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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 $$\begin{pmatrix} 0 & t_{\rm eff} \\ t_{\rm eff} & \epsilon\end{pmatrix}$$

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The desired result can, for instance, be obtained by using the Feshbach projection operator technique. This method may be some overkill for the simple problem at hand. Anyway, here it is. $$ H=\left( \begin{array}{ccc} h_{1} & 0 & h_{13} \\ 0 & h_{2} & h_{23} \\ h_{31} & h_{32} & h_{3}% \end{array}% \right) ,\;P=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0% \end{array}% \right) ,\;Q=1-P=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1% \end{array}% \right) $$ $$ P[z-H]^{-1}P=[z-PHP-PHQ[z-QHQ]^{-1}QHP]^{-1}P $$ \begin{eqnarray*} PHP &=&\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0% \end{array}% \right) \left( \begin{array}{ccc} h_{1} & 0 & h_{13} \\ 0 & h_{2} & h_{23} \\ h_{31} & h_{32} & h_{3}% \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0% \end{array} \right) =\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0% \end{array} \right) \left( \begin{array}{ccc} h_{1} & 0 & 0 \\ 0 & h_{2} & 0 \\ h_{31} & h_{32} & 0% \end{array} \right) =\left( \begin{array}{ccc} h_{1} & 0 & 0 \\ 0 & h_{2} & 0 \\ 0 & 0 & 0% \end{array}% \right) \\ PHQ &=&\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0% \end{array}% \right) \left( \begin{array}{ccc} h_{1} & 0 & h_{13} \\ 0 & h_{2} & h_{23} \\ h_{31} & h_{32} & h_{3}% \end{array}% \right) \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1% \end{array}% \right) =\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0% \end{array}% \right) \left( \begin{array}{ccc} 0 & 0 & h_{13} \\ 0 & 0 & h_{23} \\ 0 & 0 & h_{3}% \end{array}% \right) =\left( \begin{array}{ccc} 0 & 0 & h_{13} \\ 0 & 0 & h_{23} \\ 0 & 0 & 0% \end{array}% \right) \\ QHP &=&\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ h_{31} & h_{32} & 0 \end{array} \right) \\ QHQ &=&\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} h_{1} & 0 & h_{13} \\ 0 & h_{2} & h_{23} \\ h_{31} & h_{32} & h_{3} \end{array} \right) \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) =\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 0 & 0 & h_{13} \\ 0 & 0 & h_{23} \\ 0 & 0 & h_{3} \end{array} \right) =\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & h_{3} \end{array} \right) \end{eqnarray*} \begin{eqnarray*} \lbrack z-QHQ]^{-1} &=&Q[z-h_{3}]^{-1}Q \\ PHQ[z-QHQ]^{-1}QHP &=&\left( \begin{array}{ccc} 0 & 0 & h_{13} \\ 0 & 0 & h_{23} \\ 0 & 0 & 0 \end{array} \right) Q[z-h_{3}]^{-1}Q\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ h_{31} & h_{32} & 0 \end{array} \right) \\ &=&\left( \begin{array}{ccc} 0 & 0 & h_{13}[z-h_{3}]^{-1} \\ 0 & 0 & h_{23}[z-h_{3}]^{-1} \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ h_{31} & h_{32} & 0 \end{array} \right) \\ &=&\left( \begin{array}{ccc} h_{13}[z-h_{3}]^{-1}h_{31} & h_{13}[z-h_{3}]^{-1}h_{32} & 0 \\ h_{23}[z-h_{3}]^{-1}h_{31} & h_{23}[z-h_{3}]^{-1}h_{32} & 0 \\ 0 & 0 & 0 \end{array} \right) \end{eqnarray*} \begin{eqnarray*} z-PHP-PHQ[z-QHQ]^{-1}QHP \\=z-\left( \begin{array}{ccc} h_{1} & 0 & 0 \\ 0 & h_{2} & 0 \\ 0 & 0 & 0 \end{array} \right) -\left( \begin{array}{ccc} h_{13}[z-h_{3}]^{-1}h_{31} & h_{13}[z-h_{3}]^{-1}h_{32} & 0 \\ h_{23}[z-h_{3}]^{-1}h_{31} & h_{23}[z-h_{3}]^{-1}h_{32} & 0 \\ 0 & 0 & 0 \end{array} \right) \\ =z-\left( \begin{array}{ccc} h_{1}+h_{13}[z-h_{3}]^{-1}h_{31} & h_{13}[z-h_{3}]^{-1}h_{32} & 0 \\ h_{23}[z-h_{3}]^{-1}h_{31} & h_{2}+h_{23}[z-h_{3}]^{-1}h_{32} & 0 \\ 0 & 0 & 0 \end{array} \right) \end{eqnarray*} Forgetting about the vanishing third components $$ \lbrack z-PHP-PHQ[z-QHQ]^{-1}QHP]^{-1}\rightarrow \mathcal{G}% _{P}(z)=[z-\left( \begin{array}{cc} h_{1}+h_{13}[z-h_{3}]^{-1}h_{31} & h_{13}[z-h_{3}]^{-1}h_{32} \\ h_{23}[z-h_{3}]^{-1}h_{31} & h_{2}+h_{23}[z-h_{3}]^{-1}h_{32}% \end{array}% \right) ]^{-1} $$ Next $$ h_{13}=h_{23}=h_{31}=h_{32}=t,\;h_{3}=\Delta ,\;h_{1}=0,\;h_{2}=\epsilon $$ $$ \mathcal{G}_{P}(z)=[z-\left( \begin{array}{cc} h_{1}+h_{13}[z-h_{3}]^{-1}h_{31} & h_{13}[z-h_{3}]^{-1}h_{32} \\ h_{23}[z-h_{3}]^{-1}h_{31} & h_{2}+h_{23}[z-h_{3}]^{-1}h_{32}% \end{array}% \right) ]^{-1}=[z-\left( \begin{array}{cc} t^{2}[z-\Delta ]^{-1} & t^{2}[z-\Delta ]^{-1} \\ t^{2}[z-\Delta ]^{-1} & \epsilon +t^{2}[z-\Delta ]^{-1}% \end{array}% \right) ]^{-1} $$ The new eigenvalues are the poles of $\mathcal{G}_{P}(z)$, and are found by solving the eigenvalue problem $$ z\mathbf{f}-\left( \begin{array}{cc} t^{2}[z-\Delta ]^{-1} & t^{2}[z-\Delta ]^{-1} \\ t^{2}[z-\Delta ]^{-1} & \epsilon +t^{2}[z-\Delta ]^{-1}% \end{array}% \right) \mathbf{f}=0 $$ or \begin{eqnarray*} zf_{1} &=&t^{2}[z-\Delta ]^{-1}f_{1}+t^{2}[z-\Delta ]^{-1}f_{2} \\ zf_{2} &=&t^{2}[z-\Delta ]^{-1}f_{1}+(\epsilon +t^{2}[z-\Delta ]^{-1})f_{2} \end{eqnarray*} or \begin{eqnarray*} (z[z-\Delta ]-t^{2})f_{1}-t^{2}f_{2} &=&0 \\ t^{2}f_{1}-\{(z-\epsilon )[z-\Delta ]-t^{2}\}f_{2} &=&0 \end{eqnarray*} The rest is simple algebra. It not only supplies the eigenvalues but also the corresponding eigenvectors. Warning: I usually make mistakes in my calculations but the principle is (hopefully) clear.

