# What is a stoquastic Hamiltonian?

Recently, I've often read about the term 'stoquastic Hamiltonian'. But I couldn't find a precise definition anywhere. I found that the Ising-Hamiltonian is a stoquastic Hamiltonian, but that does not help a lot...

Can anyone give a precise definition?

Edit: Here's a paper where such a 'stoquastic Hamiltonian' is mentioned, The Complexity of Stoquastic Local Hamiltonian Problems (2007).

• Also, it looks like Bravyi mentions the answer to your question on page 2 of the PDF in the arXiv link; can you explain what is lacking in the definition given there? Commented Nov 15, 2017 at 11:19
• Do you mean Stochastic?
– alfC
Commented Dec 7, 2017 at 22:19
• @alfC I thought so too, but the linked paper does actually say "stoquastic." (It also gives a definition in the very first sentence of the abstract, so it's hard to see the need for this question.) Commented Jun 9, 2019 at 6:32
• This has got to be my least favorite physics coinage of all time! Commented Jun 9, 2019 at 6:43
• @knzhou You can always use "sign-problem free" instead. Commented Jun 9, 2019 at 7:20

Hamiltonians where all the off-diagonal elements in the standard basis are real and non-positive are called "stoquastic."

One important thing you need to note is that the notion of stoquasticity is basis dependent. That is the single most tricky part in the definition, and once you are OK with that, the idea should be fairly simple.

To keep things simple, let's just consider a quantum spin system with spin-1/2 i.e. qubits. Now, we need to fix one basis for defining "stoquastic", and here, let's just choose the nicest case of the $$z$$-basis (aka computational basis).

A Hamiltonian $$H$$ is stoquastic if and only if, when you write it down as a matrix in the fixed basis ($$z$$-basis for now), all off-diagonal entries of that matrix are non-positive.

This is it! For example, a two-qubit Hamiltonian for the transverse field Ising model may look like $$$$H=JZ_iZ_j-h(X_i+X_j),$$$$ which in an explicit matrix form (with $$z$$-basis!) will look like $$$$H=\begin{pmatrix} J & -h & -h & 0\\ -h& -J & 0 & -h\\ -h& 0 & -J & -h\\ 0 & -h & -h & 0 \end{pmatrix},$$$$ so you can see that it's stoquastic whenever $$h>0$$. Note that the form of the Hamiltonian will change when we use another basis, and that's closely tied with the basis-dependency of stqoasticity. For example, if we choose the $$x$$-basis instead, the condition for stoquasticity will become $$J<0$$.

It could actually be a bit confusing, since some people define stoquasticity as the given Hamiltonian $$H$$ to be admitting some local basis transformation so that it satisfies the above "non-positive" condition. This sort of lack of consensus in definitions is a common thing in newly developing fields, and is part of physics I think (15 years ago, no one really knew what quantum spin liquids are!). While this local basis transformation definition also makes sense, I think it keeps things easier to just define the notion as a basis-dependent concept like I did first; in the paper you linked, the authors argues this point but only briefly, and I feel that's causing trouble too.

For example, in the Hamiltonian above, if we have $$h<0$$, the Hamiltonian is nonstoquastic (by my definition), but obviously the physics doesn't change. In the light of basis transformation, we can see that all we need to do is to apply a basis rotation by conjugating with $$Z_i$$ and $$Z_j$$. This will leave the $$ZZ$$ interaction untouched (since $$Z_k Z_i Z_k = Z_i$$), but will flip the $$X$$ terms ($$Z_iX_iZ_i=-X_i$$) and make the Hamiltonian stoquastic again. This is the reason why one may want to define the idea of stoquasticity "up to local basis transformation", but IMO it keep things simpler if we just define the concept not allowing any basis transformation and simply say "well, that Hamiltonian is easily stoquastifiable by a local basis transformation" for this kind of example.

Finally, I see some comments that are basically saying "isn't it just the same as sign-problem free?", and I'd like to comment on that. The short answer is "that's almost correct, but not exactly", and this also connects to the above point. For example, consider the antiferromagnetic Heisenberg Hamiltonian for two qubits: $$$$H=X_iX_j + Y_iY_j +Z_iZ_j = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 2 & 0\\ 0& 2 & -1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix},$$$$ where I again use the computational basis. This Hamiltonian, without basis transformation, is nonstoquastic because it has a +2 off-diagonal element. However, if you do standard world-line or stochastic-series expansion type quantum Monte Carlo algorithms, it is sign-problem free because of the periodic boundary condition of the imaginary time (cf. Marshall's sign rule). This example shows that when a Hamiltonian is stoquastic it is necessarily sign-problem free, but the converse does not always hold.

Actually, the cases with the Heisenberg model with Marshall's sign rule (like this example) will also always have a simple local basis transformation that turns the Hamiltonian stoquastic, so this again perhaps motivates the definition allowing local transformations. However, ultimately, deciding whether a given Hamiltonian can be made stoquastic by a local basis transformation is known to be $$\mathsf{NP}$$-hard, so I think it's better to stay away from "curability with transformations". It's not ideal to have a definition that results in saying "For a given Hamiltonian, it is NP-hard to know if it is stoquastic" IMO. Furthermore, it will need to have another definition for "locality" which comes with additional baggage. Also, sign-problem freeness can be a bit different for fermionic spin systems with determinant Monte Carlo, so I think it's good to have a precise definition just about the Matrix.