.gitattributes

+# GitHub syntax highlighting
+pixi.lock linguist-language=YAML linguist-generated=true

.gitignore

@@ -3,4 +3,7 @@
 site
 *.pyc
 __pycache__
-*.swp
\ No newline at end of file
+*.swp
+.cache# pixi environments
+.pixi
+*.egg-info

.gitlab-ci.yml

-image: gitlab.kwant-project.org:5005/qt/research-docker
+variables:
+  FF_USE_FASTZIP: "true"
+  CACHE_COMPRESSION_LEVEL: "fastest"
 
 build lectures:
-  before_script:
-    - pip install -Ur requirements.txt
-  script:
-    - mkdocs build
-  artifacts:
+  image: ghcr.io/prefix-dev/pixi:0.28.0
+  cache:
+    key: "$CI_JOB_NAME"
     paths:
-      - site
-    expire_in: 1 week
-  needs: []
-
-compile mathjax:
-  image: gitlab.kwant-project.org:5005/qt-restricted/registry/mathjax-jsdom
+        - .pixi
   script:
-    - find site -name "*.html" | xargs -L1 mathjax-jsdom-mod
+    - pixi run build
   artifacts:
     paths:
       - site
     expire_in: 1 week
-  needs:
-    - build lectures
+  needs: []
 
 .prepare_deploy: &prepare_deploy
+  image: eeacms/rsync
   only:
     - branches@solidstate/lectures
   before_script:
@@ -42,12 +37,12 @@ compile mathjax:
     ## Create the SSH directory and give it the right permissions
     - mkdir -p ~/.ssh
     - chmod 700 ~/.ssh
-    - ssh-keyscan tnw-tn1.tudelft.net >> ~/.ssh/known_hosts
+    - ssh-keyscan qt4.tudelft.net >> ~/.ssh/known_hosts
     - chmod 644 ~/.ssh/known_hosts
   script:
-    - "rsync -rv site/* solidstate@tnw-tn1.tudelft.net:$DEPLOY_PATH"
+    - "rsync -rv site/* uploader@qt4.tudelft.net:$DEPLOY_PATH"
   needs:
-    - compile mathjax
+    - build lectures
 
 deploy delft version:
   <<: *prepare_deploy
@@ -79,12 +74,13 @@ undeploy test version:
     DEPLOY_PATH: "test_builds/$CI_COMMIT_REF_NAME"
   script:
     - mkdir empty/
-    - "rsync -rlv --delete empty/ solidstate@tnw-tn1.tudelft.net:$DEPLOY_PATH"
+    - "rsync -rlv --delete empty/ uploader@qt4.tudelft.net:$DEPLOY_PATH"
   environment:
     name: $CI_COMMIT_REF_NAME
     action: stop
 
 merge into local:
+  image: bitnami/git
   only:
     - master@solidstate/lectures
   script:

docs/10_xray_solutions.md

@@ -7,110 +7,79 @@ from matplotlib import pyplot as plt
 import numpy as np
 from math import pi
 ```
+
 # Solutions for lecture 10 exercises
 
 ## Warm-up exercises
-#### Question 1.
-Hint: you can make use of the [scalar triple product](https://en.wikipedia.org/wiki/Triple_product#Scalar_triple_product).
-#### Question 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.
-#### Question 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.
-#### Question 4.
-No, an increase of the unit cell size cannot create new diffraction peaks (see lecture).
-
-## Exercise 1: Equivalence of direct and reciprocal lattice
-
-#### Question 1.
-Solution
-
-$$
-V^*=\left|\mathbf{b}_{1} \cdot\left(\mathbf{b}_{2} \times \mathbf{b}_{3}\right)\right| = \frac{2\pi}{V}\left| (\mathbf{a}_{2} \times \mathbf{a}_{3}) \cdot\left(\mathbf{b}_{2} \times \mathbf{b}_{3}\right)\right| = \frac{(2\pi)^3}{V}
-$$
-
-In the second equality, we used the reciprocal lattice vector definition (see notes). 
-In the third equality, we used the identity:
 
-$$
-(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d})-(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c})
-$$
+1. The reciprocal lattice of a 1D lattice defined by $R = n_1 a$ is $G = m_1 2\pi/a$, with $n_1, m_1 \in \mathbb{Z}$. The Wigner Seitz cell defines the first Brillouin zone, and is constructed by connecting nearest-neighbor reciprocal lattice points and drawing perpendicular bisectors.
+2. We expect $\mathbf{a_i}\cdot\mathbf{b_j} = 2\pi\delta_{ij}$.
+3. If $\mathbf{k}-\mathbf{k'}\neq \mathbf{G}$, then the argument of the exponent in the sum  $\sum_\mathbf{R}\mathrm{e}^{i\left((\mathbf{k'}-\mathbf{k})\cdot\mathbf{R}-\omega t\right)}$
+represents a phase that depends on the location $\mathbf{R}$ of the real-space lattice point.
+Because we sum over all lattice points, each argument has a different phase. Summing over all these phases results in a total amplitude of 0, resulting in no intensity peaks.
+4. No, there is a single atom, and thus only one term in the structure factor. Therefore the structure factor is non-zero.
+5. No, an increase of the unit cell size cannot create new diffraction peaks (see lecture). Even though the increase leads to extra reciprocal lattice points, the structure factor will cancel their contributions to the scattered wave amplitudes.
 
-#### Question 2.
-Because the relation between direct and reciprocal lattice is symmetric, so are the expressions for the direct lattice vectors through the reciprocal ones:
+## Exercise 2*. The reciprocal lattice of the bcc lattice
 
-$$
-\mathbf{a}_{i} \epsilon_{ijk} = \frac{2\pi}{V^*} (\mathbf{b}_{j} \times \mathbf{b}_{k})
-$$
+1. A possible set of BCC primitive lattice vectors is:
 
-where $\epsilon_{ijk}$ is the [Levi-Civita tensor](https://en.wikipedia.org/wiki/Levi-Civita_symbol#Three_dimensions)
+    \begin{align*}
+    \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).
+    \end{align*}
 
-#### Question 3.
-One set of the BCC primitive lattice vectors is given by:
+    We expect that the volume of the primitive unit cell equals $a^3/2$ because the conventional unit cell has volume $a^3$ and contains two lattice points. We confirm this by calculating the volume of the primitive unit cell using $|\mathbf{a_1}\times\mathbf{a_2}\cdot\mathbf{a_3}| = a^3/2$.
 
-$$
-\mathbf{a_1} = \frac{a}{2} \left(-\hat{\mathbf{x}}+\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\
-\mathbf{a_2} = \frac{a}{2} \left(\hat{\mathbf{x}}-\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\
-\mathbf{a_3} = \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}}-\hat{\mathbf{z}} \right).
-$$
+2. The corresponding set of reciprocal lattice vectors
 
-From this, we find the following set of reciprocal lattice vectors:
+    \begin{align*}
+    \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),
+    \end{align*}
 
-$$
-\mathbf{b_1} = \frac{2 \pi}{a} \left(\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\
-\mathbf{b_2} = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{z}} \right) \\
-\mathbf{b_3} = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right),
-$$
+3. The reciprocal lattice forms an FCC lattice. Given the lattice vectors, the volume of the conventional unit cell is $(4\pi/a)^3$.
 
-which forms a reciprocal FCC lattice.
-The opposite relation follows directly from our previous result.
-
-#### Question 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.
+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!
+We see that each lattice point contains 12 neirest neighbours and thus the Wigner-Seitz cell contains 12 sides! 
+The volume of the 1st Brillouin zone is the same as the volume of any other primitive unit cell. Therefore, it is given by $\mathbf{b_1}\cdot(\mathbf{b_2}\times\mathbf{b_3})| = \tfrac{1}{4} (4\pi/a)^3$. This is one quarter of the conventional unit cell calculated in subquestion 3, which is as expected because the conventional unit cell of the fcc lattice contains 4 lattice points.
 
-## Exercise 2: Miller planes and reciprocal lattice vectors
+5. The bcc and fcc lattices are reciprocal to each other.
 
-#### Question 1.
-Hints
+## Exercise 2: Miller planes and reciprocal lattice vectors
 
-??? hint "First small hint"
+1. To prove this, we can show that $\mathbf{G}$ is orthogonal to two non-parallel vectors in the Miller plane. We therefore first find two vectors in the Miller plane, such as $\mathbf{v_1} = \mathbf{a_1}/h-\mathbf{a_2}/k$ and $\mathbf{v_2} = \mathbf{a_2}/h - \mathbf{a_3}/l$. We then show that $\mathbf{G}\cdot \mathbf{v_{1,2}}=0$
 
-    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? 
+2. We compute the distance between two parallel planes by projecting any vector connecting the two planes onto a unit vector perpendicular to the planes. Using the result of subquestion 1, the unit vector is
 
-??? hint "Second small hint"
+    $$
+    \hat{\mathbf{n}} = \frac{\mathbf{G}}{|\mathbf{G}|}
+    $$
 
-    Whats the best vector operation to show orthogonality between two vectors? 
+    For lattice planes, there is always a plane intersecting the zero lattice point (0,0,0). As such, we can project the vector $\mathbf{a_1}/h$ onto $\mathbf{\hat{n}}$, yielding a distance:
 
-#### Question 2.
-One can compute the normal to the plane by using result from Subquestion 1:
+    $$
+    d = \hat{\mathbf{n}} \cdot \frac{\mathbf{a_1}}{h} = \frac{2 \pi}{|\mathbf{G}|}
+    $$
 
-$\hat{\mathbf{n}} = \frac{\mathbf{G}}{|G|}$
+3. Since $\rho=d / V$, we must maximize $d$ and therefore find the smallest $|\mathbf{G}|$. Using the primitive reciprocal lattice vectors derived in the previous question, we observe that all $\{\mathbf{b_i}\}$ have the same length and we cannot combine them to obtain an even shorter vector. Therefore, we have that the {100} family of planes (in terms of the BCC primitive lattice vectors) has the highest density of lattice points. In terms of the conventional lattice vectors, this is the {110} family of planes.
 
-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:
+4. To identify one of the Miller planes of the previous subquestion in a sketch of the real space lattice, we first note that the normal to these planes points in the direction of the $\{\mathbf{b_i}\}$. In addition, we use that the planes are parallel to 2 out of 3 primitive lattice vectors, and intersect the third at the end. To sketch the plane, it may help to make a sketch of the projections of the $\{\mathbf{a_i}\}$ onto the $xy$ plane.
 
-$ d = \hat{\mathbf{n}} \cdot \frac{\mathbf{a_1}}{h} = \frac{2 \pi}{|G|} $
+5. (100) does not correspond to a family of lattice planes because they do not contain the lattice point in the center of the bcc unit cell. (200) on the other hand, does.
 
-#### Question 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).
+6. The structure factor is zero except when $h+k+l$ is even. Therefore, the shortest valid reciprocal lattice vector has indices $(hkl)=(110)$, which is indeed the same family of planes as that found in 4 (use a sketch)
 
 ## Exercise 3: X-ray scattering in 2D
 
-#### Question 1.
-See figure
-
 ```python
+b_y, b_x = 18.4, 13.4
 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)
+    y = np.repeat(np.linspace(-b_y*(N//2),b_y*(N//2),N),N)
+    x = np.tile(np.linspace(-b_x*(N//2),b_x*(N//2),N),N)
 
     plt.figure(figsize=(5,5))
 
@@ -125,55 +94,76 @@ def reciprocal_lattice(N = 7, lim = 40):
 reciprocal_lattice()
 ```
 
-#### Question 2.
-Since we have elastic scattering, we obtain
-$|\mathbf{k}| = |\mathbf{k}'| = \frac{2 \pi}{\lambda} = 37.9 nm^{-1}$
-#### Question 3.
-See figure
+1. See figure above
+2. Since we have elastic scattering, we have $|\mathbf{k}| = |\mathbf{k}'| = \frac{2 \pi}{\lambda} = 37.9 nm^{-1}$
+3. We can draw the (210) Miller plane using its intersections with the lattice vectors as described in the lecture notes. We plot the scattering triangle in the figure below
 
 ```python
 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.arrow(
+    b_x*2, b_y, -b_x*2, -b_y, 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.arrow(-6,37.4,6+b_x*2,-37.4+b_y,color='k',zorder=11,head_width=2,length_includes_head=True)
 plt.annotate('$\mathbf{k\'}$',(15,30),fontsize=14, ha='center',color='k');
 ```
 
-## Exercise 4: Structure factors
+4.
+Since there is only 1 atom in the basis, there are no missing peaks due to a structure factor. We will get diffraction peaks at angles given by Bragg's law $\sin2\theta = \lambda/d_{hkl} = \lambda |\mathbf{G_{hkl}}|/2\pi$. We see that the shortest reciprocal lattice vector gives the smallest angle. Therefore, as a function of increasing $\theta$, we will see peaks at $(hkl)= (10) \quad (01) \quad (11) \quad (20) \quad  (21) \quad (02)$, where we took into account that $|\mathbf{b_1}|<|\mathbf{b_2}|$.
 
-#### Question 1.
-$S(\mathbf{G}) = \sum_j f_j e^{i \mathbf{G} \cdot \mathbf{r_j}} = f(1 + e^{i \pi (h+k+l)})$
-#### Question 2.
-Solving for $h$, $k$, and $l$ results in 
+## Exercise 4: Analyzing a 3D power diffraction spectrum
 
-$$
-S(\mathbf{G}) = \begin{cases}
-    2f, \: \text{if $h+k+l$ is even}\\
-    0, \: \text{if $h+k+l$ is odd}.
-\end{cases}
-$$
+1. The structure factor is $S(\mathbf{G}) = \sum_j f_j e^{i \mathbf{G} \cdot \mathbf{r_j}} = f(1 + e^{i \pi (h+k+l)})$
 
-Thus if $h+k+l$ is odd, diffraction peaks dissapear
+2. Solving for $h$, $k$, and $l$ results in 
 
-#### Question 3.
-Let $f_1 \neq f_2$, then
+    $$
+    S(\mathbf{G}) = \begin{cases}
+        2f, \: \text{if $h+k+l$ is even}\\
+        0, \: \text{if $h+k+l$ is odd}.
+    \end{cases}
+    $$
 
-$$
-S(\mathbf{G}) = \begin{cases}
-f_1 + f_2, \text{if $h+k+l$ is even}\\
-f_1 - f_2, \text{if $h+k+l$ is odd}
-\end{cases}       
-$$
+    Thus if $h+k+l$ is odd, diffraction peaks are absent even though the Laue condition is satisfied. The reason is that the Laue condition is based on the reciprocal lattice vectors constructed from the conventional unit cell instead of a primitive unit cell.
 
+3. Let $f_1 \neq f_2$, then
 
-#### Question 4.
-Due to bcc systematic absences, the peaks from lowest to largest angle are:
+    $$
+    S(\mathbf{G}) = \begin{cases}
+    f_1 + f_2, \text{if $h+k+l$ is even}\\
+    f_1 - f_2, \text{if $h+k+l$ is odd}
+    \end{cases}       
+    $$
+
+4. Due to the systematic absences of peaks caused by the structure factor, the peaks from lowest to largest angle are:
 $(110),(200),(211), (220), (310)$
 
-#### Question 5.
-$a = 2.9100$Å
+5. We use $d_{hkl} = \lambda /(2\sin\theta)$. We can for instance read off from the graph that $\theta = 32$ deg. for the $(hkl) =(200)$ peak, which gives $d_{200} = 0.145$ nm, and therefore the side-length of the conventional unit cell is $a=0.29$ nm.
+
+## Extra exercise 2: Equivalence of direct and reciprocal lattice
+
+1. We get
+
+    $$
+    V^*=\left|\mathbf{b}_{1} \cdot\left(\mathbf{b}_{2} \times \mathbf{b}_{3}\right)\right| = \frac{2\pi}{V}\left| (\mathbf{a}_{2} \times \mathbf{a}_{3}) \cdot\left(\mathbf{b}_{2} \times \mathbf{b}_{3}\right)\right| = \frac{(2\pi)^3}{V}
+    $$
+
+    In the second equality, we used the reciprocal lattice vector definition (see notes). 
+    In the third equality, we used the identity:
+
+    $$
+    (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d})-(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c})
+    $$
+
+2. 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](https://en.wikipedia.org/wiki/Levi-Civita_symbol#Three_dimensions)

docs/11_nearly_free_electron_model.md

@@ -12,7 +12,7 @@ pi = np.pi
 ```
 
 # The nearly free electron model & Bloch theorem
-_(based on chapter 15 of the book)_  
+_(based on chapter 15 of the book)_
 
 !!! success "Expected prior knowledge"
 
@@ -20,15 +20,15 @@ _(based on chapter 15 of the book)_
 
     - Write down the dispersion and wavefunction of an electron in free space (solving the Schrödinger equation).
     - Describe how the periodicity of a band structure (= dispersion) is related to the reciprocal lattice.
-    - Write down a Fourier series representation of a periodic function. 
+    - Write down a Fourier series representation of a periodic function.
     - Diagonalize a 2x2 matrix (i.e., find its eigenvalues and eigenfunctions).
-    
+
 !!! summary "Learning goals"
 
     After this lecture you should be able to:
 
     - Derive the electron band structure when the interaction with the lattice is weak using the **Nearly free electron model**.
-    - Describe what momentum states of particles in a crystal may couple through the periodic lattice potential.  
+    - Describe what momentum states of particles in a crystal may couple through the periodic lattice potential.
     - Formulate a general way of computing the electron band structure - the **Bloch theorem**.
     - Recall that in a periodic potential, all electron states are Bloch waves.
 
