I encounter a difficulty in comouting the time evolution of occupation number. I want to compute the time evolution of occupation number of Aubry-Andre model to show that there exists Anderson localization under some parameter. The Aubry-Andre model is described as follow:

$$\begin{equation} H = -t \sum_{i}^{L} (c^{\dagger}_{i} c_{i+1} + h.c.) + \sum_{i}^{L}\lambda_{i} c^{\dagger}_{i} c_{i}~,~ \lambda_{i} = \cos( 2 \pi \sigma i) \end{equation}$$ where $\sigma^{-1}$ is the golden ratio. We can diagonalize the single particle Hamiltonian and turn to the following form: \begin{align} H = \sum_{ij}^{L} c^{\dagger}_{i}h_{ij} c_j = \sum_{k} \big( c^{\dagger}_{i} V_{ik} \big) E_{k} \big( V^{\dagger}_{kj} c_{j} \big) = \sum_{k} E_{k} d^{\dagger}_{k} d_{k} ~~,~~ d_{k} = \sum_{i}V^{\dagger}_{ki}c_{i} \end{align} where $V$ is the set of eigenstates that diagonalize $h_{ij}. $Having this setup, we can focus on the time evolution occupation number
$$\begin{align} \langle n_{i}(t) \rangle = \langle \psi | U^{\dagger}(t) c^{\dagger}_{i} c_{i} U(t) | \psi \rangle ~~,~~ U(t) = \exp(-i H t) \end{align}$$ My idea is that first turning the $c^{\dagger}$ into $d^{\dagger}$ such that $$\begin{align} \langle \psi | U(t)^{\dagger} c^{\dagger}_{i} c_{i} U(t) | \psi\rangle = \sum_{\mu \nu} V_{i \mu} V_{\nu i}^{\dagger} \langle \psi| U^{\dagger}(t)d^{\dagger}_{\mu} d_{\nu} U(t) | \psi \rangle \end{align}$$ Since $U(t) = \exp(-iHt) = \exp(-i t \sum_{k}E_{k}d^{\dagger}_{k}d_{k})$, I expect that the time evolution operator of $d^{\dagger}_{k} = \exp(-iE_{k} t) d^{\dagger}_{k}$ such that: $$\begin{align} \langle \psi | U(t)^{\dagger} c^{\dagger}_{i} c_{i} U(t) | \psi\rangle = \sum_{\mu \nu} V_{i \mu} V_{\nu i}^{\dagger} e^{i(E_{\mu} - E_{\nu})t} \langle \psi| d^{\dagger}_{\mu} d_{\nu}| \psi \rangle \end{align}$$ The main difficulty is that I do not know how to evaluate the term $\langle \psi| d^{\dagger}_{\mu} d_{\nu}| \psi \rangle $. My purpose is to perform Anderson localization. So I would set the initial state $|\psi \rangle$ as a single-particle state localize at certain site. But I have no idea how to evaluate the correlation matrix $\langle \psi| d^{\dagger}_{\mu} d_{\nu}| \psi \rangle $? Could anyone give me some suggestion on it?


1 Answer 1


It's not necessary to diagonalize the single particle Hamiltonian. The Aubry-Andre model has a self-duality: Fourier transforming the single particle Hamiltonian results in a dual Hamiltonian that has the same form, but with rescaled hopping and disorder terms. The Fourier transform also maps localized states to delocalized states, and vice versa. Using these properties you can show that if there is no localization for a given set of parameters, then the dual Hamiltonian must exhibit localization.

Numerically, the correlation matrix can be evaluated by expanding the single site initial state as $| \psi \rangle = c_j^{\dagger} |0\rangle$, where $j$ is the initial site, and expanding in the eigenbasis as $$ \langle \psi| d_{\mu}^{\dagger} d_{\nu} |\psi \rangle = \sum_{\alpha\beta} V_{\alpha j}^{\dagger} V_{j\beta} \langle 0| d_{\alpha} d_{\mu}^{\dagger} d_{\nu} d_{\beta}^{\dagger} | 0 \rangle = \sum_{\alpha\beta} V_{\alpha j}^{\dagger} V_{j\beta} \delta_{\alpha\mu}\delta_{\nu \beta}, $$ which leaves you with $$ \langle n_i(t) \rangle = \sum_{\mu\nu} V_{i\mu} V_{\mu j}^{\dagger} V_{j\nu} V_{\nu i}^{\dagger} e^{i(E_{\mu} - E_{\nu})t}. $$

  • $\begingroup$ Thank you for the information. I did not know any about the duality of Aubry-Andre model and it seems interesting that it can tell us the localization. In fact I want to numerically show that there exist a localization with a correctly chosen parameter so that I encounter the difficulty of evaluating the correlation matrix $\endgroup$
    – Ricky Pang
    Apr 13, 2022 at 15:54
  • $\begingroup$ You can obtain the correlation matrix by expanding the initial state $|\psi\rangle$ in the eigenbasis. I have edited my original answer to show this. $\endgroup$ Apr 14, 2022 at 9:09

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