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...@@ -55,6 +55,7 @@ In the last lecture, we introduced crystallographic terminology in order to be a ...@@ -55,6 +55,7 @@ In the last lecture, we introduced crystallographic terminology in order to be a
In this lecture, we will 1) study how real-space lattices give rise to lattices in reciprocal space (with the goal of understanding dispersion relations) and 2) consider how to probe crystal structures using X-ray diffraction experiments. In this lecture, we will 1) study how real-space lattices give rise to lattices in reciprocal space (with the goal of understanding dispersion relations) and 2) consider how to probe crystal structures using X-ray diffraction experiments.
## Recap: the reciprocal lattice in one dimension ## Recap: the reciprocal lattice in one dimension
In [lecture 7](7_tight_binding.md) we discussed the reciprocal space of a simple 1D lattice with lattice points $x_n = na$, where $n$ is an integer and $a$ is the spacing between the lattice points. To obtain the dispersion relation, we considered waves of the form In [lecture 7](7_tight_binding.md) we discussed the reciprocal space of a simple 1D lattice with lattice points $x_n = na$, where $n$ is an integer and $a$ is the spacing between the lattice points. To obtain the dispersion relation, we considered waves of the form
$$ $$
e^{ikx_n} = e^{ikna}, \quad n \in \mathbb{Z}, e^{ikx_n} = e^{ikna}, \quad n \in \mathbb{Z},
...@@ -71,12 +72,13 @@ The set of points $G=2\pi m/a$ forms the **reciprocal lattice** in $k$-space. ...@@ -71,12 +72,13 @@ The set of points $G=2\pi m/a$ forms the **reciprocal lattice** in $k$-space.
Let us now generalize this idea to describe reciprocal lattices in three dimensions. Let us now generalize this idea to describe reciprocal lattices in three dimensions.
## Extending to three dimensions ## The reciprocal lattice in three dimensions
We start from a lattice in real space: We start from a lattice in real space:
$$ $$
\mathbf{R}=n_1\mathbf{a}_1+n_2\mathbf{a}_2+n_3\mathbf{a}_3, \quad \{n_1, n_2, n_3\} \in \mathbb{Z}, \mathbf{R}=n_1\mathbf{a}_1+n_2\mathbf{a}_2+n_3\mathbf{a}_3, \quad \{n_1, n_2, n_3\} \in \mathbb{Z},
$$ $$
where the $\{\mathbf{a}_i\}$ are the primitive lattice vectors. The reciprocal lattice is: where the $\{\mathbf{a}_i\}$ are primitive lattice vectors. The reciprocal lattice is:
$$ $$
\mathbf{G}=m_1\mathbf{b}_1+m_2\mathbf{b}_2+m_3\mathbf{b_3}, \quad \{m_1, m_2, m_3\} \in \mathbb{Z}, \mathbf{G}=m_1\mathbf{b}_1+m_2\mathbf{b}_2+m_3\mathbf{b_3}, \quad \{m_1, m_2, m_3\} \in \mathbb{Z},
$$ $$
...@@ -105,8 +107,9 @@ Because $n_i$ and $m_j$ are both integers, the exponent evaluates to 1. ...@@ -105,8 +107,9 @@ Because $n_i$ and $m_j$ are both integers, the exponent evaluates to 1.
The relation $\mathbf{a_i}\cdot\mathbf{b_j}=2\pi\delta_{ij}$ implies that if we write the $\{\mathbf{a_i}\}$ as rows of a matrix, the reciprocal lattice vectors are $2\pi$ times the columns of the inverse of that matrix. The relation $\mathbf{a_i}\cdot\mathbf{b_j}=2\pi\delta_{ij}$ implies that if we write the $\{\mathbf{a_i}\}$ as rows of a matrix, the reciprocal lattice vectors are $2\pi$ times the columns of the inverse of that matrix.
### 2D example: triangular lattice ### Example: the reciprocal lattice of a 2D triangular lattice
To gain extra intuition into the properties of the reciprocal lattice, we study an example.
To gain extra intuition for the properties of the reciprocal lattice, we study an example.
In the previous lecture we studied the triangular lattice shown in the figure below. In the previous lecture we studied the triangular lattice shown in the figure below.
