14
$\begingroup$

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?

$\endgroup$
1
4
$\begingroup$

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...

$\endgroup$
2
  • $\begingroup$ I agree it definitely simplifies the form quite a bit. My hang up was that they describe this as "the lowest order correction". I would think any sort of perturbative or series expansion would include the whole matrix at any order. Does the diagonal matrix actually fit into any systematic expansion by order? Or is the phrasing here misleading and it's just a convenient form for the equation without a physical justification or mathematical bound on the error? $\endgroup$
    – Tyberius
    May 30 '20 at 13:58
  • $\begingroup$ I wouldn't get too hung up about the phrase "lowest order" here. We are used to hearing this phrase used to describe the first term of a Taylor series (for example), but in this case it seems they just mean: "the simplest". We can think of H = D + V where D is diagonal and V is some small off-diagonal perturbation, then maybe expand as a series, but I'm not sure how valuable the "simplicity" of the next order correction would be. Without more information, I'm comfortable with guessing that "lowest-order correction" is just used to mean "simplest". $\endgroup$ May 30 '20 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.