# Calculation of effective Hamiltonian

I'm stuck with calculating the effective Hamiltonian of a model system. The energy-dependence of the effective Hamiltonian is confusing me a lot. My question is how to treat this dependence theoretically and numerically.

In principle, the model system I'm investigating is large but I'm even stuck with the easy three-level system. Hence, I will sketch you shortly what I have derived so far for this easy model. It would be very kind if you could help me with the problem.

The Hamiltonian reads $$H=\begin{pmatrix} \epsilon_1 & 0 & t \\ 0 & \epsilon_2 & t \\ t & t & \epsilon_3 \end{pmatrix}$$. I would like to calculate the effective Hamiltonian by considering the levels with energy $$\epsilon_2$$ and $$\epsilon_3$$ as part 'b'. Then the effective Hamiltonian reads $$H_{eff}(E)= \epsilon_1 - \begin{pmatrix} 0 & t \end{pmatrix} \big( \begin{pmatrix} \epsilon_2 & t \\ t & \epsilon_3 \end{pmatrix} - \begin{pmatrix} E & 0 \\ 0 & E \end{pmatrix} \big)^{-1} \begin{pmatrix} 0 \\ t \end{pmatrix}$$. Moreover, the effective Hamiltonian has to fulfill the condition $$H_{eff}(E) = E$$. I'm not sure how to proceed with calculating the effective Hamiltonian. Of course, I'm able to invert a two-by-two matrix. However, the model system I want to treat is larger than the three-level system and hence, I cannot explicitly perform the matrix subtraction.

Question:

In general, the equation $$H_{eff}(E) = E$$ is very complex to solve. Is it possible to calculate all three eigenvalues $$E_1$$, $$E_2$$, $$E_3$$ of the whole Hamiltonian $$H$$ and just plug it into this equation? Is the effective Hamiltonian then the sum as $$H_{eff}=H_{eff}(E_1) + H_{eff}(E_2) + H_{eff}(E_3)$$ or does just the action of $$H_{eff}$$ depend on the energy of the ket state, e.g. $$H_{eff} |E_1> = H_{eff}(E_1)|E_1>$$ where $$|E_1>$$ eigenstate with energy $$E_1$$? Is there an iterative approach that can tackle the problem of solving $$H_{eff}(E) = E$$ for any arbitrary model system?

• If you can solve the Hamiltonian exactly, then there is no point to use an effective Hamiltonian. I suppose that for your large system such a solution is impossible, so obtaining an effective Hamiltonian involves an approximation. E.g., if it is the effective Hamiltonian for level 1, you may be justified setting $E=\epsilon_1$. Often it boils down to second order perturbation corrections. Commented Apr 14, 2020 at 5:38
• Thank you for your comment. Yes, it is impossible for the large system. However, how does the effective Hamiltonian read for the three-level system without any approximations? Is it $H(E_1)$ or a linear combination as $H(E_1) + H(E_2)$? Commented Apr 14, 2020 at 8:06