Theme update

TODO:

• fix plotly aspect ratio (see here

Closes #119 (closed), #105 (closed)

docs/10_xray.md

@@ -24,15 +24,12 @@ import matplotlib.patches as patches
 import matplotlib.transforms as mtransforms
 import matplotlib.image as img 
 
-import plotly.offline as py
 from plotly.subplots import make_subplots
 import plotly.graph_objs as go
 
 from common import draw_classic_axes, configure_plotting
 
 configure_plotting()
-
-py.init_notebook_mode(connected=True)
 ```
 
 _based on chapters 13.1-13.2 & 14.1-14.2 of the book_  
@@ -67,9 +64,11 @@ Now our goal is twofold, we will:
 ## Reciprocal lattice motivation 1D case
 In [lecture 7](7_tight_binding.md) we discussed the reciprocal space of a simple 1D lattice. 
 To obtain the dispersion relation we considered waves of the form
+
 $$
 e^{ikx_n} = e^{ikna}, \quad n \in \mathbb{Z},
 $$
+
 where $x_n = na$ is the 1D lattice point.
 We then observed that these waves with wave vectors $k$ and $k+G$, where $G=2\pi m/a$ with integer $m$, are exactly the same: 
 
@@ -263,35 +262,45 @@ fig.update_layout(
 fig.update_xaxes(range=[-plot_range, plot_range], row=1, col=2, visible=False)
 fig.update_yaxes(row=1, col=2, scaleanchor="x", scaleratio=1)
 fig.update_yaxes(row=1, col=1, scaleanchor="x", scaleratio=1)
-fig.show()
-
+fig
 ```
 
 To find the reciprocal lattice vectors, we use the relation
+
 $$
 \mathbf{a_i}\cdot\mathbf{b_j}=2\pi\delta_{ij},
 $$
+
 which gives us the following equations:
+
 $$
 \mathbf{a}_1\cdot\mathbf{b}_2=\mathbf{a}_2\cdot\mathbf{b}_1=0,
 $$
+
 and
+
 $$
 \mathbf{a}_1\cdot\mathbf{b}_1=\mathbf{a}_2\cdot\mathbf{b}_2=2\pi.
 $$
+
 We substitute $\mathbf{a_i}\cdot\mathbf{b_i} = \lvert \mathbf{a_i} \rvert \lvert \mathbf{b_i} \rvert \cos(\theta_i)$:
+
 $$
 \lvert \mathbf{a}_1 \rvert \lvert \mathbf{b}_1 \rvert =\frac{2\pi}{\cos(\theta_1)} \:\: \text{and} \:\: \lvert \mathbf{a}_2 \rvert \lvert \mathbf{b}_2 \rvert =\frac{2\pi}{\cos(\theta_2)},
 $$
+
 where $\theta_i$ is the angle between the vectors $\mathbf{a}_i$ and $\mathbf{b}_i$.
 To find the angles $\theta_1$ and $\theta_2$, we use the orthogonality relations above and the fact that the angle between $\mathbf{a}_1$ and $\mathbf{a}_2$ is $60^\circ$.
 From this we conclude that $\theta_1 = \theta_2 = 30^\circ$.
 Because $\lvert \mathbf{a}_1 \rvert = \lvert \mathbf{a}_2 \rvert = a$, we find
+
 $$
 \lvert \mathbf{b}_1 \rvert = \lvert \mathbf{b}_2 \rvert = \frac{4\pi}{a\sqrt{3}}. 
 $$
+
 Unsurprisingly, we find that the lengths of the reciprocal lattice vectors are equal and inversely proportional to the lattice constant $a$.
 With $\lvert \mathbf{b}_2 \rvert$ and $\mathbf{a}_1 \perp \mathbf{b}_2$, we easily find 
+
 $$
 \mathbf{b}_2 = \frac{4\pi}{a\sqrt{3}} \mathbf{\hat{y}}.
 $$
@@ -304,9 +313,11 @@ $$
 
 ??? Question "Is the choice of a set of reciprocal lattice vectors unique? If not, which other ones are possible?"
     There are many equivalent ways to choose lattice vectors of the reciprocal lattice. In the example above we could as well use
+
     $$
     \mathbf{b}_1 = \frac{4\pi}{a\sqrt{3}} \left(-\frac{\sqrt{3}}{2} \mathbf{\hat{x}} + \frac{1}{2}\mathbf{\hat{y}} \right) \quad \text{and} \quad \mathbf{b}_2 = -\frac{4\pi}{a\sqrt{3}} \mathbf{\hat{y}}.
     $$
+
     There is however only one choice that satisfies the relations $\mathbf{a_i}\cdot\mathbf{b_j}=2\pi\delta_{ij}$.
 
