## Scattering in crystals ### Coherent scattering from crystal, assumes $b$ and $R$ uncorrelated As stated at the last part in [[Coherent and incoherent scattering#Decompose scattering cross-section for many particles]], for the scattering treatment we assumed that scattering length of atoms are uncorrelated with their positions. Only under this assumption, we may extract $\langle b\rangle^{2}$ out from the phase term containing real space $\mathbf{R}$, and this gives, $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}_{\text{coh}} = \langle b \rangle^2 \sum_{i,j} e^{iq \cdot (\mathbf{R}_i - \mathbf{R}_j)}$ For a crystal with a Bravais lattice, every unit cell gives the same contribution, so we may change the displacement vector difference to $\mathbf{r} = \mathbf{R}_{i}-\mathbf{R}_{j}$, while $\mathbf{R}_{i}$, $\mathbf{R}_{j}$ are **(scattering) center of unit cells**. Under this consideration, each unit cell is one scattering source, and $\mathbf{r}$ is the translational vectors between cells. In this way, we should multiply $N$, which is the number of atoms per unit cell, $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}_{\text{coh}} = N \langle b \rangle^2 \sum_{\mathbf{r}} e^{i\mathbf{q} \cdot \mathbf{r}}$ Real space lattice vector $\mathbf{R}$: $\mathbf{R} = n_1\mathbf{a}_1 + n_2\mathbf{a}_2 + n_3\mathbf{a}_3, \quad n_i \in \mathbb{Z}$ Reciprocal lattice vector $\mathbf{G}$: $\mathbf{G} = h\mathbf{b}_1 + k\mathbf{b}_2 + l\mathbf{b}_3, \quad h, k, l \in \mathbb{Z}$ Write wavevector difference (here we may call it diffraction vector) $\mathbf{q}$ as $\mathbf{b}$, we have $\mathbf{q} = q_1\mathbf{b}_1 + q_2\mathbf{b}_2 + q_3\mathbf{b}_3, \quad q_i \in \mathbb{R}$ Now we know that, under Bragg condition, the scattering happens at $\mathbf{R}\cdot \mathbf{q}=2\pi m$, where $m\in \mathbb{Z}$. And $\mathbf{R}\cdot \mathbf{q}$ is $\mathbf{R} \cdot \mathbf{q} = 2\pi(n_1q_1 + n_2q_2 + n_3q_3)$ This means all $q_{i} \in \mathbb{Z}$ for all $n_{i}$. Therefore, we have $\mathbf{R} \cdot \mathbf{G} = 2\pi(n_1h + n_2k + n_3l)$ >[!Note]- >A quick recap: the Bragg condition, written in wavelength form >$n\lambda=2\mathrm{d}\sin \theta$ >is the same as the vector form >$\mathbf{R}\cdot \mathbf{q}=2\pi m$ >This is due to the intrinsic property of reciprocal space, that $b_{i}\cdot a_{j}=2\pi \delta_{ij}$. And the diffraction vector $\mathbf{q}$ has the value $q=\frac{4\pi}{\lambda} \sin\theta$. >Writing $\mathbf{R}$ in $a_{i}$ and $\mathbf{q}$ in $b_{j}$, we get $\mathbf{R}\cdot \mathbf{q}$, and their value, $R$ and $q$ would contribute to $d$ and $\lambda$, eventually reduced to $n\lambda=2\mathrm{d}\sin \theta$. With $\mathbf{R}$ and $\mathbf{G}$ relation obtained, the phase term can be expressed with $\mathbf{G}$, i.e., $\sum_{\mathbf{r}} e^{i\mathbf{q} \cdot \mathbf{r}} =\frac{(2\pi)^3}{V} \sum_{\mathbf{G}} \delta(\mathbf{q} - \mathbf{G})$ $V$ is the volume of the crystal, and $\sum_{\mathbf{G}} \delta(\mathbf{q} - \mathbf{G})$ is the reciprocal lattice vector. Insert this back to $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}_{\text{coh}}$, we have $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}_{\text{coh}} = N \frac{(2\pi)^3}{V}\langle b \rangle^2 \sum_{\mathbf{G}} \delta(\mathbf{q} - \mathbf{G})$ This gives how differential scattering cross-section change with scattering vector $\mathbf{q}$. >[!Notice] >One may immediately notice that above result does not include **the height of the peaks**, or the **intensity**, only the positions of peaks. >This is due to decoupling $b_{i}$ and $R_{i}$.