the pump pulse is expressed in terms of the transverse vector potential [@footnote0], $$\begin{aligned}
 \label{pulseShape}
\textbf{A}_{\textbf{q}}(t)=\textbf{A}_{\textrm{p}}  e^{-\left(\frac{2\sqrt{\ln{2}}t}{\tau_{\textrm{p}}}\right)^2}\left(\delta_{\textbf{q},\textbf{q}_{\textrm{p}} }e^{-i\omega_{\textrm{p}}  t}+\delta_{\textbf{q},-\textbf{q}_{\textrm{p}} }e^{i\omega_{\textrm{p}}  t}\right) .\end{aligned}$$ Thereby, the coupling of the pump pulse to the superconductor is described by $H_{\text{em}}=H_{\text{em}}^{(1)}+H_{\text{em}}^{(2)}$, where

$$\begin{aligned}
H_{\text{em}}^{(1)}&=\frac{e\hbar}{2m}\sum_{\textbf{k},\textbf{q},\sigma}(2\textbf{k}+\textbf{q})\textbf{A}_{\textbf{q}}(t)c^{\dagger}_{\textbf{k}+\textbf{q},\sigma}c^{\phantom{\dagger}}_{\textbf{k},\sigma},\\
H_{\text{em}}^{(2)}&=\frac{e^2}{2m}\sum_{\textbf{k},\textbf{q},\sigma}\left(\sum_{\textbf{q}^{\prime}}\textbf{A}_{\textbf{q}-\textbf{q}^{\prime}}(t)\textbf{A}_{\textbf{q}^{\prime}}(t)\right)c^{\dagger}_{\textbf{k}+\textbf{q},\sigma}c^{\phantom{\dagger}}_{\textbf{k},\sigma} .\end{aligned}$$

We stress that the wave vector ${\bf q}_{\textrm{p}}$ must be kept finite in order to desribe the effect of the pump pulse correctly. If it were set to zero, all linear couplings of the electromagnetic field to the fermions are gone. This can be most clearly seen if we consider the metallic case without superconductivity. The single-band model does not allow any direct excitation process $\Delta {\bf q}=0$ but only indirect ones. Thus, neglecting the finiteness of the wave vector prevents any excitation in the linear regime. Moreover, certain effects such as the lowering of the order parameter $\Delta(t)$ and the Pauli blocking would be ill described if we set ${\bf q}_{\textrm{p}}=0$ [@Papen07].

The absorption spectrum of the nonequilibrium state is measured by a probe pulse, which follows the pump pulse after a certain delay time $\delta t$. The probe pulse has the same shape as the pump pulse, Eq. , but much weaker intensity. We consider both negative 