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I'm reading these lecture notes on Anderson localization, and I cannot understand how the resonant regions contribute to the divergence of the resolvent expansion (sections 3.1 and 3.2). The relevant Hamiltonian is

$$ H=H_0+gT$$

where $$H_0=\sum_{i}\epsilon_i |i\rangle\langle i|,\quad T=-\sum_{\langle i,j\rangle}(|i\rangle\langle j|+|i\rangle\langle j|)$$

$i$,$j$ are sites on a cubic lattice, $\langle i,j\rangle$ are nearest neighbor. The author defines the resolvent as

$$ G(E)=\frac{1}{E-H}, \quad E\notin \sigma(H)$$

where $\sigma(H)$ denotes the spectrum of $H$. Also, call $G_0(E)=\frac{1}{E-H_0}$. After some algebra one can arrive at the series

$$ G(E)=G_0(E)+\sum_{n=1}^\infty (G_0T)^nG_0 $$

In this basis we can express $G(E)$ as a sum over walks from the starting point to the ending point: each $G_0$ contributes with a term like $\frac{1}{E-e_k}$ and $T$ makes us "walk" around the lattice.

$$ \langle i|G(E)|j\rangle=\frac{1}{E-\epsilon_i}+\sum_{n=1}^\infty (-g)^n\sum_{\substack{\pi:i\to j\\|\pi|=n}}\prod_{s=1}^n \frac{1}{E-\epsilon_{\pi(s)}}$$

The author later (beginning of section 3.2, page 13) says that if there are neighboring sites such that $\frac{g}{\epsilon_i-\epsilon_j}\geq 1$, then the series diverges because it contains terms of the form $$ \left(\frac{g}{\epsilon_i-\epsilon_j}\right)^m $$

I cannot see any such terms. I see $ \left(\frac{g}{E-\epsilon_i}\frac{g}{E-\epsilon_j}\right) $, but this does not cause divergences. What am I missing? Where does the divergence come from?

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  • $\begingroup$ Well, they are comparable to your highlighted terms, aren't they? $\endgroup$ – daydreamer Oct 14 '20 at 9:38
  • $\begingroup$ @daydreamer what do you mean? $\endgroup$ – user2723984 Oct 14 '20 at 9:52
  • $\begingroup$ Well take the last terms you mentioned: if the greater than one condition holds for each of the terms, then it won't matter if powers of the same thing happen or just products of stuff that grow at a similar pace. Perhaps he handwaved that by using a power expression $\endgroup$ – daydreamer Oct 14 '20 at 10:58
  • $\begingroup$ @daydreamer I'm not sure what you mean, the last term I mentioned is different because it doesn't have $\epsilon_i-\epsilon_j$ in the denominator, but $E-\epsilon_i$ $\endgroup$ – user2723984 Oct 14 '20 at 14:18
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    $\begingroup$ @daydreamer and if that were the case, there would be divergence problems regardless of whether $\epsilon_i \sim \epsilon_j$, because it suffices to have $E\sim \epsilon_i$ to cause divergences regardless of what the other energies do $\endgroup$ – user2723984 Oct 14 '20 at 14:26
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I think I got it. Seems that at the beginning of page 11 the author explicitly states as an hypothesis that what is being studied is the energies over the vicinities of $\epsilon_i$. Therefore, each of the Es are close to some epsilon... Hence my intuition-driven comment

" Well take the last terms you mentioned: if the greater than one condition holds for each of the terms, then it won't matter if powers of the same thing happen or just products of stuff that grow at a similar pace. Perhaps he handwaved that by using a power expression "

seems to hold

Am I missing something?enter image description here

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    $\begingroup$ Thanks! I cannot find what you're talking about at the beginning of page $11$. $\endgroup$ – user2723984 Oct 14 '20 at 15:02
  • $\begingroup$ Image attached. You're welcome! Thanks for the question $\endgroup$ – daydreamer Oct 14 '20 at 15:04
  • $\begingroup$ I guess you mean that there is a typo and it should be $E_\alpha=\epsilon_\alpha+o(1)$? $\endgroup$ – user2723984 Oct 14 '20 at 15:06
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    $\begingroup$ I cannot find anything in the passage you highlighted that says that the author examines energies close to $\epsilon_i$. Perhaps you are inferring this by the fact that we expand around the point where the spectrum is composed by the $\epsilon_i$, and so we are interested in the resolvent for energies close to $\epsilon_i$, but still if this was the case, as I commented earlier, then these divergences would be a problem regardless of the difference $\epsilon_j-\epsilon_i$, because you would always have terms like $E-\epsilon_i\ll 1, E\sim \epsilon_i$ $\endgroup$ – user2723984 Oct 14 '20 at 15:15
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    $\begingroup$ and the point should be that these resonances cause the divergences $\endgroup$ – user2723984 Oct 14 '20 at 15:16

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