Imaginary Frequency in Bosonic Hamiltonian I'm doing some calculations for my Thesis involving a Bosonic Hamiltonian of the form:
\begin{equation}
H=\sum_{\vec{k}}\alpha\ a^{+}_{\vec{k}}a^{+}_{-\vec{k}} + \beta\  a^{+}_{\vec{k}}a^{-}_{\vec{k}} +\gamma\  a^{-}_{\vec{k}}a^{+}_{\vec{k}}  + \delta\  a^{-}_{\vec{k}}a^{-}_{-\vec{k}}
\end{equation}
where $a^{\pm}$ are creation and annihilation operators and the Greek letters are just real coefficients that may be functions of $\vec{k}$. This Hamiltonian is hermitian only if $\alpha=\delta$. After diagonalizing this using a Bogoliuvov transformation 
\begin{align*}
b^{\pm}_{\vec{k}} = \cosh(\phi)\ a^{\pm}_{\vec{k}}-\sinh(\phi)\ a^{\mp}_{-\vec{k}}
\end{align*}
One obtains the usual result
\begin{equation}
H=E_0+\sum_{\vec{k}}\omega(\vec{k})b^+_{{\vec{k}}}b^-_{\vec{k}}=E_0+ \sum_{\vec{k}}\omega(\vec{k})\ \hat{n}_{\vec{k}}
\end{equation}
Where $E_0$ is a constant that doesn't matter at all and $\omega(\vec{k})$ is
\begin{equation}
\omega(\vec{k})=(\beta+\gamma)\sqrt{1-\left(\frac{2\alpha}{\beta +\gamma}\right)^2}
\end{equation}
It is quite obvious that when $\frac{\beta+\gamma}{2}<\alpha$ the frequency becomes imaginary. This condition, back in the Hamiltonian, means that the non-diagonal terms weight more than the diagonal ones. This is because the condition for the angle $\phi$ to diagonalize $H$ is $\tanh (2\phi) = \frac{2 \alpha}{\beta + \gamma}$. This means that when $2\alpha>\beta + \gamma$ we can't perform the Bogoliubov transformation (This is why the frequency was imaginary). So the question remains:
How do I diagnalize $H$ when $2\alpha>\beta + \gamma$? 
 A: Diagonalization is not meaningful if $2\alpha > \beta + \gamma$ because the Hamiltonian becomes unphysical (unbounded below) in this case.
To see this, let's consider a simpler Hamiltonian:
\begin{equation}
H = 2A\, a^+ a^- + B\, (a^+ a^+ + a^- a^-),
\end{equation}
where the canonical commutation relation $[a^-, a^+] = 1$ holds. Next, let's construct the "position" and "momentum" operators as
\begin{equation}
x = \frac{a^+ - a^-}{\sqrt{2}},\quad p = \frac{i(a^+ - a^-)}{\sqrt{2}}.
\end{equation}
(Note that $[x,p] = i$ is satisfied.) Inverting the above relations gives
\begin{equation}
a^\mp = \frac{1}{\sqrt{2}} (x \pm ip).
\end{equation}
Then, the Hamiltonian can be written as
\begin{equation}
\begin{split}
H &= A\, (x^2 + p^2 - 1) + B\,(x^2 - p^2)\\
&=(A+B)\,x^2 + (A-B)\,p^2 - A.
\end{split}
\end{equation}
One should have $A \ge 0$ and $-A \le B \le A$ for the above Hamiltonian to be bounded below.
To perform the same analysis on the particular Hamiltonian OP considered, let's define a new set of ladder operators $c_{\vec{k}}^\pm$ and $d_{\vec{k}}^\pm$ as follows:
\begin{equation}
\begin{split}
&a_{\vec{k}}^\pm = \frac{1}{\sqrt{2}} (c_{\vec{k}}^\pm + d_{\vec{k}}^\pm),\\
&a_{-\vec{k}}^\pm = \frac{1}{\sqrt{2}} (c_{\vec{k}}^\pm - d_{\vec{k}}^\pm).
\end{split}
\end{equation}
Then, OP's Hamiltonian can be represented as a sum over unordered pairs $\{\vec{k},-\vec{k}\}$ of the following:
\begin{equation}
\begin{split}
H_{\vec{k}} &= \alpha\, c_{\vec{k}}^+c_{\vec{k}}^+ + \beta\, c_{\vec{k}}^+c_{\vec{k}}^- + \gamma\, c_{\vec{k}}^-c_{\vec{k}}^+ + \delta\, c_{\vec{k}}^-c_{\vec{k}}^- + (c \rightarrow d),\\
&= \alpha\, (c_{\vec{k}}^+c_{\vec{k}}^+ + c_{\vec{k}}^-c_{\vec{k}}^-) + (\beta+\gamma)\, c_{\vec{k}}^+c_{\vec{k}}^- + (c \rightarrow d) + 2\gamma
\end{split}
\end{equation}
where $(c\rightarrow d)$ denotes the terms obtained by replacing $c_{\vec{k}}^\pm$ by $d_{\vec{k}}^\pm$. Identifying $\alpha = B$ and $\beta + \gamma = 2A$, one can see that the Hamiltonian becomes unbounded below if $2\alpha > \beta + \gamma$.
