# What's the “effective potential” for photons in $X$-ray diffraction?

The slickest way to introduce $X$-ray diffraction is to invoke scattering theory in quantum mechanics. One treats the incoming photon as just another particle in a scattering problem; by Fermi's golden rule and the Born approximation, the scattering rate from $\mathbf{k}$ to $\mathbf{k}'$ is $$\Gamma(\mathbf{k}', \mathbf{k}) \propto |\langle \mathbf{k}'| V | \mathbf{k} \rangle|^2$$ where $V$ is the potential experienced by the photon. Since $V$ has the periodicity of the lattice, it follows that the scattering rate vanishes unless $\mathbf{k} - \mathbf{k}'$ is a reciprocal lattice vector.

However, most sources do not say much about this "photon potential" or justify its form. For example, David Tong's lecture notes simply avoid the issue:

Firing a beam of particles — whether neutrons, electrons or photons in the X-ray spectrum — at the solid reveals a characteristic diffraction pattern. [...] Our starting point is the standard asymptotic expression describing a wave scattering off a central potential.

Tong declines to comment on the potential at all. Steve Simon's solids textbook says a bit more:

If we think of the incoming wave as being a particle, then we should think of the sample as being some potential $V(r)$ that the particle experiences. [...] X-rays scatter from the electrons in a system. As a result, the scattering potential is proportional to the electron density.

That is, the potential is the electron density, and nuclei don't contribute because they are heavier.

This sounds plausible, but I have no idea how to derive this from first principles. From the QFT side, I imagine such a calculation would start from the QED interaction $$\mathcal{L}_{\text{int}} = \bar{\psi} \gamma^\mu A_\mu \psi.$$ Treating the field $\psi$ as a classical, static background field we have $$\mathcal{L}_{\text{int}} = j^\mu A_\mu = \rho A_0$$ since the components $j^i$ are zero. But I'm not sure how this is supposed to be a "potential for the photon"; it seems to only affect one component of $A_\mu$. And it's unclear where the mass of the fermion is going to come in, to make the electrons count more than the nuclei.

On the other end, I suppose one could start from classical electromagnetism. In that case we're talking about Thomson scattering, and heavier particles indeed contribute less. The challenge is then exporting quantum mechanical ideas, such as partial waves and the Born approximation, to this classical context. Maybe this is manageable, but I've never seen this done anywhere either.

The derivation of X-ray diffraction peaks from quantum scattering theory is very slick, but how is it justified in detail?

## This question has an open bounty worth +50 reputation from KF Gauss ending in 7 days.

The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.

Looking for an explanation that connects the non-relativistic QM and relativistic QFT derivations for x-ray diffraction

• Guess what you're looking for is a two-point function. In that case, it may be helpful to look at the derivation of the CJT effective action. – flippiefanus Aug 27 '18 at 4:26
• I think the answer to this depends on what you are scattering from. There are a number of regimes where some direct approximations can be made, the QED interaction you suggest seems a bit simplistic in general, since a variety of scattering processes can happen. Are you interested in a specific case? E.g. Bragg scattering? Laue Scattering? Resonant scattering? The Parratt regime? Also note that cross-sections are typically measured for single atoms and then multi-particle interference is calculated from that. Particularly for resonant scattering there are analytical expressions though. – Wolpertinger Aug 30 '18 at 17:03
• The potential $V$ is the potential of the matter the x-ray is scattering from. You have to know something about that matter in order to actually perform the calculation. And even then, it is very likely you will have to make myriad approximations to successfully complete the mathematics. As far as I know, there is no "first principles" reason for matter to be configured the way it is, it was all discovered through experimentation. In fact, our understanding of warm dense matter is pretty poor compared to, say, cold dense matter. – Finncent Price Aug 31 '18 at 15:38
• I don't think your interaction term is correct, that term would describe the absorption of a photon, rather than the scattering of a photon into another photon (e.g. diffraction). – KF Gauss Sep 1 '18 at 17:25
• @user157879 It’s the basic QED interaction. You get scattering at second order from it as usual. – knzhou Sep 1 '18 at 17:27