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I'm reading about the complexity classes related to quantum computation, currently I'm studying QMA class. A language is in QMA(c,s) if there exists a polynomial time verifier and polynomial $p(n)$ such that

if $x\in L \implies$ there exists quantum state $\psi$ such that the probability V accepts is greater than $c$

if $x \notin L \implies$ For all quantum states $\psi$ the probability that V accepts is less than $s$.

Usually it is remarked that $c$ and $s$ have the following property $c-s = \frac{1}{poly(n)}$. This last condition is what confuses me. What is the intuition behind imposing this condition on the difference of $c$ and $s$? Not understanding this makes me feel I haven't fully understood QMA class definition.

Something similar happens with the problem called Local Hamiltonian problem. Here two constants are defined such that $a-b = \frac{1}{poly(n)}$. If someone could explain the intuition behind this and if it is related to the previous question that would be great. Thanks!

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While this question deals with the difference between $\mathsf{BPP}$ and $\mathsf{PP}$, it should still be useful for you to read through. The "short answer" is that the "gap" between $c$ and $s$ allows for standard amplification techniques to be used, which end up showing that $\mathsf{QMA}(c,s) = \mathsf{QMA}(2^{-n},1-2^{-n})$ (provided $c - s$ meets the condition you wrote down).

The "case study" of $\mathsf{BPP}$ vs $\mathsf{PP}$ should be interesting overall for your question, as the only difference between them is the presence / lack of the $c - s \geq 1 / \mathsf{poly}(n)$ condition. It's known that: $$\mathsf{BQP}\subseteq\mathsf{QMA}\subseteq \mathsf{PP}\subseteq\mathsf{PSPACE}$$ But: $$\mathsf{BPP}\subseteq\mathsf{BQP}$$ This is quite a large difference. I'm unsure of what is known about $\mathsf{QMA}(c,s)$ where $c - s$ is negligible (meaning $n^{-\omega(1)}$), where amplification techniques can't be used.

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  • $\begingroup$ Thanks! The same argument you have provided here applies for the Local Hamiltonian Problem, right? $\endgroup$ – Pam Dec 23 '18 at 10:34

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