I was wondering if the complexity of a quantum circuit that solves a problem that is in NP implies P=NP?


In short, no — as things stand, if there were a polynomial-time algorithm to solve an $\mathsf{NP}$-complete problem on a(n idealised) quantum computer (with reasonably low error probabilities), this would not imply that $\mathsf{P} = \mathsf{NP}$. The question of "what problems are in $\mathsf{P}$" is a very different one from "what problems can you efficiently solve in principle on an idealised quantum computer".

The problems that are in $\mathsf{P}$, are not actually "the problems that can be solved in practise"; it is the set of decision problems that can be solved, in polynomial time, very specifically on a classical computer (more precisely: a deterministic Turing machine, or any model of computation which is polynomial-time equivalent to a deterministic Turing machine).

It is, perhaps, possible that there is a classical algorithm to efficiently simulate an idealised quantum computer, i.e., to simulate polynomial-time uniform quantum circuit families which produce outcomes in {0,1}. But we have no particular evidence that this is the case, and certainly no proof. So, if we discovered an $\mathsf{NP}$-complete problem L which happened to be solvable efficiently (and with low enough probability of error) by quantum computers — that is, if we discovered such a problem L, which belonged to the class $\mathsf{BQP}$ of problems solvable by a quantum computer in that sense — this would not necessarily provide us with any way to solve the problem in polynomial time using deterministic Turing machines (or in other words to show that L is also in $\mathsf{P}$), nor even any evidence in that direction.

More generally, there are no other formal results or other evidence to the effect that $\mathsf{NP} \subseteq \mathsf{BQP}$ only if $\mathsf{NP} = \mathsf{P}$. So, even if we could solve $\mathsf{NP}$-complete problems on a quantum computer, this wouldn't provide us with any proof (or formal reason to believe) that $\mathsf{P} = \mathsf{NP}$.

(It would, however, be an extremely exciting and surprising result.)

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    $\begingroup$ Pedantically, although this excellent answer directly answers the question of whether "a quantum circuit that solves a problem that is NP-complete implies P=NP?" (it does not), to the OP's original question about whether "a quantum circuit that solves a problem that is merely in NP" has any bearing on whether P=NP, this also does not, if only because $\mathsf{P}\subseteq\mathsf{NP}$, and thus the OP's considered problem in NP could also have been in P. $\endgroup$
    – Mark S
    Aug 26 '20 at 22:47
  • $\begingroup$ @MarkS: that is also quite correct. $\endgroup$ Aug 26 '20 at 23:52

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