1 / 11

When Qubits Go Analog A Relatively Easy Problem in Quantum Information Theory

When Qubits Go Analog A Relatively Easy Problem in Quantum Information Theory. Scott Aaronson (MIT).

reeves
Télécharger la présentation

When Qubits Go Analog A Relatively Easy Problem in Quantum Information Theory

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. When Qubits Go AnalogA Relatively Easy Problem in Quantum Information Theory Scott Aaronson (MIT)

  2. Erik Demaine(motivated by a computational genetics problem): “Suppose a PSPACE machine can flip a coin with bias p an unlimited number of times. Can it extract an exponential amount of information (or even more) about p?” Me: “I’m sure whatever the answer is, it’s obvious...” Didn’t seem too likely there could be superpowerful“Advice Coins”

  3. Indeed, Hellman-Cover (1970) proved the following... Suppose a probabilistic finite automaton is trying to decide whether a coin has bias ½ or ½+. Then even if it can flip the coin an unlimited number of times, the automaton needs (1/) states to succeed with probability (say) 2/3. Implies PSPACE/coin = PSPACE/poly poly(n) advice bits another poly(n) advice bits Bias=0.000000000000000110101111101

  4. Yet quantum mechanics nullifies the Hellman-Cover Theorem! Halt with probability ~2/100 at each time step, by measuring along the third dimension Expected difference in final angle after halting, in p=½ vs. p=½+ cases: 1 radian Standard deviation in angle: Keep flipping the coin. Whenever the coin lands heads, rotate /100 radians counterclockwise. Whenever it lands tails, rotate /100 radians clockwise. Theorem: For any >0, can distinguish a coin with bias p=½ from a coin with bias p=½+ (with bounded error) using a single qutrit of memory.

  5. So, could BQPSPACE/coin=ALL? Theorem: No. Proof: Let’s even let the machine run infinitely long; it only has to get the right answer in the limit Let 0 = superoperator applied to our memory qubits whenever coin lands heads, 1 = superoperator when it lands tails Then induced superoperator at each time step: p = coin’s bias We’re interested in a fixed-point of p: a mixed state p such that

  6. Fixed-Points of Superoperators Studied by [A.-Watrous 2008] in the context of quantum computing with closed timelike curves Our result there:BQPCTC = PCTC = PSPACE Quantum computers with CTCs have exactly the same power as classical computers with CTCs, namely PSPACE (or: “CTCs make time and space equivalent as computational resources”) 0.000000000000000110101111101

  7. Fixed-point p of a superoperator p can be expressed in terms of degree-2srational functions of p, where s is the number of qubits (Proof: Use Cramer’s Rule on 2s2s matrices) Let ax(p) be the probability that the PSPACE machine accepts, on input x{1,...,N} and an advice coin with bias p Then ax(p) is a degree-2s rational function of p By calculus, a degree-2s rational function can cross the origin (or the line y=½) at most 22s times Key Point 0.000000000000000110101111101

  8. ax(p) p To specify p, well enough to decide whether ax(p)½ for any x: suffices to say how many reals 0<q<p there are such that someax(p) crosses the line y=½ at q This takes log2(22sN)=s+1+log2(N) bits So, coin can specify distinct functions

  9. New candidate problem we should use for this: “Fourier Checking” OK, how about a harder problem? • Given oracle access to two Boolean functions,f,g:{-1,1}n{-1,1}n • Promised that either • f and g are both uniformly random, or • f,g were chosen by picking a random unit vector and letting f(x)=sgn(vx), g(x)=sgn(Hnvx) • Problem:Decide which Is there an oracle relative to which BQPPH?

  10. then measure in the Hadamard basis On the other hand, I conjecture the Fourier Checking problem is not in PH I can show that any poly(n) bits of f(x) and g(x) are close to uniformly random. I conjecture that this suffices to put the problem outside PH (“Generalized Linial-Nisan Conjecture”) I claim this problem is in BQP

  11. PROVE QUANTUM LOWER BOUNDS NO WE CAN’T

More Related