Trying to wrap my head around the following situation.
Consider first a case that I understand well: Let's assume a three level system where the lowest two levels are degenerate and individually couple to a third level, higher in energy by some large $\Delta$. The Hamiltonian may be written as $$\begin{pmatrix}0 & 0 &t \\0 & 0 & t \\ t & t & \Delta\end{pmatrix}$$
If we want to integrate out the higher energy level, we can use the partition method / downfolding to get an effective coupling $t' = -t^2/\Delta$ to arrive at an effective Hamiltonian for the two-level system, $$\begin{pmatrix}0 & t' \\ t' & 0\end{pmatrix}$$
The denominator $\Delta$ comes from a term $E - H_1$ in the perturbation expansion where $H_1$ is the Hamiltonian for the higher-energy sector.
So far, so good, because both states in the low-energy sector have the same energy.
But what if the energies aren't the same?
Let's consider the Hamiltonian $$\begin{pmatrix} 0 & 0 & t \\ 0 & \epsilon & t \\ t & t & \Delta \end{pmatrix}$$ with a small $\epsilon$. However, then I cannot simply set $E = 0$ in the denominator for the perturbation expansion.
I know that I can still use Rayleigh-Schroedinger perturbation theory to calculate the energy corrections, but what if I don't want to do that and rather want an effective Hamiltonian for those two low-energy states?
Just to be clear: I don't want to "solve" the Hamiltonian per se, and rather want to arrive at an effective Hamiltonian for the low-energy subspace, which will look something like $$\begin{pmatrix} 0 & t_{\rm eff} \\ t_{\rm eff} & \epsilon\end{pmatrix}$$
