# QMA definition of difference between probabilities intuition

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!

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.