# Divergences in resolvent expansion of Anderson hamiltonian

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?

• Well, they are comparable to your highlighted terms, aren't they? Oct 14, 2020 at 9:38
• @daydreamer what do you mean? Oct 14, 2020 at 9:52
• 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 Oct 14, 2020 at 10:58
• @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 $E-\epsilon_i$ Oct 14, 2020 at 14:18
• @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 Oct 14, 2020 at 14:26

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? • Thanks! I cannot find what you're talking about at the beginning of page $11$. Oct 14, 2020 at 15:02
• Image attached. You're welcome! Thanks for the question Oct 14, 2020 at 15:04
• I guess you mean that there is a typo and it should be $E_\alpha=\epsilon_\alpha+o(1)$? Oct 14, 2020 at 15:06
• 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 $E-\epsilon_i\ll 1, E\sim \epsilon_i$ Oct 14, 2020 at 15:15
• and the point should be that these resonances cause the divergences Oct 14, 2020 at 15:16