The left panel shows the real-space lattice with primitive lattice vectors $\mathbf{a}_1 = a \mathbf{\hat{x}}$ and $\mathbf{a}_2 = a/2\mathbf{\hat{x}} + \sqrt{3}a/2 \mathbf{\hat{y}}$. The right panel shows the corresponding reciprocal lattice and its primitive lattice vectors $\mathbf{b}_1$ and $\mathbf{b}_2$. The left panel shows the real-space lattice with primitive lattice vectors $\mathbf{a}_1 = a \mathbf{\hat{x}}$ and $\mathbf{a}_2 = a/2\mathbf{\hat{x}} + \sqrt{3}a/2 \mathbf{\hat{y}}$. The right panel shows the corresponding reciprocal lattice and its primitive lattice vectors $\mathbf{b}_1$ and $\mathbf{b}_2$.
...@@ -294,7 +297,7 @@ $$ ...@@ -294,7 +297,7 @@ $$
There is however only one choice that satisfies $\mathbf{a_i}\cdot\mathbf{b_j}=2\pi\delta_{ij}$. There is however only one choice that satisfies $\mathbf{a_i}\cdot\mathbf{b_j}=2\pi\delta_{ij}$.
### 3D lattice example ### Constructing a 3D reciprocal lattice
Let us now consider a more involved example of the 3D lattice. Let us now consider a more involved example of the 3D lattice.
The explicit expression for the reciprocal lattice vectors in terms of their real-space counterparts is: The explicit expression for the reciprocal lattice vectors in terms of their real-space counterparts is:
...@@ -345,7 +348,8 @@ $$ ...@@ -345,7 +348,8 @@ $$
{\mathcal F}_\mathbf{k}\left[\rho(\mathbf{r})\right]=\int \mathrm{d}\mathbf{r}\ \mathrm{e}^{i\mathbf{k}\cdot\mathbf{r}} \rho(\mathbf{r}) = \sum_\mathbf{G}\delta(\mathbf{k}-\mathbf{G}). {\mathcal F}_\mathbf{k}\left[\rho(\mathbf{r})\right]=\int \mathrm{d}\mathbf{r}\ \mathrm{e}^{i\mathbf{k}\cdot\mathbf{r}} \rho(\mathbf{r}) = \sum_\mathbf{G}\delta(\mathbf{k}-\mathbf{G}).
$$ $$
## Periodicity of the reciprocal lattice ## The importance of the 1st Brillouin zone
We have now seen how the structure of the reciprocal lattice is directly determined by the structure of the real-space lattice. An important reason to study the reciprocal lattice is that we are often interested in understanding the dispersion relation of electronic or vibrational modes in a material. And because waves with wavevectors differing by a reciprocal lattice vector $\mathbf{G}$ are identical, we only need to understand the dispersion in a single primitive unit cell of the reciprocal lattice. But what unit cell to choose? We learned that the choice of a primitive unit cell is not unique. We have now seen how the structure of the reciprocal lattice is directly determined by the structure of the real-space lattice. An important reason to study the reciprocal lattice is that we are often interested in understanding the dispersion relation of electronic or vibrational modes in a material. And because waves with wavevectors differing by a reciprocal lattice vector $\mathbf{G}$ are identical, we only need to understand the dispersion in a single primitive unit cell of the reciprocal lattice. But what unit cell to choose? We learned that the choice of a primitive unit cell is not unique.
A general convention in reciprocal space is to use the Wigner-Seitz cell, which is also called the **1st Brillouin zone**. A general convention in reciprocal space is to use the Wigner-Seitz cell, which is also called the **1st Brillouin zone**.
...@@ -356,9 +360,9 @@ In the previous lecture we already discussed how to construct Wigner-Seitz cells ...@@ -356,9 +360,9 @@ In the previous lecture we already discussed how to construct Wigner-Seitz cells
![](figures/brillouin_mod.svg) ![](figures/brillouin_mod.svg)
## Diffraction ## Determining crystal structures using diffraction experiments
### Reciprocal lattice: Laue conditions ### The Laue condition
Another reason to understand the reciprocal lattice is that it manifests directly in diffraction experiments. Such experiments are some of our most powerful tools for determining the crystal structure of materials. Another reason to understand the reciprocal lattice is that it manifests directly in diffraction experiments. Such experiments are some of our most powerful tools for determining the crystal structure of materials.
