Frequency difference of absorbed and emitted photons? In a transparent medium, how does the frequency of incoming light compare to the exiting light? I thought it was the same, but doesn't there have to be an energy difference in two electron states in the atoms of the medium that is exactly equal to the energy of the photon? I feel like an exact match would be unlikely. Is there a litlle energy lost in the process that lowers the exit frequency by a slight amount?
 A: In an ideal transparent medium, without any dipoles, then there is no frequency difference between the incident light ray and the transmitted ray, as no scattering or absorption can take place. 
Now materials that are transparent may have very different atomic structures, e.g. crystalline, glassy, but the important factors in light scattering are the inter-atomic distances, wavelength of the light, electronic energy levels of the material, atomic vibrations etc. If any of the mentioned factors is not fulfilled in a material, then scattering takes place (e.g. Raman inelastic scattering, red-blue light shift). One should also note that in solids, gaseous or liquid mediums, there are further different complications to consider, like diffuse reflection, Rayleigh, Thomson scatterings, etc.
Finally to take the famous example of glasses, consisting of silica in amorphous form, the electrons have no available energy levels above them in range of the visible light, or if they do, there's no appreciable absorption because of their amorphous structure, all of which makes them "ideal" transparent material. Ideal in quotation marks, because again there are other complications like difference of density in the material which can create scattering centers.
Lastly, a transparent medium, unlike a translucent one, follows the snell's law of refraction in change of medium (change of index of refraction), which means that for one thing the frequency does not change, the light is transported through the medium and allows for image formation as the photons are not scattered at the interfaces.
A: There doesn't have to be a difference between two energy levels that is the same as the photon energies of the incoming and exiting light. The molecule of the medium can be in a virtual energy state that doesn't correspond to an electronic state, for a very short time, if the product of the energy and time $\Delta E \Delta t \leq \hbar/2$, due to the uncertainty principle. Only the initial and final states have to be actual energy levels of the molecule. This corresponds to a photon being absorbed and almost immediately re-emitted.
