One-way quantum verification The theory of cluster-state computation is well-established by now, showing that any BQP circuit can be modified so it uses only single qubit quantum gates, possibly classically controlled, provided ample supply of a state known as the "cluster state" - which is a simple to produce stablizer state.

My question is: is a similar notion known for quantum verification - i.e.
can one replace QMA circuits with classically controlled 1-qubit gates, possibly using some "special state"?
At least initially, I'm unclear on why the cluster state can even work in this case.
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).