# How to diagonalize the BCS Hubbard Hamiltonian using the Bogoliubov transformation?

How do I diagonalize the following BCS (Bardeen-Cooper-Schrieffer) Hubbard Hamiltonian: $$H= \sum\limits_{k \in [-\frac{π}{2}, +\frac{π}{2}[} \begin{bmatrix}c^\dagger_k & c^\dagger_{k+\pi} \end{bmatrix}\hat{\cal H}\begin{bmatrix}c_k \\ c_{k+\pi} \end{bmatrix} \space\space\space\space\space with\space\space\space\space\space \hat{\cal H}= \begin{bmatrix} \epsilon_k & v \\ v & -\epsilon_k \end{bmatrix}$$ using the following Bogoliubov transformation: $$a^\dagger_k=u_k c^\dagger_k + v_k c^\dagger_{k+\pi}\\ b^\dagger_k=v_k c^\dagger_k - u_k c^\dagger_{k+\pi}\\ with \space\space\space\space u^2+v^2=1$$ The result should be $$H= \sum\limits_{k \in [-\frac{π}{2}, +\frac{π}{2}[} E_k (b^\dagger_{k}b_{k}-a^\dagger_{k}a_{k}) \space\space\space with \space\space\space E_k=\sqrt{\epsilon_k^2+v^2}$$

I tried to reverse the Bogoliubov transformations to find the transformation for $c_k$, $c^\dagger_k$,$c_{k+\pi}$, $c^\dagger_{k+\pi}$ and insert it, but this gives only a sum of 16 terms involving linear combinations of $a^\dagger_ka_k,b^\dagger_kb_k,a^\dagger_kb_k,b^\dagger_ka_k$. And the terms $\epsilon_k$ and v only show up linearly, so I don't see how a term like $\sqrt{\epsilon_k^2+v^2}$ should show up.

At first, you should recognize that the eigenvalues of Hamiltonian is nothing but $\sqrt{\epsilon_k^2+\nu^2}$, according to $|\lambda I-H|=0$.
Then, you can find eigenvectors expressed by $c_k$ and $c_{k+\pi}$ also $c_k^\dagger$ and $c_{k+\pi}^\dagger$, which is nothing but the Bogoliubov transformation.
Usually in textbook, people write down Bogoliubov transformation without any explanation, just some mathematical expression. So you don't know why you should mix $c_k$ and $c_{k+\pi}$ also $c_k^\dagger$ and $c_{k+\pi}^\dagger$, and what will do next step.
For your case, you have expressed Hamiltonian by $a$ and $b$, then you should find proper values of $u$ and $v$ to make Hamiltonian satisfy the diagonal form, in which equations are not difficult to solve.