# Can inelastic scattering still give rise to diffraction?

I am having a conceptual difficulty reconciling inelastic events and diffraction, particularly whether or not you can have inelastic diffraction.

Here is my thought experiment that I am working through (see cartoon picture below). Imagine you have a line of atoms spatially separated by a periodic spacing $$a$$, and that each atom has two internal states $$\vert 0 \rangle$$ and $$\vert 1 \rangle$$ with energy separation $$\Delta \omega$$. There is no coupling between atoms, so the ground state of the whole system is $$\vert 0 \cdots 0\rangle$$. Likewise, the excited states are all binary combinations of $$1$$'s and $$0$$'s (for example $$\vert 1,0 \cdots 0\rangle$$, $$\vert 0,0,1 \cdots 0\rangle$$ or $$\vert 0,1 \cdots 0\rangle$$ and so on...).

Now let's say you send an planewave made of photons (or neutrons, doesn't really matter) with momentum $$k$$ and energy $$\omega$$ onto this array of atoms and look at the scattered waves. The incident wave will be (elastically) diffracted by transferring momentum $$k-k'=q=2\pi n/a$$ to the atoms, which gives rise to diffraction peaks. It is important that while the momentum of the incident waves changes, the energy does not (i.e. elastic scattering).

But now consider the scenario where the incident wave of photons inelastically scatters, promoting a single atom from $$\vert 0 \rangle$$ to $$\vert 1 \rangle$$. These photons will lose energy, so $$\omega-\omega'=\Delta \omega$$. My question is, will these inelastic scattered photons that excite atoms from $$\vert 0\rangle$$ to $$\vert 1 \rangle$$ still form a diffraction pattern because the atoms are periodically spaced by $$a$$? one could distinguish these waves with the elastic scattered ones by using a energy-discriminating detector, for example.

Intuitively, I would expect the inelastic scattered photons to still form a diffraction pattern, since the atoms form a perfectly spaced array after all. However, if I look at Fermi's golden rule for scattering, it tells me that the different inelastic scattering processes don't add coherently (i.e. the phase doesn't matter between different inelastic scattered events even if they involve the same energy and momentum). Apparently this will mean that the inelastic photons do not form a diffraction pattern. How do I reconcile these two viewpoints?

$$\Gamma(\Delta\omega) \propto \sum_f \vert \langle f \vert \hat{M} \vert i\rangle\vert^2 \delta(\omega_{fi} - \Delta\omega)$$

• I am not sure what do you mean exactly by "diffraction pattern" and what is the role of Fermi's golden rule in that. However, inelastic diffraction is a thing. In acousto-optic events, the energy of the photons scattered by the phonons is not conserved. Nov 12, 2020 at 16:39
• By "diffraction pattern" I just mean whether the inelastic light will change direction to form a periodic pattern in the far field, or whether they will only lose energy and nothing else. Fermi's golden rule is the standard method to describe scattering in the first Born approximation, which is usually quite accurate away from a resonance or in highly non-perturbative high-field limits. Nov 12, 2020 at 20:06
• Inelastic neutron diffraction is an established technique. One can for example measure crystal-field excitations of rare earth elements. And also the dispersion relations of phonons and magnons.
– user137289
Nov 16, 2020 at 20:49
• @Pieter for crystal field excitations, which I think are mostly decoupled from each other, can you form a diffraction pattern? From S. Mcgrew's answer, it seems the most important quality is whether one could identify which atom was excited after scattering. If he is right, you should get a diffraction pattern with delocalized excitations (phonon and magnon), since you can't identify which atom is excited. On the other hand for crystal fields you are able to identify the atom excited (not sure about this to be honest), so S. Mcgew's answer would predict no diffraction in that case. Nov 17, 2020 at 4:40

If there is a way to measure the scattering particles after a single scattering event and determine thereby which particle did the scattering, then there should be no diffraction pattern. However, if there are multiple simultaneous scattering events, e.g., because of high photon flux or careful construction of the experiment, there may be some diffraction because then the scattering particle may NOT be identifiable.

This was an interesting question that really made me think, so thanks for asking it. Your question ultimately maps onto the problem of diffraction from a solid that possesses a zero-momentum optical phonon. (Note: it is easy to generalize what I say below to non-zero-momentum phonons as well, but in that case it is obvious that the scattered particles would get redirected to regions of the screen between diffraction peaks. Therefore, it's not important to consider that case here.)

The answer then has two parts: one is that the inelastic scattering will decrease the intensity of the diffracted peaks in a way reminiscent of the Debye-Waller factor in a solid, but will not wash away the diffraction peaks completely. Depending on how large the matrix element is in the Fermi golden rule expression, the diffraction peaks will correspondingly decrease in intensity. The analogy to the solid with q=0 optical phonons would be that the larger the amplitude of the phonons, the more substantial the decrease in the Bragg peak intensity.

