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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.

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  • $\begingroup$ atoms and photons are quantum mechanical entities they do not follow the classical mechanics that the answer you are quoting assumes a priori . $\endgroup$
    – anna v
    Sep 7, 2021 at 9:13
  • $\begingroup$ @annav what I like about that answer is the suggestion that all the collisions are really inelastic. Do soft photons commongly occur in particle physics experiments? $\endgroup$ Sep 7, 2021 at 9:17
  • $\begingroup$ In particle physics the particles taking part in the interaction are countable. In photon atom scattering, three things may happen at the center of mass photon-atom 1) elastic scattering 2)inelastic where the whole energy/momentum of photon is absorbed by the atom 3) the atom is ionized, an electron kicked out,, and also a lower energy photon goes off conserving energy and momentum . "soft photons" have no meaning. $\endgroup$
    – anna v
    Sep 7, 2021 at 10:25

3 Answers 3

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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.

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  • $\begingroup$ +1 This certainly goes in the right direction - please do not delete it. But I still would like to link it with an absorption by an atom: would it still work for a bound electron, transitioning between two levels? Is it still the same in non-relativistic limit (non-relativistic for the atom/electron, not the photons of course)? My my background is in condensed matter and quantum optics, so I am coming to it from different direction - I think natural band width implies soft photons. $\endgroup$ Sep 7, 2021 at 14:51
  • $\begingroup$ I think it should work for bound electrons too, the amplitude $\mathcal{M}(\gamma N+e^- \rightarrow N+e^-)$ ($N$ is the nucleus) has in it the amplitude I used in this answer (modulo extraction of the polarization vector of the photon which becomes a propagator). The only part which may be problematic is the calculation of the cross-section from the amplitude since one has to use different polarization bi-spinors. But since the zeroth order, the first order, and the Bremsstrahlung cross-sections all three involve terms of the form $u \overline{u}$, it should be OK. $\endgroup$ Sep 7, 2021 at 15:23
  • $\begingroup$ For the non relativistic limit, I don't know :/ $\endgroup$ Sep 7, 2021 at 15:23
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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.

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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.

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  • $\begingroup$ The point about the reference frame for the Doppler shifts is useful. My second bullet refers to the natural linewidth - this however means that atom would absorb photons that do not match exactly its transition frequency, due to its coupling to the vacuum photon field. Hence my conjecture about low-energy photons. The third bullet refers to the finite experiment duration (not to finite lifetime, although it may be awkwardly formulated): the sharp atomic resonances exist only in Fermi Golden rule or scattering theory, where the duration of the experiment is effectively taken to infinity. $\endgroup$ Sep 13, 2021 at 11:56
  • $\begingroup$ @RogerVadim The fact that the atomic transition energy does not match the photon energy is not a contradiction (and therefore does not require soft photons to "fix"). The states you are comparing are eigenstates of $H_\mathrm{atom}+H_\text{field-only}$, but this is not the energy/Hamiltonian operator of the system (because it's missing the coupling) and therefore it is not conserved. It's tempting to thing of the eigenvalues of $H_\mathrm{atom}$ and $H_\text{field-only}$ as "energies", but it is very important to keep in mind that this is only an approximation. $\endgroup$ Sep 13, 2021 at 12:03
  • $\begingroup$ Indeed, you are right about this one. $\endgroup$ Sep 13, 2021 at 12:06
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    $\begingroup$ One useful instance where this is very clear is in the application of the Rotating-Wave Approximation to derive the Jaynes-Cummings model from the Rabi model, where it is often argued that the counter-rotating terms must be discarded because they "don't conserve energy". Of course, this is complete gibberish: only the full Hamiltonian of the system is the energy; the counter-rotating terms fail to conserve the excitation number, which in some occasions is useful as it is approximately conserved (but nothing more). The same happens for your soft photons. $\endgroup$ Sep 13, 2021 at 12:06

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