@@ -38,7 +38,7 @@ Let's summarize what we learned about electrons so far:
 * Electrons on isolated atoms live in discrete orbitals ([lecture 5](5_atoms_and_lcao.md))
 * When orbitals hybridize we get *LCAO* or *tight-binding* band structures ([lecture 7](7_tight_binding.md))
 
-In this lecture, we will analyze how electrons behave in solids using the *nearly-free electron model*. This model considers electrons as plane waves (as in the free electron model) that are weakly perturbed by the periodic potential associated with the atoms in a solid. This approach is opposite to that of the tight-binding model, where our starting point was that the electrons are strongly bound to the individual atoms and we included hopping to other atoms as a small effect. Perhaps surprisingly, we will find that the nearly-free electron model gives very similar results to the tight binding model: it also leads to the formation of energy bands, and these bands are separated by *band gaps* - regions in the band structure where there are no allowed energy states.  
+In this lecture, we will analyze how electrons behave in solids using the *nearly-free electron model*. This model considers electrons as plane waves (as in the free electron model) that are weakly perturbed by the periodic potential associated with the atoms in a solid. This approach is opposite to that of the tight-binding model, where our starting point was that the electrons are strongly bound to the individual atoms and we included hopping to other atoms as a small effect. Perhaps surprisingly, we will find that the nearly-free electron model gives very similar results to the tight binding model: it also leads to the formation of energy bands, and these bands are separated by *band gaps* - regions in the band structure where there are no allowed energy states.
 
 ## Nearly free electron model
 
@@ -80,6 +80,7 @@ ax.set_ylabel("$E$")
 ax.set_ylim(-.5, max_en + 5)
 ax.set_xticks(pi * np.arange(-3, 4))
 ax.set_xticklabels(fr"${i}\pi$".replace("1", "") if i else "$0$" for i in range(-3, 4))
+ax.set_yticks([])
 
 draw_classic_axes(ax, xlabeloffset=4)
 ```
@@ -88,7 +89,7 @@ In this figure, the orange curves represent the nearly-free electron dispersion,
 
 ### Analyzing the avoided crossings
 
-_Remark: An avoided crossing is an important concept in quantum mechanics that can be analyzed using **perturbation theory**. You will only learn this theory later in QMIII, so we will need to postulate some important facts here._
+_Remark: An avoided crossing is an important concept in quantum mechanics that is captured by the **perturbation theory**._
 
 To analyze what happens near the crossings, we first neglect the lattice potential and consider the free-electron dispersion near the crossing at $k=\pi/a$ in 1D. Near this crossing, we see that two copies of the dispersion come together (one copy centered at $k=0$, the other at $k=2\pi/a$). We call the corresponding plane-wave eigenfunctions $|k\rangle$ and $|k'\rangle =|k-2\pi/a\rangle$. We now express the wavefunction near this crossing as a linear superposition $|\psi\rangle = \alpha |k\rangle + \beta |k'\rangle$. Note that this wave function is very similar to that used in the LCAO model, except there we used linear combinations of the orbitals $|1\rangle$ and $|2\rangle$ instead of the plane waves $|k\rangle$ and $|k'\rangle$.
 
@@ -125,19 +126,19 @@ $$ E(\delta k) = E_0 \pm \sqrt{v^2\hbar^2\delta k^2 + |W|^2}$$
 
 #### Calculating the magnitude of the gaps
 
-We will now show that $W=\langle k | V |k' \rangle$ represents a Fourier component of the lattice potential. To see this, we express the lattice potential (which is periodic with $V(x)=V(x+a)$) as a Fourier series 
+We will now show that $W=\langle k | V |k' \rangle$ represents a Fourier component of the lattice potential. To see this, we express the lattice potential (which is periodic with $V(x)=V(x+a)$) as a Fourier series
 
 $$
 V(x) = \sum_{n=-\infty}^{\infty} V_n e^{2\pi i n x/a}
 $$
 
-and recall that such a series has Fourier components $V_n$ given by 
+and recall that such a series has Fourier components $V_n$ given by
 
 $$
 V_n = \frac{1}{a}\int_0^a e^{- i n 2\pi x /a} V(x) dx
 $$
 
-Calculating $W$, we find 
+Calculating $W$, we find
 
 $$
 W = \langle k | V | k' \rangle = \frac{1}{a}\int_0^{a} e^{-i k x} V(x) e^{i k'x}  dx = \frac{1}{a}\int_0^a e^{-i 2\pi x /a} V(x) dx = V_1
@@ -150,10 +151,10 @@ where we have used that $k-k' =2\pi/a$ because we are analyzing the first crossi
 
 Everything we did can also be applied to the higher-energy crossings seen in the figure above. We note that all crossings occur between parabola's that are shifted by integer multiples of reciprocal lattice vectors $n 2\pi/a$. The first crossing corresponds to $n=1$, and we found that the magnitude of the gap is given by $V_1$. Similarly, $V_2$ determines the gap between the second and third bands, $V_3$ for the crossing between third and fourth, etc.
 
-The key conclusion is that the lattice potential couples plane-wave states that differ by integer multiples of reciprocal lattice vectors. This coupling alters the band structure most strongly where the free-electron eigenenergies cross, opening up gaps of which the magnitudes are determined by the Fourier components of the lattice potential.  
+The key conclusion is that the lattice potential couples plane-wave states that differ by integer multiples of reciprocal lattice vectors. This coupling alters the band structure most strongly where the free-electron eigenenergies cross, opening up gaps of which the magnitudes are determined by the Fourier components of the lattice potential.
 
 ??? question "Suppose the lattice potential is $V(x)=A\cos(2\pi/ax)$. At what locations in the dispersion does $V(x)$ lead to the formation of gaps?"
-    Hint: The Fourier series of $V(x)$ is $V(x)=A(e^{i2\pi/ax}+e^{-i2\pi/ax})/2$, so the only non-zero Fourier components are $V_1=V_{-1} = A/2$.  
+    Hint: The Fourier series of $V(x)$ is $V(x)=A(e^{i2\pi/ax}+e^{-i2\pi/ax})/2$, so the only non-zero Fourier components are $V_1=V_{-1} = A/2$.
 
 
 ## General description of a band structure in a crystal - Bloch theorem
@@ -203,8 +204,8 @@ $$
 
 which shows that each Bloch wave can be written as a sum over plane waves that differ by a reciprocal lattice vector.
 
-??? question "Does the tight-binding wavefunction $|\psi\rangle = \sum_n e^{ikna}(\phi_0|n,1\rangle+\psi_0|n,2\rangle)$ (see exercise 2 in Lecture 8) satisfy the Bloch theorem? What part of $|\psi\rangle$ describes $u(x)$ in this case? Try to describe in words how this Bloch wave is built up."
-    
+!!! question "Does the tight-binding wavefunction $|\psi\rangle = \sum_n e^{ikna}(\phi_0|n,1\rangle+\psi_0|n,2\rangle)$ (see exercise 2 in Lecture 8) satisfy the Bloch theorem? What part of $|\psi\rangle$ describes $u(x)$ in this case? Try to describe in words how this Bloch wave is built up."
+
 ### Repeated vs reduced vs extended Brillouin zone
 
 There are several common ways to **plot** the same dispersion relation (no difference in physical information).
@@ -238,6 +239,7 @@ ax.set_ylabel("$E$")
 ax.set_ylim(-.5, max_en + 5)
 ax.set_xticks(pi * np.arange(-1, 2))
 ax.set_xticklabels(r"$-\pi$ $0$ $\pi$".split())
+ax.set_yticks([])
 
 draw_classic_axes(ax, xlabeloffset=4)
 ```
@@ -267,6 +269,7 @@ ax.set_ylabel("$E$")
 ax.set_ylim(-.5, max_en + 5)
 ax.set_xticks(pi * np.arange(-3, 4))
 ax.set_xticklabels(fr"${i}\pi$".replace("1", "") if i else "$0$" for i in range(-3, 4))
+ax.set_yticks([])
 
 draw_classic_axes(ax, xlabeloffset=4)
 ```
@@ -279,65 +282,47 @@ draw_classic_axes(ax, xlabeloffset=4)
 ### Exercise 1: Bloch's theorem
 Suppose we have a crystal with lattice vectors $\mathbf{a}_ 1$, $\mathbf{a}_ 2$, and $\mathbf{a}_ 3$.
 
-#### Question 1.
-What can be said about the symmetry of the Hamiltonian $\hat{H}$ of this crystal?
-#### Question 2.
-Now define the translation operator $\hat{T}_{\alpha,\beta,\gamma}$ so that 
-
-$$
-\hat{T}_{\alpha,\beta,\gamma} \psi(\mathbf{r}) = \psi(\mathbf{r} - \alpha \mathbf{a}_1 - \beta \mathbf{a}_2 - \gamma \mathbf{a}_3),
-$$
-
-where $\alpha$, $\beta$, $\gamma$ are integers. Show that $\hat{T}_{\alpha,\beta,\gamma}$ and $\hat{H}$ commute.
-
-#### Question 3.
-Show that the Bloch wavefunctions defined in the lecture notes are eigenfunctions of $\hat{T}_{\alpha,\beta,\gamma}$. What are the corresponding eigenvalues? What does this say about the eigenfunctions of $\hat{H}$.
-#### Question 4.
-By applying $\hat{H}$ to the Bloch wavefunction, show that the Schrödinger equation can be rewritten as
-
-$$
-\left[ \frac{\mathbf{\hat{p}}^2}{2m} + \frac{\hbar}{m} \mathbf{k} \cdot \mathbf{\hat{p}} + \frac{\hbar^2 \mathbf{k}^2}{2m} + V(\mathbf{r}) \right] u_{n,\mathbf{k}}(\mathbf{r}) = E_{n,\mathbf{k}} u_{n,\mathbf{k}}(\mathbf{r}),
-$$
-
+1. What can be said about the symmetry of the Hamiltonian $\hat{H}$ of this crystal?
+2. To describe the translation through the crystal in terms of the lattice vectors, we define the translation operator $\hat{T}_{\alpha,\beta,\gamma}$ in such a way that $$
+\hat{T}_{\alpha,\beta,\gamma} \psi(\mathbf{r}) = \psi(\mathbf{r} - \alpha \mathbf{a}_1 - \beta \mathbf{a}_2 - \gamma \mathbf{a}_3)
+, $$
+where $\alpha$, $\beta$, $\gamma$ are integers.
+Show that $\hat{T}_{\alpha,\beta,\gamma}$ and $\hat{H}$ commute.
+3. Show that the Bloch wavefunctions defined in the lecture notes are eigenfunctions of $\hat{T}_{\alpha,\beta,\gamma}$. What are the corresponding eigenvalues? What does this say about the eigenfunctions of $\hat{H}$.
+4. By applying $\hat{H}$ to the Bloch wavefunction, show that the Schrödinger equation can be rewritten as $$
+\left[ \frac{\mathbf{\hat{p}}^2}{2m} + \frac{\hbar}{m} \mathbf{k} \cdot \mathbf{\hat{p}} + \frac{\hbar^2 \mathbf{k}^2}{2m} + V(\mathbf{r}) \right] u_{n,\mathbf{k}}(\mathbf{r}) = E_{n,\mathbf{k}} u_{n,\mathbf{k}}(\mathbf{r})
+, $$
 where $\mathbf{\hat{p}} =-i\hbar\nabla$.
-
-#### Question 5.
-What is $u_{n,\mathbf{k}}(\mathbf{r})$ in case of free electrons? Is your answer consistent with the equation above?
+5. What is $u_{n,\mathbf{k}}(\mathbf{r})$ in case of free electrons? Is your answer consistent with the equation above?
 
 ### Exercise 2: the central equation in 1D
-Let's consider a 1D crystal with a period $a$. Let $k_0$ be any wave number of an electron in the first Brillouin zone.
+Let's consider a 1D crystal with a periodicity of $a$.
+Let $k_0$ be any wave number of an electron in the first Brillouin zone.
 
-#### Question 1.
-What $k_n$ are equivalent to $k_0$ in this crystal?
-#### Question 2.
-First, we assume that the electrons with these $k_n$ are free. In that case, what are the wavefunctions $\phi_n(x)$ and energies $E_n$ of these electrons?
-#### Question 3.
-Make a sketch of the dispersion relation using a repeated Brillouin zone representation. Indicate some $k_n$ and $E_n$ as well as the first Brillouin zone in your sketch.
+1. Which $k_n$ are equivalent to $k_0$ in this crystal?
 
-We will now introduce a weak periodic potential $V(x) = V(x+na)$ in our system. This causes coupling between eigenstates $\left| \phi_n\right>$ in the free electron case. In order to find the right eigenstates of the system with that potential, we need an 'LCAO-like' trial eigenstate given by
-
-$$
-\left|\psi\right> = \sum_{n=-\infty}^{\infty}C_n \left|\phi_n\right>
-$$
+2. First, we assume that the electrons with these $k_n$ are free. In that case, what are the wavefunctions $\phi_n(x)$ and energies $E_n$ of these electrons?
 
-#### Question 4.
-Using the trial eigenstate above and the Schrödinger equation, show that 
+3. Make a sketch of the dispersion relation using a repeated Brillouin zone representation.
+    Indicate some $k_n$ and $E_n$ as well as the first Brillouin zone in your sketch. We will now introduce a weak periodic potential $V(x) = V(x+na)$ to the system, which causes coupling between eigenstates $\left| \phi_n\right>$ of the free electrons. In order to find the right eigenstates of the system with that potential, we use an 'LCAO-like' trial eigenstate $$ \left|\psi\right> = \sum_{n=-\infty}^{\infty}C_n \left|\phi_n\right> $$
+    
+4. Using this trial eigenstate and the Schrödinger equation, show that
 
-$$
-E C_m = \varepsilon_m C_m+\sum_{n=-\infty}^{\infty} V_{n}C_{m-n},
-$$
+    $$ 
+    E C_m = \varepsilon_m C_m+\sum_{n=-\infty}^{\infty} C_{n}V_{m-n},
+    $$
 
-where $V_n$ are the Fourier components of the potential defined [above](#physical-meaning-of-w). Find an expression for $\varepsilon_m$. _**NB:** This equation is also known as the central equation (in 1D)._
+    where $V_n$ are the Fourier components of the potential defined [above](#physical-meaning-of-w).
+    Find an expression for $\varepsilon_m$. _**NB:** This equation is also known as the central equation (in 1D)._
 
-??? hint
-    - Apply $\left<\phi_m\right|$ to the Schrödinger equation.
-    - To evaluate $\left<\phi_m\right| \hat{H} \left| \phi_n\right>$, it may be helpful to separate the kinetic energy and potential energy of the Hamiltonian.
+    ??? hint
+        - Apply $\left<\phi_m\right|$ to the Schrödinger equation.
+        - To evaluate $\left<\phi_m\right| \hat{H} \left| \phi_n\right>$, it may be helpful to separate the kinetic energy and potential energy of the Hamiltonian.
 
-#### Question 5.
-Why is the dispersion relation only affected near $k=0$ and at the edge of the Brillouin zone (see also figures [above](#repeated-vs-reduced-vs-extended-brillouin-zone))?
+5. Why is the dispersion relation only affected near $k=0$ and at the edge of the Brillouin zone (see also figures [above](#repeated-vs-reduced-vs-extended-brillouin-zone))?
 
-??? hint
-    To answer this question, only consider consider two free electron wavefunctions in the Hamiltonian and ignore all the others. Between what two of free electron wavefunctions does the coupling give significant contribution to the energy levels of the free electron wavefunctions?
+    ??? hint
+        To answer this question, only consider consider two free electron wavefunctions in the Hamiltonian and ignore all the others. Between what two of free electron wavefunctions does the coupling give significant contribution to the energy levels of the free electron wavefunctions?
 
 ### Exercise 3: the tight binding model vs. the nearly free electron model
 Consider a 1D crystal with a periodic potential given by delta peaks:
@@ -349,36 +334,43 @@ $$
 where $\lambda>0$. In this exercise, we will find the band structure of this crystal in two ways:
 
 - By means of the nearly free electron model explained in this lecture.
-- By means of the tight binding model explained in [lecture 7](/7_tight_binding).
+- By means of the tight binding model explained in [lecture 7](7_tight_binding.md).
 
-#### Question 1.
-We first find the band structure using the nearly free electron model. To this end, we consider the effect of the potential on the free electron wavefunctions given by $\psi_1(x) \propto e^{ikx}$ and $\psi_2(x) \propto e^{i[k-2\pi/a]x}$ on the interval $k=[0,\pi/a]$. Derive a dispersion relation of the lower band using the Schödinger equation and the trial eigenstate
+We first find the band structure using the nearly free electron model. 
 
-$$
-\Psi(x) = \alpha \psi_1(x) + \beta \psi_2(x).
-$$
+1. To this end, we consider the effect of the potential on the free electron wavefunctions given by $\psi_1(x) \propto e^{ikx}$ and $\psi_2(x) \propto e^{i[k-2\pi/a]x}$ on the interval $k=[0,\pi/a]$. Derive a dispersion relation of the lower band using the Schödinger equation and the trial eigenstate 
+
+    $$ 
+    \Psi(x) = \alpha \psi_1(x) + \beta \psi_2(x). 
+    $$
+
+    ??? hint
+
+        Using the Schrödinger equation and the trial eigenstate, first derive a 2×2 eigenvalue problem given by
+
+        $$
+        E \begin{pmatrix}\alpha \\ \beta\end{pmatrix} = \begin{pmatrix}\varepsilon_0(k)+V_0 & V_1^* \\ V_1 & \varepsilon_0(k - 2\pi/a) + V_0\end{pmatrix} \begin{pmatrix}\alpha \\ \beta\end{pmatrix}.
+        $$
 
-??? hint
-    Using the Schrödinger equation and the trial eigenstate, first derive a 2×2 eigenvalue problem given by
+        What are $\varepsilon_0(k)$, $V_0$ and $V_1$?
 
+2. Make a sketch of the lower band.
+
+3. We now use a tight binding model approach to derive the dispersion relation. 
+    We know from that the corresponding dispersion is 
     $$
-    E \begin{pmatrix}\alpha \\ \beta\end{pmatrix} = \begin{pmatrix}\varepsilon_0(k)+V_0 & V_1^* \\ V_1 & \varepsilon_0(k - 2\pi/a) + V_0\end{pmatrix} \begin{pmatrix}\alpha \\ \beta\end{pmatrix}.
+    E = \varepsilon_0 - 2 t \cos (ka). 
     $$
+    Find an expression for $\varepsilon_0=\left<n\right| \hat{H} \left|n\right>$ and $-t=\left<n-1\right| \hat{H} \left| n \right>$, using the bound state wavefunction around a single $\delta$-peak, centered at site $n$: $$
+    |n\rangle = \kappa e^{- \kappa | x-na | }
+    , $$
+    where $\kappa = -\frac{m \lambda}{\hbar^2}$.
 
-    What are $\varepsilon_0(k)$, $V_0$ and $V_1$?
-
-#### Question 2.
-Make a sketch of the lower band.
-#### Question 3.
-We now use the tight binding model, where we know that the dispersion relation can be described by
+    ??? hint
 
-$$
-E = \varepsilon_0 - 2 t \cos (ka).
-$$
+        To ease the calculating $\epsilon_0$ and $t$, calculate them for $| n = 0 \rangle $ and $ | n = 1 \rangle $.
 
-Find an expression for $\varepsilon_0=\left<n\right| \hat{H} \left|n\right>$ and $-t=\left<n-1\right| \hat{H} \left| n \right>$, where $|n\rangle$ is the wavefunction of a single $\delta$-peak well at site $n$. You may make use of the results obtained in [exercise 2 of lecture 5](/5_atoms_and_lcao/#exercise-2-application-of-the-lcao-model) or [look up the wavefunction](https://en.wikipedia.org/wiki/Delta_potential).
+        You may also make use of the results obtained in [exercise 2 of lecture 5](5_atoms_and_lcao.md#exercise-2-application-of-the-lcao-model) or found on the [wikipedia](https://en.wikipedia.org/wiki/Delta_potential).
 
-#### Question 4.
-Compare the bands obtained in exercise 1 and 2: what are the minima and bandwidths (difference between maximum and minimum) of those bands?
-#### Question 5.
-For what $a$ and $\lambda$ is the nearly free electron model more accurate? And for what $a$ and $\lambda$ is the tight binding model more accurate?
+4. Compare the bands obtained in exercise 1 and 2: what are the minima and bandwidths (difference between maximum and minimum) of those bands?
+5. For what $a$ and $\lambda$ is the nearly free electron model more accurate? And for what $a$ and $\lambda$ is the tight binding model more accurate?

docs/11_nearly_free_electron_model_solutions.md

@@ -55,9 +55,9 @@ All $k_n$ that differ by an integer multiple of $2\pi/a$ from $k_0$ have the exa
 \phi_n(x)=\frac{1}{\sqrt{\Omega}} \exp\left[i \left(k_0+\frac{2\pi n}{a}\right)x \right]
 \end{equation*}
 
-\begin{equation}
+\begin{equation*}
 E_n=\frac{\hbar^2}{2m}\left(k_0+\frac{2\pi n}{a}\right)^2
-\end{equation}
+\end{equation*}
 
 ### Question 3.
 ```python
@@ -91,7 +91,7 @@ def dispersions(N = 5):
     plt.xlabel('$k$',fontsize=19)
     plt.ylabel('$E$',fontsize=19)
     plt.xticks((-2*pi, -pi, 0 , pi,2*pi),('$-2\pi/a$','$-\pi/a$','$0$','$\pi/a$','$2\pi/a$'),fontsize=15)
-    plt.yticks((1,27,54),('$E_0$','$E_1$','$E_{-1}$'))
+    plt.yticks((1,27,54),('$E_0$','$E_{-1}$','$E_{1}$'))
     