...@@ -398,7 +402,8 @@ $$ ...@@ -398,7 +402,8 @@ $$
to get constructive interference. In other words, we can only get constructive interference at very specific angles, as determined by the structure of the reciprocal lattice. This requirement is known as the _Laue condition_. to get constructive interference. In other words, we can only get constructive interference at very specific angles, as determined by the structure of the reciprocal lattice. This requirement is known as the _Laue condition_.
As a result, the interference pattern produced in diffraction experiments is a direct measurement of the reciprocal lattice! As a result, the interference pattern produced in diffraction experiments is a direct measurement of the reciprocal lattice!
### Structure factor ### The structure factor
Above we assumed that the unit cell contains only a single atom. Above we assumed that the unit cell contains only a single atom.
What if the basis contains more atoms though? What if the basis contains more atoms though?
In the figure below we show a simple lattice which contains multiple atoms in the unit cell. In the figure below we show a simple lattice which contains multiple atoms in the unit cell.
...@@ -432,7 +437,7 @@ In diffraction experiments, the intensity of the scattered wave is $I \propto A^ ...@@ -432,7 +437,7 @@ In diffraction experiments, the intensity of the scattered wave is $I \propto A^
Therefore, the intensity of a scattered wave depends on the structure factor $I \propto S(\mathbf{G})^2$. Therefore, the intensity of a scattered wave depends on the structure factor $I \propto S(\mathbf{G})^2$.
Because the structure factor depends on the form factors and the positions of the basis atoms, by studying the visibility of different diffraction peaks we may learn the locations of atoms within the unit cell. Because the structure factor depends on the form factors and the positions of the basis atoms, by studying the visibility of different diffraction peaks we may learn the locations of atoms within the unit cell.
### Non-primitive unit cell ### The Laue condition and structure factor for non-primitive unit cells
Laue conditions allow scattering as long as the scattering wave vector is a reciprocal lattice vector. Laue conditions allow scattering as long as the scattering wave vector is a reciprocal lattice vector.
However if we consider a non-primitive unit cell of the direct lattice, the reciprocal lattice contains more lattice points, seemingly leading to additional interference peaks. However if we consider a non-primitive unit cell of the direct lattice, the reciprocal lattice contains more lattice points, seemingly leading to additional interference peaks.
...@@ -477,14 +482,15 @@ $$ ...@@ -477,14 +482,15 @@ $$
We now see that the reciprocal lattice points with nonzero amplitude exactly form the reciprocal lattice of the FCC lattice. We now see that the reciprocal lattice points with nonzero amplitude exactly form the reciprocal lattice of the FCC lattice.
### Powder Diffraction ### Powder diffraction
The easiest way to do diffraction measurements is to take a crystal, shoot an X-ray beam through it and measure the direction of outgoing waves. The easiest way to do diffraction measurements is to take a crystal, shoot an X-ray beam through it and measure the direction of outgoing waves.
However growing a single crystal may be hard because many materials are polycrystalline However growing a single crystal may be hard because many materials are polycrystalline
A simple alternative is to perform **powder diffraction**. A simple alternative is to perform **powder diffraction**.
By crushing the crystal into a powder, the small crystallites are now orientated in random directions. Crushing the crystal into a powder results in many small crystallites that are oriented in random directions.
This improves the chances of fulfilling the Laue condition for a fixed direction incoming beam. This improves the chances of fulfilling the Laue condition for a fixed direction incoming beam.
The experiment is illustrated in the figure above. The experiment is illustrated in the figure below.
The result is that the diffracted beam exits the sample via concentric circles at discrete **deflection angles** $2 \theta$. The result is that the diffracted beam exits the sample via concentric circles at discrete **deflection angles** $2 \theta$.
```python ```python
...@@ -530,8 +536,7 @@ plt.ylim([-1,0.5]) ...@@ -530,8 +536,7 @@ plt.ylim([-1,0.5])
plt.axis('off'); plt.axis('off');
``` ```
To deduce the values of $\theta$ of a specific crystal, let us put the Laue condition into a more practical form.
In order to deduce the values of $\theta$ of a specific crystal, let us put the Laue condition into a more practical form.
We first take the modulus squared of both sides: We first take the modulus squared of both sides:
\begin{align} \begin{align}
...@@ -539,7 +544,6 @@ We first take the modulus squared of both sides: ...@@ -539,7 +544,6 @@ We first take the modulus squared of both sides:
G^2 &= 2k^2-2\mathbf{k'} \cdot \mathbf{k}, G^2 &= 2k^2-2\mathbf{k'} \cdot \mathbf{k},
\end{align} \end{align}
where we used $|\mathbf{k'}| = |\mathbf{k}|$. where we used $|\mathbf{k'}| = |\mathbf{k}|$.
We then substitute the Laue condition $\mathbf{k'} = \mathbf{k}+\mathbf{G}$: We then substitute the Laue condition $\mathbf{k'} = \mathbf{k}+\mathbf{G}$:
...@@ -582,8 +586,8 @@ $$ ...@@ -582,8 +586,8 @@ $$
Bragg's law allows us to obtain atomic distances in the crystal $d_{hkl}$ through powder diffraction experiments! Bragg's law allows us to obtain atomic distances in the crystal $d_{hkl}$ through powder diffraction experiments!
## Summary ## Summary
* We described how to construct a reciprocal lattice from a real-space lattice. * We described how to construct a reciprocal lattice from a real-space lattice.
* Points in reciprocal space that differ by a reciprocal lattice vector are equivalent. * Points in reciprocal space that differ by a reciprocal lattice vector are equivalent.
* Diffraction experiments reveal information about crystal structure. * Diffraction experiments reveal information about crystal structure.
...@@ -591,72 +595,78 @@ Bragg's law allows us to obtain atomic distances in the crystal $d_{hkl}$ throug ...@@ -591,72 +595,78 @@ Bragg's law allows us to obtain atomic distances in the crystal $d_{hkl}$ throug
* Structure factor: describes the contribution of the atoms in a unit cell to the diffraction pattern. * Structure factor: describes the contribution of the atoms in a unit cell to the diffraction pattern.
* Powder diffraction and relating its experimental results to the crystal structure via Bragg's law. * Powder diffraction and relating its experimental results to the crystal structure via Bragg's law.
## Exercises ## Exercises
### Warm up exercises ### Warm up exercises*
1. Calculate $\mathbf{a}_1 \cdot \mathbf{b}_1$ and $\mathbf{a}_2 \cdot \mathbf{b}_1$ using the definitions of the reciprocal lattice vectors given in the lecture. Is the result what you expected?
2. Why is the amplitude of a scattered wave zero if $\mathbf{k'}-\mathbf{k} \neq \mathbf{G}$? 1. Study the 1D phonon dispersion relation that was plotted in the Tight-binding model lecture. Identify the reciprocal lattice points and the first Brillouin zone. Confirm that the first Brillouin zone is the Wigner Seitz unit cell of the reciprocal lattice.
3. Suppose we have a unit cell with a single atom in it. 2. Use the [scalar triple product](https://en.wikipedia.org/wiki/Triple_product#Scalar_triple_product) and the definitions of the 3D reciprocal lattice vectors given in the lecture to calculate $\mathbf{a}_1 \cdot \mathbf{b}_1$ and $\mathbf{a}_2 \cdot \mathbf{b}_1$. Discuss if the result is as expected.
Can any intensity peaks dissapear as a result of the structure factor? 3. Why is the amplitude of a scattered wave zero if $\mathbf{k'}-\mathbf{k} \neq \mathbf{G}$?
4. Can increasing the unit cell in real space introduce new diffraction peaks due to reciprocal lattice having more points? 4. Suppose we have a unit cell with a single atom in it. Can any intensity peaks dissapear as a result of the structure factor?
5. Can increasing the unit cell in real space introduce new diffraction peaks due to reciprocal lattice having more points?
### Exercise 1*: The reciprocal lattice of the bcc and fcc lattices
In this lecture, we studied how to construct the reciprocal lattice from a real-space lattice. We will now zoom in the properties of the fcc and bcc lattices, which are two of the most common lattices encountered in crystal structures. We analyze the reciprocal lattices and the shape of the first Brillouin zone. This helps us understanding e.g. the periodicity of 3D band structures and the Fermi surface database shown in the Attic.
### Exercise 1: Equivalence of direct and reciprocal lattice We consider a bcc lattice of which the conventional unit cell has a side length $a$ and volume $V=a^3$.
The volume of a primitive cell of a lattice with lattice vectors $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$ [equals](https://en.wikipedia.org/wiki/Parallelepiped#Volume) $V = |\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)|$. 1. Write down the primitive lattice vectors of the [BCC lattice](https://solidstate.quantumtinkerer.tudelft.nl/test_builds/lecture_9/9_crystal_structure/#body-centered-cubic-lattice). Calculate the volume of the primitive unit cell. Is the volume the expected fraction of $V$?
#### Question 1. 2. Calculate the reciprocal lattice vectors associated with the primitive lattice vectors you found in the previous subquestion.
Find the volume of a primitive unit cell $V^* = \left| \mathbf{b}_1 \cdot (\mathbf{b}_2 \times \mathbf{b}_3) \right|$ of the corresponding reciprocal lattice.
#### Question 2.
Derive the expressions for the lattice vectors $\mathbf{a}_i$ through the reciprocal lattice $\mathbf{b}_i$.
??? hint 3. Sketch the reciprocal lattice. Which type of lattice is it? What is the volume of its conventional unit cell?
Make use of the vector identity $\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = \mathbf{B}(\mathbf{A}\cdot\mathbf{C}) - \mathbf{C}(\mathbf{A}\cdot\mathbf{B})$
#### Question 3. 4. Describe the shape of the 1st Brillouin zone. How many sides does it have? (Note that the Brillouin zones are sketched in the Fermi surface periodic table in the Attic). Calculate the volume of the 1st Brillouin zone and check if it is the expected fraction of the volume you found in the previous subquestion.
Write down the primitive lattice vectors of the [BCC lattice](https://solidstate.quantumtinkerer.tudelft.nl/test_builds/lecture_9/9_crystal_structure/#body-centered-cubic-lattice) and calculate its reciprocal lattice vectors.
Which type of lattice is the reciprocal lattice of a BCC crystal? 5. Based on the insight gained in this question, argue what lattice is the reciprocal lattice of the _fcc_ lattice.
#### Question 4.
Determine the shape of the 1st Brillouin zone.
### Exercise 2: Miller planes and reciprocal lattice vectors ### Exercise 2: Miller planes and reciprocal lattice vectors
Miller indices are central to the description of the various planes in crystals. In this question we will analyze the Miller indices and their associated reciprocal lattice vectors, and show that the distance between Miller planes follows from the length of these vectors. We also highlight the convenience of using the conventional unit cell for describing cubic crystal structures.
Consider a family of Miller planes $(hkl)$ in a crystal. Consider a family of Miller planes $(hkl)$ in a crystal.
#### Question 1. 1. Prove that the reciprocal lattice vector $\mathbf{G} = h \mathbf{b}_1 + k \mathbf{b}_2 + l \mathbf{b}_3$ is perpendicular to the Miller plane $(hkl)$.
Prove that the reciprocal lattice vector $\mathbf{G} = h \mathbf{b}_1 + k \mathbf{b}_2 + l \mathbf{b}_3$ is perpendicular to the Miller plane $(hkl)$.
??? hint
Choose two vectors that lie within the Miller plane and are not parallel to each other.
2. Show that the distance between two adjacent Miller planes $(hkl)$ is $d = 2\pi/|\mathbf{G}_\textrm{min}|$, where $\mathbf{G}_\textrm{min}$ is the shortest reciprocal lattice vector perpendicular to these Miller planes.
3. In exercise 1, you derived the reciprocal lattice vectors of the BCC lattice from a set of primitive lattice vectors. Use these vectors to find the family of Miller planes that has the highest density of lattice points $\rho$. Use that $\rho = d/V$, where $V$ is the volume of the primitive unit cell and $d$ is the distance between adjacent planes derived in the previous subquestion. Formulate the Miller plane indices with respect to the primitive lattice vectors.
4. Make a sketch of the BCC structure and identify a Miller plane with the highest density of lattice points. Hint: it may help to make a sketch of the projections of the real-space lattice vectors $\{\mathbf{a_i}\}$ onto the $xy$ plane to identify which plane the Miller indices correspond to.
5. For cubic crystal structures, the interpretation of Miller indices strongly simplifies by using the lattice vectors of the conventional instead of a primitive unit cell. However, when doing so not all reciprocal lattice vectors correspond to a family of lattice planes. The reason is that such a family, by definition, must contain all lattice points.
??? hint Consider the reciprocal lattice vector $\mathbf{G} = h\mathbf{b_1} +k\mathbf{b_2} + l\mathbf{b_3}$, constructed from the conventional unit cell, such that $\mathbf{b_1}=2\pi/a\mathbf{\hat{x}}$, $\mathbf{b_2}=2\pi/a\mathbf{\hat{y}}$, $\mathbf{b_3}=2\pi/a\mathbf{\hat{z}}$. Do the indices $(hkl) = (100)$ correspond to a family of lattice planes? And $(200)$?. Discuss why (not).
Choose two vectors that lie within the Miller plane and are not parallel to each other.
#### Question 2. 6. As in the previous subquestion, here we still consider the reciprocal lattice constructed from the conventional unit cell. It turns out that, to understand which sets of $(hkl)$ indices can describe families of lattice planes, we can use the structure factor. Calculate (or recall from a previous question) the structure factor for the BCC lattice and discuss which sets of $(hkl)$ indices describe Miller planes. From this, identify the Miller planes with the highest density of lattice points and check if you got the same result as in subquestion 4.
Show that the distance between two adjacent Miller planes $(hkl)$ of any lattice is $d = 2\pi/|\mathbf{G}_\textrm{min}|$, where $\mathbf{G}_\textrm{min}$ is the shortest reciprocal lattice vector perpendicular to these Miller planes.
#### Question 3.
Find the family of Miller planes of the BCC lattice that has the highest density of lattice points. To solve this problem use that the density of lattice points per unit area on a Miller plane is $\rho = d/V$. Here $V$ is the volume of the primitive unit cell and $d$ is the distance between adjacent planes given in 2.2.
### Exercise 3: X-ray scattering in 2D ### Exercise 3: X-ray scattering in 2D
*(adapted from ex 14.1 and ex 14.3 of "The Oxford Solid State Basics" by S.Simon)* *(adapted from ex 14.1 and ex 14.3 of "The Oxford Solid State Basics" by S.Simon)*
Consider a two-dimensional crystal with a rectangular lattice and lattice vectors $\mathbf{a}_1 = (0.468, 0)$ nm and $\mathbf{a}_2 = (0, 0.342)$ nm (so that $\mathbf{a}_1$ points along $x$-axis and $\mathbf{a}_2$ points along $y$-axis). Using x-ray scattering, we can infer information on the crystal structure of a material. Here we visualize the geometry of this procedure by analyzing an elementary 2D crystal structure.
#### Question 1. Consider a two-dimensional crystal with a rectangular lattice and primitive lattice vectors $\mathbf{a}_1 = d_1\mathbf{\hat{x}}$ and $\mathbf{a}_2 = d_2\mathbf{\hat{y}}$, where $d_1=0.47$ nm and $d_2=0.34$ nm. We conduct an X-ray scattering experiment using monochromatic X-rays with wavelength $\lambda = 0.166$ nm. The wavevectors of the incident and reflected X-ray beams are $\mathbf{k}$ and $\mathbf{k'}$ respectively.
Sketch the reciprocal lattice of this crystal.
#### Question 2. 1. Calculate the reciprocal lattice vectors and sketch both the real- and the reciprocal lattice of this crystal.
Consider an X-ray diffraction experiment performed on this crystal using monochromatic X-rays with wavelength $0.166$ nm. By assuming elastic scattering, find the magnitude of the wave vectors of the incident and reflected X-ray beams. 2. Consider an X-ray diffraction experiment performed on this crystal using monochromatic X-rays with wavelength $\lambda = 0.166$ nm. By assuming elastic scattering, find the magnitude $k$ of the wavevectors of the incident and reflected X-ray beams.
#### Question 3. 3. In the sketch of the real-space lattice of subquestion 1, indicate a (210) Miller plane. Indicatet the associated reciprocal lattice vector $\mathbf{G}$ in the sketch of the reciprocal lattice. Also sketch the "scattering triangle" formed by the vectors $\mathbf{k}$, $\mathbf{k'}$, and $\mathbf{G}$ corresponding to diffraction from (210) planes.
