# Why are diagonal Hamiltonians considered classical?

I've been following UT QML course (http://localhost:8888/tree/UNI/PHD/UT-QML) and during their lecture on the Ising hamiltonian, they point out that the hamiltonian of an Ising model without a transverse field commutes

$$H=-\sum_{} J_{ij} \sigma^Z_i \sigma^Z_{j} - \sum_i h_i \sigma^Z_i.$$

and To make this a quantum Ising model, we need to add a term that does not commute with the rest of the terms. A transverse field is such, which is an on-site interaction just like the external field.

$$H=-\sum_{} J_{ij} \sigma^Z_i \sigma^Z_{j} - \sum_i h_i \sigma^Z_i - \sum_i g_i \sigma^X_i.$$

I am confused by the following. What do they actually mean by a commuting hamiltonian? and second, what is the relationship between a hamiltonian commuting and its 'quantumness'?

In their written assignment, the difference between a quantum and non-quantum hamiltonian is just that one is a diagonal operator and the other is not.

• localhost:8888/tree/UNI/PHD/UT-QML is not a particularly useful URL. Mar 23 at 20:38
• Your URL doesn't work. Mar 25 at 16:55

## Classical Hamiltonians

By the spectral theorem, for every Hamiltonian there exists a basis in which it is diagonal. Thus, it is not correct to say that diagonal Hamiltonians are classical since this would apply to all Hamiltonians.

A Hamiltonian $$H$$ which is diagonal in a product basis $$\mathcal{B}$$ is sometimes described as classical, because the evolution of any initial state in $$\mathcal{B}$$ driven by such $$H$$ does not give rise to quantum effects such as superposition and entanglement. Moreover, a measurement in $$\mathcal{B}$$ at any point during the evolution yields a single, deterministic result and does not entail wavefunction collapse.

Note that the designation of certain bases as product bases and of certain Hamiltonians as classical depends on the choice of partitioning of a composite system into subsystems.

## Commuting tensor factors

The following characterization of classical Hamiltonians is useful.

Theorem Let $$H$$ be a Hermitian operator on a two-body Hilbert space $$\mathcal{H}_A\otimes\mathcal{H}_B$$. Then $$H$$ is diagonal in a product basis if and only if it can be written as $$H=\sum_{i=0}^{k-1}A_i\otimes B_i$$ for some integer $$k$$ and Hermitian operators $$A_i$$ and $$B_i$$ such that $$A_i$$ commutes with $$A_j$$ for every $$i,j=0,\dots,k-1$$ and $$B_i$$ commutes with $$B_j$$ for every $$i,j=0,\dots,k-1$$.

Proof Suppose $$H$$ is diagonal in the product basis $$|\psi_\alpha\rangle\otimes|\phi_\beta\rangle$$ with $$\alpha=0,\dots,m-1$$ and $$\beta=0,\dots,n-1$$ where $$m=\dim\mathcal{H}_A$$ and $$n=\dim\mathcal{H}_B$$. The spectral theorem guarantees that $$|\psi_\alpha\rangle\otimes|\phi_\beta\rangle$$ are orthogonal. Therefore, $$|\psi_\alpha\rangle$$ are orthogonal and $$|\phi_\beta\rangle$$ are orthogonal. Moreover, we can choose both bases to be orthonormal. Then, the eigendecomposition of $$H$$ is

$$H=\sum_{\alpha=0}^{m-1}\sum_{\beta=0}^{n-1}\lambda_{\alpha\beta}|\psi_\alpha\rangle\langle\psi_\alpha|\otimes|\phi_\beta\rangle\langle\phi_\beta|$$

where $$\lambda_{\alpha\beta}$$ is the eigenvalue of $$H$$ associated with eigenvector $$|\psi_\alpha\rangle\otimes|\phi_\beta\rangle$$. Let $$k=mn$$. Any $$i=0,\dots,k-1$$ can be written as $$i=\alpha n + \beta$$ for unique $$\alpha=0,\dots,m-1$$ and $$\beta=0,\dots,n-1$$. Define $$A_{\alpha n + \beta}=\lambda_{\alpha\beta}|\psi_\alpha\rangle\langle\psi_\alpha|$$ and $$B_{\alpha n + \beta}=|\phi_\beta\rangle\langle\phi_\beta|$$. Then

\begin{align} H &= \sum_{\alpha=0}^{m-1}\sum_{\beta=0}^{n-1}\lambda_{\alpha\beta}|\psi_\alpha\rangle\langle\psi_\alpha|\otimes|\phi_\beta\rangle\langle\phi_\beta| \\ &=\sum_{\alpha=0}^{m-1}\sum_{\beta=0}^{n-1}A_{\alpha n+\beta}\otimes B_{\alpha n+\beta} \\ &= \sum_{i=0}^{k-1}A_i\otimes B_i\tag1 \end{align}

and $$A_i$$ commutes with $$A_j$$ for every $$i,j=0,\dots,k-1$$ and $$B_i$$ commutes with $$B_j$$ for every $$i,j=0,\dots,k-1$$.

