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

Question

How can a Hermitian Hamiltonian $H = b^\dagger b^\dagger + b b$, with $b$ boson, cannot be diagonlized?

Given a Hamiltonian $$\hat H = b^\dagger b^\dagger + b b \tag{1}$$ with $b, b^\dagger$ boson operators and $[b, b^\dagger] = 1$, this Hamiltonian is obviously Hermitian ${\hat H}^\dagger = {\hat H}$.

Let's compute the eigenvalue of this Hamiltonian, we need to do Bogoliubov transformation,

$$b = u a + v a^\dagger\tag{2}$$ $$b^\dagger = u^* a^\dagger + v^* a$$

The commutation relation requires: $$[b,b^\dagger] = (|u|^2 - |v|^2) [a, a^\dagger]$$ that is $$|u|^2 - |v|^2 = 1 \tag{3}$$

Take (2) into (1): $$H = (u^* a^\dagger + v^* a)(u^* a^\dagger + v^* a）+ (u a + v a^\dagger)(u a + v a^\dagger)$$ $$H = ((u^*)^2 + v^2) a^\dagger a^\dagger + ((u)^2 + (v^*)^2) a a + (v^* u^* + v u) (a a^\dagger + a^\dagger a)$$

The diagonalizing of $H$ requires $$(u^*)^2 + v^2 = 0 \tag{4}$$

$(3)\implies$

$$u = \cosh r e^{i \theta_1}$$ $$v = \sinh r e^{i \theta_2} \tag{5}$$ Take $(5)$ into $(4)$, we get $$\cosh^2 r e^{-2 i \theta_1} + \sinh^2 r e^{ 2 i \theta_2 } = 0\tag{6}$$ i.e. $$\cosh^2 r \cos 2\theta_1 + \sinh^2 r \cos 2\theta_2 = 0$$ $$\cosh^2 r \sin 2\theta_1 - \sinh^2 r \sin 2\theta_2 = 0$$

We can prove there is no real solutions of above equations.

How is it possible a Hermitian Hamiltonian (especially a quadratic Hamiltonian) cannot be diagonalized?

My questions:

1. Is there any loophole in above argument? I can't believe it is possible. If $H$ can be diagonalized, how to get the eigenvalue and eigenvector of this specific Hamiltonian?

2. If it's really true that $H$ cannot be diagonalized, please give me the reason why it contradicts the common sense like Hermitian matrix/operator can be diagonalized with real eigenvalue, quadratic Hamiltonian can be exactly solvable, etc.

Answer 1

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.

Answer 2

The formula (1) doesn't define the Hamiltonian it only gives a formal expression. The operator corresponding to this formal expression is not a bounded one. To define an unbounded operator in the Hilbert space of infinite dimension, one has to specify its definition area. Only in this way can one judge if the operator is Hermitian or not, does it have a self-adjoint extension or not.

I actually should admit that I don't know if there is a Hermitian operator, corresponding to (1). In case if it exists, it is possible it has no discrete spectrum. Some Hermitian operators do not have normalizable eigenvectors. Then your conclusion about non-diagonalizability of the Hamiltonian by the Bogolyubov transformation might be a consequence of a discrete spectrum absence. Otherwise, a new bosons' vacuum would be a normalizable eigenvector.