 dispersions(5)
 ```
@@ -99,34 +99,34 @@ dispersions(5)
 
 First the kinetic term,
 
-\begin{equation}
-\left\langle\phi_m|\hat{K}|\phi_n\right\rangle=C_m\frac{\hbar^2k_m^2}{2m}
-\end{equation}
+\begin{equation*}
+\left\langle\phi_m|\hat{K}|\psi\right\rangle=C_m\frac{\hbar^2k_m^2}{2m}
+\end{equation*}
 
 And the potential term,
 
-\begin{align}
+\begin{align*}
 \left\langle\phi_m|V(x)|\psi\right\rangle=\sum_{n=-\infty}^{\infty}C_n\int_0^a V(x) e^{-i\frac{2\pi}{a}(m-n)x}dx
-\end{align}
+\end{align*}
 
 then relabel indices and combine both expressions to find the final answer and expression for $\varepsilon_m$ (which is the free electron dispersion).
 
 ### Question 5.
 From the expression for the energy, it is clear that the difference with respect to the free electron model is given by the Fourier component $V_{m-n}$, describing the coupling between two states $m$ and $n$. The question becomes: when does this term contribute significantly? For that we look at two orthogonal states $\phi_n$ and $\phi_m$, and construct the Hamiltonian in the basis ($\phi_n$,$\phi_m$),  
 
-\begin{equation}
+\begin{equation*}
 \hat{H}=
 \begin{pmatrix}
 \dfrac{\hbar^2 k_n^2}{2m} & V_{n-m}\\
 V_{m-n} & \dfrac{\hbar^2 k_m^2}{2m}
 \end{pmatrix}
-\end{equation}
+\end{equation*}
 
 The eigenvalues of this fellow are 
 
-\begin{equation}
+\begin{equation*}
 E=\frac{\hbar^2(k_n^2+k_m^2)}{4m}\pm\sqrt{\frac{\hbar^4}{16m^2}(k_n^2-k_m^2)^2+|V_{n-m}|^2}.
-\end{equation}
+\end{equation*}
 
 This clearly displays that only if $|k_n|\approx |k_m|$, the band structure will be affected (given that the potential is weak, and therefore small). This nicely demonstrates how an avoided crossing arises.
 
@@ -155,10 +155,10 @@ See the lecture notes.
 ### Question 3.
 We split the Hamiltonian into two parts $H=H_n+H_{\overline{n}}$, where $H_n$ describes a particle in a single delta-function potential well, and $H_\hat{n}$ is the perturbation by the other delta functions:
 
-\begin{align}
+\begin{align*}
 H_n = & \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} - \lambda\delta(x-na) \\
 H_\overline{n} = & - \lambda \sum_{m\neq n}\delta(x-ma)
-\end{align}
+\end{align*}
 
 such that $H_n|n\rangle = \epsilon_0|n\rangle = -\hbar^2\kappa^2/2m |n\rangle$ with $\kappa=m\lambda/\hbar^2$. We can now calculate 
 
@@ -188,10 +188,10 @@ $$
 
 and
 
-\begin{align}
+\begin{align*}
 \langle n-1|H_\overline{n}|n\rangle = & -\kappa \lambda \sum_{m \neq 0 }\int e^{-\kappa|x-a|} \delta(x-ma)  e^{-\kappa|x|} \\
 =& -\kappa \lambda \sum_{m \neq 0 } e^{-\kappa a|m-1|} e^{-\kappa a |m|} =-\kappa \lambda(e^{ka}+e^{-ka}) \sum_{m=1}^{m=\infty} e^{-2\kappa a m}
-\end{align}
+\end{align*}
 
 In the limit $\kappa a \gg 1$ (i.e., where the distance between the delta functions is large compared to the width of the isolated orbitals), the onsite energy becomes
 

docs/12_band_structures_in_higher_dimensions.md

 ```python tags=["initialize"]
 import numpy as np
 import plotly.graph_objs as go
-
-pi = np.pi
+from matplotlib import pyplot as plt 
+from math import pi
 ```
 
 # Tight binding and nearly free electrons
 _(based on chapter 16 of the book)_  
 
+
+!!! success "Expected prior knowledge"
+
+    Before the start of this lecture, you should be able to:
+
+    - Apply perturbation theory to understand the band structure at crossings
+    - Derive the dispersion relation for a tight-binding model
+    - Recall the notion of the Fermi energy
+    - Calculate the Fermi surface, i.e. through the Fermi wavevector $k_F$, for two and three-dimensional systems
+    - Construct the unit cell of a crystal
+
 !!! summary "Learning goals"
 
     After this lecture you will be able to:
@@ -16,19 +27,26 @@ _(based on chapter 16 of the book)_
     - describe how the light absorption spectrum of a material relates to its band structure.
 
 ## Band structure
-
-How are material properties related to the band structure?
+In nature we see that some materials conduct electricity (conductors), some don't (insulators), and some do under specific circumstances (semiconductors).
+The motion of electrons is described by the band structure (dispersion relation).
+So far we have derived the dispersion relation for different type of models, such as the tight-binding model in which the electrons are bound by a strong potential, or when the electrons are perturbed by a small periodic potential as was the case in the NFEM.
+We have not yet discussed how the band structure does affect the electrical properties of a material.
 
 For a material to be a conductor, there should be available electron states at the Fermi level. Otherwise all the states are occupied, and all the currents cancel out.
 
-A band structure of a 1D material may look similar to this:
+Suppose we have a band structure of a 1D material similar to one below:
 
 ![](figures/band_structure_sketch.svg)
 
-We see several **energy bands** that may be separated by a **band gap** or overlapping.
+We see several energy bands that may either be separated by a **band gap** or overlap.
 
 When the Fermi level lies in the band gap, the material is called a semiconductor (or dielectric or insulator). When the Fermi level lies within a band, it is a conductor (metal).
 
+??? Question "Suppose the Fermi energy lies in the band gap. What feature about the band gap determines if a material is insulating or a semiconductor?"
+
+    Its size
+
+
 ### A simple requirement for insulators
 
 In an insulator every single band is either completely filled or completely empty.
@@ -47,18 +65,25 @@ We come to the important rule:
 
 > Any material with an odd number of electrons per unit cell is a metal.
 
-If the material has an even number of electrons per unit cell it may be a semiconductor, but only if the bands are not overlapping (see the figure above). For example: Si, Ge, Sn all have 4 valence electrons. Si (silicon, band gap 1.14 eV) and Ge (germanium, band gap 0.67 eV) are semiconductors, Sn (tin) is a metal. **Interesting feature: the heaviest material is a metal, why?**
+If the material has an even number of electrons per unit cell it may be a semiconductor, but only if the bands are not overlapping (see the figure above), or it could be a metal. To illustrate this, let us consider the three elements Si (silicon), Ge (germanium) and Sn (tin). All three of these have 4 valence electrons. Both Si (band gap 1.14 eV) and Ge (band gap 0.67 eV) are semiconductors, while Sn (tin) is a metal.
+
+??? Question "Why is Sn not a semiconductor but a metal?"
+
+    It has overlapping energy bands
 
 
 ## Fermi surface using a nearly free electron model
 
-Sequence of steps (same procedure as in 1D, but harder because of the need to imagine a 2D dispersion relation):
+In the end, we are also interested in studying the electrical properties, and thus the Fermi surface, in higher dimensions as well. 
+One could either do this through a tight-binding model approach (which is very similar to the 1D case), or through the NFEM (same procedure as in 1D, but harder because of the need to imagine a higher dispersion relation). 
+Let us illustarte the procedure in 2D.
+To derive the corresponding Fermi surface using the NFEM we use the sequence of steps:
 
 1. Compute $k_f$ using the free electron model (remember this is our starting point).
 2. Plot the free electron model Fermi surface and the Brillouin zones.
 3. Apply the perturbation where the Fermi surface crosses the Brillouin zone (due to avoided level crossings).
 
-The resulting band structure looks like this (in the extended Brillouin zone scheme):
+By doing so, the resulting band structure looks like the following (in the extended Brillouin zone scheme):
 
 
 ```python
@@ -129,12 +154,12 @@ go.Figure(
 )
 ```
 
-Observe that the top of the first band is above the bottom of the lowest band. Therefore if $V$ is sufficiently weak, the material can be conducting even with 2 electrons per unit cell!
+Observe that the top of the first band is above the bottom of the second band. Therefore if $V$ is sufficiently weak, the material can be conducting even with 2 electrons per unit cell!
 
 A larger $V$ makes the Fermi surface more distorted and eventually makes the material insulating.
 Let's compare the almost parabolic dispersion of the nearly free electron model with a tight-binding model in 2D.
 
-We now have a dispersion relation $E = E_0 - 2t(\cos k_x a + \cos k_y a)$, which looks like this:
+For a tight-binding model we obtain the dispersion $E = E_0 - 2t(\cos k_x a + \cos k_y a)$:
 
