One-way quantum verification
It is possible to restrict the QMA verifier to single-qubit measurements and classical pre- and postprocessing (with randomness) while keeping QMA-completeness. 

To see why, take any class of $k$-local QMA-complete Hamiltonians on qubits. By adding a constant of order $\mathrm{poly}(n)$ and rescaling with a $1/\mathrm{poly}(n)$ factor, the Hamiltonian can be brought into the form
$$
H=\sum_i w_i h_i\ ,
$$
where $w_i&gt;0$, $\sum_i w_i=1$, and $h_i = \tfrac12(\mathrm{Id}\pm P_i)$, where $P_i$ is a product of Paulis. Estimating the smallest eigenvalue of $H$ up to accuracy $1/\mathrm{poly}(n)$ is still QMA-hard.

We can now build a circuit which only uses single-qubit measurements which, given a state $|\psi\rangle$, accepts with probability $1-\langle\psi|H|\psi\rangle$ (which by construction is between $0$ and $1$).  To this end, first randomly pick one of the $i$'s according to the distribution $w_i$. Then, measure each of the Paulis in $P_i$, and take the parity $\pi$ of the outcomes, which is now related to $\langle\psi|h_i|\psi\rangle$ via
$$
\langle\psi|h_i|\psi\rangle = \tfrac12(1\pm (-1)^\pi)\in\{0,1\}\ .
$$
The circuit now outputs $1-\langle\psi|h_i|\psi\rangle$, and the output is therefore distributed according to $\langle\psi|H|\psi\rangle$.  

This is, if we picked a yes-instance of the (QMA-complete) local Hamiltonian problem, there is a state $|\psi\rangle$ such that this verifier will accept with some probability $\ge a$, while otherwise any state will be rejected with probability $\le b$, with $a-b&gt;1/\mathrm{poly}(n)$.  The variant of QMA where the verifier is restricted to one-qubit measurements is therefore QMA-complete for some $1/\mathrm{poly}(n)$ gap.  Finally, this version of QMA can be amplified using just the conventional amplification techniques for QMA, which finally proves it is QMA-complete independent of the gap (within the same range as QMA).