2
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ What is your question? $\endgroup$
    – Anyon
    Commented Jun 21, 2022 at 14:32
  • 1
    $\begingroup$ Hi, my question is specifically, how to calculate last equation $\endgroup$
    – Jimi
    Commented Jun 21, 2022 at 16:03

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Hi, that is wonderful. It works exactly as I expected. Many thanks for your great and clear answer $\endgroup$
    – Jimi
    Commented Jun 22, 2022 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.