# How is it possible a quadratic Hermitian Hamiltonian $H = b^\dagger b^\dagger + b b$, with $b$ boson, cannot be diagonalized?

Essentially, you are trying to prove that your initial Hamiltonian can be rewritten as $$H = r a^\dagger a + sI\tag{1}$$ where $$r$$ and $$s$$ are reals and $$a^\dagger$$ and $$a$$ satisfy bosonic ccrs and there is a vacuum vector for $$a$$.

This is a much stronger condition than diagonalizability!

However, it is possible to prove that the requirement above cannot be satisfied. In fact, it immediately arises from (1) that $$H$$ is either bounded below or above depending on the sign of $$r$$, because

$$\langle \psi| H \psi \rangle = r \langle a \psi| a \psi \rangle + s \langle \psi |\psi \rangle = r ||a\psi||^2 + s||\psi||^2$$ Instead (with some technical hypothesis on the domain and some other technicality as Stone-von Neumann's theorem) one can prove that $$b^\dagger b^\dagger + bb$$ is not bounded below nor above. Therefore operators $$a$$ and $$a^\dagger$$ cannot exist.

To prove that the initial Hamiltonian cannot be bounded, observe that defining $$X = b + b^\dagger$$ and $$P = i(b-b^\dagger)$$, these operators satisfy standard ccr up to a real factor and $$H$$ is proportional to $$P^2-X^2$$ which is unbouded below and above (here technicalities should be used). In fact, this is the Hamiltonian of a repulsive oscillator so it is unbouded below, but a unitary transformation which therefore preserves the spectrum swaps $$X$$ and $$P$$ changing the sign of $$H$$, hence $$H$$ is also unbounded above.

Finally, all that does not mean that $$H$$ is not "diagonalizable". $$H$$ is at least Hermitian, presumably selfadjoint on some domain (it depends on details) thus it admits a PVM over the reals. However, the spectrum is continuos as it happens for the repulsive oscillator, instead of discrete.