# Approximation of Dyson's Equation

I've been trying to learn about Green's function in the context of computational chemistry by reading Szabo and Ostlund's Modern Quantum Chemistry.

I've reached a section about the one particle many body Green's function and I'm confused about an approximation the authors use to obtain the lowest order correction to the ionization potential/electron affinity.

They write, in regards to solving the Dyson equation,

$$\det\left[E\mathbf{1}-\epsilon_i-\mathbf{\Sigma}(E)\right]=0 \tag{7.42}$$ When $$\mathbf{\Sigma}(E)=0$$, the roots occur at $$\mathbb{\epsilon}_i$$'s. To find the lowest order correction to these Koopman's theorem results, let us ignore the off-diagonal elements of $$\mathbf{\Sigma}(E)$$. Then Eq. ($$7.42$$) simplifies to $$\prod_i(E-\epsilon_i-\Sigma_{ii}(E))\tag{7.43}$$

where $$\epsilon_i$$ are orbital energies and $$E$$ is the self-energy. What I don't understand is, what justifies ignoring the off diagonal elements? Or rather, what error is introduced by this approximation? I would think the lowest order correction would take the entire second order self energy, including off-diagonal elements.

They actually later show the results of some example calculations and have results for just the diagonal, second order self energy and the full matrix second order result. Is there something about the self energy that makes taking just the diagonal reasonable?

• May 19 '20 at 8:39

I think the justification is just that this is the simplest approach, since $$E$$ is a scalar and $$\boldsymbol{\epsilon}$$ is a vector. If you only take the diagonal part of $$\Sigma$$, you see that the secular determinant vanishes whenever $$E = \epsilon_i + \Sigma_{ii}$$. If you included the off-diagonal elements, you would essentially have to diagonalize $$\boldsymbol{\Sigma}$$ and you would lose this nice simple interpretation...