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Multilinear Formulas and Skepticism of Quantum Computing Scott Aaronson, UC Berkeley http://www.cs.berkeley.edu/~aaronson

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Outline (1)Four objections to quantum computing (2)Sure/Shor separators (3)Tree states (4)Result: QECC states require n (log n) additions and tensor products (5)Experimental (!) proposal (6)Conclusions and open problems

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Four Objections TheoreticalPractical Physical(A): QCs cant be built for fundamental reason (B): QCs cant be built for engineering reasons Algorithmic(C): Speedup is of limited theoretical interest (D): Speedup is of limited practical value

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(A): QCs cant be built for fundamental reasonLevins arguments (1) Analogy to unit-cost arithmetic model (2) Error-correction and fault-tolerance address only relative error in amplitudes, not absolute (3) We have never seen a physical law valid to over a dozen decimals (4) If a quantum computer failed, we couldnt measure its state to prove a breakdown of QMso no Nobel prize The present attitude is analogous to, say, Maxwell selling the Daemon of his famous thought experiment as a path to cheaper electricity from heat

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Responses (1) Continuity in amplitudes more benign than in measurable quantitiesshould we dismiss classical probabilities of order 10 -1000 ? (2) How do we know QMs successor wont lift us to PSPACE, rather than knock us down to BPP? (3) To falsify QM, would suffice to show QC is in some state far from e iHt |. E.g. Fitch & Cronin won 1980 Physics Nobel merely for showing CP symmetry is violated Real Question: How far should we extrapolate from todays experiments to where QM hasnt been tested?

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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)

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Alternatives to QM Roger PenroseGerard t Hooft (+ King of Sweden) Stephen Wolfram

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

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My View: Any good argument for why quantum computing is impossible must answer this questionbut I havent seen any that do What Ill Do: - Initiate a complexity theory of (pure) quantum states, that studies possible Sure/Shor separators - Prove a superpolynomial lower bound on tree size of states arising in quantum error correction - Propose an NMR experiment to create states with large tree size

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Classes of Pure States Classical Vidal Circuit AmpP MOTree OTree TSH Tree P 1 2 1 2

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Tree size TS(| ) = minimum number of unbounded- fanin + and gates, |0 s, and |1 s needed in a tree representing |. Constants are free. Permutation order of qubits is irrelevant. Example: + |0 1 |1 2 ++ |0 1 |1 1 |0 2 |1 2

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Motivation: If we accept | and |, we almost have to accept | | and | + |. But can only polynomially many tensorings and summings take place in the multiverse, because of decoherence? Tree States: Families such that TS(| n ) p(n) for some polynomial p Will abuse and refer to individual states

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Example Tree State = equal superposition over n-bit strings of parity i

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Trees involving +,, x 1,…,x n, and complex constants, such that every vertex computes a multilinear polynomial (no x i multiplied by itself) Given let MFS(f) be minimum number of vertices in multilinear formula for f Multilinear Formulas + -3ix1x1 x1x1 x2x2 Theorem: If then

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Theorem: Any tree state has a tree of polynomial size and logarithmic depth Proof Idea: Follows Brents Theorem (1974), that any function with a poly-size arithmetic formula has a formula of polynomial size and logarithmic depth Depth Reduction

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is an orthogonal tree state if it has a polynomial-size tree that only adds orthogonal states Theorem: Any orthogonal tree state can be prepared by a poly-size quantum circuit Proof Idea: If we can prepare | and |, clearly can prepare | |. To prepare | + | where | =0: let U|0 n =|, V|0 n =|. Then Add OR of 2 nd register to 1 st register

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Why Its Not Obvious: Theorem: If is chosen uniformly under the Haar measure, then with 1-o(1) probability, no state | with TS(| )=2 o(n) satisfies | | | 2 15/16 Proof Idea: Use Warrens Theorem from real algebraic geometry

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Theorem: Proof Idea: Guess and verify trees; use Goldwasser- Sipser approximate counting Evidence that TreeBQP BQP? Class of problems solvable by a quantum computer whose state at every time is a tree state. (1-qubit intermediate measurements are allowed.) BPP TreeBQP BQP TreeBQP