The second part of the answer is whether you will observe this decrease in the intensity. If you suppose that you have an energy discriminating detector that could tell the difference between the elastically and inelastically scattered particles, then you definitely would see a decrease in intensity because of the aforementioned analogy to the Debye-Waller effect.

Beyond the Debye-Waller effect, which I think is what your question is getting at, is if you will see a "diffraction pattern" if we reject the elastically scattered particles and only keep the inelastic ones. The answer would still be yes. Again, the analogy to the solid is apt here. Since the scattering particle can transfer momentum and energy to the solid only at a specific momentum values, those of the reciprocal lattice vectors, you would still get intensity at the points on the screen where you formerly had diffraction peaks in the elastic limit. Stated otherwise, you can still see a zero-momentum phonon in an inelastic scattering experiment at all reciprocal lattice points.

Now that we've considered the elastic and inelastic channels separately, I think it's relatively easy to see that you'll still have a diffraction pattern. The last question is whether the elastically and inelastically scattered particles will add coherently or incoherently. I need to think further about that, but my gut tells me that they'll add incoherently since they would arrive at the screen at different times (i.e. you would be able to tell whether the particle scattered elastically or inelastically, and therefore they cannot interfere.)

All of this is to say, yes, you'll still have diffraction.

• How does this reconcile with the answer by S. McGrew? Nov 15, 2022 at 12:16

Elastic and inelastic scattering
Part of the problem is what we call elastic and inelastic scattering. In the context of atoms, such as the experiment described in the question, elastic scattering is the Rayleigh scattering: an atom absorbs a photon and then re-emits a photon of the same energy, but with different direction and phase. This randomization of phase is already sufficient to wash out the diffraction picture - it will simply average out, since the fields interfering in every point are random.

Inelastic scattering in these context is the Raman scattering suggested by the question: an atom absorbs a photon and emits a photon of lower frequency: $$\omega ' = \omega - \omega_{10}$$ Just as in the case of Rayleigh scattering, these photons will have random directions and phases, but they will also have different frequencies. Again, the photons will interfere, but due to the randomness introduced by the scattering, the result will be some homogeneous average, rather than a diffraction picture.

Elastic and inelastic processes
Neither of these two situations is described by the Fermi golden rule. Although the two processes described above are termed elastic and inelastic, we haven't included yet any irreversible thermal effects and/or energy loss. Indeed, if we considered a wave function for the system photons+atoms, the energy of the system would be conserved and we could potentially expect restoration of the coherence, known as wave function revival in the context of Jaynes-Cummings model (simplest model of a photon coupled to a two-level atom). Irreversible loss of overall coherence could result only when the number of atoms/photons tends to infinity, i.e. in the thermodynamic limit.

A thermodynamic limit is implied in the Fermi golden rule, since all its derivations invariably involve transitions into some kind of continuous spectrum, so that they go only in one direction, rather than exhibiting the Rabi oscillations. This point is however often made rather vaguely, since the rule is introduced in introductory quantum mechanics courses. It is usually reviewed again (sometimes without naming it) in discussions of spontaneous emission and the context of lasers, or elsewhere.

Dephasing and decoherence
We considered above two situations:

• the disappearance of the interference pattern due to phase randomization (while overall processes are still elastic)
• truly inelastic processes involving energy loss

These two types of situations are often referred to as dephasing and decoherence.

• Interesting, this answer really challenges what I learned so far, thank you Vadim. But I don't believe Fermi's golden rule implies a thermodynamic limit, it is simply the first order term of time-dependent perturbation theory in the limit of $t>>1/\Delta E$. You need to include higher order perturbative terms to realize Rabi oscillations, just as you would for approximating a sine wave by polynomials Nov 13, 2020 at 19:06
• Another thing that confuses me about your answer, Fermi's golden rule is known to adequately describe diffraction and inelastic scattering in the context of inelastic neutron scattering and x-ray scattering. Why do you say that it cannot describe elastic or inelastic scattering? Or do you simply mean that Fermi's golden rule is limited to very weak/perturbative scattering so it is incomplete? Nov 13, 2020 at 19:07
• @KFGauss the condition that you cited is just one of the ways how thermodynamic limit may appear ther - btw, I mentioned precisely this when talking about collapse and revival in Jaynes-Cummings midrl. Nov 13, 2020 at 19:52
• Regarding the interference: the interference appears in the Fermi golden rule, when the matrix element is a sum of several amplitudes. This is not the same use that you suggest in your question. Nov 13, 2020 at 19:53