# What is the number of fermions in Kitaev honeycomb model?

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One way to solve the Kitaev honeycomb model: $$H = J_x \sum_{\textrm{x links}, <ij> } \sigma^x_i \sigma^x_j + J_y \sum_{\textrm{y links}, <ij> } \sigma^y_i \sigma^y_j + J_z \sum_{\textrm{z links}, <ij> } \sigma^z_i \sigma^z_j.$$

is to represent the spin operators by Majorana fermions (arXiv:cond-mat/0506438) which makes the Hamiltonian in a (sort of) quadratic form in Majorana fermions. This Hamiltonian can in turn be written in a quadratic form in some Dirac fermions: $$H = \sum_{ij} t_{ij} a^{\dagger}_i a_j + H.C.$$

Now, it is my understanding that to find the ground state of the system, we must fill the lowest lying states by $a_i$ fermions. Now, my question is: what is number of $a_i$ fermions, $N$ in this system (which is a conserved quantity)? and how is it related to physics of the original spin model?

This post imported from StackExchange Physics at 2015-03-01 12:40 (UTC), posted by SE-user Hamed
asked Feb 27, 2015

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