10_xray_solutions.md

from matplotlib import pyplot as plt 
import numpy as np
from math import pi

Solutions for lecture 10 exercises

Warm-up exercises

??? hint "Small hint"

You can make use of the [scalar triple product](https://en.wikipedia.org/wiki/Triple_product#Scalar_triple_product).

If \mathbf{k}-\mathbf{k'}\neq \mathbf{G}, then the argument of the exponent has a phase factor dependent on the real-space lattice points. Because we sum over each of these lattice points, each argument has a different phase. Summing over all these phases results in an average amplitude of 0, resulting in no intensity peaks.

No, there is a single atom, and thus only one term in the structure factor. This results in only a single exponent being present in the structure factor, which is always nonzero.

No, an increase of the unit cell size cannot create new diffraction peaks (see lecture).

Exercise 1: Equivalence of direct and reciprocal lattice

V^*=\left|\mathbf{b}_{1} \cdot\left(\mathbf{b}_{2} \times \mathbf{b}_{3}\right)\right| = \frac{2\pi}{V}\left| (\mathbf{a}_{2} \times \mathbf{a}_{3}) \cdot\left(\mathbf{b}_{2} \times \mathbf{b}_{3}\right)\right| = \frac{(2\pi)^3}{V}

In the second equality, we used the reciprocal lattice vector definition (see notes). In the third equality, we used the identity:

(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d})-(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c})

Because the relation between direct and reciprocal lattice is symmetric, so are the expressions for the direct lattice vectors through the reciprocal ones:

\mathbf{a}_{i} \epsilon_{ijk} = \frac{2\pi}{V^*} (\mathbf{b}_{j} \times \mathbf{b}_{k})

where \epsilon_{ijk} is the Levi-Civita tensor

One set of the BCC primitive lattice vectors is given by: \mathbf{a_1} = \frac{a}{2} \left(-\hat{\mathbf{x}}+\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\ \mathbf{a_2} = \frac{a}{2} \left(\hat{\mathbf{x}}-\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\ \mathbf{a_3} = \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}}-\hat{\mathbf{z}} \right).

From this, we find the following set of reciprocal lattice vectrs:

\mathbf{b_1} = \frac{2 \pi}{a} \left(\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\ \mathbf{b_2} = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{z}} \right) \\ \mathbf{b_3} = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right),

which is forms a reciprocal FCC lattice. The opposite relation follows directly from our previous result.

Because the 1st Brillouin Zone is the Wigner-Seitz cell of the reciprocal lattice, we need to construct the Wigner-Seitz cell of the FCC lattice. For visualization, it is convenient to look at FCC lattice introduced in the previous lecture and count the neirest neighbours of each lattice point. We see that each lattice point contains 12 neirest neighbours and thus the Wigner-Seitz cell contains 12 sides!

Exercise 2: Miller planes and reciprocal lattice vectors