# Valid Bogoliubov transformation?

Is it possible to diagonalize the Hamiltonian: $H=c_1 c_2^\dagger+c_2 c_1^\dagger +\delta(c_1c_2+c_2^\dagger c_1^\dagger )$ using the Bogoliubov canonical transformation, where the $c_1,c_2$ are standard fermion anihilation operators...?

I tried $\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix}$ $=$ $\begin{bmatrix} x &-y \\ y &x \end{bmatrix}$ $\begin{bmatrix} d_{1} \\ d_{2}^\dagger \end{bmatrix}$ and

$\begin{bmatrix} c_{1} \\ c_{2}^\dagger \end{bmatrix}$ $=$ $\begin{bmatrix} x &-y \\ y &x \end{bmatrix}$ $\begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix}$

At the end I allways get terms like: $x^2 d_1 d_2 ^\dagger -y^2 d_2^\dagger d_1 +x^2 d_1 d_2 -y^2 d_2^\dagger d_1^\dagger+h.c.$. In order for the Hamiltonian to be diagonal I require: $x^2 d_1 d_2 ^\dagger -y^2 d_2^\dagger d_1 +x^2 d_1 d_2 -y^2 d_2^\dagger d_1^\dagger+h.c.=0$ The problem is that there is no connection between let's say: $d_1d_2^\dagger$ and $d_2^\dagger d_1^\dagger$ in terms of anticommutation relations. Can I try something like this: $\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix}$ $=$ $\begin{bmatrix} x &-y \\ y &x \end{bmatrix}$ $\begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix}$ or does this transformation not preserve the canonical anticommutation relations(CAR)?