I was doing a calculation in Giamarchi's Introduction to Many Body Physics, chapter 3, on BCS theory and second quantization, and ran into some confusion with the BCS Hamiltonian. The pdf is here for your reference: http://dpmc.unige.ch/gr_giamarchi/Solides/Files/many-body.pdf
The main confusion comes with eqn. 3.154. Here, the BCS Hamiltonian is given by
$$ H=\sum_k \left( A(k)(\beta_k^\dagger \beta_k-\alpha_k^\dagger \alpha_k)+\Delta(\alpha_k^\dagger \beta_k+\beta_k^\dagger \alpha_k)\right)+ \sum_k\xi(k) $$
Where $\xi(k)$ and $A(k)$ are functions of $k$ and $\alpha_k$ and $\beta_k$ are fermionic operators. Now, I know that the tight-binding Hamiltonian with a periodic potential is given by eqn. 3.128: $$ H=\sum_{k}\bigg( A(k)(\beta_k^\dagger \beta_k-\alpha_k^\dagger \alpha_k)+V(\alpha_k^\dagger \beta_k+\beta_k^\dagger \alpha_k)\bigg) $$ The solution to this is easy to solve with the Bogoliubov transformation, and is given by
$$ E(k)=\sqrt{\xi(k)^2+V^2} $$ I was able to derive this without any problems. However, my question is this: would the solution of the BCS Hamiltonian then be
$$ E_{BCS}(k)=\sqrt{\xi(k)^2+V^2}+\xi(k) $$
Or would it be identical to the tight-binding Hamiltonian? Would the eigenvectors as given by the Bogoliubov transformation change as well?