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Let's argue within non-relativistic QT for simplicity (and it's entirely justified for not too high ##Z##). Due to translation invariance the total momentum of an atom is conserved, i.e., the center-of-mass motion separates from the relative motion of the electrons and the atomic nucleus. The bound energy eigenstates of the latter define the intrinsic states of the atom. Indpendently from this the atom can move with any momentum relative to your (arbitrarily chosen) reference frame.sol47739 said:How is that consistent with the Blackbody spectrum if an atom in its ground state can move around?
If you know have the atom in some excited state you can always Galilei-boost to its rest frame, i.e., to the frame, where the total momentum of this atom is 0. Now it will at some random time relax to a lower state by spontaneously emitting a photon (due to the vacuum fluctuations of the em. field). This photon has a momentum ##\vec{p}_{\gamma}=\hbar \vec{k}##, and since momentum is conserved (due to spatial translation invariance of the closed system consisting of the atom + (quantized) radiation field) the total momentum of the atom after emission of the photon must be ##\vec{P}_{\text{atom}}=-\vec{p}_{\gamma}##.
This should be in any textbook on atomic physics. I've no time to look for a specific reference though ;-)).