# Photon scattering final state

I'm looking through the linked paper on x-ray scattering. The Section $$5$$ says that the expression $$(67)$$ (which contains an error but the editor mentions the correct variant at the beginning of the paper) describes the final state of the system after scattering. It has the form $$\left| F \right\rangle = \left| {\Psi _0^{{N_{el}}}} \right\rangle {\widehat a^\dagger }_{{k_F},{\lambda _F}}\left| {{N_{EM}} - 1} \right\rangle,$$ where $$\left| {\Psi _0^{{N_{el}}}} \right\rangle$$ is the electronic ground state of the molecule with $$N_{el}$$ electrons,

$$\left| {{N_{EM}} - 1} \right\rangle$$ is the Fock state containing $$N_{EM}-1$$ photons in the mode ($${\bf k_{in}}, \lambda_{in})$$), and

$${\widehat a^\dagger }_{{k_F},{\lambda _F}}$$ creates a photon in the mode ($${\bf k_{F}}, \lambda_{F})$$)

I wonder why the field final state contains the creation operator? I thought that the final state should have been written like this $$\left| F \right\rangle = \left| {\Psi _0^{{N_{el}}}} \right\rangle \left| {{N_{EM}} - 1} \right\rangle$$ because the field lost one photon. May it mean just that a field loses the photon in mode ($${\bf k_{in}}, \lambda_{in})$$) and creates a photon in mode $$({\bf{k}_F},{\lambda _F})$$ after scattering?

I thought that the final state should have been written like this $$\left| F \right\rangle = \left| {\Psi _0^{{N_{el}}}} \right\rangle \left| {{N_{EM}} - 1} \right\rangle$$ because the field lost one photon.
May it mean just that a field loses the photon in mode ($${\bf k_{in}}, \lambda_{in})$$ and creates a photon in mode $$({\bf{k}_F},{\lambda _F})$$ after scattering?