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Consider a photon incident upon a atom. Dependent on the electronic makeup of the atom, the frequency of the light, and/or group velocity of incident photon, we might see:

  1. Various elastic/inelastic scattering processes (Thomson Scattering, Rayleigh Scattering, etc.)
  2. Absorption and spontaneous Emission
  3. Absorption and transferal into vibrational motion of the atom
  4. Interaction of photon with nuclear factors (proton(s)/neutron(s))

So, what is the actual difference between the phenomena of absorption and then emission of light (or even the absorption and transferal into vibrational atomic motion) and the various scattering phenomena?

Reasoning is that in any scattering, the incident photon interacts with the electron of an atom and is absorbed, and either:

  1. An indistinguishable photon with same frequency is emitted with a change in angle in terms of elastic scattering.
  2. Or is absorbed and a photon with changed frequency is emitted in terms of inelastic scattering.

But in both scattering the incident photon is absorbed.

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  • $\begingroup$ I always wondered how a "half transparent mirror" works but did not work on the question enough to ask it. Your question looks quite related. $\endgroup$ Commented Jun 13 at 23:59

4 Answers 4

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When considering Thompson scattering processes and absorption+spontaneous emission, indeed both the initial state (photon+atom) and the final state (photon and atom with same energy) appear to be the same. However, there are some important differences. First of all, scattering processes are not isotropic, but follow a dipole pattern $\sim \sin(\theta)^2$. In scattering processes the direction of the initial photon/ EM wave breaks the spherical symmetry, after an absorption the symmetry is restored for the excited atom. Spontaneous emission is spherically symmetric.

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  • $\begingroup$ How is the symmetry restored for an excited atom if the resulting orbital's orientation is correlated with the incident photon's polarization? $\endgroup$
    – Ruslan
    Commented Jun 14 at 11:20
  • $\begingroup$ Scattering in general does not imply the final state is the same. It usually isn't, as the secondary wave has different directions and sometimes even different frequency than the primary wave. $\endgroup$ Commented Jun 14 at 16:15
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I'm not entirely sure if this answers your question, but by quantizing the incoming radiation field, the interaction between an atom and the field is found involve three types of processes, where a different number of photons is created or annihilated: (1) single-photon process corresponding to "normal" absorption or emission (2) two-photon processes, and (3) zero-photon processes (simultaneous creation and annihilation), corresponding to scattering. One can then interpret the mathematics in various ways (e.g., Feynman diagrams).

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  • $\begingroup$ This is also my point $\endgroup$
    – Nikos M.
    Commented Jul 7 at 14:14
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I'm unhappy that the existing answers came close to the essence of your question but missed, so here goes:

Consider an isolated atom in an empty universe. Then, as is popularly stated in introductory literature, this atom cannot absorb photons if the photons are not of the exact energy differences needed to excite the electron to higher energy levels of the same atom. This means that there might not be the possibility of absorption and re-emission that you are even thinking about. Then, it can only scatter the photons, and in so doing, it is clear that scattering necessarily has to be different from absorption and re-emission.

There is some tiny time delay for scattering, but it is bounded. It cannot delay for too long. Absorption and re-emission, however, can have arbitrarily long delay. This means that the phase relationship is protected in scattering but is destroyed in absorption and re-emission. Again, they must be different.

Note that this phase relationship is quite important. For example, if you want a mirror reflection, then you have to protect this phase relationship. Otherwise, you would get a white reflection that you ought not to be able to discern what the bright object giving off the light to be. Of course, this is not anywhere near as simple, because in a mirror reflection, we are talking about a lot of electrons oscillating in phase, and this might have a bit of colour changing, say red-shifting, but the general idea I'm trying to impart is there.

Now, for an isolated atom, we can find the centre-of-momentum frame of the photon and atom. Then we can talk about elastic v.s. inelastic collisions. In an elastic collision, the incoming photon (and the atom) in the centre-of-momentum frame can only rotate the momentum vector, since the energy stays the same.