Conversely, suppose that $$H$$ can be written as $$H = \sum_{i=0}^{k-1}A_i\otimes B_i$$ with Hermitian operators $$A_i$$ and $$A_j$$ commuting for every $$i,j=0,\dots,k-1$$ and Hermitian operators $$B_i$$ and $$B_j$$ commuting for every $$i,j=0,\dots,k-1$$. The operators $$A_i$$ share a common eigenbasis, as do operators $$B_i$$. Let $$|\psi_\alpha\rangle$$ with $$\alpha=0,\dots,m-1$$ denote the shared eigenbasis of operators $$A_i$$ and $$|\phi_\beta\rangle$$ with $$\beta=0,\dots,n-1$$ the shared eigenbasis of operators $$B_i$$. We can write $$A_i=\sum_{\alpha=0}^{m-1} \mu_{i,\alpha}|\psi_\alpha\rangle\langle\psi_\alpha|$$ and $$B_i=\sum_{\alpha=0}^{n-1}\nu_{i,\beta}|\phi_\beta\rangle\langle\phi_\beta|$$ and

\begin{align} H &= \sum_{i=0}^{k-1}A_i\otimes B_i \\ &= \sum_{i=0}^{k-1}\left( \sum_{\alpha=0}^{m-1} \mu_{i,\alpha}|\psi_\alpha\rangle\langle\psi_\alpha|\right)\otimes \left(\sum_{\beta=0}^{n-1}\nu_{i,\beta}|\phi_\beta\rangle\langle\phi_\beta|\right) \\ &= \sum_{\alpha=0}^{m-1} \sum_{\beta=0}^{n-1} \left(\sum_{i=0}^{k-1} \mu_{i,\alpha}\nu_{i,\beta}\right) |\psi_\alpha\rangle\langle\psi_\alpha|\otimes |\phi_\beta\rangle\langle\phi_\beta| \\ &= \sum_{\alpha=0}^{m-1} \sum_{\beta=0}^{n-1} \lambda_{\alpha\beta} |\psi_\alpha\rangle\langle\psi_\alpha|\otimes |\phi_\beta\rangle\langle\phi_\beta| \tag2 \end{align}

where we defined $$\lambda_{\alpha\beta} = \sum_{i=0}^{k-1} \mu_{i,\alpha}\nu_{i,\beta}$$. Equation $$(2)$$ exhibits eigendecomposition of $$H$$ in the product basis $$|\psi_\alpha\rangle\otimes|\phi_\beta\rangle$$ completing the proof. $$\square$$

Remark 1: The theorem applies to many-body operators as well. This can be seen by suitable generalization of the proof above.

Remark 2: Note that in the above characterization of classical Hamiltonians it is key that tensor factors on each subsystem commute pairwise across all terms. It is insufficient for the whole terms on the composite system to commute pairwise. An example of a non-classical Hamiltonian with commuting terms is

\begin{align} H &= -\frac12\sigma_1^X \sigma_2^X - \frac12\sigma_1^Z \sigma_2^Z \\ &= -|\phi^+\rangle\langle\phi^+| + |\psi^-\rangle\langle\psi^-| \end{align}

whose non-degenerate ground state is $$|\phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$$. Another related example of a non-classical Hamiltonian with commuting terms is the Toric Code.

(Thanks to @NorbertSchuch who pointed out an error in an earlier version of this section and provided the above counterexample.)

The theorem has two consequences for the Hamiltonians in the question. Namely, it implies that the first Hamiltonian is classical and it justifies the addition of a transversal field in the second one as a valid attempt to make it non-classical. Its non-classical character can be confirmed by diagonalization.

## Terminology

Commutation is a relation between two operators, not a property of an operator. However, the expression "commuting Hamiltonian" is used sometimes as a shortcut for "Hamiltonian with pairwise commuting terms".

• For "classical", being diagonal in any product basis is sufficient. Also, commuting does not imply classical, see e.g. the toric code. Mar 23 at 23:21
• @NorbertSchuch Good catch! Thanks! Fixed both errors. Mar 24 at 1:53
• I still feel that the "Commuting tensor factor" part contains more misleading (and probably wrong) parts than not. A single two-body term can always be describes as classical. The question is whether there is one local (not necessarily single-site) basis which makes all Hamiltonian term diagonal. You can't judge this from a single term. In fact, all systems described by two-body Hamiltonian have ground states which are "almost" classical (in the sense that their ground space is spanned by products.) I feel it would be safer off to just say that commuting does not imply classical ... Mar 24 at 10:35
• ... and illustrate this with the toric code. As for the terminology, it might be poor, but it is used by reputable people ;) Mar 24 at 10:35
• @NorbertSchuch I did not mean to imply that we can judge classical character of a Hamiltonian from a single term, but I concede the text was not very clear or rigorous. Made it so now. I think the above characterization of classical Hamiltonians is interesting, useful and correct. In particular, it explains why we need a transverse field to make the Ising model exhibit quantum features. Mar 24 at 19:08

While Adam's very detailed answer is probably emaculate, it's a bit long so for people that want a shorter answer, I'll give a much more compact alternative. This is not at all to challenge or try to refute Adam's answer at all.

