Let's say I have a Hilbert space $\mathcal{H}$ (either finite-dimensional or with a countably infinite basis) with a specified Hamiltonian $\hat{H}$, representing some quantum system. Under what conditions can I factorize into tensor product components, i.e. find Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ such that

$\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2 $

and Hamiltonians $\hat{H}_1$ and $\hat{H}_2$ such that

$\hat{H} = \hat{H}_1 \otimes 1 + 1 \otimes \hat{H}_2$ ?

The idea is to figure out whether a quantum system can be factored into two non-interacting subsystems.

For simple cases such as trying to decompose a 4D Hilbert space into two qubit systems I can write out equations for matrix elements of $\hat{H}$ in terms of matrix elements of $\hat{H}_1$ and $\hat{H}_2$ in an arbitrary basis to see when they are solvable, but I am wondering if there is a more general theory for larger systems with some physical intuition behind it.

(Edit: I should also point out that the choice of basis matters in deciding whether you can do this or not, so I assume that in order for this question to be well-defined you must provide some extra structure. For example, as I understand it this paper discusses what structure is needed for a factorization of a Hilbert space into tensor products, and suggests that one should pick out a collection of subalgebras of observables satisfying certain axioms in order to define a basis-independent "tensor product structure").

I thought I ran across a paper which discussed this sort of problem when browsing through the literature for information about a different topic several months ago, but unfortunately I can't remember enough information about the paper I saw to find it again or even the right keywords to search for in case "tensor product factorization" is not the right terminology for what I am trying to do.

I would also be interested in learning about answers to generalizations of this question for which the two subsystems are allowed to interact with one another but only weakly, with something like $\hat{H} = \hat{H}_1 \otimes 1 + 1 \otimes \hat{H}_2 + \hat{H}_{\textrm{int}}$

for an interaction $\hat{H}_{\textrm{int}}$ that is "small" in some sense.


2 Answers 2


This is possible iff the spectrum of the Hamiltonian (including multiplicities of eigenvalues) can be written as the sum of two other spectra. So it doesn't depend on a basis.

  • $\begingroup$ I have a bit of a problem with the formulation of your answer. In my understaning, the spectrum of a linear operator does not depend on a basis. Never. So what does your last sentence mean? Moreover: Do you have a reference to a proof? $\endgroup$ May 28, 2020 at 21:21

The answer by jjcale gets it right. But let me elaborate a bit. You've basically asked:

Given an abstract Hamiltonian, when can we factorize the Hilbert space such that Hamiltonian is non-interacting with respect to that choice of degrees of freedom?

This is an interesting question because we're asking which Hamiltonians have a particularly simple description in terms of non-interacting parts.

You can explicitly characterize the Hamiltonians where this is possible, by specifying the necessary and sufficient conditions on their spectra: that's what jjcale answered. Of course, only very special Hamiltonians will have this property.

Meanwhile, you might ask the more general question:

Given an abstract Hamiltonian, when can we factorize the Hilbert space such that Hamiltonian is locally interacting with respect to that choice of degrees of freedom?

The answer to this more general question is again that only special Hamiltonians have this property. That is, only very special Hamiltonians allow a factorization of the Hilbert space such that the Hamiltonian looks like a sum of local interaction terms. The more interesting question then becomes: when a Hamiltonian does allow such a factorization, is it unique? The answer is essentially "yes, most of the time."

Yes, here I'm making a plug: my co-authors and I address this question in "Locality from the Spectrum."

So when you do have an abstract Hamiltonian that allows a tensor factorization of the Hilbert space such that the Hamiltonian looks local, this factorization is usually unique, and gives you something like a preferred basis where the Hamiltonian is local.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.