Electromagnetic waves carry linear momentum. So when they strike a solar sail (or any reflective surface), the object experiences radiation pressure and net energy gain.
The reflected EM waves must then have less energy. Does this result in a redshifted wavelength or a decrease in expected photon number? I'm inclined to say it lowers photon number (or in classical E&M, a decrease in field amplitude). Curious if there is an authoritative statement on the matter.
It’s a redshift.
Suppose you only have a single photon to begin with. It reflects from a mirror, imparting $\hbar \omega /c$ momentum. The mirror’s kinetic energy increases. Then you still have one photon but its energy is lower. Work out the math, and voila! It’s been redshifted (twice, from both the absorption and emission events, with the relevant velocity in each case being the average velocity of the mirror before and after the event).
In terms of quantum mechanics, photon is scattered away by net electrons field of surface atoms. This implies that due to momentum conservation, almost all photon momentum is passed back from surface electric field to photon. So energy looses of photon (if any) due to field recoil would be very very small.
More interesting case is where individual atom absorbs or emits photon. In such case atom experiences recoil, given by : $$ E_R = \frac{{E_\gamma}^{2}}{2 Mc^2} $$, Where $E_R$ is nucleus recoil kinetic energy, $E_\gamma$ photon energy and $M$ nucleus mass.
If free nucleus (not in lattice) isn't stationary and has mean kinetic energy $\overline E_k > 0$, then due to Doppler shift, this will induce spectral lines broadening of emitted photon, given by : $$ \overline {\nu} = \nu_{0} \sqrt{\frac{2 \overline E_k}{Mc^2}}$$
