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The free fermion Hamiltonian for the 1D quantum Ising model is $$H = -J\sum_i (c_{i}^{\dagger }c_{i+1} +c_{i+1}^{\dagger }c_{i}+c_{i}^{\dagger }c_{i+1}^{\dagger }+c_{i+1}c_{i}-2gc_{i}^{\dagger }c_{i} +g)$$ where the sum is over $i$, the site index.

My question is - what are the basis states of this system? I read somewhere that the matrix representation of the Hamiltonian of this form has $L \times L$ elements where $L$ are the total sites. I don't see how that is. (As opposed to the spin states where the matrix would be $2^{L}\times 2^{L}$)

I eventually want to find the ground state but do not want to resort to a Fourier transform to the momentum states, so I want the matrix in the fermionic basis states form only.

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The $2L$-by-$2L$ matrix is the one that is diagonalized by means of a Bogoliubov transformation. After a suitable such transformation $H$ ends up as $\sum_m E_m b_m^\dagger b_m$. I'm afraid that some Fourier stuff will be necessary. You start by writing $$ H= (c^\dagger_i,c_j) \left(\matrix{ A_{ik} & B_{il}\\ C_{jk} & D_{jl}}\right) \left(\matrix{ c_k \\ c^\dagger_l}\right). $$

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  • $\begingroup$ But @mike, eventually I will want to introduce disorder (not for the Ising model, but the Kitaev model) such that k is not a good quantum number. $\endgroup$
    – user2578520
    Commented Mar 20, 2017 at 4:56
  • $\begingroup$ Well, for a finite chain, it always possible to do the $2L$-by$2L$ Bogoliubov diagonalization numerically. I think its actually easier for Fermions because diagonlizaing the Hamitonian can be though of as conjugating an element of the Lie algebra of ${\rm SO}(2N)$ into its maximal torus. For Bosons the linear algebra involves the non-compact {\rm Sp}(2N, {\mathbb R})$, and there are all sorts of different cases. $\endgroup$
    – mike stone
    Commented Mar 20, 2017 at 15:28

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