## Anderson Localizaion

Today Zhenghe explained Anderson Localizaion to me.
(update according to Zhenghe’s comment)

Consider a Schrodinger operator $H_{f,\theta}:\ell^2(\mathbb{Z})\to\ell^2(\mathbb{Z})$. Let $\Sigma_{pp}(H_{f,\theta})\overset{\triangle}{=}\overline{EV(H_{f,\theta})}$ and $\ell^2(\mathbb{Z})_{pp}$ be the subspace spanned by all eigenvectors of $E\in EV(H_{f,\theta})$.

The Schrodinger operator $H_{f,\theta}$ is said to displays Anderson Localization if

1. the eigenvectors span the whole space: $\ell^2(\mathbb{Z})=\ell^2(\mathbb{Z})_{pp}$ (evidently this is stronger than $\Sigma(H_{f,\theta})=\Sigma_{pp}(H_{f,\theta})$).

2. for each $E\in EV(H_{f,\theta})$ with the nontrivial eigenvector $\phi_E\in\ell^2(\mathbb{Z})$ there exist $\epsilon>0, C\ge1$ and $n_E\in\mathbb{Z}$ such that $|\phi_E(n)|\le~C\cdot e^{-\epsilon |n-n_E|}$ for each $n\in \mathbb{Z}$.

Note that for $SL(2,\mathbb{R})$ cocycle, if an e.v. $E$ has some $\phi$ which decays exponentially, then this e.v. $E$ must be simple.

It is quite interesting. The exponential decayed eigenvector corresponds to an orbit of a Schrodinger cocycle. Hence there exists a vector that is exponentially contracted under both forward and backward iterates. This resembles the case that a vector lies in the homoclinic tangency of a hyperbolic fixed point. This is an obstacle of uniform hyperbolicity. So positive Lyapunov exponent could only coexists with nonunifom hyperbolicity.

Hi Pengfei, I am sorry I made some mistake in the first part of defintion of Anderson localization, which actually has to be replaced by $\l^2(Z)=\l^2(Z)_{pp}$, which is stronger than $\Sigma=\Sigma_{pp}$. Also part 2 you posted here is not correct, it’s actually for all $E\in EV(H_{f,\theta})$, not just exists one and the decay inequality is not exact. I am going to post it in my blog sometime later. By the way, in this case eigenvalues are always simple, which follows easily from the fact cocycles are SL(2, R)-valued. Thank you for your interests:)