 ### 3D lattice example
@@ -351,6 +362,7 @@ Therefore, we rewrite this into the form
 $$
 {\mathcal F}_{k}\left[\rho(x)\right]=\frac{2\pi}{|a|}\sum_{m} \delta\left(k-G\right).
 $$
+
 Therefore, we see that the Fourier transform is non-zero only at reciprocal lattice points.
 In other words, Fourier transforming a real-space lattice yields a reciprocal lattice! 
 The above result generalizes directly to three dimensions:
@@ -359,7 +371,6 @@ $$
 {\mathcal F}_\mathbf{k}\left[\rho(\mathbf{r})\right]=\int \mathrm{d}\mathbf{r}\ \mathrm{e}^{i\mathbf{k}\cdot\mathbf{r}} \rho(\mathbf{r}) = \sum_\mathbf{G}\delta(\mathbf{k}-\mathbf{G}).
 $$
 
-
 ## Periodicity of the reciprocal lattice
 In order to describe a reciprocal lattice, we need to define a primitive unit cell in reciprocal space.
 Previously, we learned that the choice of a primitive unit cell is not unique.
@@ -397,13 +408,11 @@ $$
 
 As a result of the travel distance, the phase difference is: 
 
-$$
 \begin{align}
 \Delta \phi &= \lvert\mathbf{k} \rvert(\Delta x_1+\Delta x_2)\\
 &= \lvert\mathbf{k}\rvert \lvert\mathbf{R}\rvert(\cos(\theta)+\cos(\theta'))\\
 &= \mathbf{k'}\cdot \mathbf{R} - \mathbf{k}\cdot \mathbf{R} = (\mathbf{k'} - \mathbf{k}) \cdot \mathbf{R}.
 \end{align}
-$$
 
 However, that is only a phase difference between waves scattered off of two atoms.
 To find the outgoing wave's amplitude, we must sum over scattered waves from each and every atom in the lattice:
@@ -440,12 +449,10 @@ In order to keep track of the atoms, we define $\mathbf{r}_j$ to be the location
 The distance $\mathbf{r}_j$ is defined with respect to the lattice point from which we construct the unit cell.
 In order to calculate the amplitude of the scattered wave, we must sum not only over all the lattice points but also over the atoms in a single unit cell:
 
-$$
 \begin{align}
 A &\propto \sum_\mathbf{R} \sum_j f_j \mathrm{e}^{i\left(\mathbf{G}\cdot(\mathbf{R}+\mathbf{r}_j)-\omega t\right)}\\
 &= \sum_\mathbf{R}\mathrm{e}^{i\left(\mathbf{G}\cdot\mathbf{R}-\omega t\right)}\sum_j f_j\ \mathrm{e}^{i\mathbf{G}\cdot\mathbf{r}_j}
 \end{align}
-$$
 
 where $f_j$ is the scattering amplitude off of a single atom, and it is called the *form factor*.
 The form factor mainly depends on the chemical element, nature of the scattered wave, and finer details like the electrical environment of the atom.
@@ -454,7 +461,7 @@ The second part gives the amplitude of the scattered wave, and it is called the
 
 $$
 S(\mathbf{G})=\sum_j f_j\ \mathrm{e}^{i\mathbf{G}\cdot\mathbf{r}_j}.
-$$    
+$$
 
 In diffraction experiments, the intensity of the scattered wave is $I \propto A^2$
 Therefore, the intensity of a scattered wave depends on the structure factor $I \propto S(\mathbf{G})^2$.
@@ -478,23 +485,21 @@ As a demonstration of how it happens, let us compute the structure factor of the
 The basis of the conventional FCC unit cell contains four identical atoms.
 With respect to the reference lattice point, these attoms are located at
 
