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......@@ -5,78 +5,100 @@ from math import pi
```
# Solutions for lecture 10 exercises
## Warm-up exercises
1.
??? hint "Small hint"
You can make use of the [scalar triple product](https://en.wikipedia.org/wiki/Triple_product#Scalar_triple_product).
2.
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.
3.
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.
4.
No, a change to the unit cell can cause intensity peaks to dissapear.
## Exercise 1: Equivalence of direct and reciprocal lattice
### Subquestion 1
1.
$$
V^*=\left|\mathbf{b}_{1} \cdot\left(\mathbf{b}_{2} \times \mathbf{b}_{3}\right)\right| = \frac{2\pi}{V}\left| (\mathbf{a}_{2} \times \mathbf{a}_{3}) \cdot\left(\mathbf{b}_{2} \times \mathbf{b}_{3}\right)\right| = \frac{(2\pi)^3}{V}
$$
In the second equality, we used the reciprocal lattice vector definition (see notes). In the third equality, we used the identity:
In the second equality, we used the reciprocal lattice vector definition (see notes).
In the third equality, we used the identity:
$$
(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d})-(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c})
$$
### Subquestion 2
2.
$$
\mathbf{a}_{i} \epsilon_{ijk} = \frac{2\pi}{V^*} (\mathbf{b}_{j} \times \mathbf{b}_{k})
$$
whereas $\epsilon_{ijk}$ is the [Levi-Civita tensor](https://en.wikipedia.org/wiki/Levi-Civita_symbol#Three_dimensions)
### Subquestion 3
BCC primitive lattice vectors are given by:
3.
One set of the BCC primitive lattice vectors is given by:
$$
\mathbf{a_1} = \frac{a}{2} \left(-\hat{\mathbf{x}}+\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\
\mathbf{a_2} = \frac{a}{2} \left(\hat{\mathbf{x}}-\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\
\mathbf{a_3} = \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}}-\hat{\mathbf{z}} \right)
\mathbf{a_3} = \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}}-\hat{\mathbf{z}} \right).
$$
using definition of reciprocal lattice vector (see notes), one can show:
From this, we find the following set of reciprocal lattice vectrs:
$$
\mathbf{b_1} = \frac{2 \pi}{a} \left(\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\
\mathbf{b_2} = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{z}} \right) \\
\mathbf{b_3} = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right)
\mathbf{b_3} = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right),
$$
which is FCC primitive lattice vectors. Using the result in Subquestion 2, the vice versa result is trivial
### Subquestion 4
which is forms a reciprocal FCC lattice.
Using the result in Subquestion 2, the vice versa result is trivial
Brillouin zone is most easily given by the Wigner Seitz unit cell (see notes) by constructing planes in the midpoint between a lattice point and a nearest neighbor. Our lattice is FCC, so the reciprocal lattice is BCC. Since BCC has 8 nearest neighbors (see interactive figure in last week's [Exercise 1: Diatomic crystal](https://solidstate.quantumtinkerer.tudelft.nl/9_crystal_structure/#exercise-1-diatomic-crystal)), there will be 8 planes. A polyhedron which has 8 faces is an octahedron.
4.
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](https://solidstate.quantumtinkerer.tudelft.nl/9_crystal_structure/#face-centered-cubic-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
### Subquestion 1
1.
??? 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?
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?
### Subquestion 2
2.
One can compute the normal to the plane by using result from Subquestion 1:
$\hat{\mathbf{n}} = \frac{\mathbf{G}}{|G|}$
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:
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|} $
### Subquestion 3
Since $\rho=d / V$, we must maximize $d$. To do that, we must minimize $|G|$ (Subquestion 2). We must therefore use the smallest possible reciprocal lattice vector which means {100} family of planes (in terms of FCC primitive lattice vectors).
3.
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
### Subquestion 1
1.
```python
def reciprocal_lattice(N = 7, lim = 40):
y = np.repeat(np.linspace(-18.4*(N//2),18.4*(N//2),N),N)
......@@ -95,14 +117,11 @@ def reciprocal_lattice(N = 7, lim = 40):
reciprocal_lattice()
plt.show()
```
### Subquestion 2
$k = \frac{2 \pi}{\lambda} = 37.9 nm^{-1}$
### Subquestion 3
Note that $|k| = |k'| = k $ since elastic scatering
2.
Since we have elastic scattering, we obtain
$|\mathbf{k}| = |\mathbf{k}'| = \frac{2 \pi}{\lambda} = 37.9 nm^{-1}$
3.
```python
reciprocal_lattice()
# G vector
......@@ -116,38 +135,36 @@ plt.arrow(-6,37.4,6+13.4*2,-37.4+18.4,color='k',zorder=11,head_width=2,length_in
plt.annotate('$\mathbf{k\'}$',(15,30),fontsize=14, ha='center',color='k')
plt.show()
```
## Exercise 4: Structure factors
### Subquestion 1
$S_\mathbf{G} = \sum_j f_j e^{i \mathbf{G} \cdot \mathbf{r_j}} = f(1 + e^{i \pi (h+k+l)}) =$
1.
$S(\mathbf{G}) = \sum_j f_j e^{i \mathbf{G} \cdot \mathbf{r_j}} = f(1 + e^{i \pi (h+k+l)})$
2.
Solving for $h$, $k$ and $l$ results in
$$
S(\mathbf{G}) =
\begin{cases}
2f & \text{if $h+k+l$ is even}\\
0 & \text{if $h+k+l$ is odd}
\end{cases}
42, \: \text{if $h+k+l$ is even}\\
0, \: \text{if $h+k+l$ is odd}.
\end{cases}
$$
Thus if $h+k+l$ is odd diffraction peaks dissapear
where we used sum over the basis of BCC in $j$.
### Subquestion 2
See when $S_G$ is zero
### Subquestion 3
$S_\mathbf{G} =$
3.
Let $f_1 \neq f_2$, then
$$
S(\mathbf{G}) =
\begin{cases}
f_1 + f_2 & \text{if $h+k+l$ is even}\\
f_1 - f_2 & \text{if $h+k+l$ is odd}
f_1 + f_2, \mathrm{if $h+k+l$ is even}\\
f_1 - f_2, \mathrm{if $h+k+l$ is odd}
\end{cases}
$$
### Subquestion 4
4.
Due to bcc systematic absences, the peaks from lowest to largest angle are:
$(110),(200),(211), (220), (310)$
### Subquestion 5
5.
$a = 2.9100 \unicode{xC5}$
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