# What is a free fermion model?

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

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

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?

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lcv's answer to this question physics.stackexchange.com/q/2014/2451 seems relevant, see physics.stackexchange.com/questions/2014/… – 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