# Un-equal time correlation via non-interacting tight-binding Hamiltonian

Let's assume we have a model, which is initially defined by the tight-binding Hamiltonian with a random on-site energy $$f_n$$, as follows:

$$H^i=-J\sum_n^{L-1}\left(a_n^\dagger a_{n+1}+h.c\right)+\sum_n^Lf_na_n^\dagger a_n$$

The above single-particle Hamiltonian can be diagonalized in real space with basis change:

$$\eta_m^\dagger=\sum_n U_{nm}a_m^\dagger$$

So, we have $$H^i=\sum_m\epsilon_m\eta_m^\dagger\eta_m$$, and the many-body ground state can be constructed by creating a particle up to a filling point:

$$\mid\Psi_0^i \rangle=\Pi_m^N\eta_m^\dagger\mid0\rangle$$

where $$\mid 0 \rangle$$ is the vaccum state, and $$N\le L$$ is the number of particles.

Now, we do a global quench and turn the random on-site term off:

$$H^f=-J\sum_n^{L-1}\left(a_n^\dagger a_{n+1}+h.c\right)$$

This Hamiltonian is also diagonalizable in both momentum and real space, so all energy and corresponding eigenstates are available. From now on, the dynamic of the model is driven by $$H^f$$. We are interested in calculating the unequal time correlation defined by:

$$\left<\Psi_o^i|a_r(t)a_{r'}^\dagger(0)|\Psi_o^i\right>=\left<\Psi_o^i|e^{iH^ft}a_r(0)e^{-iH^ft}a_{r'}^\dagger(0)|\Psi_o^i\right>$$

The question is, how to calculate numerically the above equation?

• What is your question? Commented Jun 21, 2022 at 14:32
• Hi, my question is specifically, how to calculate last equation
– Jimi
Commented Jun 21, 2022 at 16:03

For fermionic Gaussian states, we can use the covariance matrix to completely characterize the state, $$C_{mn} = \langle \Psi|a^{\dagger}_{m} a_{n}|\Psi\rangle.$$ This also applies to mixed state. The correlation matrix is a semi-positive Hermitian matrix, with eigenvalues between 0 and 1.
Given a set of occupying orbitals labelled by $$\eta^{\dagger}_k = \sum_{m=1}^L U_{km}a^{\dagger}_m,$$ where $$1\leq k\leq N$$, the correlation matrix is simply $$C_{mn} = \sum_{k=1}^N U^{*}_{km} U_{kn}$$.
Now back to your question. The first step is to compute the coefficients $$V_{rs}$$, where $$a_r(t) = \sum_{s} V_{rs}a_s.$$ (Note also that operators with no time labelled means the operator at $$t=0$$). This can be computed via a matrix exponential $$V = e^{-i H^{f}_{\mathrm{mat}} t},$$ where $$H^{f}_{\mathrm{mat}}$$ is simply a tridiagonal matrix with off-diagonal elements $$-J$$. (This is exactly the first-quantized form of $$H^{f}$$). In your case, $$V$$ can be computed analytically by fully diagonalizing $$H^{f}_{\mathrm{mat}}$$ using Fourier modes. More generally, If $$J$$ also varies with position, then the matrix exponential must be computed numerically.
Then the correlation function is simply $$\langle a_r(t) a^{\dagger}_s(0)\rangle = \sum_{s'} V_{rs'} \langle a_{s'}a^{\dagger}_s\rangle = \sum_{s'} V_{rs'} (C_{ss'}+\delta_{ss'}),$$ where in the last equality I have used the canonical commutation relation to switch the creation and annihilation operators.