What is the definition of a quantum integrable model?

To be specific: given a quantum Hamiltonian, what makes it integrable?

  • $\begingroup$ There is a precise definition of quantum integrability for explicitly time-dependent Hamiltonians: arxiv.org/abs/1711.09945 $\endgroup$
    – user228933
    Apr 22, 2019 at 4:48

2 Answers 2


Quantum integrability basically means that the model is Bethe Ansatz solvable. This means that we can, using the Yang-Baxter relation, get a so-called "transfer matrix" which can be used to generate an infinite set of conserved quantities, including the Hamiltonian of the system, which, in turn, commute with the Hamiltonian. In other words, if we can find a transfer matrix which satisfies the Yang-Baxter relation and also generates the Hamiltonian of the model, then the model is integrable.

Please note that, oddly enough, a solvable system is not the same thing as an integrable system. For instance, the generalized quantum Rabi model is not integrable, but is solvable (see e.g. D. Braak, Integrability of the Rabi Model, Phys. Rev. Lett. 107 no. 10, 100401 (2011), arXiv:1103.2461).

A nice introduction to integrability and the algebraic Bethe Ansatz is this set of lectures by Faddeev in Algebraic aspects of the Bethe Ansatz (Int. J. Mod. Phys. A 10 no. 13 (1995) pp. 1845-1878, arXiv:hep-th/9404013)

  • $\begingroup$ This sound very interesting. Could you maybe describe the implications of the existence of this transfer matrix? How does it help? $\endgroup$
    – ungerade
    Nov 29, 2016 at 17:25

If we deal with quantum finite-dimensional systems without spin, the definition is this: if we have a system with $n$ degrees of freedom whose (quantum) Hamiltonian is given by an operator $H$, then this system is called integrable if there exist $n$ independent operators $K_i$ such that $K_1=H$ and $[K_i,K_j]=0$ for all $i,j=1,\dots,n$. All operators, of course, are assumed to be (formally) self-adjoint.

The matter of how one should interpret the word "independent" here is a bit tricky. Linear independence is not sufficient, and we should at least require the functional independence of classical limits of $K_j$ for all $j=1,\dots,n$.

For further details see e.g. Definition 6 in this paper by Miller, Post and Winternitz and references therein.

  • 5
    $\begingroup$ This definition is useless because in any system there will always be $n$ independent operators that commute with $H$ and among themselves - the projectors onto the eigenstates. That's just a restatement that the Hamiltonian can be diagonalized. So clearly integrability is not the same as diagonalizability, and some notion of locality or whatever is required of your operators $K_i$. $\endgroup$
    – nervxxx
    Oct 1, 2014 at 17:55

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