Two bands interaction: self-energy and effective low energy Hamiltonian.

Consider the following model of two-bands Hamiltonian with interaction between the bands:

$$ H = \left( \begin{array}{cc} H_{11} &H_{12} \\ H_{12}^{\dag} & H_{22} \end{array} \right)\,, $$

where $H_{11}$ and $H_{22}$ are some low and high energy bands that we know how to diagonalize and $H_{12}$ is some interaction between them. Since we know the spectrum of $H_{11}$, we can write down the GF. Here it is:

$$ G_{11}^{(0)} = \frac{1}{H_{11} - \epsilon^{0}} $$

With the interaction with band $H_{22}$, the new GF for this sub-band is of course going to adjust, like this:

$$ G_{11}(\epsilon) = \frac{1}{H_{11} - \epsilon - \Sigma(\epsilon)} \,, $$

where $\epsilon$ is the new spectrum and $\Sigma$ is called Self-energy. What might they be? We can "symbolically" look for the new GF by solving the following:

$$ \left\{ \begin{array}{rl} (H_{11}-\epsilon) G_{11}(\epsilon) + H_{12} G_{12}^{\dag}(\epsilon) & = 1 \\ H_{12}^{\dag} G_{11}(\epsilon) + (H_{22}-\epsilon) G_{12}^{\dag}(\epsilon) & = 0 \end{array} \right.\,. $$

"Solve" for $G_{11}$, and we get:

$$ G_{11}(\epsilon) = \frac{1}{H_{11} - \epsilon - H_{12} (H_{22}-\epsilon)^{-1} H_{12}^{\dag}}\,. $$

From this we identify the self-energy as:

$$ \Sigma(\epsilon) = H_{12} (H_{22} - \epsilon)^{-1} H_{12}^{\dag}\,, $$

This is of course very much useless since $\epsilon$ is the perturbed energy, which we have not found. Of course we expect $\epsilon = \epsilon_0 + E$ with $E \ll \epsilon_0$, so:

$$ G_{11} = \frac{1}{H_{11} - \epsilon_{0} - \Sigma(\epsilon_{0}) - E \left[ 1 + H_{12} (H_{22} - \epsilon_{0})^{-2} H_{12}^{\dag} \right] + O\left(E^2\left(H_{22} - \epsilon_{0}\right)^{-3}\right)}$$

At this point, the earlier assumption that the spectrum of $H_{11}$ and $H_{22}$ is far away and decoupled kicks in and makes $(H_{22} - \epsilon_{0})$ small. So we have

$$ \Sigma(\epsilon) \simeq \Sigma(\epsilon_{0}) - H_{12} (H_{22} -\epsilon_{0})^{2} H_{12}^{\dag} E\,, $$

up to $O\left(E^2\left(H_{22} - \epsilon_{0}\right)^{-3}\right)$ correction. But we really want to massage the GF back to some simpler form like $G_{11}^{0}$. The trick is to factor out the bracket that multiples $E$, we let $Z$ such that:

\begin{eqnarray*} Z^{-1} &=& 1 + H_{12} \left( H_{22} - \epsilon_{0} \right)^{-2} H_{12}^{\dag} \\ Z^{\frac{1}{2}} &\simeq& 1 - \frac{1}{2} H_{12} \left( H_{22} - \epsilon_{0} \right)^{-2} H_{12}^{\dag} \end{eqnarray*}

Using this back in $G_{11}$, we have:

$$G_{11} = Z^{1/2} \frac{1}{H_{\mbox{Eff}} - \epsilon_0 - E} Z^{1/2} \,,$$

where the effective Hamiltonian $H_{\mbox{Eff}}$ is:

\begin{eqnarray*} H_{\mbox{Eff}} - \epsilon_0 &=& Z^{1/2} \left[ H_{11} - \epsilon_{0} - \Sigma(\epsilon_{0}) \right] Z^{1/2} \\ &\simeq& H_{11} - \epsilon_{0} - \Sigma(\epsilon_{0}) - \frac{1}{2} \left\{ H_{12} \left(H_{22} - \epsilon_{0}\right)^{-2} H_{12}^{\dag}, H_{11} - \epsilon_{0} \right\}\,. \end{eqnarray*}

$\left\{A,B\right\}$ is the anti-commutator. $Z$ plays the role of residue for the GF, i.e. DOS of the perturbed spectrum.