If, instead, there is some change in energy, then the atom is in some way excited to a different energy state.

In this centre-of-momentum frame, there is no reason why the photon cannot be exactly the frequency needed to excite the atom to an higher energy state. If it is, then the photon can be absorbed, and then after an arbitrarily long time, it can be re-emitted.

If you know about Feynman diagrams, absorption alone is a phenomenon that happens with just one factor of the fine structure constant, whereas scattering without absorption requires two factors. Yes, emission will also require another factor of the fine structure constant, but the point I'm trying to make here is that, because the fine structure constant is small, this means that pure scattering is a far lower probability of happening event, whereas absorption is far higher probability. Because of this, when white light (to a good approximation, sunlight) passes through a gas (e.g. our atmosphere), the probability of absorption near the excitation energies of molecules of the gas is far greater than the pure scattering probability. This is why we get absorption lines.

To clarify, inelastic scattering is slightly different from absorption and re-emission because the incoming photon does not have to be of the exact energy difference of the molecule.

I have also taken care to talk about the centre-of-momentum frame. This automatically takes into account the recoil of the molecule, which can then translate into the vibration phonon mode if inside a solid or something like that. I have deliberately neglected the nuclear factors; those are yet more complications in this already complicated answer.

But the answer is definitely yes, scattering, elastic and inelastic, are different from absorption and re-emission, and the difference is for our own convenience. If you do not make the distinction, then discussion will become very annoying.

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  • $\begingroup$ Nice attempt to argue why scattering is not same as absorption/remission if a photon is not of such energy as existing excitation levels of a system but fails for precisely the same reasons. If the system cannot absorb some energy in some way even if it is different than its excitation levels, then scattering cannot happen as well, because if we agree that scattering can absorb some energy in another way, same can be said for absorption/remission. $\endgroup$
    – Nikos M.
    Commented Jul 5 at 20:55
  • $\begingroup$ Furthermore that some cases of absorption and remission can take arbitrarily long time is irrelevant to the fact of the cases of absorption/remission that take arbitrarily short time, and are the ones we can use to model various cases of scattering. $\endgroup$
    – Nikos M.
    Commented Jul 5 at 21:00
  • $\begingroup$ Finally quantum.mechanical calculations of scattering processes are same as quantum mechanical calculations for suitable pairs of absorption/emission processes (eg in terms of Feynman diagrams). $\endgroup$
    – Nikos M.
    Commented Jul 6 at 19:32
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So, what is the actual difference between the phenomena of absorption and then emission of light (or even the absorption and transferal into vibrational atomic motion) and the various scattering phenomena?

"Scattering" refers to matter creating a secondary wave (or a modified flow of particles, in the naive "photon-shot" model), most often with different characteristics (direction, center of origin, frequency) than the primary wave. There need not be any delay between the primary wave hitting the system, and the secondary being created; both happen at the same time. During scattering, some energy may get absorbed and reradiated, but this can often be neglected, especially at high enough frequencies, if most incoming energy gets redirected via scattering, and almost none gets dissipated after absorption. This is the case e.g. for Thomson scattering, and it is called "elastic" because of this negligible change in radiation energy.

"Absorption and emission" refers rather to energy, or to energy quanta $h\nu$, than to EM field/wave. Also, sometimes people mean there is some delay between the matter absorbing radiation, and emitting it; for example, often spontaneous emission takes a lot of time compared to time needed to get to the excited state. When this delay is important, or when substantial part of absorbed energy is not radiated back, the process is very much different from the elastic scattering above, and can be called inelastic scattering, or absorption and subsequent emission.

Some types of scattering, like the Raman scattering, can be sometimes seen described as "due to" virtual absorption/virtual emission of a photon. This is just an imperfect language to describe what Schroedinger's equation predicts about emitted radiation. There is no substantial delay between absorption and emission, scattering happens very fast, so it is an inelastic scattering, very similar to Compton scattering (both are inelastic because emitted radiation frequency changes).

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