The thermal radiation associated with some object is typically described in terms of the "black-body" spectrum for a given temperature, given by the Planck formula. This formula is based on an idealization of an object that absorbs all frequencies of radiation equally, but it works fairly well provided that the object whose thermal spectrum you're interested in studying doesn't have any transitions with resonant frequencies in the range of interest. As the typical energy scale of atomic and molecular transitions is somewhere around an eV, while the characteristic energy scale for "room temperature" is in the neighborhood of 1/40 eV, this generally isn't all that bad an assumption-- if you look in the vicinity of the peak of the blackbody spectrum for an object at room temperature, you generally find that the spectrum looks very much like a black-body spectrum.
How does this arise from the interaction between light of whatever frequency and a gas of atoms or molecules having discrete internal states? The thing to remember is that internal states of atoms and molecules aren't the only degree of freedom available to the systems-- there's also the center-of-mass motion of the atoms themselves, or the collective motion of groups of atoms.
The central idea involved with thermal radiation is that if you take a gas of atoms and confine it to a region of space containing some radiation field with some characteristic temperature, the atoms and the radiation will eventually come to some equilibrium in which the kinetic energy distribution of the atoms and the frequency spectrum of the radiation will have the same characteristic temperature. (The internal state distribution of the atoms will also have the same temperature, but if you're talking about room-temperature systems, there's too little thermal energy to make much difference in the thermal state distribution, so we'll ignore that.) This will come about through interactions between the atoms and the light, and most of these interactions will be non-resonant in nature. In terms of microscopic quantum processes, you would think of these as being Raman scattering events, where some of the photon energy goes into changing the motional state of the atom-- if you have cold atoms and hot photons, you'll get more scattering events that increase the atom's kinetic energy than ones that decrease it, so the average atomic KE will increase, and the average photon energy will decrease. (Or, in more fully quantum terms, the population of atoms will be moved up to higher-energy quantum states within the box, while the population of higher-energy photon modes will decrease.)
For thermal radiation in the room temperature regime, of course, the transitions in question are so far off-resonance that a Raman scattering for any individual atom with any particular photon will be phenomenally unlikely. Atoms are plentiful, though, and photons are even cheaper, so the total number of interactions for the sample as a whole can be quite large, and can bring both the atomic gas and the thermal radiation bath to equilibrium in time.
I've never seen a full QFT treatment of the subject, but that doesn't mean much. The basic idea of the equilibration of atoms with thermal radiation comes from Einstein in 1917, and there was a really good Physics Today article (PDF) by Dan Kleppner a few years back, talking about just how much is in those papers.
Looks like you are already familiar with the classical explanation but are still curious about the quantum version of it.
2.phase difference between absorbed and emitted light
Yeah, this is essentially the lowest order contribution to the phase shift in the photon-electron scattering. Here is the sloppy way to visualize it continuously (this is basically the 'classical EM wave scattering' point of view): you can imagine that the "kinetic energy" (-> frequency) of the "photon" increases as it approaches the atom's potential well and then it goes back to its normal frequency upon leaving the atom. This translates to a net increase in the phase ($(n-1)\omega/c$).
- "drift velocity" of photons ( they aren't the same photons, they are re-emitted all the time)
By "drift velocity" do you mean a pinball-like, zigzag motion of the photon? This won't contribute that much because it requires more scattering (basically it is a higher order process).
And also, I still don't really understand about the detail of the absorption-emission process.
Yes the absorption will still occur in all range of the frequency. The hamiltonian of the atom will be modified by the field (by $- p \cdot E$ where p is the dipole moment of the atom and E is the electric field component of the light). This will give us the required energy level to absorb the photon momentarily, which will be re-emitted again by stimulated+spontaneous emission.
edit: clarification, the term 'energy level' is misleading, since the temporarily 'excited' atom is not in an actual energy eigenstate.
See the diagram here: http://en.wikipedia.org/wiki/Raman_scattering
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Regarding the general question posed in the title, yes, it is possible in general to have photon emission (or absorption) without electrons changing energy levels. For example, in nuclear magnetic resonance (NMR), EM radiation (photons) are absorbed and re-emitted with changes in nuclear spin state. NMR spectroscopy relies on splitting between nuclear spin states due to a large applied magnetic field. This splitting causes absorption and stimulated emission of EM radiation with energy equal to the splitting. So NMR is one example of photon emission without changes in electronic energy levels playing a direct role.
However, with regards to your specific examples, electrons are directly involved. In the vibrational modes, the resonant frequencies are determined by the stiffnesses of the molecular bonds. Those stiffnesses in turn depend on the electronic structure of the molecule. The question is, how much does the electronic energy change when the bond lengths and angles are changed by small amounts? This can be calculated using electronic structure packages like Gaussian and ABINIT. The nuclei are also involved; the masses of the nuclei also factor into the resonant frequencies. The picture is similar for (dihedral) rotation of chemical bonds.