# Exact diagonalization by Bogoliubov transformation

 P: 16 Hello all, I am developing a model of multiple gaps in a square lattice. I simplified the associated Hamiltonian to make it quadratic. In this approximation it is given by, $$H = \begin{pmatrix} \xi_\mathbf{k} & -\sigma U_1 & -U_2 & -U_2\\ -\sigma U_1 & \xi_{\mathbf{k}+(\pi,\pi)} & 0 & 0\\ - U_2 & 0 & \xi_{\mathbf{k}+(\pi/2,0)} & 0\\ - U_2 & 0 & 0 & \xi_{\mathbf{k}+(0,\pi/2)} \end{pmatrix}$$ And my Nambu operator is given by, $$ψ_\mathbf{k} = \begin{pmatrix} c_{\mathbf{k},\sigma} \\ c_{\mathbf{k}+(\pi,\pi),\sigma} \\ c_{\mathbf{k}+(\pi/2,0),\sigma} \\ c_{\mathbf{k}+(0,\pi/2),\sigma} \end{pmatrix}$$ I tried to diagonalized by making three Bogoliubov transformations, the first to diagonalize the upper right submatrix of H, and then the two others (a sort of nested transformations). But I get a lengthy result, what I would like to know if there is a smart transformation which allows me to write $$H = A_1^\dagger A_2^\dagger A_3^\dagger D A_3 A_2 A_1$$ or simply $$H = U^\dagger D U$$ Or the only way is to use just brute force? Thanks