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Transactions of the American Mathematical Society

ISSN 1088-6850(online) ISSN 0002-9947(print)



The Schrödinger equation with a quasi-periodic potential

Author: Steve Surace
Journal: Trans. Amer. Math. Soc. 320 (1990), 321-370
MSC: Primary 34B25; Secondary 47E05, 81C05, 82A42
MathSciNet review: 998358
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Abstract: We consider the Schràdinger equation

$\displaystyle - \frac{{{d^2}}} {{d{x^2}}}\psi + \varepsilon (\cos x + \cos (\alpha x + \vartheta ))\psi = E\psi $

where $ \varepsilon$ is small and $ \sigma$ satisfies the Diophantine inequality

$\displaystyle \vert p + q\alpha \vert \geq C/{q^2}{\text{for}}p{\text{,}}q \in {\mathbf{Z}},q \ne 0.$

. We look for solutions of the form

$\displaystyle \psi (x) = {e^{iKx}}q(x) = {e^{iKx}}\sum {{\psi _{mn}}{e^{inx}}} {e^{im(\alpha x + \vartheta )}}$

. If we try to solve for $ \psi = {\psi _{mn}}$ we are led to the Schràdinger equation on the lattice $ {{\mathbf{Z}}^2}$

$\displaystyle H(K)\psi = (\varepsilon \Delta + V(K))\psi = E\psi $

where $ \Delta$ is the discrete Laplacian (without diagonal terms) and $ V(K)$ is some potential on $ {{\mathbf{Z}}^2}$ . We have two main results:

(1) For $ \varepsilon$ sufficiently small, $ H(K)$ has pure point spectrum for almost every $ K$.

(2) For $ \varepsilon$ sufficiently small, the operator

$\displaystyle - {d^2}/d{x^2} + \varepsilon (\cos x + \cos (\alpha x + \vartheta ))$

has no point spectrum.

To prove our results, we must get decay estimates on the Green's function $ {(E - H)^{ - 1}}$. The decay of the eigenfunction follows from this. In general, we must keep track of small divisors which can make the Green's function large. This is accomplished by a KAM (Kolmogorov, Arnold, Moser) type of multiscale perturbation analysis.

References [Enhancements On Off] (What's this?)

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