Friction in photon absorption/emission process Question: Is photon emission/absorption by an atom always accompanied by emission of soft photons (i.e. photons of very low energy)?
On the one hand, we can consider a scattering problem where at $t=-\infty$ we have an atom in its ground state and photon with the frequency exactly matching the atom axcitation energy: $\Delta=\epsilon_e-\epsilon_g=\hbar\omega$. We can calculate the probability/cross-section that at $t=+\infty$ the atom is in its excited state.
On the other hand, in practice we never encounter such a situation. In particular:

*

*there is always energy mismatch between a photon and an atom (e.g., due to the atom thermal motion)

*atom is coupled to the vacuum photon modes, which results in broadening of the transition

*the absorption happens over finite time

So in practice some energy is always lost in the form of low-energy photons, i.e., transferred into heat.
Background: The question is inspired by this answer which states that no collisions are elastic.
 A: I don't know if this is what you are seeking, so let me know in the comments if I am misunderstanding your question, and I'll delete this answer. The example is here for future people who will eventually want to know why it is, or why it should be the case.
Yes, the emission or absorption of photons is accompanied by the emission of soft photons. In QFT soft photons are very common because soft bremsstrahlung precisely cancels out IR divergences of UV integrals. Here is an example:
Let be an electron, not bound to a nucleus because the calculations are easier, that absorbs a photon (Because this calculation is a part of a real calculation taking into account the propagation of the photon, this one is considered off-shell here). The amplitude matrix at the zeroth order of perturbation theory is:
\begin{equation}
\mathcal{M}^{(n),0}_{\rho \sigma}(\gamma_k+e^-_p \rightarrow e^-_q)=-ie\overline{u}_\rho(p) \gamma^\mu u_\sigma(q) \epsilon^{(n)}_\mu(k)
\end{equation}
Where $k=q-p$. At one loop this amplitude becomes:
\begin{align}
\mathcal{M}^{(n),0+1}_{\rho \sigma}(\gamma_k+e^-_p \rightarrow e^-_q)=-ie\overline{u}_\rho(p)\left[ \gamma^\mu+\gamma^\mu F_1(k^2)-\frac{1}{4m_e}k_\alpha [\gamma^\alpha ; \gamma^\mu] F_2(k^2) \right] u_\sigma(q) \epsilon^{(n)}_\mu(k)
\end{align}
Where $F_1$ and $F_2$ are respectively the electric charge form factor and the magnetic momentum form factor. In its expression, only $F_1$ is IR divergent. Let us introduce a fictive mass to the photon, $m_\gamma$. Suppose that $m_e^2 \ll k^2$, the electric charge form factor becomes:
\begin{equation}
F_1(k^2) \stackrel{m_e^2 \ll k^2}{\approx}-\lim_{m_\gamma \rightarrow 0}\frac{\alpha}{2\pi} \ln \left( \frac{-k^2}{m_e^2} \right) \ln\left( \frac{-k^2}{m_\gamma^2} \right)
\end{equation}
The logarithms are called "Sudakov's double logarithms". So far, the cross-section of the studied case is:
\begin{equation}
\frac{d\sigma^{0+1}}{d\Omega}=\lim_{m_\gamma \rightarrow 0}\frac{d\sigma^{0}}{d\Omega}\left[ 1-\frac{\alpha}{\pi} \ln \left( \frac{-k^2}{m_e^2} \right) \ln\left( \frac{-k^2}{m_\gamma^2} \right)+\mathcal{O}(\alpha^2)\right]
\end{equation}
Now, let us introduce two phenomena: one for the emission of a soft photon before the absorption, and one for the emission of a soft photon after. After some calculations and approximations, one arrives at:
\begin{align}
\frac{d\sigma^{\text{Brem}}}{d\Omega}&=\frac{d\sigma^{0}}{d\Omega} \frac{2\alpha}{\pi}\int_0^{E_\Lambda}\frac{1}{|\vec{l}|}d|\vec{l}|\ln \left( \frac{-k^2}{m_e^2} \right)
\\ &=\frac{d\sigma^{0}}{d\Omega} \frac{\alpha}{\pi}\ln \left( \frac{E_\Lambda^2}{m_\gamma^2} \right) \ln \left( \frac{-k^2}{m_e^2} \right)
\end{align}
Where $E_\Lambda$ is some cut-off in the impulsion of emitted soft-photons. Summing the cross-sections yields:
\begin{equation}
\frac{d\sigma^{0+1+\text{Brem}}}{d\Omega}=\frac{d \sigma^0}{d\Omega} \left[ 1-\frac{\alpha}{\pi} \ln \left( \frac{-k^2}{m_e^2} \right)\ln \left( \frac{-k^2}{E_\Lambda^2} \right) +\mathcal{O}(\alpha^2)\right]
\end{equation}
Which is indeed IR finite! One can argue "Yes but what about the IR divergences in $\mathcal{O}(\alpha^2)$?" In fact, the calculation should be done at all orders of perturbation to cancel out all the IR divergences. So at an infinite number of loops, the total cross-section is:
\begin{equation}
\frac{d\sigma^\infty}{d\Omega}=\frac{d\sigma^0}{d\Omega}\exp \left[ -\frac{\alpha}{\pi} \ln \left( \frac{-k^2}{m_e^2} \right)\ln \left( \frac{-k^2}{E_\Lambda^2} \right) \right]
\end{equation}
Where an infinite number of soft photons have been emitted. And for people who don't trust my example, the KLN theorem should be sufficient.
Once again sorry if this is not what you are seeking.
A: Atomic emission and absorption are single photon processes. Atomic transitions can also occur by multiple photon absorption or emission but the probability for these is low. Any excess energy appears as atomic kinetic energy.
Note that such processes are inelastic, as kinetic energy is not conserved.
A: The absorption and emission of photons during bound-bound transitions in atoms is perfectly well described by single-photon physics, with no soft photons involved.
The fancy QFT mathematics in Jeanbaptiste's answer go beyond my expertise, but they deal with bremsstrahlung-like processes with unbound electrons, and it is missing the wrangling of QED required to describe bound states. In any case, QFT is not required to describe atomic transitions unless you are doing spectroscopy at high levels of precision – and, even then, you're still calculating small corrections to the energy of the (single) photon involved.
More particularly, the specific concerns that you posed do not justify your conclusion that "in practice some energy is always lost in the form of low-energy photons", which is not itself the same as "transferred into heat".
Im more detail:


*

*there is always energy mismatch between a photon and an atom (e.g., due to the atom thermal motion)


There can be an energy mismatch between the energy of the transition in the laboratory frame and the Doppler-shifted energy in the atom's rest frame, and this Doppler shift is perfectly easy to account for, as it is purely kinematic.
There is also a nontrivial dynamical effect in that the absorption or emission of a photon delivers a nonzero kick to the atom's centre of mass. This can be completely accounted for within standard atomic physics (I explained the details in this Q&A), and the result is simply a shift of the transition energy. In other words, the transition remains a single-photon process, and the only effect is a change in the energy of the photon.


*

*atom is coupled to the vacuum photon modes, which results in broadening of the transition

*the absorption happens over finite time


These two statements are simply Fourier transforms of each other. The atomic bound "eigenstates" are only eigenstates of the atom-only Hamiltonian, but they are not eigenstates of the full Hamiltonian of the system. (Otherwise, they would not decay!) Instead, once you account for coupling to the electromagnetic fields, they become resonances, with a finite width and a finite lifetime. But the coupling is still a single-photon one, and no soft photons are required to explain this.
