# The support of the ground state of stoquastic Hamiltonian is connected

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

• Please add details what you mean by "connected" (and "support"). Commented Aug 5 at 8:13
• @NorbertSchuch 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 ） Commented Aug 5 at 8:23
• So you just want to know if probability distributions are connected? Then what is the relation to stoquastic Hamiltonians? (And why not $p(x)\ge0$? Those are also allowed ground states.) --- One issue is that it seems quite unclear whether you want to connect different probability distributions (my thought), or whether you rather care about the $x$ where a given $p(x)$, for a given stoquastic Hamiltonian, is non-zero ... I think being specific about that would help a lot in clarifying, at least for me. Commented Aug 5 at 9:15
• Now assuming the latter, doesn't the Ising model disprove your claim? Commented Aug 5 at 9:17
• arxiv.org/abs/0806.1746 could be of interest (they implement a random walk on the space of $x$, so maybe they say sth about local moves.) Commented Aug 5 at 9:26

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.

• I know little about Ising model and many-body theory so could you give reference why the classical Ising model has two ground states? (The question is from this link:arxiv.org/abs/2404.07281v1) Definition 28, Fact H.1., where they say the ground state is unique. Commented Aug 5 at 10:55
• Do they mean that they only consider the case when the ground state is unique, then the support is connected? Commented Aug 5 at 11:03
• @qmww987 If the ground state is unique, please add this to your question. As it stands, the question is unclear and incomplete. Also, if this comes from some paper, give the reference immediately, it wiill help people to answer your question in context Commented Aug 5 at 11:07
• @qmww987 For the unique gs case, this is simply part of the Perron-Frobenius theorem, see updated answer. Commented Aug 5 at 11:13
• Can you give reference to the last paragraph, why $p(x)>0$ for all $x$? I didn't find this statement in the wiki en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem Commented Aug 5 at 12:33

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.