When we use scattering radiation to probe critical phenomena, we have the usual Bragg relation for constructive interference $$|\vec{k}|=\frac{4\pi}{\lambda}sin\frac{\theta}{2}$$ where $\vec{k}$ is the change in wavevector after scattering, $\lambda$ is the wavelength of the radiation and $\theta$ is the angle of deflection.

The scattering intensity is taken to proportional to the Fourier transform of the real space correlation function $$I(\theta)\propto g(\vec{k})=\int d\vec{r} g(\vec{r})e^{i\vec{k}\cdot \vec{r}}.$$ My question is that why the intensity of scattering is related to the Fourier transform like this. Also, this resembles the so-called atomic form factor inside the structure factor for a crystal. For your reference, the structure factor is defined to be $$S(\vec{k})=\sum_{j=1}^n f_j(\vec{k})e^{i\vec{k}\cdot \vec{d}_j}$$ where the summation is over all the atoms, whose position are given by $\vec{d}_j$, in a particular basis (e.g. $n=2$ for a honeycomb lattice). The $f_j(\vec{k})$ in the above expression is the atomic factor, which again is directly proportional to the Fourier transform of the charge density due to the atom at $\vec{d}_j$, i.e. $$f_j(\vec{k})\propto \int d\vec{r} \rho (\vec{r})e^{i\vec{k}\cdot \vec{r}}$$ Therefore, both of the two cases are related to scattering amplitude, but the former is regarding the Fourier transform of the correlation function, which can be quite different for different systems, and the latter is regarding the transform of the charge density.

So again why is the Fourier transforms related to the scattering intensity and are the two cases interrelated somehow?


1 Answer 1


A single photon with 3-momentum $p=\hbar k$ incident on the system is scattered to an outgoing state with momentum $\hbar k'$. If the interaction operator between the photon and the sample is $U$, and the initial and final states are labelled by $|k\rangle$ and $|k'\rangle$ respectively (we ignore photon polarization/spin), then the amplitude for this process is related to the matrix element $$ \langle k'|U|k\rangle=\int d^3x\langle k'|x\rangle U(x)\langle x|k\rangle =\int d^3xe^{-2\pi ix\cdot(k-k')}U(x). $$ The cross section is then proportional to $|\langle k'|U|k\rangle|^2$, which can be written $$ \int d^3x'd^3x\bigg[U(x)U(x')\exp\big(-2\pi i(x-x')(k-k')\big)\bigg] $$ The above steps are valid for a single `snapshot' of a system with statistical fluctuations. We are really bombarding our sample with many photons over a relatively long time, during which the sample fluctuates randomly. Taking the ensemble average, we simply replace $U(x)U(x')$ by $\langle U(x)U(x')\rangle=g_U(x-x')$.
For simple Bragg-like scattering, all atoms are identical and the interaction $U(x)$ is proportional to the density $\rho(x)$, leading to the relationship between correlation function $g_\rho(r)$ and scattering intensity $I(\theta)$ that you mention above. Although much of the fluctuating matter is neutral, it is made up of charged matter that can be excited into oscillating dipoles and made to radiate by incoming radiation. To make the relationship between $I$ and $g$ exact (at least to the level of Fermi's golden rule), one could just define $g(r)=g_U(r)$, the correlation function for whatever $U(x)$ happens to be.

This brings us to the second part of your question, regarding the structure factor $S(\vec k)$. More generally, we decompose the unit cell of a crystal into positive and negative charged 'atoms', since the interaction $U(x)$ depends primarily on the local charge density. In this case, $$ U(x)\propto\sum_i \rho_i(x), $$ and since we are working with a crystal, $\langle U(x)U(x')\rangle =U(x)U(x')$. Hence, to find the scattering intensity we will need to calculate $$ \int d^3xd^3x'U(x)U(x')\exp(-2\pi i(x-x')(k-k'))=\\ \sum_{i,j}\sum_{K,K'}\int d^3xd^3x'\rho_{iK}\rho_{jK'}e^{-2\pi i(x-x')(k-k')-2\pi i (Kx+K'x')} \\ =\sum_{i,j}\sum_{K,K'}\rho_{iK}\rho_{jK'}\delta(\Delta k-K)\delta(\Delta k+K')\\ =V^2\sum_{i,j}\rho_{i,\Delta k}\rho_{j,-\Delta k}=V^2 |S(\Delta k)|^2 $$ (the third line is justified as long as $\Delta k=k-k'$ is in the reciprocal lattice).


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