# Bogoliubov transformation, same eigenvalues

I have been reading about the Bogoliubov transformation of creation and annihilation operators: \begin{align} b & = u\,a + v\,a^\dagger\\ b^\dagger & = u^* a^\dagger+v^*a \end{align}

where $a,a^\dagger$ are the original operators and $b,b^\dagger$ are the "new" creation and annihilation operators.

My question: Why does the transformation preserve the eigenvalues of a hamiltonian containing $a$ and $a^\dagger$? Is there an easy way to explain this? I have looked at similar question, but still don't get it...

Similar questions:

Why must the Bogoliubov transform preserve anticommutation relations?

Bogoliubov transformation is not unitary transformation, correct?

Why would it change the eigenvalues of the hamiltonian? It is exactly the same hamiltonian, you've just re-expressed it in a form which is easier to understand. When you simplify a hamiltonian in the form \begin{align} H & = A\, a^\dagger a + B \, a^\dagger a^\dagger+ B^*a a + C \\ & = \omega b^\dagger b \end{align} for some suitable values of the parameters $u,v$, and $\omega$, the second equals sign really is an equals sign, i.e. it is the very same operator, and the only thing that's changed is that now you have a clearer canonical form to assign it. And, since it's the same operator, acting in the same way on the same space, the eigenvalues simply cannot change.