  We show that the functional analogue of QMA$\cap$coQMA, denoted
F(QMA$\cap$coQMA), equals the complexity class Total Functional QMA (TFQMA). To
prove this we need to introduce alternative definitions of QMA$\cap$coQMA in
terms of a single quantum verification procedure. We show that if TFQMA equals
the functional analogue of BQP (FBQP), then QMA$\cap$coQMA = BQP. We show that
if there is a QMA complete problem that (robustly) reduces to a problem in
TFQMA, then QMA$\cap$coQMA = QMA. These results provide strong evidence that
the inclusions FBQP$\subseteq$TFQMA$\subseteq$FQMA are strict, since otherwise
the corresponding inclusions in BQP$\subseteq$QMA$\cap$coQMA$\subseteq$QMA
would become equalities.
