Bogoliubov transformation with a slight twist

This is an eigenvalue problem.

Let's assume your Bogoliubov transformation is of the form: $(a_k,b_k)^T=X(c_k,d_k)^T$. What this transformation do is let your Hamiltonian become: $H_k=w_1c_k^\dagger c_k+w_2 d_k^\dagger d_k$, with the anti-commute relation holds for new field operators $c_k$ and $d_k$.

Now you can check that $X$ is just the matrix where its columns are just the Normalized Eigenvectors of your original matrix.

I would just like to point out that the given Hamiltonian does not require a Bogoliubov transformation to be diagonalized, since it is of the form of a single-particle operator (nevertheless in second quantization) i.e. does not contain 'off-diagonal' terms of the form $a a$,...

You can simply diagonalize it by diagonalizing the coupling matrix.

@leongz: although this matrix is also Hermitian for the true Bogoliubov case, you will generally get the wrong answer for the eigenenergies and modes if you diagonalize it. The resulting modes would not be bosonic, i.e. it would not be a canonical transformation. You can obtain the right answer (which is much more powerful than the typical ansatz for the Bogoliubov operators) by diagonalizing $\Sigma H$, where $\Sigma$ is the pseudonorm on the sympletic space you're working on. Note however, that this matrix is not always Hermitian (and not always diagonlaizable - but this is physical: one bosonic mode is missing for each Goldstone mode).