Diagonalisation of a quadratic Hamiltonian for bosons

Question

As far as I know, every quadratic Hamiltonian can be diagonalised by a suitable basis change. Let me consider the basic squeezing Hamiltonian $$ H = \gamma (a^2+{a^\dagger}^2) $$ defined for the bosonic creation and annihilation operators $a$, $a^\dagger$. If I follow the standard approach of the Bogoliubov transformation, namely looking for a $b=u a + v a^\dagger$ such that $H=\omega b^\dagger b$, I obtain a system of equations containing $|u|^2+|v|^2=0$ which is meaningless.

(For details about the math you can look here https://physics.stackexchange.com/a/224301/160397 with $E+\Delta=0$.)

Where am I misunderstanding things? If every quadratic Hamiltonian can be diagonalised, and the way to do it is a Bogoliubov transformation, what is the problem here?

EDIT: this question is not about the math, as the underlying calculation is straightforward. The question can be expressed as: what is wrong with the Bogoliubov transformation here? Why is the transformation to diagonalise H not of that form?

Answer 1

It is not true for every Hamiltonian. The issue with yours is that it is not positive. The easiest way to see this is to think classically using $a=\frac{x+ip}{\sqrt 2}$: $$ H=\gamma(x^2-p^2) $$ (in the quantized version, you’d have an extra constant term). As you can see, the origin is not the minimum but rather a saddle point so you cannot use a change of variables to get $H$ of the form: $$ H=\omega b^\dagger b $$ See a similar issue in Vacuum state of Bogoliubov quasi-particles (continued). As explained there, your case $E+\Delta=0$ corresponds to the case where your Hamiltonian ceases to be positive definite.

Hope this helps.

Answer to comment

I’ve never mentioned diagonalisation. What I wrote is that you cannot use a Boguliubov transformation: $b=ua+va^\dagger$ to write the hamiltonian in the form: $H=\omega b^\dagger b$. This is because the latter is positive definite, while the former is not.

The closest analogue would be the change of variable using the previously defined $x,p$: $$ y=\frac{x-p}{\sqrt2} \quad q = \frac{p+x}{\sqrt2} $$ so that: $H=\gamma yq$. Importantly, $y,q$ are real and canonically conjugate.

Actually, the core of the discussion has little to do with quantum mechanics and has more to do with classical Hamiltonian mechanics.

Answer 2

You need positive definiteness. The essential result is Williamson's theorem..