 ```python
 momenta = np.linspace(-pi, pi, 100)
@@ -172,16 +197,26 @@ go.Figure(
 ```
 
 ### Light absorption
+So far we have only discussed the effect of the Fermi energy and the band structure on the electrical properties of a material. 
+We know from nature that photons of external light sources can be reflected, transmitted, or absorbed by a material. Absorption, requires energy to transfer from the photon to electrons. In a filled band there are no available states where energy could be transferred (that's why insulators may be transparent).
 
-Photons of external light can be reflected, transmitted, or absorbed by the material. Absorption, in turn requires energy transfer from the photon to electrons. In a filled band there are no available states where energy could be transferred (that's why insulators may be transparent).
-
-When transition between two bands becomes possible due to photons having high energy, the absorption increases in a step-like fashion, see the sketch below for germanium.
+When transition between two bands becomes possible due to photons having a high energy, the absorption increases in a step-like fashion, see the sketch below for germanium.
 
+<figure markdown>
 ![](figures/adsorption.svg)
+  <figcaption> </figcaption>
+</figure>
+
+
 
 Here $E'_G\approx 0.9eV$ and $E_G\approx 0.8 eV$. The two visible steps are due to the special band structure of Ge:
 
+<figure markdown>
 ![](figures/direct_indirect.svg)
+  <figcaption> </figcaption>
+</figure>
+
+
 
 This band structure has two band gaps: the *direct* band gap $E'_G$ at $k=0$, and *indirect* band gap $E_G$ at any $k$. In Ge $E_G < E'_G$, and therefore it is an *indirect band gap semiconductor*. Silicon also has an indirect band gap. Direct band gap materials are for example GaAs and InAs.
 
@@ -197,48 +232,64 @@ A joint absorbtion of a photon and a phonon collision may excite an electron acr
 ## Exercises
 
 ### Exercise 1: 3D Fermi surfaces
-Using the [periodic table](../fermi_surfaces) of the Fermi surfaces, answer the following questions:
+Using the [periodic table](fermi_surfaces.md) of the Fermi surfaces, answer the following questions:
 
-#### Question 1.
-Find 4 elements that are well described by the nearly-free electron model and 4 that are poorly described by it.
-#### Question 2.
-Is the Fermi surface of lithium or potassium better described by the free electron model? What about nearly-free electron model? Why?
-#### Question 3.
-Do you expect a crystal with a simple cubic lattice and monovalent atoms to be conducting?
-#### Question 4.
-What Fermi surface shape would you expect the [NaCl crystal](https://en.wikipedia.org/wiki/Sodium_chloride) to have? Explain your answer using both the atomic valences and the optical properties of this crystal.
+1. Find 4 elements that are well described by the nearly-free electron model and 4 that are poorly described by it.
+2. Is the Fermi surface of lithium better described by the free electron model or the NFEM? What about potassium? Why?
+3. Do you expect a crystal with a simple cubic lattice and monovalent atoms to be conducting?
+4. Na has 1 valence electron and Cl has 7. What Fermi surface shape would you expect the [NaCl crystal](https://en.wikipedia.org/wiki/Sodium_chloride) to have? Explain your answer using **a.** the atomic valences, and **b.** the optical properties of this crystal.
 
 ### Exercise 2: Tight-binding in 2D
 
 Consider a rectangular lattice with lattice constants $a_x$ and $a_y$.
 Suppose the hopping parameters in the two corresponding directions to be $-t_1$ and $-t_2$.
-Consider a single orbital per atom and only nearest-neighbour interactions.
-
-#### Question 1.
-Write down a 2D tight-binding Schrödinger equation (expand to 2D the results of 1D).
-#### Question 2.
-Formulate the Bloch ansatz for the wave function.
-#### Question 3.
-Calculate the dispersion relation of this model.
-#### Question 4.
-What Fermi surface shape would this model have if the atoms are monovalent?
-#### Question 5.
-What Fermi surface shape would it have if the number of electrons per atom is much smaller than 1?
+We limit ourselves to a single orbital per atom and only nearest-neighbour interactions.
+
+1. Write down a 2D tight-binding Schrödinger equation (use your result of 1D).
+2. Formulate the Bloch ansatz for the wave function.
+3. Calculate the dispersion relation of this model.
+4. The dispersion is plotted below. 
+
+    What Fermi surface shape would this model have if the atoms are monovalent?
+
+5. What Fermi surface shape would it have if the number of electrons per atom is much smaller than 1?
+
+
+```python
+    def dispersion2D(N=100, kmax=pi, e0=2):
+        
+        # Define matrices with wavevector values
+        kx = np.tile(np.linspace(-kmax, kmax, N),(N,1))
+        ky = np.transpose(kx)
+
+        # Plot dispersion
+        plt.figure(figsize=(6,5))
+        plt.contourf(kx, ky, e0-np.cos(kx)-np.cos(ky))
+
+        # Making things look ok
+        cbar = plt.colorbar(ticks=[])
+        cbar.set_label('$E$', fontsize=20, rotation=0, labelpad=15)
+        plt.xlabel('$k_x$', fontsize=20)
+        plt.ylabel('$k_y$', fontsize=20)
+        plt.xticks((-pi, 0 , pi),('$-\pi/a$','$0$','$\pi/a$'), fontsize=17)
+        plt.yticks((-pi, 0 , pi),('$-\pi/a$','$0$','$\pi/a$'), fontsize=17)
+
+    dispersion2D()
+```
 
 ### Exercise 3: Nearly-free electron model in 2D
 _(based on exercise 15.4 of the book)_
 
 Suppose we have a square lattice with lattice constant $a$, with a periodic potential given by $V(x,y)=2V_{10}(\cos(2\pi x/a)+\cos(2\pi y/a))+4V_{11}\cos(2 \pi x/a)\cos(2 \pi y/a)$.
 
-#### Question 1.
-Use the nearly-free electron model to find the energy of state $\mathbf{q}=(\pi/a, 0)$.
+1. Use the nearly-free electron model to find the energy of state $\mathbf{q}=(\pi/a, 0)$.
+
+    ??? hint
+        This is analogous to the 1D case: the states that interact have $k$-vectors $(\pi/a,0)$ and $(-\pi/a,0)$; ($\psi_{+}\sim e^{i\pi x /a}$ ; $\psi_{-}\sim e^{-i\pi x /a}$).
 
-??? hint
-    This is analogous to the 1D case: the states that interact have $k$-vectors $(\pi/a,0)$ and $(-\pi/a,0)$; ($\psi_{+}\sim e^{i\pi x /a}$ ; $\psi_{-}\sim e^{-i\pi x /a}$).
+2. Let's now study the more complicated case of  state $\mathbf{q}=(\pi/a,\pi/a)$. How many $k$-points have the same energy? Which ones?
+3. Write down the nearly free electron model Hamiltonian near this point.
+4. Find its eigenvalues.
 
-#### Question 2.
-Let's now study the more complicated case of  state $\mathbf{q}=(\pi/a,\pi/a)$. How many $k$-points have the same energy? Which ones?
-#### Question 3.
-Write down the nearly free electron model Hamiltonian near this point.
-#### Question 4.
-Find its eigenvalues.
+    ??? hint
+        Make use of the symmetries of the matrix Hamiltonian

docs/12_band_structures_in_higher_dimensions_solutions.md

@@ -86,12 +86,15 @@ Four in total: $(\pm\pi/a,\pm\pi/a)$.
 ### Question 3.
 
 Define a basis, e.g.
-\begin{align}
-    \left|0\right\rangle &= (\pi/a,\pi/a) \\
-    \left|1\right\rangle &= (\pi/a,-\pi/a) \\
-    \left|2\right\rangle &= (-\pi/a,-\pi/a) \\
-    \left|3\right\rangle &= (-\pi/a,\pi/a)
-\end{align}
+\begin{equation*}
+    \begin{align*}
+        \left|0\right\rangle &= (\pi/a,\pi/a) \\
+        \left|1\right\rangle &= (\pi/a,-\pi/a) \\
+        \left|2\right\rangle &= (-\pi/a,-\pi/a) \\
+        \left|3\right\rangle &= (-\pi/a,\pi/a)
+    \end{align*}
+\end{equation*}
+
 The Hamiltonian becomes
 
 $$

docs/13_semiconductors.md

@@ -53,11 +53,11 @@ In this lecture, we shall grasp the basics of semiconductors by learning how to
 
 Before proceeding further, let us remind ourselves of important band structure properties.
 
-* Group velocity $v=\hbar^{-1}\partial E(k)/\partial k$. 
+* Group velocity $v=\hbar^{-1}dE(k)/dk$. 
 Descibes how quickly electrons move within the lattice. 
 * Effective mass $m^* = \hbar^2\left(d^2 E(k)/dk^2\right)^{-1}$. 
 Tells us how hard it is to *accelerate* the particles and is related to the curvature of the band.
-* Density of states $g(E) = \sum_{\textrm{FS}} (dn/dk) \times (dk/dE)$. 
+* Density of states $g(E) = \sum (dn/dk) \times (dk/dE)$, where the sum runs over all states at the same energy. 
 The amount of states per infinitesimal interval of energy at given energy. 
 The quantity is vital in order to calculate any bulk property of the material such as conductivity, heat capacity, etc.
 
@@ -78,10 +78,10 @@ We distinguish three different band filling types: filled, empty and partially f
 Despite being two opposite extreme extreme cases, filled and empty bands are very similar.
 For example, both filled and empty bands carry no electric current:
 
-\begin{align}
+\begin{align*}
 j = 2e \frac{1}{2\pi} \int_{-\pi/a}^{\pi/a} v(k) dk = 2e \frac{1}{2\pi \hbar} \int_{-\pi/a}^{\pi/a} \frac{dE}{dk} \times dk = \\
 2e \frac{1}{2\pi \hbar} [E(\pi/a) - E(-\pi/a)] = 0.
-\end{align}
+\end{align*}
 
 On the other hand, a filled band has an equal number of electrons going forwards and backwards which thus cancel and lead to zero current. 
 Similar results apply to many other physical quantities such as heat capacity and magnetization. 
@@ -102,7 +102,11 @@ The same approach applies to band filling.
 Instead of describing a lot of electrons that are present in an almost filled band, we focus on those that are absent.
 The absence of an electron is called a **hole**: a state of a completely filled band with one particle missing.
 
+<figure markdown>
 ![](figures/holes.svg)
+  <figcaption> </figcaption>
+</figure>
+
 
 In this schematic we can either say that 8×2 electron states are occupied (the system has 8×2 electrons counting spin), or 10×2 hole states are occupied.
 Electron and hole pictures correspond to two different, but equivalent ways of describing the occupation of a band.
@@ -126,8 +130,8 @@ $$
 
 which is the Fermi distribution of particles with energy $E_h= -E$ and $E_{F,h} = -E_F$.
 
-The **momentum** $p_h$ of a hole should give the correct total momentum of a partially filled band if one sums momenta of all holes.
-Therefore $p_h = -\hbar k$, where $k$ is the wave vector of the electron.
+The **momentum** $p_h$ of a hole should give the correct total momentum of a partially filled band if one sums momenta of all electrons.
+Therefore $\mathbf{p}_h = -\hbar \mathbf{k}$, where $\mathbf{k}$ is the wavevector of the missing electron.
 
 Similarly, the total **charge** should be the same regardless of whether we count electrons or holes, so holes have a positive charge $+e$ (electrons having $-e$).
 
@@ -177,26 +181,27 @@ pyplot.xlabel('$k$')
 pyplot.ylabel('$E$')
 pyplot.ylim(-1.5, 3)
 pyplot.yticks([E_C_1 - .15, E_V_1 +.15], "$E_C$ $E_V$".split())
+pyplot.xticks([])
 pyplot.legend(lines, ['Conduction band', 'Valence band'])
 draw_classic_axes(ax)
 ```
 
 Or in other words
 
-\begin{align}
+\begin{align*}
 E_e &= E_c + \frac{\hbar^2k^2}{2m_e},\\
 E_h &= E_{v,h} + \frac{\hbar^2k^2}{2m_h} = -E_{v} + \frac{\hbar^2k^2}{2m_h}.
-\end{align}
+\end{align*}
 
 Here $E_c$ is the energy of an electron at the bottom of the conduction band and $E_v$ is the energy of an electron at the top of the valence band.
 Observe that because we are describing particles in the valence band as holes, $m_h > 0$ and $E_h > -E_v$.
 
 The corresponding density of states of the two types of particles is
 
-\begin{align}
+\begin{align*}
 g(E) &= (2m_e)^{3/2}\sqrt{E-E_c}/2\pi^2\hbar^3,\\
 g_h(E_h) &= (2m_h)^{3/2}\sqrt{E_h+E_v}/2\pi^2\hbar^3.
-\end{align}
+\end{align*}
 
