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.

Post a comment or leave a trackback: Trackback URL.


  • zhenghe  On May 7, 2010 at 5:40 pm

    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:)

    • Pengfei  On May 7, 2010 at 9:56 pm

      Thanks! I made several corrections according to your suggestion.

Leave a Reply

Please log in using one of these methods to post your comment: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: