One topic in crystallography that I've found a surprising dearth of information on is what the fundamental interaction behind the interaction of the X-ray and atom. Pretty much every book just treats them as classical waves instead of explaining the quantum mechanics behind it. By this I don't mean that interference that leads to diffraction, but what actually happens when the X-ray interacts with the electron cloud.

The only description I've found so far is in Glusker, Lewis and Rossi that talks about:

"When X rays hit an atom, the rapidly oscillating electric field of the radiation sets the electrons of the atom into oscillation about their nuclei. This oscillation has the same frequency as that of the incident radiation. The result is that the electron acts as an oscillating dipole that serves as a source of secondary radiation with the frequency of the incident beam."

and goes on to note that this is coherent scattering and there is no wavelength shift.

Which doesn't make a ton of sense, isn't oscillating about the nuclei an old pre-quantum explanation of orbitals? Is this saying that this is similar to fluorescence/phosphorescence where the photo is absorbed, then another emitted? That is what most of the other grad students I've talked to have assumed, but if that was the case I'd expect a Stokes shift, the direction of emitted radiation to be random, and I don't see why you'd get a $180^\circ$ phase shift.

The other explanation I've heard is that you are actually sloshing electrons around within their orbitals, with the same frequency as the X-ray, similar to when you are dealing with optical properties, but I'm not sure what the mechanism for the interaction would be in this case.

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    $\begingroup$ I'm voting to close this question as off-topic because I think it's better located at Physics.SE. $\endgroup$ – hBy2Py Feb 22 '17 at 20:10
  • $\begingroup$ @hBy2Py Really? SCXRD is usually considered to be a part of chemistry. $\endgroup$ – Canageek Feb 22 '17 at 20:14
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    $\begingroup$ If it were a question on the interpretation of XRD spectra, then I agree it should be sited here (since there's no Materials.SE yet, AFAIK). For the theory of it, though -- I would think you're likely to get better answers at Physics.SE. But, there may be some physical chemists here that'll be able to help out. I will not be upset at all if the close vote fails! $\endgroup$ – hBy2Py Feb 22 '17 at 20:20
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    $\begingroup$ Don't close, I think we can deal with it. $\endgroup$ – Ivan Neretin Feb 22 '17 at 21:01
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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. I have retained the discussion about closing the question. If there's any relevant info, you can still find it there. Even better, edit them into the question. $\endgroup$ – orthocresol Feb 22 '17 at 22:32

The description given in your question pretty much explains things. I would put it in slightly different words which may or may not help, which is that the x-ray photon induces a dipole moment in the atom (a field induced dipole) which then radiates and so the x-ray is scattered. This is pretty much what happens in scattering by visible photons and we must assume that the photon is distant in energy from any transitions. (In Compton scattering part of the x-ray energy is imparted to the electron, so the frequency of the scattered x-ray is changed).

The field induced dipole moment is proportional to the polarisability which for a single bound electron is $$ \alpha = \frac{e^2}{m}\left ( \frac{1}{\omega_0^2-\omega^2} \right )$$

where $\omega_0$ is the resonance frequency of the bound electron and $\omega$ the x-ray frequency and so $\omega >> \omega_0$ which in the limit gives $$\alpha \rightarrow -\frac{e^2}{m\omega^2} $$ Thus the calculation reduces to understanding the QM calculation for the polarisability. See for example Atkins & Friedmann 'Molecular Quantum Mechanics' Ch $12$.

( By considering that the x-ray is a plane wave the intensity of scattered radiation relative to that incident is given by $$ \frac{I}{I_0}= \frac{e^4}{m^2c^4R^2}P(\theta)$$

where R is the distance from the scattering centre to the detector, m the electron mass and $\theta$ the scattering angle and $P(\theta)=(1+\cos^2(\theta))/2$ is called the polarisation factor. This formula is sometimes named after Thomson. The $m^{-2}$ shows why significant scattering occurs only from electrons and not from protons. The book by Flygare 'Molecular Structure and Dynamics' section 8.5 gives a detailed description of x-ray scattering in general and details such as calculating atomic scattering factors. )

  • $\begingroup$ Thanks, that makes things make perfect sense, except for "which then radiates": I get the dipole polarization, you've explained that well, but I'd like a little more detail on this. Thank you. $\endgroup$ – Canageek Feb 23 '17 at 19:02
  • $\begingroup$ All I can add is that the oscillating induced dipole radiates the scattered radiation at the same frequency as the initial photon. $\endgroup$ – porphyrin Feb 24 '17 at 9:03

Explaining the diffraction of a photon off of a crystal lattice quantum mechanically is the same as finding what momenta can be transferred to the photon by the crystal lattice. The simplest way to model this, as Ivan noted, is to describe the momentum states of a particle in this kind of potential as these are the momenta which can be transferred to the photon. Each of the barriers in this is interpreted as one of the atoms in the crystal lattice. Generally, a 1D approach is taken, but this is only because it is explanatory, and extension to 3D is not much harder.

In the interpretation I'm going to give, which I'm mostly reiterating from this link, the lattice itself should be finite as we would never really see diffraction off of an infinite lattice. Thus, the spatial wavefunction is simply a sum of periodic functions such as,$$\Psi(x)=\sum_{i=1}^N\cos(2\pi n\cdot\frac{x}{L})$$where $L$ is the distance between two atoms and $n$ is an integer. It's as simple as Fourier transforming into momentum space and finding the resulting probability density. This density is then interpreted as the diffraction pattern. The Fourier transform takes the form of $$\Phi(p)=\frac{1}{\sqrt{2\pi}}\int_{\frac{-L}{2}}^{\frac{L}{2}}\Psi(x)^2e^{ipx}dx$$

This certainly isn't the most detailed description one could give, but it is probably gonna be the theoretical groundwork for any interpretation of x-ray diffraction. It is also quite illuminating. To quote from the link earlier, the interpretation of this diffraction pattern can be stated as,

According to the quantum mechanical interpretation of diffraction, when this one‐dimensional electron density distribution interacts with an X‐ray source the individual photons are temporarily localized simultaneously at all lattice sites. The uncertainty principle requires that spatial localization leads to a delocalization (scattering) of the momentum distribution and the appearance of interference fringes because the spatial localization occurs at multiple sites.

I find that to be quite an illuminating interpretation as it is very similar to the classical idea of the photon making the electrons slosh around, but with the added detail that the electrons were already sloshing around some to begin with.

One can also take the approach of treating this as a particle-particle collision, which the link above also works through. This is more detailed, but you arrive at the same answer. The fact that you can treat the system in this way and get to the right answer basically confirms the original interpretation that the amplitudes of the momentum space wavefunction are a good representation of the diffraction pattern because these are the only momenta that could be transferred to the photon.

To explain how going to the 3-D case is not very hard, we would simply change our spatial wavefunction to be a product of periodic functions x, y, and z. Think particle in a 3-D box wavefunctions. The Fourier transform is almost the same as before because the triple integral could be separated into three identical 1-dimensional integrals. Of course, you don't really wanna do this integral for any more than like two atoms. Oddly enough if you used infinite atoms it'd actually be easier because you could use Bloch's theorem.


There is a well-known exercise in the basic courses on quantum mechanics: calculate a wavefunction of a 1D system with a rectangular potential barrier in the middle. The results are then interpreted as scattering of a wave on the said barrier: some part of it gets reflected, and some penetrates (probably tunnels) the barrier and continues on its way.

Well, the scattering of X-rays on an atom is a similar thing, only somewhat more complicated: we have a 3D setting, with spherical waves in addition to plane waves, and with an atom instead of the rectangular barrier. Other than that, the problem is pretty much the same.

  • $\begingroup$ I'd think that model is close. But that model would show a Stokes shift which I really don't think exists. So the diffracted x-ray just doesn't transfer momentum to the crystal, the crystal can also transfer momentum to the x-ray. So there will be a slight angle dependency for the exit x-ray and at the primary angle the energy of the x-rays would be slightly broadened. $\endgroup$ – MaxW Feb 22 '17 at 21:10
  • $\begingroup$ Also I'd expect that if the incidence x-rays were in a perfect line that the exit-rays would be sort of a 2-D cone. More elliptical in shape than a pure cone if that makes sense. $\endgroup$ – MaxW Feb 22 '17 at 21:27
  • $\begingroup$ The exit rays are spherical waves. It is the interference that shapes them into a series of individual spots. The very idea of a cone is simply not there. $\endgroup$ – Ivan Neretin Feb 22 '17 at 21:29
  • $\begingroup$ @ IvanNeretin - I understand that is the classical interpretation. But if the x-ray is going one way then changes direction there must be a momentum transfer. The difference in energy is so small that it doesn't make a difference practically. But a true quantum mechanical solution would take the momentum into account. $\endgroup$ – MaxW Feb 22 '17 at 21:35
  • $\begingroup$ Yeah, there should be some correction to that effect. $\endgroup$ – Ivan Neretin Feb 22 '17 at 21:41

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