 ??? question "A photon gives a single electron enough energy to move from the valence band to the conduction band. How many particles does this process create?"
     Two: one electron and one hole.
@@ -226,6 +231,7 @@ ax.set_xlabel('$E$')
 ax.set_ylabel('$g$')
 ax.set_xticks([E_V, E_C, E_F])
 ax.set_xticklabels(['$E_V$', '$E_C$', '$E_F$'])
+ax.set_yticks([])
 ax.legend()
 draw_classic_axes(ax, xlabeloffset=.2)
 ```
@@ -233,20 +239,21 @@ draw_classic_axes(ax, xlabeloffset=.2)
 We know that by itself, the semiconductor should have no charge, and therefore the total numbers of electrons and holes must be equal.
 Since increasing the Fermi level increases the number of electrons and reduces the number of holes, we will use the charge neutrality condition to determine where the Fermi level is situated.
 
-**The key algorithm of describing the state of a semiconductor:**
+!!! tip "The key algorithm of describing the state of a semiconductor"
 
-1. Compute the density of states of all types of particles.
-2. Calculate the number of electrons in the conduction band and holes in the valence band, assuming a certain value of $E_F$
-3. Write down the charge balance condition: the difference between electrons and holes should equal the total charge of the semiconductor.
-4. Apply approximations to simplify the equations (this is important!).
-5. Find $E_F$ and concentrations of electrons and holes
+    1. Compute the density of states of all types of particles.
+    2. Calculate the number of electrons in the conduction band and holes in the valence band, assuming a certain value of $E_F$
+    3. Write down the charge balance condition: the difference between electrons and holes should equal the total charge of the semiconductor.
+    4. Apply approximations to simplify the equations (this is important!).
+    5. Find $E_F$ and concentrations of electrons and holes
+    
 
 Applying the first two steps of the algorithm:
 
-\begin{gather}
+\begin{gather*}
 n_h = \int_{-E_v}^\infty f_h(E_h) g_h(E_h) dE_h = \int_{-E_v}^\infty\frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}\sqrt{E_h+E_v}\frac{1}{e^{(E_h+E_F)/k_BT}+1}dE_h,\\
 n_e = \int_{E_c}^\infty f(E)g_e(E)dE = \int_{E_c}^\infty\frac{(2m_e)^{3/2}}{2\pi^2\hbar^3} \sqrt{E-E_c}\frac{1}{e^{(E-E_F)/k_BT}+1}dE.
-\end{gather}
+\end{gather*}
 
 Note that whenever calculating the hole dependent quantities, we replace all the relevant physical quantities with their hole equivalents.
 Since the hole energy is opposite $E_h = -E$, we replace the Fermi energy $E_F \to -E_F$ and the bottom of the valence band by $E_v \to -E_v$ in the integration limits.
@@ -257,7 +264,7 @@ Therefore, the fourth step assumes that the Fermi level is far from both bands $
 As a result, the Fermi-Dirac distribution is approximately similar to Boltzmann distribution:
 
 $$
-f(E)_{e/h} \approx \exp\left[-(E_{e/h}\pm E_F)/k_BT\right].
+f(E)_{e/h} \approx \exp\left[(-E_{e/h}\pm E_F)/k_BT\right].
 $$
 
 Now we can move to the last step and calculate $n_e$ and $n_h$:
@@ -300,11 +307,13 @@ An extra observation: regardless of where $E_F$ is located, $n_e n_h = N_C N_V e
 
 $n_i$ is the **intrinsic carrier concentration**, and for a pristine semiconductor $n_e = n_h = n_i$.
 
-> The equation
->
-> $$n_e n_h = n_i^2$$
->
-> is the **law of mass action**. The name is borrowed from chemistry, and describes the equilibrium concentration of two reagents in a reaction $A+B \leftrightarrow AB$. Here electrons and hole constantly split and recombine.
+!!! note "The law of mass action"
+
+    The equation
+
+    $$n_e n_h = n_i^2$$
+
+    is the **law of mass action**. The name is borrowed from chemistry, and describes the equilibrium concentration of two reagents in a reaction $A+B \leftrightarrow AB$. Here electrons and hole constantly split and recombine.
 
 ### Conduction
 
@@ -312,10 +321,10 @@ Earlier, we deduced that empty and filled bands provide no current.
 We finish the analysis by considering partially filled bands of an intrinsic (pristine) semiconductor.
 To calculate the current, we utilize the Drude model and sum the electron and hole contributions:
 
-\begin{gather}
+\begin{gather*}
 j = -n_e e v_e + n_h e v_h \\
  -m_e v_e /\tau_e = -eE;\quad -m_h v_h /\tau_h = eE.
-\end{gather}
+\end{gather*}
 
 We see that despite opposite velocity signs for electrons and holes, they carry electric current in the same direction:
 
@@ -329,44 +338,26 @@ This, however only means that the electron current is opposite of the hole curre
 
 ## Exercises
 
-### Exercise 1: Energy, mass, velocity and cyclotron motion of electrons and holes
-
-#### Question 1.
-Consider the top of the valence band of a semiconductor (see [above](#semiconductors-materials-with-two-bands)). Does an electron near the top of the valence band have a positive or a negative effective mass? Does the electron's energy increase or decrease as $k$ increases from 0? Does the electron have a positive or negative group velocity for $k>0$?
-#### Question 2.
-Answer the same last 3 questions for a hole in the valence band.
-#### Question 3.
-We now consider an electron in a 2D semiconductor near the bottom of the conduction band described by an energy dispersion $E=E_{G}+\frac{\hbar^2}{2m^*}(k_x^2+k_y^2)$. The electron's velocity is given by $\mathbf{v}=\nabla_\mathbf{k} E/\hbar = \frac{1}{\hbar}(\frac{\partial E}{\partial k_x}\mathbf{\hat{x}} + \frac{\partial E}{\partial k_y}\mathbf{\hat{y}})$. Suppose we turn on a magnetic field $B$ in the $z$-direction. Write down the equation of motion for this electron (neglecting collisions). What is the shape of the motion of the electron? What is the characteristic 'cyclotron' frequency of this motion? What is the direction of the Lorentz force with respect to $\nabla_\mathbf{k} E$? 
-#### Question 4.
-Suppose we now consider a hole near the bottom of the conduction band and turn on a magnetic field $B$ in the $z$-direction. Is the direction of the circular motion (i.e., the chirality) of the hole the same as that of the electron? Would the chirality change if we instead consider a hole (or electron) near the top of the valence band?
+Exercise with an asterisk (*) are considered to be at the basic/essential level
 
-### Exercise 2: The Hall coefficient when both electrons and holes are present in the system
+### Warm-up exercises*
 
-#### Question 1.
-Recall from the [Drude model](/3_drude_model/) that electrons give rise to a negative Hall coefficient. Explain why the Hall coefficient is positive if holes are the charge carriers in a material.
-#### Question 2.
-Recall that the Hall coefficient $R_H$ describes the change of the transverse resistivity $\rho_{xy}$ due to a change in $B$: $\rho_{xy} = \frac{E_y}{J_x}=R_H B$. When both electrons and holes are present in a system, the Hall coefficient is given by
+In this set of warm-up exercises we consider a two-dimensional semiconductor.
 
-$$
-R_H = \frac{n_h \mu_{h}^{2}-n_e \mu_{e}^{2}}{e\left(n_h \mu_{h}+n_e \mu_{e}\right)^{2}}.
-$$
+1. Write down an integral expression for the number of electrons in the conduction band. How does the dimensionality come in?
+2. Write down an integral expression for the number of holes in the valence band.
+3. Make a sketch of the density of states (=band structure) of this semiconductor. 
+4. Solve the integrals of questions 1 and 2 assuming the Fermi energy is in the gap and far away from the band edges compared to $k_bT$.
+5. Find the Fermi energy
 
-What would be the Hall coefficient if the electrons and holes have equal concentrations and equal mobilities?
-
-### Exercise 3. Holes and electrons in a 1D tight-binding energy band
-Recall that the dispersion relation of a [1D single orbital tight binding chain](/7_tight_binding/) is given by
+### Exercise 1: Energy, mass, velocity and cyclotron motion of electrons and holes
 
-$$E(k)=\varepsilon + 2t \cos(ka),$$
+1. Consider the top of the valence band of a semiconductor (see [above](#semiconductors-materials-with-two-bands)). Does an electron near the top of the valence band have a positive or a negative effective mass? Does the electron's energy increase or decrease as $k$ increases from 0? Does the electron have a positive or negative group velocity for $k>0$?
+2. Answer the same last 3 questions for a hole in the valence band.
+3. We now consider an electron in a 2D semiconductor near the bottom of the conduction band described by an energy dispersion $E=E_{G}+\frac{\hbar^2}{2m^*}(k_x^2+k_y^2)$. The electron's velocity is given by $\mathbf{v}=\nabla_\mathbf{k} E/\hbar = \frac{1}{\hbar}(\frac{\partial E}{\partial k_x}\mathbf{\hat{x}} + \frac{\partial E}{\partial k_y}\mathbf{\hat{y}})$. Suppose we turn on a magnetic field $B$ in the $z$-direction. Write down the equation of motion for this electron (neglecting collisions). What is the shape of the motion of the electron? What is the characteristic 'cyclotron' frequency of this motion? What is the direction of the Lorentz force with respect to $\nabla_\mathbf{k} E$? 
+4. Suppose we now consider a hole near the bottom of the conduction band and turn on a magnetic field $B$ in the $z$-direction. Is the direction of the circular motion (i.e., the chirality) of the hole the same as that of the electron? Would the chirality change if we instead consider a hole (or electron) near the top of the valence band?
 
-where $a$ is the lattice constant and $\varepsilon$ and $t$ are tight binding parameters.
-#### Question 1.
-What is the group velocity and effective mass of this band for holes compared to that of electrons?
-#### Question 2.
-Give an integral expression of the hole concentration in this band given the chemical potential $\mu$ and temperature $T$.
-#### Question 3.
-Show that the sum of the electron and hole concentration in this band is constant as a function of the temperature.
-
-### Exercise 4: a 1D semiconductor
+### Exercise 2*: a 1D semiconductor
 Suppose we have a 1D semiconductor with a conduction band described by
 
 $$E_{cb} = E_G - 2 t_{cb} [\cos(ka)-1],$$
@@ -377,16 +368,31 @@ $$E_{vb} = 2 t_{vb} [\cos(ka)-1].$$
 
 Furthermore, the chemical potential is set at $0 < \mu < E_G$.
 
-#### Question 1.
-Derive an expression for the group velocity and effective mass for electrons in the conduction bands and holes in the valence band.
+1. Make a sketch of the band structure.
+2. Derive an expression for the group velocity and effective mass for electrons in the conduction bands and holes in the valence band.
+3. Assume that the Fermi level is far away from both bands. That is, $|E_G - \mu| \gg k_B T$ and $\mu\gg k_B T$. In that case, it is acceptable to approximate the bands for low $k$. Why is it acceptable? Write down an approximate expression of these bands.
+4. Write down an expression for the density of states _per unit length_ for both bands using the approximated expressions. Compare with the actual density of states per unit length. 
+5. Calculate the electron density in the conduction band and the hole density in the valence band.
+6. What would the chemical potential $\mu$ be in case of an intrinsic semiconductor?
+
+### Exercise 3*. Holes and electrons in a 1D tight-binding energy band
+Recall that the dispersion relation of a [1D single orbital tight binding chain](7_tight_binding.md) is given by
+
+$$E(k)=\varepsilon + 2t \cos(ka),$$
+
+where $a$ is the lattice constant and $\varepsilon$ and $t$ are tight binding parameters.
+
+1. What is the group velocity and effective mass of this band for holes compared to that of electrons?
+2. Give an integral expression of the hole concentration in this band given the chemical potential $\mu$ and temperature $T$.
+3. Show that the sum of the electron and hole concentration in this band is constant as a function of the temperature.
+
+### Exercise 4*: The Hall coefficient when both electrons and holes are present in the system
+
+1. Recall from the [Drude model](3_drude_model.md) that electrons give rise to a negative Hall coefficient. Explain why the Hall coefficient is positive if holes are the charge carriers in a material.
+2. Recall that the Hall coefficient $R_H$ describes the change of the transverse resistivity $\rho_{xy}$ due to a change in $B$: $\rho_{xy} = \frac{E_y}{J_x}=R_H B$. When both electrons and holes are present in a system, the Hall coefficient is given by
 
-Assume that the Fermi level is far away from both bands. That is, $|E - \mu| \gg k_B T$. In that case, it is acceptable to approximate the bands for low $k$.
+    $$
+    R_H = \frac{n_h \mu_{h}^{2}-n_e \mu_{e}^{2}}{e\left(n_h \mu_{h}+n_e \mu_{e}\right)^{2}}.
+    $$
 
-#### Question 2.
-Why is it acceptable? Write down an approximate expression of these bands.
-#### Question 3.
-Write down an expression for the density of states _per unit length_ for both bands using the approximated expressions. Compare with the actual density of states per unit length. 
-#### Question 4.
-Calculate the electron density in the conduction band and the hole density in the valence band.
-#### Question 5.
-What would the chemical potential $\mu$ be in case of an intrinsic semiconductor?
+    What would be the Hall coefficient if the electrons and holes have equal concentrations and equal mobilities?

docs/13_semiconductors_solutions.md

@@ -9,98 +9,158 @@ from math import pi
 ```
 # Solutions for lecture 13 exercises
 
-## Exercise 1: Energy, mass, velocity and cyclotron motion of electrons and holes
+## Warm-up exercises*
 
-#### Question 1.
+In this set of warm-up exercises we consider a two-dimensional semiconductor.
 
-Electrons near the top of the valence band have a negative effective mass, their energy decreases as $k$ increases from 0, and they have a negative group velocity for $k>0$.
-
-#### Question 2.
+#### Question 1.
 
-Holes near the top of the valence band have a positive effective mass, their energy increases as $k$ increases from 0, and they have a negative group velocity for $k>0$.
+The dimensionality of the system enters the integral implicitly with the density of states, which in 2D and expressed per unit volume is a constant:
 
-#### Question 3.
+$$
+g_{2 D}(E)=\frac{2_s}{2 \pi} \frac{m}{\hbar^2}
+$$
 
-The equation of motion for an electron near the bottom of the conduction band is:
 
 $$
-m^* \frac{d\mathbf{v}}{dt} = -e\mathbf{v} \times \mathbf{B}
+n_e = \int_{E_c}^\infty f(E)g_e(E)dE = \int_{E_c}^\infty\frac{m_e}{\pi\hbar^2}\frac{1}{e^{(E-E_F)/k_BT}+1}dE.
 $$
 
-and when replacing we get two coupled equations:
-
-\begin{align}
-\dot{k_x} &= -\frac{e}{m^*}B k_y\\
-\dot{k_y} &= +\frac{e}{m^*}B k_x
-\end{align}
+#### Question 2. 
 
-The solution to this equation is circular motion of cyclotron frequency of $\omega_c = \frac{eB}{m^*}$, where the Lorentz
-force is perpendicular to $\nabla_\mathbf{k} E$.
+Following the derivation in the lecture notes, whenever calculating the hole-dependent quantities, replace all relevant physical quantities with their hole equivalents.
+    
+1. For example, state the integral in terms of the electron energy scale using $E \equiv E_e $ and $E_F \equiv E_{F,e}$ with $f_n(E) = 1 -f(E)$:
+    $$
+    n_h = \int_{-\infty}^{E_v} f_h(E)g_h(E)dE  = \int_{-\infty}^{E_v}\frac{m_h}{\pi\hbar^2}\left(\frac{1}{e^{-(E-E_F)/k_BT}+1}\right)dE
+    $$    
+2. Next, using $E_h = -E$, change the integration element and the corresponding integration limits, and the energy variable:
+    $$
+    n_h = \int_{-E_v}^\infty\frac{m_h}{\pi\hbar^2}\frac{1}{e^{(E_h+E_F)/k_BT}+1}dE_h,
+    $$
+    
+    remembering that $E_{F,h} = - E_F$.
 
-#### Question 4.
+#### Question 3.
 
-A hole near the bottom of the conduction band will have the same chirality as an electron.
-The chirality would be just the opposite if we would consider the valence band (for both electrons and holes).
+The sketch of the density of states (=band structure) for a 2D semiconductor: 
 
+```python
+from matplotlib import pyplot
 
+import numpy as np
+from scipy.optimize import curve_fit
+from scipy.integrate import quad
+
+from common import draw_classic_axes, configure_plotting
+
+configure_plotting()
+
+# Band structure parameters.
+E_V, E_C, E_F = -1.2, 1.8, .4
+E_D, E_A = E_C - .7, E_V + .5
+m_h, m_e = 1, .5
+
+default_colors = pyplot.rcParams['axes.prop_cycle'].by_key()['color']
+blue, red = default_colors[0], default_colors[3]
+E = np.linspace(-3, 3, 1000)
+fig, ax = pyplot.subplots()
+
+n_F = 1/(np.exp(2*(E - E_F)) + 1)
+# define g_e and g_h as step functions at E_C and E_V, respectively:
+g_e = m_e * np.heaviside(E - E_C, 0)
+g_h = m_h * np.heaviside(E_V - E, 0)
+ax.plot(E, g_e, label="$g_e$", c=blue)
+ax.plot(E, g_h, label="$g_h$", c=red)
+ax.fill_between(E, 10 * g_e * n_F, 0, alpha=.7, label="$n_e$", color=blue)
+ax.fill_between(E, 10 * g_h * (1-n_F), 0, alpha=.7, label="$n_h$", color=red)
+ax.plot(E, n_F, label="$n_F$", linestyle='dashed', c='k')
+ax.axvline(E_F, color='gray', linestyle='dotted', linewidth=1)
+ax.set_ylim(bottom=0,top=1.4)
+
+ax.set_xlabel('$E$')
+ax.set_ylabel('$g$')
+ax.set_xticks([E_V, E_C, E_F])
+ax.set_xticklabels(['$E_V$', '$E_C$', '$E_F$'])
+ax.set_yticks([])
+ax.legend()
+draw_classic_axes(ax, xlabeloffset=.1)
+```
 
-### Exercise 2: The Hall coefficient when both electrons and holes are present in the system
+#### Question 4.
 
-#### Question 1.
+Assuming that the Fermi energy is in the gap $(E_v < E_F < E_c)$ and far away from the band edges compared to $k_BT$;
 
 $$
-R_{H,h} = \frac{1}{ne}
-$$
-
-#### Question 2.
+E_F-E_v \gg k_BT \text{ and } E_c - E_F \gg k_BT,
+$$ 
 
-Filling in $n_e = n_h$ and $\mu_e=\mu_h$ results in $R_H = 0$.
+the Fermi-Dirac distribution can be approximated as Boltzmann distribution:
 
-### Exercise 3. Holes and electrons in a 1D tight-binding energy band
+$$
+f(E)_{e/h} \approx \exp\left[-(E_{e/h}\pm E_F)/k_BT\right].
+$$
 
-#### Question 1.
-
-For electrons we find
+and the integrals for $n_h$ and $n_e$ simplify to:
 
 $$
-m_e = -\frac{\hbar^2}{2ta^2\cos(ka)}
+n_h \approx \int_{-E_v}^\infty\frac{m_h}{\pi\hbar^2}e^{-(E_h+E_F)/k_BT}dE_h = N_V e^{(E_v-E_F)/k_BT}
 $$
 
 $$
-v_e = -\frac{2ta\sin(ka)}{\hbar},
+n_e \approx \int_{E_c}^\infty\frac{m_e}{\pi\hbar^2}e^{-(E-E_F)/k_BT}dE = N_C e^{-(E_c-E_F)/k_BT}
 $$
 
-and for holes we obtained
+where
 
 $$
-m_h = -m_e
+N_{V,C}= \frac{m_{h,e} k_B T}{\pi\hbar^2}
 $$
-
+    
+#### Question 5.
+Under the assumption of the charge balance ($n_e = n_h$), the Fermi energy can be found by dividing the results of the previous question:
 $$
-v_h = v_e.
+1 = \frac{n_h}{n_e} = \left(\frac{N_V}{N_C}\right) e^{(E_v+E_c-2E_F)/k_BT} = \left(\frac{m_h}{m_e}\right) e^{(E_v+E_c-2E_F)/k_BT}, 
+$$
+and taking the logarithm of both sides of the equation:
+$$
+E_F = \frac{E_V+E_C}{2} + \frac{k_B T}{2}\ln{\left(\frac{m_h}{m_e}\right)}.
 $$
 
-Thus the effective masses of electrons an hole will be of opposite sign, while the group velocities will be the same!
+## Exercise 1: Energy, mass, velocity and cyclotron motion of electrons and holes
+
+#### Question 1.
+
+Electrons near the top of the valence band have a negative effective mass, their energy decreases as $k$ increases from 0, and they have a negative group velocity for $k>0$.
 
 #### Question 2.
 
-The number of holes is defined as
+Holes near the top of the valence band have a positive effective mass, their energy increases as $k$ increases from 0, and they have a negative group velocity for $k>0$.
 
-$$ 
-N_h = \int_{-\varepsilon-2t}^{-\varepsilon+2t} f(E_h, E_{F,h}) g_h(E_h)d E_h,
+#### Question 3.
+
+The equation of motion for an electron near the bottom of the conduction band is:
+
+$$
+m^* \frac{d\mathbf{v}}{dt} = -e\mathbf{v} \times \mathbf{B}
 $$
 
-where $g_h(E_h)$ is the density of states of the holes in terms of hole energy $E_h$.
+and when replacing we get two coupled equations:
 
-#### Question 3.
+\begin{align*}
+\dot{k_x} &= -\frac{e}{m^*}B k_y\\
+\dot{k_y} &= +\frac{e}{m^*}B k_x
+\end{align*}
 
-??? hint "Small hint"
+The solution to this equation is circular motion of cyclotron frequency of $\omega_c = \frac{eB}{m^*}$, where the Lorentz
+force is perpendicular to $\nabla_\mathbf{k} E$.
 
-    It is convenient to write the hole integral in terms of the electron energy. 
-    
+#### Question 4.
 
+A hole near the bottom of the conduction band will have the same chirality as an electron.
+The chirality would be just the opposite if we would consider the valence band (for both electrons and holes).
 
-## Exercise 4: a 1D semiconductor
+## Exercise 2*: a 1D semiconductor
 
 ```python
 def dispersion(EG, tcb, tvb, N=100, kmax=np.pi/2):
@@ -141,7 +201,7 @@ v_{vb,h} = -\frac{2at_{vb}}{\hbar}\sin (ka)
 $$
 
 $$
-m_{vb,h} =  -\frac{\hbar^2}{2a^2t_{vb}\cos (ka)}.
+m_{vb,h} =  \frac{\hbar^2}{2a^2t_{vb}\cos (ka)}.
 $$
 
 #### Question 2.
@@ -182,18 +242,18 @@ where $E_h = -E$.
 
 The electron density in the conduction band is given by
 
-\begin{align}
+\begin{align*}
 n_e &= \int_{E_G}^{\infty} f(E)g_c(E)dE\\
 &\approx \int_{E_G}^{\infty} e^{-\beta (E-\mu)} g_c(E) dE\\
 &=\sqrt{\frac{k_B T}{\pi t_{cb}}}\frac{e^{\beta (\mu - E_G)}}{a}.
-\end{align}
+\end{align*}
 
 Analoguous, we find for the hole density
 
-\begin{align}
+\begin{align*}
 n_h &= \int_{0}^{\infty} f(E_h) g_h(E_h) dE_h\\
 &\approx \sqrt{\frac{k_B T}{\pi t_{vb}}}\frac{e^{-\beta \mu}}{a}.
-\end{align}
+\end{align*}
 
 #### Question 5.
 
@@ -204,3 +264,60 @@ $$
 \mu = \frac{1}{2}E_G + \frac{k_B T}{4}ln(\frac{t_{cb}}{t_{vb}})
 $$
 
+### Exercise 3*. Holes and electrons in a 1D tight-binding energy band
+
+#### Question 1.
+
+For electrons we find
+
+$$
+m_e = -\frac{\hbar^2}{2ta^2\cos(ka)}
+$$
+
+$$
+v_e = -\frac{2ta\sin(ka)}{\hbar},
+$$
+
+and for holes we obtained
+
+$$
+m_h = -m_e
+$$
+
+$$
+v_h = v_e.
+$$
+
+Thus the effective masses of electrons an hole will be of opposite sign, while the group velocities will be the same!
+
+#### Question 2.
+
+The number of holes is defined as
+
+$$ 
+N_h = \int_{-\varepsilon-2t}^{-\varepsilon+2t} f(E_h, E_{F,h}) g_h(E_h)d E_h,
+$$
+
+where $g_h(E_h)$ is the density of states of the holes in terms of hole energy $E_h$.
+
+#### Question 3.
+
+??? hint "Small hint"
+
+    It is convenient to write the hole integral in terms of the electron energy. 
+    
+
+
+
+
+### Exercise 4*: The Hall coefficient when both electrons and holes are present in the system
+
+#### Question 1.
+
+$$
+R_{H,h} = \frac{1}{ne}
+$$
+
+#### Question 2.
+
+Filling in $n_e = n_h$ and $\mu_e=\mu_h$ results in $R_H = 0$.
\ No newline at end of file

docs/14_doping_and_devices_solutions.md

@@ -27,13 +27,13 @@ $$ n_e - n_h + n_D - n_A = N_D - N_A $$
 Since $E_G \gg k_B T$, we can only use the law of mass action. 
 But the question offers us another piece of information - we are around $|N_D-N_A| \approx n_i$.
 That means that we are near the transition between extrinsic and intrinsic regimes.
-Because the dopants stop being ionized at very low temperatures (see next exercise), we can neglect $n_D$ and $n_A$ in this exercise, just like in the lecture.
+In this regime we can neglect $n_D$ and $n_A$ in this exercise, just like in the lecture.
 Writing $n_e n_h = n_i^2$ and $n_e - n_h = N_D - N_A$ and solving these together, we obtain
 
-\begin{align} 
+\begin{align*} 
 n_{e} = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D),\\
 n_{h} = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)
-\end{align}
+\end{align*}
 
 where $D = N_D - N_A$.
 
@@ -105,43 +105,42 @@ $$ I_s(T) \propto e^{-E_{gap}/k_BT}$$
 ### Question 2.
 This a "particle in a box" problem.
 
-\begin{align}
+\begin{align*}
 -\frac{\hbar^2}{2m_e^{\ast}} \nabla^2 \Psi_e &= (E_e-E_c)\Psi_e\\
--\frac{\hbar^2}{2m_h^{\ast}} \nabla^2 \Psi_h &= (-E_h-E_v)\Psi_h 
-\end{align}
+-\frac{\hbar^2}{2m_h^{\ast}} \nabla^2 \Psi_h &= (E_h+E_v)\Psi_h 
+\end{align*}
 
 Here and below $E_h$ is the energy of a hole in the valence band.
 
 ### Question 3.
 
-\begin{align}
+\begin{align*}
 E_e = E_c + \frac{\hbar^2}{2m_e^{\ast}} ((\frac{\pi n}{L})^2+k_x^2+k_y^2),\\
--E_h = E_v - \frac{\hbar^2}{2m_h^{\ast}} ((\frac{\pi n}{L})^2+k_x^2+k_y^2)
-\end{align}
+E_h = -E_v + \frac{\hbar^2}{2m_h^{\ast}} ((\frac{\pi n}{L})^2+k_x^2+k_y^2)
+\end{align*}
 
 ### Question 4.
 This is a 2D electron/hole gas, therefore the DOS per unit area expression is the same as in 2D parabolic dispersion:
 
-\begin{align}
+\begin{align*}
 g_e = \frac{m_e^{\ast}}{\pi\hbar^2},\\
 g_h = \frac{m_h^{\ast}}{\pi\hbar^2}
-\end{align}
+\end{align*}
 
 ### Question 5.
-L can be found here using previous Qestions..
-Setting
+$L$ can be found here using previous questions, by setting:
 
 $$
-E_e - E_h - E_c + E_v = 1 eV = \frac{\hbar^2}{2}(\frac{\pi n}{L}^2+k_x^2+k_y^2)
+E_e + E_h - E_c + E_v = 1 eV = \frac{\hbar^2}{2}(\frac{\pi n}{L}^2+k_x^2+k_y^2)
 (\frac{1}{m_e^{\ast}}+\frac{1}{m_h^{\ast}})
 $$
 
-By choosing the correct $n$, $k_x$ and $k_y$, L can be found as $6.85$ nm approx.
+By choosing the correct $n$, $k_x$ and $k_y$, $L$ can be found as $\approx 6.85$ nm 
 
 ### Question 6.
 
-For a laser one wants to fix the emission wavelength to a certain value. With
-this setup the band gap is "easy" to design (set by L, which is fixed).
+For a laser, one wants to fix the emission wavelength to a certain value. With
+this setup, the band gap is "easy" to design (set by $L$, which is fixed).
 
 ### Question 7.
 If donor impurities are put outside of the well (on both sides, for example)

docs/1_einstein_model.md

@@ -21,7 +21,6 @@ from scipy.optimize import curve_fit
 from scipy.integrate import quad
 import plotly.offline as py
 import plotly.graph_objs as go
-from ipywidgets import interact, FloatSlider, Layout
 
 from common import draw_classic_axes, configure_plotting
 
@@ -252,10 +251,10 @@ $$
 
 Substituting $n_B$ into the expression for the oscillator energy, we obtain the expectation value of the energy stored in the oscillator at temperature $T$—the *thermal energy* (which, for brevity, we will simply denote as the total energy $E$):
 
-\begin{align}
+\begin{align*}
 E(T) &= (n_B(\omega_0,T) + \frac{1}{2})\hbar\omega_0 \\
 &= \frac{\hbar\omega_0}{e^{\hbar\omega_0/k_B T}-1} + \frac{1}{2}\hbar\omega_0
-\end{align}
+\end{align*}
 
 A plot of the Bose-Einstein distribution and the expected value of the energy are shown below.
 
@@ -297,12 +296,12 @@ Because the energy in the oscillator becomes approximately constant when $k_B T\
 
 Having found an expression for $E(T)$, we now calculate the heat capacity per harmonic oscillator explicitly by using its definition:
 
-\begin{align}
+\begin{align*}
 C(T) &\equiv \frac{d E(T) }{d T}\\
 &= -\frac{\hbar\omega_0}{\left(e^{\hbar\omega_0/k_B T}-1\right)^2}\frac{d}{d T}\left(e^{\hbar\omega_0/k_B T}-1\right)\\
 &= \frac{\hbar^2\omega_0^2}{k_B T^2}\frac{ e^{\hbar\omega_0/k_B T}}{\left(e^{\hbar\omega_0/k_B T}-1\right)^2}\\
 &=k_B \left(\frac{\hbar\omega_0}{k_B T}\right)^2\frac{ e^{\hbar\omega_0/k_B T}}{\left(e^{\hbar\omega_0/k_B T}-1\right)^2}
-\end{align}
+\end{align*}
 
 We can rewrite this equation into
 

docs/1_einstein_model_solutions.md

@@ -14,6 +14,8 @@ jupyter:
     name: python3
 ---
 
+# Solutions for lecture 1 exercises
+
 ```python tags=["initialize"]
 from matplotlib import pyplot as plt
 from mpl_toolkits.axes_grid1 import make_axes_locatable
@@ -27,8 +29,8 @@ from common import draw_classic_axes, configure_plotting
 configure_plotting()
 ```
 
-# Solutions for lecture 1 exercises
-### Warm-up exercises
+## Warm-up exercises
+
 1. By definition, the particles in an ideal gas do not interact. As such, their energy is independent of their position, and only depends on the momenta of the particles. This means that only the 3 momentum degrees of freedom can store energy and thereby contribute to the heat capacity. In contrast, both the position and the momentum of atoms described by harmonic oscillators (e.g. masses on springs) determine the energy. This means that  3 momentum + 3 positional degrees of freedom can store energy and thereby contribute to the heat capacity. Finally, equipartition tells us that each degree of freedom contributes $1/2k_BT$ to the energy.
 2. Following last question's reasoning, we get $C = 2k_B$ per atom.
 3. See the plot with the slider in the lecture notes. For fun, the Einstein dist. for two different values of $T$ is plotted below
@@ -50,9 +52,37 @@ ax.set_yticklabels(['$0$','$1$', '$2$'])
 ax.legend()
 draw_classic_axes(ax, xlabeloffset=.2)
 ```
-### Exercise 1: The harmonic oscillator and the Bose-Einstein distribution¶
 
-### Exercise 2: The quantum harmonic oscillator - connection with statistical physics
+## Exercise 1: The harmonic oscillator and the Bose-Einstein distribution
+
+### Question 1
+
+- The number of phonons in the oscillator is given by the Bose-Einstein distribution $n_B(\beta\hbar\omega_0) = 1/(e^{\beta\hbar\omega_0} - 1)$.
+- In the high temperature limit, $\beta\hbar\omega_0 \ll 1$, so $\exp(\beta\hbar\omega_0) \approx 1 + \beta\hbar\omega_0$, and $n_B(\beta\hbar\omega_0) \approx 1/(\beta\hbar\omega_0)$.
+- This result has a natural interpretation: the thermal energy is $k_BT$, which requires $n = k_BT/\hbar\omega_0$ phonons to store it.
+
+### Question 2
+
+- The expectation value of the energy is given by $\langle E\rangle = \hbar\omega_0(n_B(\beta\hbar\omega_0) + 1/2)$.
+- In the high temperature limit, if we apply the first order Taylor expansion like above, we get an almost correct answer $\langle E\rangle \approx k_BT + \hbar\omega_0/2$. To get a more accurate result (not required for this problem), we need to use the second order expansion, $\exp(\beta\hbar\omega_0) \approx 1 + \beta\hbar\omega_0 + (\beta\hbar\omega_0)^2/2$, which gives $\langle E\rangle \approx k_BT$.
+- This result is consistent with the expected number of phonons in the high temperature limit.
+
+### Question 3
+
+In the high temperature limit the heat capacity is the derivative of the above expression with temperature, and it is $C = k_B$.
+
+### Question 4
+
+- At zero temperature, the harmonic oscillator contains no phonons.
+- Phonons start getting excited at $T \sim \hbar\omega_0/k_B \equiv T_E$.
+- At the low temperature limit the number of phonons is exponentially small because of the exponential in the Bose-Einstein distribution.
+- This means that both the energy and the heat capacity are exponentially small at low temperatures.
+
+### Question 5
+
+There are many ways to argue that this function cannot be Bose-Einstein distribution. For example, the Bose-Einstein distribution is an expected number of particles, and the function in question is negative.
+
+## Exercise 2: The quantum harmonic oscillator - connection with statistical physics
 
 1. The energy spectrum is $E_n = (n+1/2)\hbar\omega_0$ with $n=0,1,2,...$
 2. We calculate the partition function using 
@@ -73,7 +103,7 @@ draw_classic_axes(ax, xlabeloffset=.2)
 
   where we used $\sum_{n = 0}^{\infty}nr^n = \frac{r}{(1 - r)^2}$.
 
-### Exercise 3*: Total heat capacity of a diatomic material.
+## Exercise 3*: Total heat capacity of a diatomic material.
 
 1. We use $\omega_{1,2} = \sqrt{\frac{k}{m_{1,2}}}$.
 

docs/2_debye_model.md

@@ -122,12 +122,12 @@ where $v_s$ is the _sound velocity_ of a material.
 As discussed in the previous lecture, the quantum mechanical excitations of a harmonic oscillator are called *phonons*, and the expected number of phonons in the oscillator at temperature $T$ is given by the Bose-Einstein distribution $n_B(\beta \hbar \omega(\mathbf{k}))$. Instead of having $3N$ oscillators with the same frequency $\omega_0$ as in the Einstein model, we now have $3N$ oscillators (the vibrational modes) with frequencies depending on $\textbf{k}$ through the dispersion relation $\omega(\mathbf{k}) = v_s|\mathbf{k}|$. Apart from this crucial difference with the Einstein model, the calculation of the expectation value of the energy stored in the vibrational modes proceeds in the same way: 
 The expected value of the total energy stored in the oscillators (which, from now on, we will simply denote as the total energy $E$) is given by the sum of the energy stored in the individual oscillators. These oscillators are characterized by their wavevector $\mathbf{k}$:
 
-\begin{align}
-E &= 3 \sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\hbar \omega(\mathbf{k}) n_{B}(\hbar \omega(\mathbf{k}))\right)\\
+\begin{align*}
+E &= 3 \sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\hbar \omega(\mathbf{k}) n_{B}(\beta \hbar \omega(\mathbf{k}))\right)\\
 &= 3 \sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{ e^{\hbar\omega(\mathbf{k})/{k_BT}}-1}\right).
-\end{align}
+\end{align*}
 
-Here we used that the expected occupation number is $n_B(\hbar \omega(\mathbf{k}))$.
+Here we used that the expected occupation number is $n_B(\beta \hbar \omega(\mathbf{k}))$.
 
 ??? question "Where does the factor 3 come from?"
 
@@ -195,10 +195,10 @@ $$
 
 We can use this approximation to rewrite the total energy as an integral:
 
-\begin{align}
+\begin{align*}
 E &=  3 \sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{ e^{\hbar\omega(\mathbf{k})/k_B T}-1}\right) \\
 &\approx 3 \left(\frac{L}{2\pi}\right)^3 \int \textrm{d} \textbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{e^{\hbar\omega(\mathbf{k})/k_B T}-1}\right).
-\end{align}
+\end{align*}
 
 Where $\omega(\mathbf{k}) = v_s |\mathbf{k}|$ is the dispersion relation, and the integral goes over 3-dimensional $k$-space.
 
@@ -213,10 +213,10 @@ $$
 
 Performing the change of variables, we obtain the expression for the total energy in spherical coordinates is
 
-\begin{align}
+\begin{align*}
 E &= 3 \frac{L^3}{(2\pi)^3}\int_0^\infty  4 \pi k^2 \left(\frac{1}{2}\hbar\omega(k)+\frac{\hbar\omega(k)}{ {e}^{\hbar\omega(k)/{k_BT}}-1}\right) \textrm{d}k \\
 &=  3 \frac{L^3}{(2\pi)^3}\frac{4 \pi}{v_s^3}\int_0^\infty \omega^2 \left(\frac{1}{2}\hbar\omega+\frac{\hbar\omega}{ e^{\hbar\omega/{k_BT}}-1}\right) \textrm{d}\omega.
-\end{align}
+\end{align*}
 
 We utilized the dispersion relation $\omega(k) = v_s |k|$ and omitted the absolute value of $k$ due to the integral over $k$ only running from 0 to $\infty$ after conversion to spherical coordinates.
 The integral above can be split up into two factors.
@@ -266,10 +266,10 @@ Despite $E_{\textrm Z}$ diverging towards infinity, does not contribute to $C$.
 The integral depends on the temperature through the $e^{\hbar\omega/k_BT}$ term.
 In order evaluate the integral, we substitute $x\equiv\frac{\hbar\omega}{k_BT}$ and remove the temperature dependence of the integrand:
 
-\begin{align}
+\begin{align*}
 E &= E_Z + \frac{3L^3}{2\pi^2 v_s^3}\frac{\left(k_BT\right)^4}{\hbar^3}\int\limits_0^\infty\frac{x^3}{ {\textrm e}^x-1} {\textrm{d}}x \\
 &= E_Z + \frac{3L^3}{2\pi^2 v_s^3}\frac{\left(k_BT\right)^4}{\hbar^3}\frac{\pi^4}{15},
-\end{align}
+\end{align*}
 
 where we used the fact that the integral is equal to $\frac{\pi^4}{15}$[^4].
 As we can see, the energy scales as $T^4$.
@@ -289,11 +289,11 @@ These modes have $\hbar \omega(\mathbf{k}) \lesssim k_B T$.
     Therefore the number of excited modes is proportional to the volume of a sphere $V_{\textbf{k}} = \frac{4 \pi}{3} |\mathbf{k}|^3$, multiplied by the density of modes in $k$-space, $\left(\frac{L}{2 \pi}\right)^3$. 
     Thus the total number of excited modes is 
 
-    \begin{align}
+    \begin{align*}
     N_\textrm{modes} &= V_{\textbf{k}} \left(\frac{L}{2\pi}\right)^3\\
     &\sim \left( |\mathbf{k}| L \right)^3\\
     &\sim (k_B T L/\hbar v_s)^3
-    \end{align}
+    \end{align*}
 
     where we have substituted $|\mathbf{k}| \simeq k_B T /\hbar v_s$ and left out all numerical factors.
 
