to the lab frame and the average flow of particles can be quantified by calculating the average momentum $K(t) = \int n(k,t)\, k\, dk$. While calculating the full momentum distribution is a numerically demanding process for large particle numbers, the average momentum can be calculated efficiently as it is related to the probability current density of the system as $  K(t) =  \frac{1}{L}  \int j(x,t)\;dx\;$. For a TG gas, the probability current density is identical to that of the free Fermi-gas and is simply given as the sum of the single-particle currents corresponding to the occupied energy levels $$j(x,t) = \frac{\hbar}{m} \mathrm{Im} \sum_{n=0}^{N-1} \psi^*_n(x,t)\, \partial_x \psi_n(x,t).$$ For the time-dependent wavefunction given in Eq. , the explicit form of the average momentum can therefore be calculated as $$\begin{aligned}
K(t) &=&K_0 +K_t(t) \nonumber \\
 &=&\frac{N \hbar q}{m L} + \frac{\hbar}{m L} \sum_{n=0}^{N-1}  \sum_{j} |c_{jn}|^2 \mathrm{Im} \left[ F_{jj}\right] \nonumber \\ 
&+&\mathrm{Im} \left[\frac{\hbar}{m L} \sum_{n=0}^{N-1} \sum_{j \neq k} c_{jn} c_{kn}^* e^{-i(E_{k,q}-E_{j,q})t/ \hbar}  F_{jk}(t)\right],
\label{eq:explicitcurrent}\end{aligned}$$ where $$F_{jk}(t) = \int dx  \psi_{j,q}^*(x-vt) \, \partial_x \psi_{k,q}(x-vt).
\label{Fjk}$$ The elements $F_{jj}$ are independent of time in all cases we consider, as the two eigenfunctions are the same, which means that the time-dependence just shifts the integrand. For $j 