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The standard popular-news account of quantum computing is that a quantum computer (QC) would work by splitting into exponentially many noninteracting parallel copies of itself in different universes and having each one attempt to verify a different certificate, then at the end of the calculation, the single copy that found a valid certificate "announces" its solution and the other branches magically vanish.

People who know anything about theoretical quantum computation know that this story is absolute nonsense, and that the rough idea described above more closely corresponds to a nondeterministic Turing machine (NTM) than to a quantum computer. Moreover, the compexity class of problems efficiently solvable by NTMs is **NP** and by QCs is **BQP**, and these classes are not believed to be equal.

People trying to correct the popular presentation rightfully point out that the simplistic "many-worlds" narrative greatly overstates the power of QCs, which are not believed to be able to solve (say) **NP**-complete problems. They focus on the misrepresentation of the measurement process: in quantum mechanics, which outcome you measure is determined by the Born rule, and in most situations the probability of measuring an incorrect answer completely swamps the probability of measuring the right one. (And in some cases, such as black-box search, we can *prove* that no clever quantum circuit can beat the Born rule and deliver an exponential speedup.) If we *could* magically "decide what to measure", then we would be able to efficiently solve all problems in the complexity class **PostBQP**, which is believed to be much large than **BQP**.

But I've never seen anyone explicitly point out that there is *another* way in which the popular characterization is wrong, which goes in the other direction. **BQP** is believed to be not a strict subset of **NP**, but instead incomparable to it. There exist problems like Fourier checking which are believed to not only lie outside of **NP**, but in fact outside of the entire polynomial hierarchy **PH**. So with respect to problems like these, the popular narrative actually *under*states rather than overstates the power of QCs.

My naive intuition is that if we *could* "choose what to measure", then the popular narrative would be more or less correct, which would imply that these super-quantum-computers would be able to efficiently solve exactly the class **NP**. But we believe that this is wrong; in fact **PostBQP=PP**, which we believe to be a strict superset of **NP**.

Is there any intuition for what's going on behind the scenes that allows a quantum computer to be (in some respects) more powerful than a nondeterministic Turing machine? Presumably this "inherently quantum" power, when combined with postselection (which in a sense NTMs already have) is what makes a super-QC so much more powerful than a NTM. (Note that I'm looking for some intuition that directly contrasts NTMs and QCs with postselection, without "passing through" the classical complexity class **PP**.)

Are

PandPSPACEcounting classes? Naively it seems that yes forP, as it could be defined as the set of problems such that every path accepts, but I'm not sure aboutPSPACE. – tparker – 2018-03-30T19:25:03.2801

PSPACEis not a counting class, no. You're on the right track withP--- you must require that eithereverypath accepts oreverypah rejects (or a similarly strong requirement), or else you might end up withcoNP,coRP, or some other class not known to equalP. – Niel de Beaudrap – 2018-03-30T19:30:42.043Presumably

PHisn't a counting class either, since it's naturally formulated in terms of an alternating rather than nondeterministic Turing machine? – tparker – 2018-06-05T20:45:22.400If

BPPrequires two-thirds of the branches to accept,PPrequires half to accept, andNPonly requires one to accept, wouldn't that imply that $\mathbf{BPP} \subset \mathbf{PP} \subset \mathbf{NP}$? But in fact $\mathbf{BPP} \subset \mathbf{NP} \subset \mathbf{PP}$, and both inclusions are believed to be strict. – tparker – 2018-06-11T16:18:36.3031@tparker: You are missing some details, for instance the behaviour of NO instances in the definition of

BPP(see the post). In short, these thresholds do not map in a linear manner to broadness of instances taken to be YES instances. (And to the best of my knowledge, no such relation as $\mathsf {BPP \subseteq NP} $ is known --- at best we know $\mathsf {BPP \subseteq NP^{NP} \cap coNP^{NP}}$.) – Niel de Beaudrap – 2018-06-11T21:19:16.843