The support of the ground state of stoquastic Hamiltonian is connected

Question

A Hamiltonian $H$ is stoquastic in the standard basis if all the off-diagonal terms of the Hamiltonian are non-positive. If we choose $\beta$ small enough, all entries of $I-\beta H$ are non-negative. By the Perron-Frobenius Theorem, the eigenvector which corresponds to the largest eigenvalue (which is also the ground state of $H$) can be chosen to have non-negative entries. Suppose the ground state is unique ,and denote it as $|\psi\rangle =\sum_x \sqrt{p(x)}|x\rangle$. How to prove that the support of {$p(x)$} is connected on the boolean hypercube?

Support is the 𝑥 that 𝑝(𝑥)>0, connected means connected component in a graph （i.e there exist a path for every two points, in the boolean hypercube, there exist an edge between two points iff they differ in only one position).

The question is from Definition 28, Fact H.1. of this paper

Answer 1

In case your ground state is not unique, your claim is incorrect: The classical Ising model (on any connected graph) is stoquastic, and has two ground states |0...0> and |1...1>, which are not connected on the hypercube.

On the other hand, for a unique ground state, the Perron-Frobenius theorem states that $p(x)>0$ for all $x$, and thus, the support is the entire hypercube, and thus connected.

Answer 2

For what it's worth I also think the reference to the "Boolean hypercube" in the question is a bit of a red herring. In particular I can imagine the Hamiltonian acting stoquastically on a number of quitrits, in which case the dimension of the underlying Hilbert space is 3$^n$ and not 2$^n$. The converse of the Perron-Frobenius theorem provides that the amplitudes have full support over each of the 3$^n$ basis vectors.

Indeed, we can still embed our graph into a Hilbert space of dimension $2^n$, but the graph might not need have a number of vertices that's a power of two and we could have oracular access to the entries and edges. For example, given oracles to access entries to the adjacency matrix $A$ of some sparse unweighted, undirected, connected graph on $N\lt 2^n$ vertices having degree matrix $D$, then the Laplacian $\mathcal L=D-A$ is stoquastic, and the ground state has full support over all $N$ basis states. If $A$ is $d$-regular and aperiodic, then each $p(x)$ is $1/N$, and the stationary distribution for a random walk on the graph is uniform over all $N$ vertices.

It's only by convenience I guess that we like to work with the $n$ qubits, and we just "ignore" the $2^n-N$ basis vectors that are not on the graph - the full $2^n\times 2^n$ matrix would be block-diagonal with respect to these basis vectors, which correspond to one or more other connected components, not of the graph, but of the larger Boolean hypercube.