On the reciprocal lattice sketched in 3.1, draw the "scattering triangle" corresponding to the diffraction from (210) planes. To do that use the Laue condition $\Delta \mathbf{k} = \mathbf{G}$ for the constructive interference of diffracted beams. 4. Sketch the first 5 peaks in an x-ray powder diffraction spectrum of this crystal as a function of $\sin 2\theta$, where $\theta$ is the deflection angle. Label the peaks according the Miller indices. Make sure you have the correct order of the peaks. Are there missing peaks because of the structure factor?
### Exercise 4: Structure factors and powder diffraction ??? Hint
Use the result of exercise 2 to express the distance between Miller planes in terms of the length of the reciprocal lattice vector $d_{hkl} = 2\pi/|\mathbf{G_{hkl}}|$
#### Question 1.
Compute the structure factor $\mathbf{S}$ of the BCC lattice. ### Exercise 4: Analyzing a 3D power diffraction spectrum
#### Question 2.
Which diffraction peaks are missing? In this question, we analyze the diffraction pattern we expect for an x-ray experiment on a 3D material with a BCC crystal structure.
#### Question 3.
How does this structure factor change if the atoms in the center of the conventional unit cell have a different form factor from the atoms at the corner of the conventional unit cell? 1. Using a conventional unit cell plus a basis to construct the BCC crystal structure, calculate the structure factor $\mathbf{S}$. (assume all the atoms to be the same).
#### Question 4. 2. Which diffraction peaks are missing because of the structure factor? Discuss why they are missing in relation to the crystal structure and the conventional unit cell.
A student carried out X-ray powder diffraction on Chromium (Cr) which is known to have a BCC structure. The first five diffraction peaks are given below. Furthermore, the student took the liberty of assigning Miller indices to the peaks. Were the peaks assigned correctly? Fix any mistakes and explain your reasoning. 3. How does this structure factor change if the atom in the center of the conventional unit cell has a different form factor from the atoms at the corners?
4. A student carried out an X-ray powder diffraction experiment on chromium (Cr) which is known to have a BCC structure. The measured spectrum is shown given below. Furthermore, the student assigned Miller indices to the peaks. Were these indices assigned correctly? Fix any mistakes and explain your reasoning.
![](figures/cr_xray_exercise.svg) ![](figures/cr_xray_exercise.svg)
#### Question 5. 5. Calculate the lattice constant $a$ of the conventional chromium bcc unit cell. Use that the X-ray diffraction experiment was carried out using Cu K-$\alpha$ (wavelength $\lambda = 1.5406$ Å) radiation.
Calculate the lattice constant, $a$, of the chromium bcc unit cell. Note that X-ray diffraction was carried out using Cu K-$\alpha$ ($1.5406$Å) radiation.
docs/10_xray_solutions.md
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...@@ -7,106 +7,74 @@ from matplotlib import pyplot as plt ...@@ -7,106 +7,74 @@ from matplotlib import pyplot as plt
import numpy as np import numpy as np
from math import pi from math import pi
``` ```
# Solutions for lecture 10 exercises # Solutions for lecture 10 exercises
## Warm-up 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:
$$ 1. The reciprocal lattice of a 1D lattice defined by $R = n_1 a$ is $G = m_1 2\pi/a$, with $n_1, m_1 \in \mathbb{Z}$. The Wigner Seitz cell defines the first Brillouin zone, and is constructed by connecting nearest-neighbor reciprocal lattice points and drawing perpendicular bisectors.
(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d})-(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c}) 2. We expect $\mathbf{a_i}\cdot\mathbf{b_j} = 2\pi\delta_{ij}$.
$$ 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. ## Exercise 2*. The reciprocal lattice of the bcc lattice
Because the relation between direct and reciprocal lattice is symmetric, so are the expressions for the direct lattice vectors through the reciprocal ones:
$$ 1. A possible set of BCC primitive lattice vectors is:
\mathbf{a}_{i} \epsilon_{ijk} = \frac{2\pi}{V^*} (\mathbf{b}_{j} \times \mathbf{b}_{k})
$$
where $\epsilon_{ijk}$ is the [Levi-Civita tensor](https://en.wikipedia.org/wiki/Levi-Civita_symbol#Three_dimensions) \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. 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 $|\mathbf{a_1}\times\mathbf{a_2}\cdot\mathbf{a_3}| = a^3/2$.