"What do they actually mean by a commuting hamiltonian?"

In the specific case in your question, they are referring to whether or not the terms in the Hamilotnian commute with each other. Terms $$A$$ and $$B$$ commute if: $$AB - BA = 0$$ where in this case $$A$$ and $$B$$ are Pauli matrices so $$AB$$ means matrix multiplication.

"what is the relationship between a hamiltonian commuting and its 'quantumness'?"

If your Hamiltonian is entirely comprised of $$Z$$ matrices, it only has diagonal elements (as you already noted) and those diagonal elements are the outputs of a classical function of classical variables: $$f(z_1,z_2,z_3,...,z_n)$$ where $$z_i \in \{-1,1\}$$, $$f(b_1,b_2,b_3,...,b_n)$$ where $$b_i$$ are the classical binary variables $$\{0,1\}$$ if you make the completely reversible transformation $$b_i = \frac{1}{2}(z_i+1)$$.

Many people also believe that a quantum annealer minimizing a diagonal Hamiltonian is no more powerful than a classical comptuer because the simplest Hamiltonian that has been proven to provide as much power as other quantum computers involves both $$X$$ and $$Z$$ and the same time; although a Grover-type search over the $$2^n$$ diagonal elements can be done in $$\mathcal{O}(\sqrt{2^n})$$ time rather than $$\mathcal{O}(2^n)$$, so there is indeed some potential for speed-up, but not as much as the full powerful of quantum computing would provide.

As Adam has noted, all Hamiltonians can be diagonalized, but in general that's hard to do.

I was confused the first time Tameem Albash put his hand up at the end of one of my talks in 2016 and said "but that Hamiltonian's classical" because I was talking about the interaction between electron spins, and spin is an inherently quantum property of a particle for which there is no classical analogue. But from a computational perspective, there's certainly a difference between a Hamiltonian that's already been diagonalized vs one that's not been, and if you do quantum annealing on a Hamiltonian that's already diagonal, it won't likely be as powerful as a universal quantum computer unless perhaps the driving Hamiltonian has both $$X$$ and $$Z$$ terms in the same basis.

To pose a very simple answer to compete with all these complex (but also excellent) answers: the Ising model is a classical Hamiltonian because it is diagonal as it's written and therefore all of its eigenstates are classical states in the $$S^z$$ basis (no superposition required).

• Time evolution produces no mixing of the $$S^z$$ basis.
• All observables (that you can write in the $$S_z$$ basis) commute with $$H$$ (and all the interesting observables are in the $$S^z$$ basis).
• You can use classical techniques to find the energy, understand phase transitions, etc.

## A classical limit of a quantum model

To elaborate a bit, the Ising model can be thought of as a classical limit of a quantum model. The whole idea that you would have some localized magnetic moment (spin) that has only two possible states is a very quantum idea. The quantum version of the Ising model is the Heisenberg model: $$H = -J \sum \limits_{} \vec S_i \cdot \vec S_j \tag{1}$$ A better way of writing the Heisenberg model is this: $$H = -J \sum \limits_{} S^z_i S^z_j + \frac{\Delta}{2} \left( S^+_i S^-_j + S^-_i S^+_j \right) \tag{2}$$ Now you have an Ising-like term and the off-diagonal interactions in the XY plane are separated out explicitly. If $$\Delta=1$$, you have the full isotropic Heisenberg model, but if you start turning off the XY interactions $$\Delta \to 0$$, you will arrive at the Ising model.

Here's the interesting part: the Heisenberg model has two classical limits. One is the Ising model. The other is the classical Heisenberg model, which is essentially the limit that the total spin $$S\to\infty$$. Then you just have the same Hamiltonian, but with classical vectors of length S.

In fact, the Heisenberg model for $$S=3/2$$ is less quantum-y than the $$S=1/2$$ case. The longer you make the quantum spin vector, the more it behaves like a classical vector.

## Getting more philosophical

1. Classical doesn't mean trivial. The vanilla Ising model is pretty interesting in its own right, but throw in a competing nearest neighbor interaction, or randomized interactions, it can become extremely difficult to find the ground state. See spin glass
2. The Ising model has no intrinsic dynamics. Think about it, it just stays in whatever state it starts in. Despite this, there are tons of studies of Ising model dynamics.
3. This is a weird special case of a classical limit. For a quantum particle in a box, there is only one classical limit, and even in that case you still have dynamics. In some sense, as you take $$\Delta\to0$$, you are literally turning off the interactions that make it quantum-y! It's very similar to taking the hopping to zero in a tight-binding/Hubbard model.
• +1 and welcome to Quantum Computing Stack Exchange! We hope to see much more of you in the future, and thank you for your contributions!! That's an excellent first post. I very much liked "classical doesn't mean trivial". Mar 25 at 18:33