pulse and $$\begin{aligned}
 \label{defCurrent}
{\bf j}_{\textbf{q}_{\text{pr}}}(\delta t, t)
&&
=\frac{-e\hbar}{2mV}
\sum\limits_{\textbf{k}, \sigma}(2\textbf{k}+\textbf{q}_{\text{pr}})
\left\langle c^{\dagger}_{\textbf{k},\sigma}c^{\phantom{\dagger}}_{\textbf{k}+\textbf{q}_{\text{pr}},\sigma}\right\rangle (\delta t, t) 
\nonumber\\
&&
\; \quad- \frac{e^2}{m V}
\sum\limits_{\textbf{k}, \textbf{q}, \sigma}
{\bf A}_{\textbf{q}_{\text{pr}} - \textbf{q}} \left\langle c^{\dagger}_{\textbf{k},\sigma}c^{\phantom{\dagger}}_{\textbf{k}+\textbf{q} ,\sigma}\right\rangle (\delta t, t)    . \;\end{aligned}$$ Formally, the current depends on two times, namely the delay time $\delta t$ between pump and probe pulse and the actual time $t$ at which it is measured or computed, respectively. For the numerical calculations we neglect the second term in Eq.  since it only results in a constant offset of the imaginary part of the conductivity spectra. Then, the pump-probe conductivity $\sigma ( \delta t,  \omega )$ is obtained from Eq.  via

\[eq:sigma\] $$\begin{aligned}
\sigma(\delta t , \omega)=\frac{j(\delta t, \omega)}{i\omega A( \delta t , \omega)},\end{aligned}$$ where $$\begin{aligned}
j ( \delta t,  \omega ) =  \int_{-\infty}^{\infty} d t \;  \hat{{\bf e}}_y  \cdot {\bf j}_{{\bf q}_{\textrm{pr}}} ( \delta t, t) \, e^{i \omega t}  \end{aligned}$$ and $$\begin{aligned}
A ( \delta t, \omega ) =  \int_{-\infty}^{\infty} d t \;  \hat{{\bf e}}_y  \cdot  {\bf A}_{{\bf q}_{\textrm{pr}} }
( \delta t, t) \, e^{i \omega t}  \end{aligned}$$

denote the Fourier transformed $y$ components of the current density ${\bf j}_{{\bf q}_{\textrm{pr}}}(\delta t, t)$ and the vector potential 