Consider the accepting POVM ($\Pi$, $I-\Pi$) of a fixed QMA machine. Let d be the number of eigenstates of $\Pi$ that are at least 2/3. Decide if d is at least some fixed s, or at most s/2.
This is a quantum version of a decision class that “approximately counts” NP witnesses. In that case, it is NP-hard and at most BPP^NP by Stockmeyer.
In the quantum case, it is QMA-hard, and it is known to be in QIP[2] (actually, qq-QAM). But is there a tighter upper bound or lower bound?
What about the complement? Is it in QMA(2)?