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Consider the operator on $\ell^2(\mathbb{Z})$ $$ H = \Delta + v. $$ Here $\Delta$ is the nearest neighbour Laplacian on $\mathbb{Z}$, $\Delta_{k, \ell} =1 $ if $|k - \ell| =1 $ and zero otherwise, while $v_\ell , \ell \in \mathbb{Z}$ are independent identically distributed random variables with common distribution $\mu$.

If supp$\mu$ is bounded than we have the phenomenon of Anderson localization; with probability one, the spectrum of $H$ is pure point and the $\ell^2$-eigenvectors are exponentially localized. Focusing on the localization of the eigenvectors, this means that if $\psi$ is an eigenvector of $H$ then there exists $A, B >0 $ such that $$ \psi(k) \leq A \exp\left(-B|k| \right) $$

I was wondering if there are explicit bounds on $B$? I've been looking for papers that discuss the ${rate}$ of decay of eigenvectors for RSO but I haven't been able to find anything.

The proof of localization that I've seen relies on the fact that for every fixed $E \in \mathbb{R}$ the Lyapunov exponent $\gamma(E)$ of the corresponding transfer matrix is non-zero almost surely. Is it true that the rate of decay of an eigenfunction with eigenvalue $E$ is bounded by $\gamma(E)$?

Thank you for any help or sources you can point me to.

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    $\begingroup$ Your last paragraph is correct, in the following sense: Fix $E$. Then $|\psi(n)|\lesssim e^{-\gamma |n|}$ almost surely. Note, however, that this on its own does not imply that almost surely, we have exponential decay for eigenvalues $E$ (this needs to be proved separately). $\endgroup$
    – Christian Remling
    Commented Jul 7, 2014 at 18:45
  • $\begingroup$ Thanks! Does this follow just from the fact that the Lyapunov exponent of the transfer matrix is non-zero? In other words, if I knew that with probability one, the Lyapunov exponent is non-zero Lebesgue almost everywhere, would this imply the exponential decay for all eigenvectors? $\endgroup$
    – Ben
    Commented Jul 7, 2014 at 18:50
  • $\begingroup$ No, positive Lyapunov exponent will not give this. By Fubini, you obtain that almost surely, for almost all $E$ you have exponential decay, but the (singular) spectral measure could be supported on the exceptional set. $\endgroup$
    – Christian Remling
    Commented Jul 7, 2014 at 18:56
  • $\begingroup$ A reference you might useful (or not, it's not exactly what you asked) is math.caltech.edu/SimonPapers/250.pdf (see Section 7 especially). $\endgroup$
    – Christian Remling
    Commented Jul 7, 2014 at 18:57
  • $\begingroup$ Right. This would just give that there is no absolutely continuous part of the spectral measure. $\endgroup$
    – Ben
    Commented Jul 7, 2014 at 18:57

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