# How to discretize a Hamiltonian?

In the famous article Physical Review B 92, 064520 (2015), a theoretical model was proposed to realize the chiral Majorana zero mode.

$$H_{\mathrm{BdG}}=\left(\begin{array}{cc}{H_{0}(\mathbf{k})-\mu} & {\Delta_{\mathbf{k}}} \\ {\Delta_{\mathbf{k}}^{\dagger}} & {-H_{0}^{*}(-\mathbf{k})+\mu}\end{array}\right)$$

Here, $$\mathcal{H}_{\mathrm{BdG}}=\sum_{\mathbf{k}} \Psi_{\mathbf{k}}^{\dagger} H_{\mathrm{BdG}} \Psi_{\mathbf{k}} / 2 ,$$ where $$\Psi_{\mathbf{k}}=\left[\left(c_{\mathbf{k} \uparrow}^{t}, c_{\mathbf{k} \downarrow}^{t}, c_{\mathbf{k} \uparrow}^{b}, c_{\mathbf{k} \downarrow}^{b}\right),\left(c_{-\mathbf{k} \uparrow}^{t \dagger}, c_{-\mathbf{k} \downarrow}^{t \dagger}, c_{-\mathbf{k} \uparrow}^{b \dagger}, c_{-\mathbf{k} \downarrow}^{b \dagger}\right)\right]^{T}.$$

Obviously, the Hamiltonian contains both $$k_x$$ and $$k_y$$.

My question is how to discretize the Hamiltonian and apply it to a square lattice ribbon (of finite size in the $$y$$ direction but periodic in the $$x$$ direction)?

A tight-binding Hamiltonian $$H = -t\sum_{j} c^{\dagger}_j c_{j+1} +{\rm h.c.}$$ in the momentum representation is $$H=-2t\sum_k\cos(k)c^{\dagger}_k c_k$$ and when we expand for small $$k$$ we get the quadratic dispersion of a free particle. In a similar manner, if we will hop with $$it$$ instead of $$t$$ we will get a $$\sin(k)$$ dispersion that for small $$k$$ will give the linear behavior you are looking for. So to answer your question - you discretize by writing a tight-binding Hamiltonian where quadratic dispresion are real tight-binding hopping and linear terms in $$k$$ are tight-binding with hopping $$i$$. Note that you must adjust the chemical potential $$\mu$$ accordingly. For example, in 1d the tight-binding Hamiltonian $$H = -t\sum_{j} c^{\dagger}_j c_{j+1} +{\rm h.c.} - \mu\sum_j c^{\dagger}_j c_j = -\sum_k[\mu+2t\cos(k)]c^{\dagger}_kc_k$$ and one chooses $$\mu=-2t$$ to get the bottom of the band to be near zero energy, which is $$\mu=0$$ for the continuous model.
HOWEVER, and this is a big however, one should proceed with care as to the phase diagram, since now you have also $$k=\pi$$ as low energy states, which in the continuous model will be high-energy states! In general it should work at least for a limited part of the phase diagram, but you should check for this specific model what you get.