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

??? hint "First small hint"

The $(hkl)$ plane intersects lattice at position vectors of $\frac{\mathbf{a_1}}{h}, \frac{\mathbf{a_2}}{k}, \frac{\mathbf{a_3}}{l}$. 
Can you define a general vector inside the $(hkl)$ plane? 

??? hint "Second small hint"

Whats the best vector operation to show orthogonality between two vectors? 

One can compute the normal to the plane by using result from Subquestion 1:

$\hat{\mathbf{n}} = \frac{\mathbf{G}}{|G|}$

Let us consider a very simple case in which we have the miller planes

$(h00)$

. For lattice planes, there is always a plane intersecting the zero lattice point (0,0,0). As such, the distance from this plane to the closest next one is given by:

$ d = \hat{\mathbf{n}} \cdot \frac{\mathbf{a_1}}{h} = \frac{2 \pi}{|G|} $

Since

$\rho=d / V$

, we must maximize

$d$

. To do that, we minimize must

$|G|$

. Therefore the smallest possible reciprocal lattice vectors are the (100) family of planes (in terms of FCC primitive lattice vectors).

Exercise 3: X-ray scattering in 2D

def reciprocal_lattice(N = 7, lim = 40):
    y = np.repeat(np.linspace(-18.4*(N//2),18.4*(N//2),N),N)
    x = np.tile(np.linspace(-13.4*(N//2),13.4*(N//2),N),N)

    plt.figure(figsize=(5,5))

    plt.plot(x,y,'o', markersize=10, markerfacecolor='none', color='k')
    plt.xlim([-lim,lim])
    plt.ylim([-lim,lim])
    plt.xlabel('$\mathbf{b_1}$')
    plt.ylabel('$\mathbf{b_2}$')
    plt.xticks(np.linspace(-lim,lim,5))
    plt.yticks(np.linspace(-lim,lim,5))
    
reciprocal_lattice()
plt.show()

Since we have elastic scattering, we obtain

$|\mathbf{k}| = |\mathbf{k}'| = \frac{2 \pi}{\lambda} = 37.9 nm^{-1}$

reciprocal_lattice()
# G vector
plt.arrow(0,0,13.4*2,18.4,color='r',zorder=10,head_width=2,length_includes_head=True)
plt.annotate('$\mathbf{G}$',(17,6.5),fontsize=14,ha='center',color='r')
# k vector
plt.arrow(-6,37.4,6,-37.4,color='b',zorder=11,head_width=2,length_includes_head=True)
plt.annotate('$\mathbf{k}$',(-8,18),fontsize=14, ha='center',color='b')
# k' vector
plt.arrow(-6,37.4,6+13.4*2,-37.4+18.4,color='k',zorder=11,head_width=2,length_includes_head=True)
plt.annotate('$\mathbf{k\'}$',(15,30),fontsize=14, ha='center',color='k')
plt.show()

Exercise 4: Structure factors

$S(\mathbf{G}) = \sum_j f_j e^{i \mathbf{G} \cdot \mathbf{r_j}} = f(1 + e^{i \pi (h+k+l)})$

Solving for

$h$

,

$k$

, and

$l$

results in

$ S(\mathbf{G}) = \begin{cases} 2f, : \text{if

$h+k+l$

is even}\ 0, : \text{if

$h+k+l$

is odd}. \end{cases} $

Thus if

$h+k+l$

is odd, diffraction peaks dissapear

Let

$f_1 \neq f_2$

, then

$ S(\mathbf{G}) = \begin{cases} f_1 + f_2, \text{if

$h+k+l$

is even}\ f_1 - f_2, \text{if

$h+k+l$

is odd} \end{cases}
$

Due to bcc systematic absences, the peaks from lowest to largest angle are:

$(110),(200),(211), (220), (310)$

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