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  • $\begingroup$ While I appreciate the effort you put in this answer, it's not quite what I was looking for: First, it has to be solved self-consistently and second, I don't see how to generalize it to the case of highly degenerate levels. What I'm looking for is the "non-degenerate" equivalent to how, for example, Heisenberg exchange is derived via 2nd order perturbation theory... $\endgroup$ – Lagerbaer May 14 '14 at 6:43
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    $\begingroup$ You asked for an effective Hamiltonian acting in the subspace associated with the two lower levels. That is what the Feshbach procedure gives you, the effective Hamiltonian being H_P=PHP+PHQ[z-QHQ]^(-1)QHP. Note that it is energy-dependent through z. Note that it is exact, no approximations are made.But it can be used to obtain a perturbation expansion if the coupling term PHP+PHQ[z-QHQ]^(-1)QHP is small. Highly degenerate levels are not a problem. But, as I said it is some overkill for the present case. $\endgroup$ – Urgje May 14 '14 at 8:55
  • $\begingroup$ Mh, I see. If the entire low-energy subspace is degenerate, then a good approximation is to set $z = E_0$ with $E_0$ the energy of the subspace without the perturbation. That way, you get an effective Hamiltonian that's energy independent. I was hoping for something similar but now with different energy levels in the subspace. $\endgroup$ – Lagerbaer May 14 '14 at 15:02

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