@@ -301,10 +301,10 @@ These modes have $\hbar \omega(\mathbf{k}) \lesssim k_B T$.
     Therefore, each mode contributes $k_B$ to the heat capacity (Equipartition theorem). 
     As a result, the heat capacity is
 
-    \begin{align}
+    \begin{align*}
     C &= N_{\textrm{modes}} k_B \\
     &\propto k_B (k_B T L/\hbar v_s)^3.
-    \end{align}
+    \end{align*}
 
 
 ## Debye's interpolation for medium $T$
@@ -338,10 +338,10 @@ $$
 Let us now compute $\omega_D$.
 We know that for a 3D system with $N$ atoms has to have exactly $3N$ phonon modes.
 
-\begin{align}
+\begin{align*}
 3N &= \int_0^{\omega_D} g(\omega)d\omega = \frac{3L^3}{2 \pi^2 v_s^3} \int_0^{\omega_D}\omega^2 {\textrm{d}}\omega\\
 &= \frac{L^3\omega_D^3}{2\pi^2v_s^3},
-\end{align}
+\end{align*}
 
 which gives us
 
@@ -419,7 +419,7 @@ What is the material-dependent parameter that plays a similar role in the Debye
 6. Derive an expression for the shortest possible wavelength in the Debye model it in terms of the interatomic distance $a$. 
 Hint: assume that the number of atoms is given by $N=V/a^3$. Discuss if the answer is reasonable.
 
-### Exercise 1*: Deriving the density of states for the linear dispersion relation of the Debye model.
+### Exercise 1*: Deriving the density of states for the linear dispersion relation of the Debye model
 In this lecture, we found that the linear dispersion considered in the Debye model yields a density of states $g(\omega)\propto \omega^2$ (in three dimensions). In this exercise, we will practice this important derivation again and extend it to 1D and 2D. 
 
 1. Write down the dispersion relation of the vibrational modes in the Debye model.

docs/2_debye_model_solutions.md

@@ -73,19 +73,19 @@ ax.legend();
 2. The distance between nearest-neighbour points in $\mathbf{k}$-space is $2\pi/L$. The density of $\mathbf{k}$-points in 1, 2, and 3 dimensions is $L/(2\pi)$, $L^2/(2\pi)^2$, and $L^3/(2\pi)^3$ respectively.
 3. Express the number of states between frequencies $0<\omega<\omega_0$ as an integral over k-space. Do so for 1D, 2D and 3D. Do not forget the possible polarizations. We assume that in $d$ dimensions there are $d_p$ polarizations.
     
-    \begin{align}
+    \begin{align*}
     N_\text{states, 1D} & = 1_p \frac{L}{2\pi} \int_0^{k_0} 2 dk \\
     N_\text{states, 2D} & = 2_p \left(\frac{L}{2\pi}\right)^2 \int_0^{k_0} 2\pi k dk \\
     N_\text{states, 3D} & = 3_p  \left(\frac{L}{2\pi}\right)^3 \int_0^{k_0} 4\pi k^2 dk 
-    \end{align}
+    \end{align*}
 
 4.   We use $k=\mathbf{k}| = \omega/v_s$ and $dk = d\omega/v_s$ to get 
 
-    \begin{align}
+    \begin{align*}
     N_\text{states, 1D} & = 1_p \frac{L}{2\pi} \int_0^{\omega_0} 2 \frac{1}{v_s} d\omega  := \int_0^{\omega_0} g_{1D}(\omega) d\omega \\
     N_\text{states, 2D} & = 2_p \left(\frac{L}{2\pi}\right)^2 \int_0^{\omega_0} 2\pi \frac{\omega}{v_s^2} d\omega := \int_0^{\omega_0} g_{2D}(\omega) d\omega \\
     N_\text{states, 3D} & = 3_p  \left(\frac{L}{2\pi}\right)^3 \int_0^{\omega_0} 4\pi \frac{\omega^2}{v_s^3} d\omega := \int_0^{\omega_0} g_{3D}(\omega) d\omega
-    \end{align}
+    \end{align*}
 
     The integral boundaries set the frequency region in which you calculate the density of states.
 
@@ -94,10 +94,10 @@ ax.legend();
 ###  Exercise 2: Debye model in 2D.
 1. The energy stored in the vibrational modes of a two-dimensional Debye solid is:
     
-    \begin{align}
-    E & = \int_0^{\omega_D}(n_B(\omega(\mathbf{k}))+\frac{1}{2})\hbar\omega(\mathbf{k}) d\mathbf{k} \\    
+    \begin{align*}
+    E & = 2_p \frac{L^2}{4\pi^2}\int_0^{\infty}(n_B(\omega(\mathbf{k}))+\frac{1}{2})\hbar\omega(\mathbf{k}) d\mathbf{k} \\    
     & = \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\beta\hbar\omega_D}\frac{x^2}{e^{x} - 1}dx + E_0
-    \end{align}
+    \end{align*}
 
   
 2. The high-$T$ limit implies $\beta \rightarrow 0$. Therefore, $n_B \approx k_B T/\hbar\omega$, and the integral becomes particularly illuminating:
@@ -106,7 +106,7 @@ ax.legend();
     E = \int_{0}^{\omega_D} \hbar\omega  n_B(\omega) g(\omega) d\omega \approx \int_{0}^{\omega_D} k_B T g(\omega) d\omega = N_\text{modes} k_B T
     $$
 
-    where we neglected the zero-point energy. In 3D, we have $N_\text{modes} = 3_p N_\text{atoms}$, so that we recover the Dulong Petit $C_v = dE/dT = 3 k_B$ per atom
+    where we neglected the zero-point energy. In 2D, we have $N_\text{modes} = 2_p N_\text{atoms}$, so that we recover the 2D law of Dulong–Petit $C_v = dE/dT = 2 k_B$ per atom.
 
 3.  In the low temperature limit, the high-energy modes are not excited so we can safely let the upper boundary of the integral go to infinity. For convenience, we write $g(\omega) = \alpha \omega$, with $\alpha = \frac{L^2}{\pi v_s^2}$. We get 
     
@@ -123,10 +123,10 @@ ax.legend();
 ###  Exercise 3: Longitudinal and transverse vibrations with different sound velocities
 1. The key idea is that the total energy in the individual harmonic oscillators (the vibrational modes) is the sum of the energies in the individual oscillators: $E = \int_0^{\omega_D}\frac{\hbar \omega g(\omega)}{e^{\beta\hbar\omega} - 1}d\omega + E_Z$, where $g(\omega) = g_\parallel(\omega) + g_\perp(\omega)$. Using 
     
-    \begin{align}
+    \begin{align*}
     g_\parallel & = 1_p  \frac{L^3}{2\pi^2} \frac{\omega^2}{v_\parallel^3}, \\
     g_\perp     & = 2_p  \frac{L^3}{2\pi^2} \frac{\omega^2}{v_\perp^3},
-    \end{align}
+    \end{align*}
 
     we get 
     
@@ -156,16 +156,16 @@ ax.legend();
 ### Exercise 4: Anisotropic sound velocities.
 In this case, the velocity depends on the direction. Note however that, in contrast with the previous exercise, the polarization does not affect the dispersion of the waves. We get
     
-\begin{align}
+\begin{align*}
 E =& 3_p\left(\frac{L}{2\pi}\right)^3 \int \frac{\hbar\omega(\mathbf{k})}{e^{\beta\hbar\omega(\mathbf{k})}-1} dk_x dk_y dk_z + E_Z \\
 =&   3_p\left(\frac{L}{2\pi}\right)^3\frac{1}{v_x v_y v_z} \int \frac{\hbar\kappa}{e^{\beta\hbar\kappa} - 1}d\kappa_x d\kappa_y d\kappa_z + E_Z,
-\end{align}
+\end{align*}
 
 where we made the substitutions $\kappa_x = k_x v_x,\kappa_y = k_y v_y, \kappa_z = k_z v_z$ so that $\omega = \kappa$. Going to spherical coordinates:
 
-\begin{align}
+\begin{align*}
 E &= 3_p \frac{3_p L^3}{2\pi^2}\frac{1}{v_x v_y v_z}\int_{0}^{\kappa_D} \frac{\hbar\kappa^3}{e^{\beta\hbar\kappa} - 1} d\kappa + E_Z  \\
 & = \frac{3_p L^3}{2\pi^2\hbar^3\beta^4}\frac{1}{v_x v_y v_z}\int_0^{\beta\hbar\kappa_D} \frac{x^3}{e^x - 1}dx + E_Z,
-\end{align}
+\end{align*}
 
 Therefore, $C = \frac{dE}{dT} = \frac{6k_B^4 L^3 T^3}{\pi^2\hbar^3}\frac{1}{v_x v_y v_z}\int_0^{\beta\hbar\kappa_D} \frac{x^3}{e^x - 1}dx$. We see that the result is similar to the one with the linear dispersion, the only difference is the factor $1/(v_x v_y v_z)$ instead of $1/v^3$.

docs/3_drude_model.md

@@ -18,6 +18,7 @@ _(based on chapter 3 of the book)_
     - Write down the equation describing the Lorentz force
     - Formulate Newton's equation of motion for a particle subject to a force, including a damping term
     - Discuss the concepts of voltage, current, resistivity, and conductivity
+    - Describe how the number of phonons scales with temperature $T$ 
 
 !!! summary "Learning goals"
 
@@ -107,7 +108,7 @@ scatter_pts = scattering_positions
 r_min = np.min(r.reshape(2, -1), axis=1) - 1
 r_max = np.max(r.reshape(2, -1), axis=1) + 1
 
-fig = pyplot.figure(figsize=(9, 6))
+fig = pyplot.figure(figsize=(9, 6), dpi=70)
 ax = fig.add_subplot(1, 1, 1)
 ax.axis("off")
 ax.set(xlim=(r_min[0], r_max[0]), ylim=(r_min[1], r_max[1]))
@@ -123,7 +124,7 @@ def frame(i):
     trajectories.set_data(*concatenated_lines)
     scatterers.set_offsets(scatter_pts[:, :i].reshape(2, -1).T)
 
-anim = animation.FuncAnimation(fig, frame, interval=100)
+anim = animation.FuncAnimation(fig, frame, frames=scatter_pts.shape[1], interval=100)
 
 pyplot.close()
 
@@ -156,11 +157,11 @@ $$
 
 To find the average velocity, we take a weighted average of these two groups of particles:
 
-\begin{align}
+\begin{align*}
 m\mathbf{v}(t+dt)  &= [m\mathbf{v}(t) + F dt]\left(1 - \frac{dt}{\tau}\right) + 0⋅\frac{dt}{\tau} \\
   &= [m\mathbf{v}(t) + \mathbf{F} dt] \left(1 - \frac{dt}{\tau}\right) \\
-                  &= m\mathbf{v}(t) + dt [\mathbf{F} - m\frac{\mathbf{v(t)}}{\tau}] - \frac{\mathbf{F}}{\tau} dt^2
-\end{align}
+                  &= m\mathbf{v}(t) + dt [\mathbf{F} - m\frac{\mathbf{v}(t)}{\tau}] - \frac{\mathbf{F}}{\tau} dt^2
+\end{align*}
 
 We now neglect the term proportional to $dt^2$ (it vanishes when $dt → 0$).
 Finally, we recognize that $\left[\mathbf{v}(t+dt) - \mathbf{v}(t)\right]/dt = d\mathbf{v}(t)/dt$, which results in
@@ -290,7 +291,7 @@ Hall-voltage measurements in an applied magnetic field are a powerful tool for d
 
 We consider a planar (two-dimensional) sample that sits in the $xy$ plane. The sample has width $W$ in the $y$-direction and length $L$ in the $x$-direction. We apply a current $I$ along the $x$-direction and a magnetic field $\bf B$ along the $z$-direction.
 
-1.  Make a sketch of the system. Indicate where on the sample you can measure the Hall voltage $V_H$. Express the Hall resistance $R_{xy} = V_H/I$ as a function of $B = |\bf B|$. Does $R_{xy}$ depend on the geometry of the sample? Also express $R_{xy}$ in terms of the Hall coefficient $R_H$.
+1.  Make a sketch of the system. Indicate where on the sample you can measure the Hall voltage $V_H$. Express the Hall resistance $R_{yx} = V_H/I$ as a function of $B = |\bf B|$. Does $R_{yx}$ depend on the geometry of the sample? Also express $R_{yx}$ in terms of the Hall coefficient $R_H$.
 
     ??? question "What is the relation between the electric field and the electric potential?"
         $V_b - V_a = -\int_{\Gamma} \mathbf{E} \cdot d\mathbf{\ell}$ if $\Gamma$ is a path from $a$ to $b$.
@@ -307,7 +308,7 @@ We consider copper, which has a density of 8960 kg/m$^3$, an atomic weight of 63
 1. From these numbers, calculate the Drude scattering time $τ$ at room temperature. 
 2. Assume that the electrons move with rms speed $v=\sqrt\frac{3k_BT}{m}$. Use this to estimate the electron mean free path $\lambda$, defined as the average distance an electron travels in between scattering events.
 3. Now assume that the mean free path is independent of temperature. How would the electrical resistivity $ρ$ depend on temperature under this assumption? Discuss how it scales with temperature and sketch $ρ(T)$.
-4. Compare your sketch of $ρ(T)$ with that in the lecture notes. In what respect do they differ? Discuss possible reasons for differences.
+4. Compare your sketch of $ρ(T)$ with what you expect if electron scattering is caused by phonons of which the density $\propto T$.
 
 ### Exercise 3: The Hall conductivity matrix and the Hall coefficient
 We apply a magnetic field $\bf B$ along the $z$-direction to a current-carrying 2D sample in the xy plane. As we have described in the lecture notes, in this situation, the electric field $\mathbf{E}$ is related to the current density $\mathbf{j}$ by the resistivity matrix:
@@ -324,6 +325,7 @@ $$
     $$
 
     allowing you to express $\mathbf{j}$ as a function of $\mathbf{E}$.
+    
 3. Sketch $\sigma_{xx}$ and $\sigma_{xy}$ as a function of the magnetic field $\bf B$.
 4. Give the definition of the Hall coefficient. What does the sign of the Hall coefficient indicate?
 
@@ -334,8 +336,8 @@ As before, we consider a 2D sample in the xy plane. We push a current through th
 
 1. Express the current density $\bf J$ as a function of the drift velocities $\mathbf{v_e}$ and $\mathbf{v_h}$ of the charge carriers. 
 2. Show that the drift velocities are given by $\mathbf{v_e} = -\mu_e(\mathbf{E} + \mathbf{v_e}\times\mathbf{B})$ and $\mathbf{v_h} = \mu_h(\mathbf{E} + \mathbf{v_h}\times\mathbf{B})$ 
-3. Assume that $E_x \gg |v_{e,y}| B_z$ and $E_x \gg |v_{h,y}| B_z$ to show that $J_x = eE_x(n_e\mu_e + n_h\mu_h)$. Discuss the two terms in this equation and why they should be added.
-4. Under the same assumption, show that $J_y = eE_x(n_e\mu_e + n_h\mu_h) + eB_zE_x(n_e\mu_e^2-n_h\mu_h^2)$. What should $J_y$ be equal to?
+3. Assume that $E_x \gg |v_{e,y}| B_z$ and $E_x \gg |v_{h,y}| B_z$ to show that $J_x = eE_x(n_e\mu_e + n_h\mu_h)$. Discuss the two terms of this last equation and why they should be added.
+4. Under the same assumption, show that $J_y = eE_y(n_e\mu_e + n_h\mu_h) + eB_zE_x(n_e\mu_e^2-n_h\mu_h^2)$. What should $J_y$ be equal to?
 5. Use the derived equations to show that the Hall electric field is
     $$ E_y = J_xB_z\frac{n_h\mu_h^2 - n_e\mu_e^2}{e(n_h\mu_h + n_e \mu_e)^2} $$
     Extract the Hall coefficient $R_H$ and discuss what determines its sign. 

docs/3_drude_model_solutions.md

@@ -5,32 +5,25 @@ search:
 # Solutions for Drude model exercises
 
 ### Exercise 1: Extracting quantities from basic Hall measurements
+1. The Hall voltage is measured across the sample width. It is given by
 
-#### Question 1.
-Hall voltage is measured across the sample width. Hence,
-
-$$
-V_H = -\int_{0}^{W} E_ydy
-$$
-
-where $E_y = -v_xB$.
-
-$R_{xy}$ = $-\frac{B}{ne}$, so it does not depend on the sample geometry.
+    $$
+    V_H = V(W)-V(0) = -\int_{0}^{W} E_ydy = -E_y W
+    $$
 
+    From the steady-state solution of the Drude equation of motion, we obtain $E_y = \rho_{yx}j_x = \frac{-B}{ne}j_x $. Using $j_x = I_x / W$ and $V_H=-E_y W$, we obtain $R_{yx} = V_H/I_x = \frac{B}{ne}$.
+    so it does not depend on the sample geometry.
 
