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  What is the definition of a quantum integrable model?

+ 4 like - 0 dislike

What is the definition of a quantum integrable model?

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

This post imported from StackExchange Physics at 2014-04-25 13:36 (UCT), posted by SE-user lagoa
asked Sep 9, 2013 in Theoretical Physics by lagoa (20 points) [ no revision ]

2 Answers

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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)

This post imported from StackExchange Physics at 2014-04-25 13:36 (UCT), posted by SE-user Bubble
answered Sep 9, 2013 by Bubble (210 points) [ no revision ]
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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.

This post imported from StackExchange Physics at 2014-04-25 13:36 (UCT), posted by SE-user just-learning
answered Feb 21, 2014 by just-learning (95 points) [ no revision ]

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