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QECC States Let C be a coset in then Codewords of stabilizer codes (Gottesman, CSS) Later well add phases to reduce codeword size Take the following distribution over cosets: choose u.a.r. (where k=n 1/3 ), then let Result:

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Razs Breakthrough Given coset C, let Need to lower-bound multilinear formula size MFS(f) LOOKS HARD Until June, superpolynomial lower bounds on MFS didnt exist Raz: n (log n) MFS lower bounds for Permanent and Determinant of n n matrix (Exponential bounds conjectured, but n (log n) is the best Razs method can show)

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Idea of Razs Method Given choose 2k input bits u.a.r. Label them y 1,…,y k, z 1,…,z k Randomly restrict remaining bits to 0 or 1 u.a.r. Yields a new function Let Show M R has large rank with high probability over choice of f R f R (y,z)M R = y {0,1} k z {0,1} k

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Intuition: Multilinear formulas can compute functions with huge rank, i.e. But once we restrict everything except y 1,…,y k, z 1,…,z k, with high probability rank becomes small Theorem (Raz):

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Lower Bound for Coset States b x A If these two k k matrices are invertible (which they are with probability > 0.288 2 ), then M R is a permutation of the identity matrix, so rank(M R )=2 k

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Corollary First superpolynomial gap between multilinear and general formula size of functions f(x) is trivially NC 1 just check whether Ax=b Determinant not known to be NC 1 best formulas known are n O(log n) Still open: Is there a polynomial with a poly-size formula but no poly-size multilinear formula?

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Inapproximability of Coset States Fact: For an N N complex matrix M=(m ij ), (Follows from Hoffman-Wielandt inequality) Corollary: With (1) probability over coset C, no state | with TS(| )=n o(log n) has | |C | 2 0.98

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Shor States Superpositions over binary integers in arithmetic progression: letting w = (2 n -a-1)/p, (= 1 st register of Shors alg after 2 nd register is measured) Conjecture: Let S be a set of integers with |S|=32t and |x| exp((log t) c ) for all x S and some c>0. Let S p ={x mod p : x S}. For sufficiently large t, if we choose a prime p uniformly at random from [t,5t/4], then |S p | 3t/4 with probability at least 3/4 Theorem: Assuming the conjecture, there exist p,a for which TS(|pZ+a )=n (log n)

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Challenge for NMR Experimenters Create a uniform superposition over a generic coset of (n 9) or even better, Clifford group state Worthwhile even if you dont demonstrate error correction Well overlook that its really (1-10 -5 )I/512 + 10 -5 |C C| New test of QM: are all states tree states? Whats been done: 5-qubit codeword in liquid NMR (Knill, Laflamme, Martinez, Negrevergne, quant-ph/0101034) TS(| ) 69

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Tree Size Upper Bounds for Coset States 0123456789101112 113 2377 3491710 4511212713 56 25493316 67152957773919 78173365121894522 891937731451851015125 9102141811613052251135728 10112345891773534332491256331 11122549971933857055452731376934 121327531052094178339935932971497537 log 2 (# of nonzero amplitudes) n#ofqubitsn#ofqubits Hardest cases (to left, use naïve strategy; to right, Fourier strategy)

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For Clifford Group States 0123456789101112 113 23711 3491725 4511214153 561325498985 67152957113153133 78173365129225233189 89193773145289369345301 910214181161321545561537413 1011234589177353705865817793541 1112254997193385769 1281131312651177 733 12132753105209417833 16651985188918411689957 log 2 (# of nonzero amplitudes) n#ofqubitsn#ofqubits

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Open Problems Exponential tree-size lower bounds Lower bound for Shor states Explicit codes (i.e. Reed-Solomon) Concrete lower bounds for (say) n=9 Extension to mixed states Separate tree states and orthogonal tree states PAC-learn multilinear formulas? TreeBQP=BPP? Non-tree states already created in solid state? Important for experiments

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Conclusions Complexity theory is relevant for experimental QIP Complexity of quantum states deserves further attention QC skeptics can strengthen their case (and help us) by proposing Sure/Shor separators QC experiments will test quantum mechanics itself in a fundamentally new way

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