This question focus on another aspect of my previous question. Consider a toy bilinear Hamiltonian consisting of two bosons $\{b_i\}_{i=1}^2$:
$$ \begin{align*} \mathsf{H}[b^\dagger,b] &= \Delta(b^\dagger_1 b^\dagger_2 + h.c.) + \lambda (b^\dagger_1 b_1 + b^\dagger_2 b_2) \\ &= [b^\dagger_1, b_2] \ H \begin{bmatrix} b_1 \\ b^\dagger_2 \end{bmatrix} - \lambda, \quad H = \begin{bmatrix} \lambda & \Delta \\ \Delta & \lambda \end{bmatrix} \end{align*} $$
Here $\lambda > 0$ and $\Delta > 0$. It can be diagonalized by a Bogoliubov transformation: define new boson particles $\{\beta_i\}_{i=1}^2$ as
$$ \begin{bmatrix} b_1 \\ b^\dagger_2 \end{bmatrix} = W \begin{bmatrix} \beta_1 \\ \beta^\dagger_2 \end{bmatrix} \ \Rightarrow \ \mathsf{H} = [\beta^\dagger_1, \beta_2] \ W^\dagger H W \begin{bmatrix} \beta_1 \\ \beta^\dagger_2 \end{bmatrix} - \lambda $$
$$ \Lambda \equiv W^\dagger H W = \begin{bmatrix} E & 0 \\ 0 & E \end{bmatrix} , \quad E = \sqrt{\lambda^2 - \Delta^2} $$
$$ W = \begin{bmatrix} u & -v \\ -v & u \end{bmatrix}, \quad u = \sqrt{\frac{\lambda + E}{2E}}, \quad v = \sqrt{\frac{\lambda - E}{2E}} $$
The ground state of $\mathsf{H}$ is the vacuum of the $\beta$ particles:
$$ \beta_i |0_\beta \rangle = 0 \quad (i = 1,2) $$
$|0_\beta \rangle$ can be expressed in terms of the original bosons $b_i$ and their vacuum $|0\rangle$: (see also this question)
$$ |0_\beta \rangle = e^Q |0\rangle, \quad Q = g b^\dagger_1 b^\dagger_2 $$
$$ g = -\frac{v}{u} = -\frac{\Delta}{\lambda+E} $$
The (squared) norm of this state is
$$ \begin{align*} \langle 0 | e^{Q^\dagger} e^Q | 0 \rangle &= \sum_{m,n=0}^\infty \frac{(g^*)^m g^n}{m!n!} \underbrace{\langle 0 | (b_1 b_2)^m (b^\dagger_1 b^\dagger_2)^n | 0 \rangle }_{= 0 \text{ if } n \ne m} \\ &= \sum_{n=0}^\infty \frac{|g|^{2n}}{(n!)^2} \langle 0 | (b_1 b_2)^n (b^\dagger_1 b^\dagger_2)^n | 0 \rangle \\ &= \sum_{n=0}^\infty |g|^{2n} = \frac{1}{1-|g|^2} = u^2 \end{align*} $$
When $\Delta < \lambda$ all thing are fine: the energy $E$ is positive, the norm of $e^Q |0\rangle$ is finite so $|0_\beta \rangle$ is well-defined. However, in other cases, curious things happen:
When $\Delta = \lambda$, the energy $E = 0$, and I suppose that Bose condensation occurs. Since now $|g| = 1$, the vacuum of $\beta$ (which seems to be $e^{-b^\dagger_1 b^\dagger_2} |0\rangle$) has an infinite norm. Something is wrong about this state.
Even worse, if $\Delta > \lambda$ the energy $E$ becomes imaginary.
Now for my question:
What should be the correct ground state when $\Delta = \lambda$, i.e. $E = 0$? Something is definitely wrong with $\exp(-b^\dagger_1 b^\dagger_2) |0\rangle$.
What is going on when $\Delta > \lambda$?