-$$
-\begin{aligned}
+\begin{align}
 \mathbf{r}_1&=(0,0,0)\\
 \mathbf{r}_2&=\frac{1}{2}(\mathbf{a}_1+\mathbf{a}_2)\\
 \mathbf{r}_3&=\frac{1}{2}(\mathbf{a}_2+\mathbf{a}_3)\\
 \mathbf{r}_4&=\frac{1}{2}(\mathbf{a}_3+\mathbf{a}_1),
-\end{aligned}
-$$
+\end{align}
+
 with $f_1=f_2=f_3=f_4\equiv f$. Let the reciprocal lattice be described by $\mathbf{G}=h\mathbf{b}_1+k\mathbf{b}_2+l\mathbf{b}_3$, where $h$, $k$ and $l$ are integers. Using the basis described above and the reciprocal lattice, we calculate the structure factor:
 
-$$
-\begin{aligned}
+\begin{align}
 S&=f\left(\mathrm{e}^{i\mathbf{G}\cdot\mathbf{r}_1}+\mathrm{e}^{i\mathbf{G}\cdot\mathbf{r}_2}+\mathrm{e}^{i\mathbf{G}\cdot\mathbf{r}_3}+\mathrm{e}^{i\mathbf{G}\cdot\mathbf{r}_4}\right)\\
 &=f\left(1+\mathrm{e}^{i(h\mathbf{b}_1\cdot\mathbf{a}_1+k\mathbf{b}_2\cdot\mathbf{a}_2)/2}+\mathrm{e}^{i(k\mathbf{b}_2\cdot\mathbf{a}_2+l\mathbf{b}_3\cdot\mathbf{a}_3)/2}+\mathrm{e}^{i(h\mathbf{b}_1\cdot\mathbf{a}_1+l\mathbf{b}_3\cdot\mathbf{a}_3)/2}\right)\\
 &=f\left(1+\mathrm{e}^{i\pi(h+k)}+\mathrm{e}^{i\pi(k+l)}+\mathrm{e}^{i\pi(h+l)}\right).
-\end{aligned}
-$$
+\end{align}
+
 Because $h$, $k$ and $l$ are integers, all exponents are either $+1$ or $-1$, and the interference is only present if
 
 $$
@@ -504,6 +509,7 @@ S =
     0, \: \mathrm{in \: all \: other \: cases}.
 \end{cases}
 $$
+
 We now see that the reciprocal lattice points with nonzero amplitude exactly form the reciprocal lattice of the FCC lattice.
 
 ### Powder Diffraction
@@ -556,8 +562,7 @@ plt.annotate('$ 2\\theta $',(-0.56,-0.44),fontsize=14)
 
 plt.xlim([-1,0.5])
 plt.ylim([-1,0.5])
-plt.axis('off')
-plt.show()
+plt.axis('off');
 ```
 
 
@@ -572,12 +577,10 @@ $$
 where we used $|\mathbf{k'}| = |\mathbf{k}|$.
 We then substitute the Laue condition $\mathbf{k'} = \mathbf{k}+\mathbf{G}$:
 
-$$
 \begin{align}
 \lvert \mathbf{G} \rvert ^2 &= 2k^2-2 \left(\mathbf{k}+\mathbf{G}\right) \cdot \mathbf{k} \\
 &= -2 \mathbf{G} \cdot \mathbf{k}.
 \end{align}
-$$
 
 Using $\mathbf{k} \cdot \mathbf{G} = \lvert \mathbf{k} \rvert \lvert \mathbf{G} \rvert \cos(\phi)$,  we obtain
 
@@ -676,4 +679,4 @@ Consider a two-dimensional crystal with a rectangular lattice and lattice vector
 3. How does this structure factor change if the atoms in the center of the conventional unit cell have a different form factor from the atoms at the corner of the conventional unit cell?
 4. A student carried out X-ray powder diffraction on Chromium (Cr) which is known to have a BCC structure. The first five diffraction peaks are given below. Furthermore, the student took the liberty of assigning Miller indices to the peaks. Were the peaks assigned correctly? Fix any mistakes and explain your reasoning.
 ![](figures/cr_xray_exercise.svg)
-5. Calculate the lattice constant, $a$, of the chromium bcc unit cell. Note that X-ray diffraction was carried out using Cu K-$\alpha$ ($1.5406 \unicode{xC5}$) radiation.
+5. Calculate the lattice constant, $a$, of the chromium bcc unit cell. Note that X-ray diffraction was carried out using Cu K-$\alpha$ ($1.5406$Å) radiation.