In the original QMA verification of the Local Hamiltonian Problem, could Merlin have cheated if he knew which term Arthur would measure?

Question

This a follow-up to a previous question about the verification procedures for Arthur to make sure the witness $|\eta\rangle$ provided by Merlin really does have energy $\lambda\le a$ (and not $\ge b$).

Recall the first protocol - Arthur chooses a random $k$-local term $H_i$ from among the $r$ such terms, acts locally only on those $k$ qubits, and rotates an ancilla answer qubit based on the decomposition of $H_i$. As an aside in the post I had commented that Arthur should choose a term at random, unknowable to Merlin, as a way to bind Merlin and make sure the witness $|\eta\rangle$ really is compliant. In @Norbert's great answer, he noted that randomness is really needed for averaging purposes.

But even still, if Merlin happened to predict what term $H_i$ will be used for the rotation, couldn't he prepare a fake witness state $|\eta'\rangle$ such that the subsystem for the $k$ qubits from which $H_i$ is composed will satisfy the local test that Arthur will apply (with the other $n-k$ qubits not being relevant?) Quoting @Norbert below, could Merlin use $|\eta'\rangle$ to convince Arthur that the ground state energy of $H$ is lower than it actually is?

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I had originally envisioned a serial process of Merlin providing a bunch of copies of $|\eta\rangle$ to Arthur, with Arthur performing his rotation of a random term for each such copy received. At each step Arthur secretly tosses his $r$-sided dice to decide which term to measure. Arthur has to make sure that Merlin didn't load the die.

After the discovery of the Marriott-Watrous trick, I guess, clearly Arthur can reuse a single witness and as such Merlin had better provide a compliant, true witness state. Alternatively perhaps if, instead of classically tossing an $r$-sided die, we have a quantum circuit choose the $r^{th}$ term and run the rotation only for that term then Merlin can't cheat quantum physics and know a-priori what term the quantum circuit will choose. Still further Arthur need not engage serially with Merlin, and could save all of his copies of $|\eta\rangle$ and run SWAP tests to make sure they are all the same, before doing his rotation based on a random $r$.

Answer 1

Given $H=\sum H_i$, if Merlin knows which $H_i$ Artur will measure, he can just prepare a ground state of $H_i$ on the qubits on which $H_i$ acts (times anything on the rest).

Labeling the ground state energy of $H_i$ by $E_i$, the energy which Artur will then estimate (or for which he will accept) is $\sum E_i$. This energy is generally lower than the ground state energy $E$ of $H$. (The class of Hamiltonians where this is not the case, i.e., where $E=\sum E_i$, are called frustration free, and they are very special.)

The possibly simplest example is a 3-qubit system with a antiferromagnetic Heisenberg interaction between qubits 1-2 and 2-3. The ground state of each $H_i$ is a singlet state between the adjacent pair of qubits (and it is unique on those two qubits), but this is not compatible with a global state (due to monogamy of entanglement) -- the true ground state energy will thus have a higher energy.