Eigenvalues of the Hamiltonian in quantum optics I have the following Hamiltonian (non-Hermitian) 
$$
H=-(iM_1 +N)A^\dagger A + (iM_2 - N) B^\dagger B - L(A^\dagger B + A B^\dagger),
$$ 
where $A$ and $B$ are the annihilation operators for two harmonic oscillators coupled via the term $L(A^\dagger B + A B^\dagger)$. This Hamiltonian has the eigenvalues 
$$\lambda_{\pm} =  N-\frac{i}{2}(M_1-M_2) \pm 2\sqrt{L-(M_1+M_2)^2}.$$
Also, $M_1$, $M_2$, $N$ and $L$ are the constants. I want to know how to obtain these eigenfrequencies/eigenvalues?
 A: This kind of problems are generally solved by an unitary transformation on the Hamiltonian. You can choose
$$
U=\begin{bmatrix}\cosh(r)&e^{-i\theta}\sinh(r)\\e^{i\theta}\sinh(r)&\cosh(r)\end{bmatrix}
$$
where the parameters $\theta$ and $r$ should be determined by taking
$$
\begin{bmatrix}A\\B\end{bmatrix}=U\begin{bmatrix}C\\D\end{bmatrix}
$$
and one has a new Hamiltonian with the operator $C$ and $D$. Then, your parameter choice is the one that sets to zero the cross terms like $C^\dagger D$ and $D^\dagger C$. In this way you are left with the diagonal form
$$
  H'=\lambda_CC^\dagger C+\lambda_DD^\dagger D+E_0. 
$$
being $E_0$ a constant. Your final result will be written as $E_{n_1,n_2}=n_1\lambda_C+n_2\lambda_D+E_0$. 
A: You can see by inspection that your Hamiltonian preserves the total number of quanta $N=n_a+n_b$.  Thus, working in a subspace spanned by 
$$
\vert n_a\rangle\vert n_b\rangle \, ,\qquad n_a+n_b=N
$$
you can easily obtain the matrix form of your Hamiltonian. 
Assuming your Hilbert space is 2-dimensional (as your problem suggests there are only two eigenvalues) would mean that $N=1$ and that a basis for your problem is $\{\vert 0\rangle \vert 1\rangle,\vert 1\rangle \vert 0\rangle\}$ so it’s then not much of a job to verify that you have the correct eigenvalues by diagonalizing the corresponding $2\times 2$ matrix using the standard action
$$
A^\dagger \vert n_a\rangle = \sqrt{n_a+1}\vert n_a+1\rangle
\qquad A\vert n_a\rangle = \sqrt{n_a}\vert n_a-1\rangle\, ,
$$
etc.
More generally there is nothing in your problem to suggest you need to restrict the Hilbert to dimension $2$ (except the statement of the 2 eigenvalues).  
You could in principle repeat the process within a $3$-dimensional Hilbert space for $N=2$ spanned by $\{\vert 2\rangle\vert 0\rangle, \vert 1\rangle \vert 1\rangle,\vert 0\rangle\vert 2\rangle\}$, and obtain the $3$ eigenvalues for this system by diagonalizing the $3\times 3$ matrix representation of $H$ in this space.  Of course you could also do the same for any $N$, obtaining an $N+1$ dimensional Hilbert space and $N+1$ eigenvalues.
A: here are some thoughts.
The Hamiltonian you are using should have some eigenfunctions, which are determined by the action of the creation-annihilation operators $A,A^{\dagger},B,B^{\dagger}$ on them. In a certain notation, one would have something like
$$A|\lambda>_A \sim \sqrt{\lambda}|\lambda-1>_A$$ 
$$A^{\dagger}|\lambda>_A \sim \sqrt{\lambda+1}|\lambda+1>_A$$
$$B|\lambda>_B \sim \sqrt{\lambda}|\lambda-1>_B$$
$$B^{\dagger}|\lambda>_B \sim \sqrt{\lambda+1}|\lambda+1>_B$$
where the subscripts indicate the difference between the eigenfunctions of A's and B's. Also, note that the pairs of A's and B's satisfy a set of commutation relations that reads
$$[A,A^{\dagger}]=[B,B^{\dagger}]=1$$
Now, these eigenfunctions are also eigenfunctions of the Hamiltonian. This means that if you act with the Hamiltonian (as a 4x4 matrix) on these eigenfunctions (as a column matrix), you can read the eigenvalues of the Hamiltonian after a trivial diagonalization. 
I really hope this helps with your problem. 
