quantum mechanics - Second quantization and Hamiltonian diagonalization

So I want to diagonalize my Hamiltonian (it is bosonic hamiltonian) which is:

$H=(E+\Delta)a^{\dagger}a + 1/2\Delta(a^{\dagger}a^{\dagger} + aa)$

My class didn't cover this material so I don't really know how to procede. I would be grateful for any literature which covers this topics and a problem book with solutions would be great too.

What I tried to do was writing my Hamiltonian in matrix form which would be: $\begin{pmatrix} 1/2 \Delta & 1/2(E+\Delta) \\ 1/2(E+\Delta) & 1/2 \Delta \\ \end{pmatrix}$

And then diagonalize it, find eigenstates etc. Is this the correct way?

Answer

Diagonalizing the Hamiltonian means you want to bring it into the form $H=\omega b^\dagger b$, and it is pretty obvious that $b$ should be a linear combination of $a$ and $a^\dagger$, and $b$ should satisfy the canonical commutation of annihilation operators, namely $[b,b^\dagger]=1, [b,b]=0$.

Now let's write $b=ua+va^\dagger$ (this is called the Bogoliubov transformation, by the way). The condition $[b,b^\dagger]=1$ leads to $|u|^2-|v|^2=1$. Let us expand out $b^\dagger b$:

$$b^\dagger b= |u|^2 a^\dagger a+ |v|^2 a a^\dagger + u^*v a^\dagger a^\dagger + uv^* aa.$$

Therefore

$$\omega(|u|^2+|v|^2)=E+\Delta, \omega u^*v = \frac{1}{2}\Delta.$$

Together with $|u|^2-|v|^2=1$, we have three equations for three variables ($u, v, \omega$). In fact, in this case one can safely assume $u$ and $v$ are both real. The rest is just algebra.