are locally unique pure quantum states also ground states of some local hamiltonian? Let $H=\sum_i H_i$ be some k-local hamiltonian with a unique ground state $|\psi>$. Then it is easily shown that $|\psi>$ is k-locally distinguishable from any other state $|\psi'>$.
Is the converse also true? 
In other words assume $|\psi>$ is a pure state such that for any other pure state $|\psi'>$ there exists a subset of qubits $K$ s.t $|K| \leq k$  for a fixed $k>1$ and 
$tr_{[n]\backslash K}(|\psi>) \neq tr_{[n]\backslash K}(|\psi'>)$
is it true that there exists some hamiltonian $H=\sum_i H_i$ where $H_i$ acts on at most k qubits s.t $|\psi>$ is the only ground state of $H$?
 A: The question, as I understand it, asks the following:
Consider a state $|\psi\rangle$ which is uniquely determined among all other pure states by its K-party reduced density matrices. Is it guaranteed that this state is the ground state of a K-local Hamiltonian?
First, we should note that not only is it the case that any unique pure ground state of a K-local Hamiltonian is uniquely determined among all other pure states by its K-party reduced density matrices, but it is also uniquely determined among all states (including mixed states) by its K-party reduced density matrices.
More concretely:
Define the set of n-particle quantum states which have the same K-RDMs as $\rho$ to be $A_k(\rho)$. Define the unique state with maximum entropy among all states in $A_k(\rho)$ to be the K-correlated state $\tilde{\rho}_k$. A space $V$ is called K-correlated if the maximally mixed state supported on $V$ is also K-correlated. It is observed in this paper that if $V$ is the ground state space of a K-local Hamiltonian, then $V$ must be K-correlated.
Consider a quantum state $|\psi\rangle$ which is uniquely determined by its K-RDMs. If there were a K-local Hamiltonian with this quantum state as its unique ground state, then the ground state space $V$ would be one dimensional, and since $V$ must be K-correlated, the state with maximum entropy which has the same K-RDMs as $|\psi\rangle$ is just $|\psi\rangle$. This implies there are also no mixed states which share the same K-RDMs as $|\psi\rangle$.
So if there existed a pure state which was uniquely determined among all pure states by its K-RDMs, but not uniquely determined among all states (including mixed states) by its K-RDMs, then it could not be the ground state of a K-local Hamiltonian.
In this paper, posted today, I, and my collaborators, show that such states do exist. Thus the answer to your question is no, the converse is not true. Even if you show that there do not exist any other pure states with the same K-RDMs as your pure state, it is possible that it can not be the ground state of some K-local Hamiltonian.
