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Consider the Anderson model given by the Hamiltonian $H \in B(l^2( \mathbb{Z}^d)) $ defined by $H = - \Delta + V$ where the potential $V$ acts on a unit vector $ \vert x \rangle  \in l^2( \mathbb{Z}^d)$ by $V \vert x \rangle = V(x) \vert x \rangle $ where $V(x)$ are i.i.d uniformly distributed in $\lbrack 0,1 \rbrack$. This system exhibits Anderson localization.

Now, if we put $d=1$ and I pick $V(2x) $ i.i.d uniformly distributed in $\lbrack 0,1 \rbrack$ and then let $V(2x+1) = V(2x)$. Does this system then still exhibit Anderson localization?

Unanswered in physics stackexchange: https://physics.stackexchange.com/questions/631193/does-anderson-localisation-occur-if-the-potential-are-equal-in-pairs

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    $\begingroup$ certainly, you have a potential with a correlation length of two lattice constants --- the eigenfunctions will be exponentially localized with a localization length that is twice as large as usual. $\endgroup$
    – Carlo Beenakker
    Apr 29, 2021 at 18:31
  • $\begingroup$ Is this physical intuition or can one prove it with some self-advoiding walk expansion? $\endgroup$
    – Frederik Ravn Klausen
    Apr 29, 2021 at 18:45
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    $\begingroup$ As Carlo already pointed out, this has to be true. For a formal proof, you'll probably have to go through the proofs in the traditional setting and make sure that these can be adapted, I don't think there will be a very easy argument. $\endgroup$
    – Christian Remling
    Apr 29, 2021 at 19:41

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