Today Zhenghe explained Anderson Localizaion to me.

(update according to Zhenghe’s comment)

Consider a Schrodinger operator . Let and be the subspace spanned by all eigenvectors of .

The Schrodinger operator is said to displays *Anderson Localization* if

1. the eigenvectors span the whole space: (evidently this is stronger than ).

2. for each with the nontrivial eigenvector there exist and such that for each .

Note that for cocycle, if an e.v. has some which decays exponentially, then this e.v. 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.

## Comments

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 , which is stronger than . Also part 2 you posted here is not correct, it’s actually for all , 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:)

Thanks! I made several corrections according to your suggestion.