Where is the missing energy in the Debye-Waller effect?

Question

X-ray/neutron scattering from crystals and liquids is well-described by the scattering theory to give the (dynamic) structure factor which is a function of momentum and energy:

$$S(\mathbf{k},\omega)= \int d\omega \, e^{i \omega t}\langle \hat{\rho}^{\dagger}(\mathbf{k}, t) \hat{\rho}(\mathbf{k}, 0)\rangle$$

Often, the energy-dependence is ignored (because you can't measure it) and one says that they measure the (static) structure factor

$$S(\mathbf{k}) = \int d\omega \,S(\mathbf{k},\omega)$$

From here, people model the temperature dependence of the structure factor by (quoting wikipedia)

$$S(\mathbf{k}, T\neq0) = S(\mathbf{k}) \, \langle\textrm{exp}(i\mathbf{k}\cdot \mathbf{u}) \rangle^2 \sim S(\mathbf{k}) \textrm{exp}(-k^2 \langle u^2 \rangle)$$

Where the term $\langle\textrm{exp}(i\mathbf{q}\cdot \mathbf{u}) \rangle^2 $ is the Debye-Waller factor that is temperature-dependent and comes from atoms moving randomly at finite temperature ($u$ is the average displacement which depend on temperature).

But, because the number of atoms hasn't changed, the total amount of scattering is conserved, and we need the following relation

$$\int d\mathbf{k}\, S(\mathbf{k}) \textrm{exp}(-k^2 \langle u^2 \rangle)= \int d\mathbf{k}\, S(\mathbf{k}) = \int d\mathbf{k} \,d\omega\, S(\mathbf{k},\omega)$$

The introduction of the Debye-Waller factor however, causes the integral to decrease, and it is commonly said that the rest of the integral comes from inelastic scattering that has a $1- \textrm{exp}(-k^2 \langle u^2 \rangle)$ dependence to exactly balance things.

My question is, where is the energy going? Specifically, where exactly in $(\mathbf{k},\omega)$-space is it located?

For example, is it going to increased Compton scattering off of the atoms, into more phonons, or both? Is it going to high or low $k$,$\omega$? Is there anisotropy? I'm having trouble locating a general answer to this, but I am sure it should have been answered decades ago. Ideally I would like to see how these processes give back the missing $1- \textrm{exp}(-k^2 \langle u^2 \rangle)$ term

Answer 1

I need to put this as an answer since it's too long. You can have an elastic broadening due to Xray-phonon elastic scattering. Take the scattering of one phonon for example. The scattering amplitude due to a phonon of wavevector k at second order is (in 1D, with N atoms and lattice paramter=1 for simplicity)

$A(q) = \sum\limits_{n=-N/2}^{n=N/2} e^{iq(n+u\cos(kn))}$

Where $u$ is the phonon amplitude (related to the number of phonon and therefore to the temperature and energy cost of this phonon). Doing a Taylor expansion to second order in $u$ and taking only into account the terms for $q$ close to $0$ (or to a bragg value) you get

$\begin{align} A(q) &\approx \sum\limits_{n=-N/2}^{n=N/2} e^{iqn}[1+iqu\cos(kn)-\frac{(qu)^2}{2}\cos^2(kn)] \\ &= \sum\limits_{n=-N/2}^{n=N/2} e^{iqn}[(1-\frac{(qu)^2}{4})+iqu\cos(kn)-\frac{(qu)^2}{4}\cos(2kn)]\end{align}$

If you continue the calculation and look at the diffracted intensity $I(q)=|A(q)|^2$, you'll see that the first term correspond to a Bragg with intensity lowered by the phonon due to the $-\frac{(qu)^2}{4}$ term, this is the Debye-Waller factor. Then the first cosine term will give small peaks at $q=\text{Bragg}\pm k$ and finally the last term is even smaller peaks at $q=\text{Bragg}\pm 2k$.

So you see that each elastic scattering on a thermal phonon will create small peaks in the Bragg tail. But the total integrated intensity will still be the same. Since the phonon at low wavevector $k$ cost a small amount of energy, they will have the most important contribution. The tail of the Bragg decreases rapidly when $q$ is far from a Bragg value.

No need for dynamical effects since the phonon is so slow regarding the Xray speed, you can make the assumption that the phonon is frozen during the elastic scattering.

I Hope it helped.