Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Title says it all really.. Why is the XX spin chain a free fermion (non-interacting) model, and the XXZ chain not?

Is it right that $\sum_l a_l^\dagger a_{l+1}$ isn't an interaction between fermions because it's creating a fermion on one site and destroying it on another? But why is $\sum_l a_l^\dagger a_l a_{l+1}^\dagger a_{l+1}$ an interaction term?

Is something like

\begin{equation} H_1 = -\sum_l (J+(-1)^lK) ( \sigma_l^x \sigma_{l+1}^x +\sigma_l^y \sigma_{l+1}^y) \end{equation}

a free fermion model? If not, why not?

Edit I don't have enough reputation to set a bounty, but if anyone could answer this question, I'd be very grateful!

Edit 2 Anyone?

share|cite|improve this question
lcv's answer to this question seems relevant, see… – Qmechanic Nov 8 '11 at 18:35
@Qmechanic Thank you for replying. As far I can see, Icv's answer just mentions free fermion models but doesn't say what they actually are, which is what I'm asking. – user6050 Nov 8 '11 at 18:50
Whenever the Hamiltonian may be written as at most bilinear polynomial of the basic fields, it's a "free theory". Free fermion models are models with at most quadratic terms in the fermions. Such Hamiltonians are solvable in terms of one-particle states that are occupied by particles which move independently of each other. Higher-than-quadratic terms in fermions are called "interacting" because they interact: energy eigenstates can't be easily obtained from free one-particle states. If you use Feynman diagrams, interactions produce vertices of the diagrams. – Luboš Motl Nov 8 '11 at 20:25
@LubošMotl Thank you. So the example Hamiltonian is a free fermion model since in fermions, it only contains $\sum_l a_l^\dagger a_{l+1}$ and $\sum_l (-1)^l a_l^\dagger a_{l+1}$ type terms. By this reasoning, even a spin chain $H_2 = - \sum_l J_l (\sigma_l^x \sigma_{l+1}^x + \sigma_l^y \sigma_{l+1}^y)$ where $J_l$ is different for each $l$ is a free fermion model. Is that right? – user6050 Nov 9 '11 at 13:05
@LubošMotl Also, if you were to expand your comment as an answer, I'd be happy to accept it. – user6050 Nov 10 '11 at 10:15

To expand on LubošMotl's comment, see the following classic paper by Lieb, Schultz and Mattis. For one-dimensional systems and nearest neighbor interactions, the spin chain that you mention as an example in the comment can be converted into a free fermionic model. See section II in the above paper for details.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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