator

Download ppt "Are Quantum States Exponentially Long Vectors? Scott Aaronson (who did and will have an affiliation) (did: IASwill: Waterloo) Distributions over n-bit."

Similar presentations