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In classical mechanics it is known that a certain model is solvable exactly (integrable) if it posses a sufficient amount of "well behaved" conserved charges.

On the other hand in quantum mechanics we say that a model described by Hamiltonian $H$ is integrable if it can be solved using the algebraic Bethe ansatz (look at this related question for a more clear explanation and references). If that is the case we can construct a transfer matrix which is able to generate a (possibly infinite) set of conserved charges $\{ \hat{Q}_i \}$ such that

\begin{equation} [H, \hat{Q}_i]=0 \quad \text{and} \quad [\hat{Q}_i,\hat{Q_j}]=0 \quad \forall i,j. \tag{1} \end{equation}

At this point it seems natural for me to make a parallel with classical mechanics and state that a quantum system is integrable if it posses a set of conserved charges as the one defined in $(1)$.

However given $H$ we can always construct a set of "well behaved" conserved charges whose dimension is the same as the dimension of the Hilbert space by simple taking the eigenstates $|\psi_i \rangle$ of $H$ and defining \begin{equation} \hat{Q}_i = |\psi_i\rangle \langle \psi_i |. \tag{2} \end{equation} To me it is obvious that this cannot imply that $H$ describes an integrable model because if that were the case then any model should be integrable. My immediate solution to this problem is to say that the charges $\hat{Q}_i$ constructed in $(2)$ are "functionally dependent" on $H$, in the sense that they were build using $H$ and they do not actually contribute any new information to solving the problem.

My questions are: Is my line of logic correct or am I making a huge mistake? And if I am correct is there a way to express the idea of "functionally dependent" in a more rigorous mathematical way?

Truth-Beauty-and Hatred
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  • Could you provide more background for the statement that a model is integrable if it has enough conserved charges? Is it a theorem? Can you provide a reference? – flippiefanus Feb 12 '24 at 02:20
  • I have edited my question so hopefully now it is more clear. – Truth-Beauty-and Hatred Feb 12 '24 at 09:07
  • WP. – Cosmas Zachos Feb 12 '24 at 12:18
  • This might be salutary. – Cosmas Zachos Feb 12 '24 at 15:39
  • Oh, after writing my answer here I realised that this question is a duplicate of https://physics.stackexchange.com/q/781632/ , where I wrote an answer too. Since there's some overlap between my answers, but they're not identical, I'll keep both – Jules Lamers Mar 18 '24 at 19:50
  • An important thing to note is that $\hat{Q}_i = |\psi_i\rangle \langle \psi_i |$ has some undesirable properties. In particular, it is generically an extremely nonlocal operator for eigenstates towards the middle of the spectrum. Thus the "sum of local operators" generated by derivatives of the transfer matrix (or even sums of quasilocal operators in the XXZ model found by differentiating with respect to a parameter that would naively make the transfer matrix non-commuting), are perhaps more natural objects to consider. – user196574 Mar 18 '24 at 20:12

1 Answers1

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Exactly this objection, and possible resolutions, are discussed in

J.-S. Caux and J. Mossel, Remarks on the notion of quantum integrability, J. Stat. Mech. (2011) P02023 (arXiv:1012.3587).

See the notion of "QI:N" and ensuing discussion therein. Here's the abstract:

We discuss the notion of integrability in quantum mechanics. Starting from a review of some definitions commonly used in the literature, we propose a different set of criteria, leading to a classification of models in terms of different integrability classes. We end by highlighting some of the expected physical properties associated to models fulfilling the proposed criteria.

Another approach, using representation theory, moves the notion of "(infinitely, or at least sufficiently) many conserved charges" into the abstract algebra itself. I believe this works as follows. Consider the Heisenberg XXX spin chain, for example, where the integrability is known to be governed by an infinite quantum group called the Yangian (of $SU(2)$, or more precisely $\mathfrak{gl}_2$). The transfer matrix $t(u)$, which is the generating function of the conserved charges, can already be defined 'abstractly', as an element of the Yangian itself (rather than in some representation like a spin chain). More precisely, one can show $[t(u),t(v)]=0$ holds not just inside any representation, but already abstractly. That is, $t(u)$ defines a 1-parameter family of commuting operators inside the Yangian. Its (infinitely many) coefficients as a (formal) power series in $u$ all commute with each other. In other words, those 'operator modes' form an abelian subalgebra of the Yangian. It can be shown (maybe one needs to introduce a small 'twist' to make it rigorous) that this is in fact a maximal abelian subalgebra of the Yangian: you can't add anything else without spoiling commutativity. In a representation, say $L$ spin 1/2 sites (tensor product of evaluation modules with equal inhomogeneities), this maximal abelian algebra provides the "infinitely many" commuting charges (though they won't be all independent anymore in this finite dimensional setting). This maximal abelian subalgebra is called the Bethe subalgebra, since it is simultaneously diagonalised by the (e.g. algebraic) Bethe ansatz.

Jules Lamers
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