# Fourier Transform of a HgTe 2D strip

I am currently trying to derive the Fourier transform of a 2D HgTe Hamiltonian, with $k_x$ PBC and vanishing boundary conditions in the y direction at 0 and L. Here is the Hamiltonian:

$$H = \sum_{k}\tilde{c_k}^{\dagger}[A\sin{k_x}\sigma_x + A\sin{k_y}\sigma_y + (M-4B+2B[\cos{k_x} + \cos{k_y}])\sigma_z]\tilde{c_k}$$

Where the $\sigma_i$ are the Pauli matrices, and $A$, $M$, and $B$ are material dependent parameters, and the $\tilde{c_k}^{\dagger}$ and $\tilde{c_k}$ are the Nambu spinors ($\tilde{c_k} = (c_k,c^{\dagger}_k)^T)$ and the c's are the fermion creation/annihilation operators. So what I want to do is Fourier transform in the $y$ direction to real space, and leave the $x$ direction in momentum space. My attempt was to split the $H$ into two parts,

$$H_x = \sum_{k_x}\tilde{c_k}^{\dagger}[A\sin{k_x}\sigma_x + (M-4B+2B[\cos{k_x})\sigma_z]\tilde{c_k}$$

$$H_y = \sum_{k_y}\tilde{c_k}^{\dagger}[ A\sin{k_y}\sigma_y + (M-4B+2B[\cos{k_y}])\sigma_z]\tilde{c_k}$$

I took the Fourier transform into real space for $H_y$ and got this:

$$H_y = \sum_i^L\tilde{c_i}^{\dagger}D\tilde{c_i} + \sum_i^{L-1}(\tilde{c_{i+1}}^{\dagger}T\tilde{c_i} + \tilde{c_i}^{\dagger}T^{\dagger}\tilde{c_{i+1}})$$

With the matrices being equal to $$D = (M-4B)\sigma_z, \quad\quad T = \left(\frac{B}{a^2}\sigma_z - \frac{A}{2ia}\sigma_y\right)$$

But now it doesn't make sense to add the two parts together...I was thinking that maybe I fourier transform the whole thing into real space, and then write it in Nambu notation and include boundary conditions on the matrix like for y direction j and x direction i $c_i = c_{L+1}$ and $c_{j-1} = 0, c_{L+1} = 0$. Could someone help point me in the right direction? I've worked on 1D matrices in PBC before but I'm not sure how it should look in 2D. Would the Nambu spinors be 4N length?