-#### Question 2.
-If hall resistance and magnetic field are known, the charge density is calculated from $R_{xy} = -\frac{B}{ne}$.
 
-As $V_x = -\frac{I_x}{ne}B$, a stronger field makes Hall voltages easier to measure.
-
-#### Question 3.
+2. If the hall resistance and the magnetic field are known, we extract the charge density using $R_{yx} = \frac{B}{ne}$. Because $V_H = \frac{B}{ne}I_x$, a stronger field yields a larger Hall voltage, making it easier to measure. Likewise, a lower charge density yields a larger Hall voltage for a given bias current, making it easier to measure.
 
+3. The longitudinal resistance is
 
-$$
-R_{xx} = \frac{\rho_{xx}L}{W}
-$$
+    $$
+    R_{xx} = \rho_{xx}\frac{L}{W}
+    $$
 
-where $\rho_{xx} = \frac{m_e}{ne^2\tau}$. Therefore, scattering time ($\tau$) is known and $R_{xx}$ depend upon the sample geometry.
+    with $\rho_{xx} = \frac{m_e}{ne^2\tau}$ the longitudinal Drude resistivity. Therefore, knowing the electron density $n$, we can extract the scattering time ($\tau$). We observe that $R_{xx}$ depends on the sample geometry ($L$ and $W$), whereas the Hall resistance $R_{yx}$ does not.
 
 <!---
 #### Question 4.
@@ -43,39 +36,52 @@ See 3_drude_model.md
 
 ### Exercise 2: Temperature dependence of resistance in the Drude model
 
-#### Question 1.
-Find electron density from $n_e =   \frac{ZnN_A}{W} $
+1. We first find the electron density using $n_e =   \frac{ZnN_A}{W}$, where $Z=1$ is the number of free electrons per copper atom, *n* is the mass density of copper, $N_A$ is Avogadro's constant, and *W* is the atomic weight of copper. Then, we find the scattering time $\tau = 2.57 \cdot 10^{-14}$ s from the longitudinal Drude resistivity $\rho = \frac{m_e}{n_e e^2\tau}$.
+2. Using $\lambda = \langle v \rangle\tau$, we get $\lambda =3$ nm.
+3. The scattering time $\tau \propto \frac{1}{\sqrt{T}}$, such that we expect $\rho \propto \sqrt{T}$.
+4. For most metals, the measured resistivity scales as $\rho \propto T$, in contrast with the $\rho \propto \sqrt{T}$ predicted by the Drude model. The linear scaling can be understood by assuming that the scattering is caused by phonons, as their number scales linearly with T at high temperature (recall that the high-temperature limit of the Bose-Einstein distribution functions is $n_B = \frac{kT}{\hbar\omega}$ leading to $\rho \propto T$). The inability to explain this linear dependence is a failure of the Drude model.
 
-where *Z* is valence of copper atom, *n* is density, $N_A$ is Avogadro constant and *W* is atomic weight. Use $\rho$ from the lecture notes to calculate scattering time.
+### Exercise 3: The Hall conductivity matrix and the Hall coefficient
 
-#### Question 2.
-$\lambda = \langle v \rangle\tau$
+1. $\rho_{xx}$ is independent of B and $\rho_{xy} \propto B$
 
-#### Question 3.
-Scattering time $\tau \propto \frac{1}{\sqrt{T}}$; $\rho \propto \sqrt{T}$
+2. The conductivities are 
 
-#### Question 4.
-In general, $\rho \propto T$ as the phonons in the system scales linearly with T (remember high temperature limit of Bose-Einstein factor becomes $\frac{kT}{\hbar\omega}$ leading to $\rho \propto T$). Inability to explain this linear dependence is a failure of the Drude model.
+    $$
+    \sigma_{xx} = \frac{\rho_{xx}}{\rho_{xx}^2 + \rho_{xy}^2} = \frac{mne^2\tau}{m^2+e^2\tau^2 B^2}
+    $$
 
-### Exercise 3: The Hall conductivity matrix and the Hall coefficient
+    $$
+    \sigma_{xy} = \frac{-\rho_{yx}}{\rho_{xx}^2 + \rho_{xy}^2} = \frac{Bne^3 \tau^2}{m^2+e^2 \tau^2 B^2}
+    $$
 
-#### Question 1.
-$\rho_{xx}$ is independent of B and $\rho_{xy} \propto B$
+3. These equations describe [Lorentzian](https://en.wikipedia.org/wiki/Spectral_line_shape#Lorentzian)-like functions.
 
-#### Question 2.
-4 (Refer to the lecture notes).
 
-#### Question 3.
+4. The Hall coefficient is $R_H = -\frac{1}{ne}$. The sign of the Hall coefficient depends on sign of the charge carriers (analyzed further in the next exercise).
 
-$$
-\sigma_{xx} = \frac{\rho_{xx}}{\rho_{xx}^2 + \rho_{xy}^2}
-$$
 
-$$
-\sigma_{xy} = \frac{-\rho_{yx}}{\rho_{xx}^2 + \rho_{xy}^2}
-$$
+### Exercise 4. Positve and negative charge carriers
+
+1. The current is the sum of the currents carried by the positive and negative charge carriers:
 
-This describes a [Lorentzian](https://en.wikipedia.org/wiki/Spectral_line_shape#Lorentzian).
+    $$ 
+    \mathbf{J} = -n_e e \mathbf{ v_e} + n_h e \mathbf{ v_h}
+    $$
+
+2. Write down the Drude equation of motion for the positive and negative charge carriers separately and use $ \mu_e = \frac{e\tau_e}{m_e}$ and $ \mu_h = \frac{e\tau_h}{m_h}$
+
+3. From the assumption, we can ignore the $(\mathbf{ v_i} \times \mathbf{B})_x$ terms. Then combining question 1 and 2 gives the result. 
+
+4. The net current in the y-direction should be zero.
+
+5. The Hall coefficient is
+
+    $$
+    R_H = \frac{n_h\mu_h^2 - n_e\mu_e^2}{e(n_h\mu_h + n_e \mu_e)^2}
+    $$
+    We observe that the sign of $R_H$ is determined by the density and the mobilty of the positive and negative charge carriers. Therefore, a measurement of the Hall voltage reveals the dominant charge carrier.
+<!---
 
 Solution extra exercise:
 
@@ -153,4 +159,4 @@ $$
 
 This represents a [cycloid](https://en.wikipedia.org/wiki/Cycloid#/media/File:Cycloid_f.gif): a circular motion around a point that moves in the $y$-direction with velocity $v_d=\frac{E_x}{B_z}$.
 
-
+-->

docs/4_sommerfeld_model.md

@@ -93,10 +93,10 @@ $$
 Here $\beta = 1/k_{B}T$, $\varepsilon$ is the energy, and $\mu$ is the _chemical potential_. The chemical potential is the characteristic energy above which $n_{F}$ goes to zero. 
 We obtain the number of electrons in the system by summing over the occupied states, as determined the by Fermi-Dirac distribution:
 
-\begin{align}
+\begin{align*}
 N &= 2 \sum_{\mathbf{k}} n_{F}(\beta(\varepsilon-\mu))\\
 &= 2 \left( \frac{L}{2 \pi} \right)^3 \int n_{F}(\beta(\varepsilon(\mathbf{k})-\mu))\mathrm{d} \mathbf{k}.
-\end{align}
+\end{align*}
 
 Just like with phonons, we replaced the discrete sum over $\mathbf{k}$ with a volume integral.
 The factor 2 here accounts for the number of distinct electron states per $\mathbf{k}$ value.
@@ -238,26 +238,26 @@ Let us calculate the density of states for a 3D system.
 Because the free-electron dispersion is isotropic[^2], we use spherical coordinates.
 The total number of states below energy $\varepsilon$ is then
 
-\begin{align}
+\begin{align*}
 N_\text{states} &\overset{\mathrm{3D}}{=}2_s \left(\frac{L}{2\pi}\right)^3\int_{\varepsilon(\mathbf{k}) < \varepsilon}\mathrm{d}{\mathbf{k}}\\
 &=2_s \left(\frac{L}{2\pi}\right)^3 4\pi\int k^2\mathrm{d}k\\
 &=\frac{V}{\pi^2}\int k^2\mathrm{d}k,
-\end{align}
+\end{align*}
 
 We rewrite this expression into an integral over energy using the dispersion relation, substituting $k=\hbar^{-1}\sqrt{2m\varepsilon}$ and $\mathrm{d}k=\hbar^{-1}\sqrt{m/2\varepsilon} d\varepsilon$:
 
-\begin{align}
+\begin{align*}
 N_\text{states} &=\frac{V}{\pi^2}\int\frac{2m \varepsilon}{\hbar^3}\sqrt{\frac{m}{2\varepsilon}}\mathrm{d}\varepsilon\\
 &=\frac{Vm^{3/2}}{\pi^2\hbar^3}\int\sqrt{2\varepsilon}\ \mathrm{d}\varepsilon \\
 & = \int g(\varepsilon) \mathrm{d}\varepsilon \\
-\end{align}
+\end{align*}
 
 We thus find the density of states:
 
-\begin{align}
+\begin{align*}
 g(\varepsilon) &= \frac{ \mathrm{d}N_\text{states}}{ \mathrm{d}\varepsilon}\\
 & =\frac{Vm^{3/2}\sqrt{2\varepsilon}}{\pi^2\hbar^3} \propto\sqrt{\varepsilon}
-\end{align}
+\end{align*}
 
 We observe that the density of states for the 3D parabolic free-electron dispersion is proportional to the square root of energy: 
 
@@ -284,17 +284,19 @@ ax.plot(E, 15*np.sqrt(E), label = '3D')
 ax.set_ylabel(r"$g(\varepsilon)$")
 ax.set_xlabel(r"$\varepsilon$")
 ax.legend()
+ax.set_xticks([])
+ax.set_yticks([])
 draw_classic_axes(ax, xlabeloffset=.2)
 ```
 
 ## Relation between the Fermi energy and the number of electrons
 The Fermi energy sets the number of electrons in the system $N$. We can see this by expressing $N$ at $T = 0$ as an integral over the density of states:
 
-\begin{align}
+\begin{align*}
 N &= \int \limits_0^{\infty} n_{F}(\beta(\varepsilon-\mu)) g(\varepsilon) \mathrm{d}\varepsilon\\
 &\overset{\mathrm{T = 0}}{=}\int \limits_0^{\varepsilon_F}g(\varepsilon)\mathrm{d}\varepsilon \\
 &\overset{\mathrm{3D}}{=} \frac{V}{3\pi^2\hbar^3}(2m\varepsilon_F)^{3/2}.
-\end{align}
+\end{align*}
 
 Solving this equation for the Fermi energy yields:
 
@@ -368,6 +370,7 @@ ax.annotate('', xy=(1, 0), xytext=(1, 1),
 ax.text(1.2, .7, r'$g(ε_F)$', ha='center')
 ax.set_xticks([1])
 ax.set_xticklabels([r'$ε_F$'])
+ax.set_yticks([])
 
 ax.set_ylabel(r"$g(ε)$")
 ax.set_xlabel(r"$ε$")
@@ -387,19 +390,19 @@ Since the base of the triangle is proportional to $k_BT$ and the height is $\sim
 
 These electrons have gained $k_BT$ of thermal energy, such that the total extra energy is
 
-\begin{align}
+\begin{align*}
 E(T) &= E(T = 0) + N_\mathrm{exc}k_BT\\
 &\approx  E(T = 0) + g(\varepsilon_F)k_B^2T^2.
-\end{align}
+\end{align*}
 
 Therefore, the electron heat capacity $C_e$ is
 
-\begin{align}
+\begin{align*}
 C_e &= \frac{ \mathrm{d}E}{ \mathrm{d}T}\\
  &\approx 2 g(ε_F)k_B^2T\\
  &\overset{\mathrm{3D}}{=} 3 Nk_B\frac{T}{T_F}\\
  &\propto T,
-\end{align}
+\end{align*}
 
 where we used $N=\frac{2}{3}\varepsilon_Fg(\varepsilon_F)$ and the Fermi temperature $T_F \equiv \varepsilon_F/k_B$.
 
@@ -460,7 +463,7 @@ The Sommerfeld model provides a good description of free electrons in alkali met
 3. Calculate the free electron density $n$ in potassium.
 4. Compare this with the actual electron density of potassium, which can be calculated by using the density, atomic mass and atomic number of potassium. What can you conclude from this?
 
-### Exercise 3: graphene
+### Exercise 3: The electron dispersion, density of states, and heat capacity of graphene
 One of the most famous recently discovered materials is [graphene](https://en.wikipedia.org/wiki/Graphene). It consists of carbon atoms arranged in a 2D honeycomb structure. 
 Unlike in metals, electrons in graphene cannot be treated as 'free'. Instead, close to the Fermi level, the dispersion relation can be approximated by a linear relation:
 $ \varepsilon(\mathbf{k}) = \pm c|\mathbf{k}|.$ Note that the $\pm$ here means that there are two energy levels at a specified $\mathbf{k}$. 
@@ -482,7 +485,18 @@ Your result should be linear with $|\varepsilon|$.
 
 4. Calculate the heat capacity $C_e$ as a function of the temperature $T$.
 
-[^1]: This is not completely true, as we will see when learning about [semiconductors](13_semiconductors)
+### Exercise 4: Two energy bands
+An 'energy band' is a range of energies within which there are states available for the particles in the system. An energy band is therefore closely related to the dispersion relation. For instance, for the free electron dispersion $\varepsilon = \hbar^2k^2/2m$, the available states (=the energy band) lie between zero and infinite energy. As we will encounter more often in our course, many materials have multiple, possibly overlapping energy bands, which are each described by their own dispersion. Here we analyze a system with two, partially overlapping, energy bands. These bands are each described by a free-electron dispersion, but they are  offset with repect to each other in energy. The goal is to calculate the density of states and electron occupation of the system.
+
+The dispersion of energy band 1 is $\varepsilon_1(\mathbf{k}) = \tfrac{\hbar^2 \mathbf{k}^2}{2m}$ and that of band 2 is $\varepsilon_2(\mathbf{k}) = \tfrac{\hbar^2 \mathbf{k}^2}{2m} + \varepsilon_0$. The Fermi energy is $E_F$, and we assume $E_F \gg \varepsilon_0$. We consider a _two-dimensional_ system.
+
+1. Sketch the two dispersions in one plot. Indicate the Fermi energy.
+2. Calculate the density of states and sketch it as a function of energy. Hint: the total density of states is obtained by adding the density of states associated with the individual bands.
+3. Express the number of electrons in the system in terms of the Fermi energy $E_F$.
+4. Express the number of electrons in the energy range $\varepsilon_a<\varepsilon<\infty$, for some energy $\varepsilon_a \geq \varepsilon_0$, as an integral over energy, for $T>0$.
+5. Assuming $\varepsilon_a - E_F \gg k_B T$, explicitly calculate the integral of the previous subquestion.
+
+[^1]: This is not completely true, as we will see when learning about [semiconductors](13_semiconductors.md)
 [^2]: An [isotropic](https://en.wikipedia.org/wiki/Isotropic_solid) material means that the material is the same in all directions.
 [^3]: The mean inter-particle distance is related to the electron density $n = N/V$ as $\langle r \rangle \propto n^{-1/3}$. The exact proportionality constant depends on the properties of the system. The Fermi wavelength sets the scale at which quantum interference effects of the electronic waves become important. In some materials (e.g. graphene) it can be on the 100 nm scale - accessible to nanofabrication techniques. Striking images of [electron interference at the atomic](https://en.wikipedia.org/wiki/Quantum_mirage) scale are visible with a scanning tunneling microscope.