(Anderson Localization)Time evolution of occupation number of free fermion model

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:

$$$$H = -t \sum_{i}^{L} (c^{\dagger}_{i} c_{i+1} + h.c.) + \sum_{i}^{L}\lambda_{i} c^{\dagger}_{i} c_{i}~,~ \lambda_{i} = \cos( 2 \pi \sigma i)$$$$ where $$\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?

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 $$\langle n_i(t) \rangle = \sum_{\mu\nu} V_{i\mu} V_{\mu j}^{\dagger} V_{j\nu} V_{\nu i}^{\dagger} e^{i(E_{\mu} - E_{\nu})t}.$$
• 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. Apr 14, 2022 at 9:09