$t=0$ our system is in an eigenstate at a given lattice depth and then suddenly rotating the external lattice potential at finite speed $v$, so that for $t>0$ we have $V_\mathrm{ext}(x_n-vt)=V_0 \cos^2(k_l (x_n-vt))$. The time-dependent many-body wavefunction after the quench can still be calculated via the Bose-Fermi mapping theorem, so that we only need to evolve the initial non-rotating single particle states according to the time-dependent Schr[ö]{}dinger equation $$i \hbar \partial_t \psi_n(x,t) = \left( -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2}+V_\mathrm{ext}(x-vt) \right) \psi_n(x,t)\;.
\label{eq:timedependentequation}$$ The time-dependent solutions can be found exactly in terms of the eigenfunctions $\psi_n(x,0)$ of the initial Hamiltonian and the eigenfunctions $\psi_{j,q}(x)$ and eigenvalues $E_{j,q}$ of the Hamiltonian in the co-rotating frame as [@Schenke2012] $$\psi_n(x,t) = e^{iqx}e^{-\frac{i \hbar q^2}{2m}t} \sum_j c_{jn} e^{-iE_{j,q}t/ \hbar} \psi_{j,q}(x-vt),
\label{eq:timedependentsol}$$ where $q=\frac{m v}{\hbar}$. The eigenfunctions in the co-rotating frame are a solution to the Schrödinger equation $$E_{j,q}\psi_{j,q}(x) = \left( -\frac{\hbar^2}{2 m}\frac{\partial^2}{\partial x^2}+V_\mathrm{ext}(x) \right) \psi_{j,q}(x),
\label{eq:comovingeq}$$ obeying the twisted boundary conditions $\psi_{j,q}(x+L)=e^{-i q L}\psi_{j,q}(x)$. The coefficients $c_{jn}$ then contain all the information about the initial state of the system and are defined as $$c_{jn}= \langle \psi_{j,q}(x)|e^{-iqx}| \psi_n(x,0) \rangle = \int dx \, \psi_{j,q}^*(x)e^{-iqx} \psi_n (x,0).$$

The applied rotation will lead to transport of particles with respect 