Decoding Quantum Puzzles: Unraveling the Complexity of Information Extraction

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