One set of the BCC primitive lattice vectors is given by:
$$ 2. The corresponding set of reciprocal lattice vectors
\mathbf{a_1} = \frac{a}{2} \left(-\hat{\mathbf{x}}+\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\
\mathbf{a_2} = \frac{a}{2} \left(\hat{\mathbf{x}}-\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\
\mathbf{a_3} = \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}}-\hat{\mathbf{z}} \right).
$$
From this, we find the following set of reciprocal lattice 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}
$$ 3. The reciprocal lattice forms an FCC lattice. Given the lattice vectors, the volume of the conventional unit cell is $(4\pi/a)^3$.
\mathbf{b_1} = \frac{2 \pi}{a} \left(\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\
\mathbf{b_2} = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{z}} \right) \\
\mathbf{b_3} = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right),
$$
which forms a reciprocal 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.
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.
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. 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. ## Exercise 2: Miller planes and reciprocal lattice vectors
Hints
??? 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 $\frac{\mathbf{a_1}}{h}, \frac{\mathbf{a_2}}{k}, \frac{\mathbf{a_3}}{l}$. 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
Can you define a general vector inside the $(hkl)$ plane?
??? 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 $(h00)$. 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.
For lattice planes, there is always a plane intersecting the zero lattice point (0,0,0).
As such, the distance from this plane to the closest next one is given by:
$ d = \hat{\mathbf{n}} \cdot \frac{\mathbf{a_1}}{h} = \frac{2 \pi}{|G|} $ 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. 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)
Since $\rho=d / V$, we must maximize $d$.
To do that, we minimize must $|G|$.
Therefore the smallest possible reciprocal lattice vectors are the (100) family of planes (in terms of FCC primitive lattice vectors).
## Exercise 3: X-ray scattering in 2D ## Exercise 3: X-ray scattering in 2D
#### Question 1.
See figure
```python ```python
def reciprocal_lattice(N = 7, lim = 40): def reciprocal_lattice(N = 7, lim = 40):
y = np.repeat(np.linspace(-18.4*(N//2),18.4*(N//2),N),N) y = np.repeat(np.linspace(-18.4*(N//2),18.4*(N//2),N),N)
...@@ -125,11 +93,9 @@ def reciprocal_lattice(N = 7, lim = 40): ...@@ -125,11 +93,9 @@ def reciprocal_lattice(N = 7, lim = 40):
reciprocal_lattice() reciprocal_lattice()
``` ```
#### Question 2. 1. See figure above
Since we have elastic scattering, we obtain 2. Since we have elastic scattering, we have $|\mathbf{k}| = |\mathbf{k}'| = \frac{2 \pi}{\lambda} = 37.9 nm^{-1}$
$|\mathbf{k}| = |\mathbf{k}'| = \frac{2 \pi}{\lambda} = 37.9 nm^{-1}$ 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
#### Question 3.
See figure
```python ```python
reciprocal_lattice() reciprocal_lattice()
...@@ -144,36 +110,57 @@ plt.arrow(-6,37.4,6+13.4*2,-37.4+18.4,color='k',zorder=11,head_width=2,length_in ...@@ -144,36 +110,57 @@ plt.arrow(-6,37.4,6+13.4*2,-37.4+18.4,color='k',zorder=11,head_width=2,length_in
plt.annotate('$\mathbf{k\'}$',(15,30),fontsize=14, ha='center',color='k'); plt.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 $(hkl)= (100) \quad (010) \quad (110) \quad (200), (020)$, where we took into account that $|\mathbf{b_1}|<|\mathbf{b_2}|$.
#### Question 1. ## Exercise 4: Analyzing a 3D power diffraction spectrum
$S(\mathbf{G}) = \sum_j f_j e^{i \mathbf{G} \cdot \mathbf{r_j}} = f(1 + e^{i \pi (h+k+l)})$
#### Question 2.
Solving for $h$, $k$, and $l$ results in
$$ 1. The structure factor is $S(\mathbf{G}) = \sum_j f_j e^{i \mathbf{G} \cdot \mathbf{r_j}} = f(1 + e^{i \pi (h+k+l)})$
S(\mathbf{G}) = \begin{cases}
2f, \: \text{if $h+k+l$ is even}\\
0, \: \text{if $h+k+l$ is odd}.
\end{cases}
$$
Thus if $h+k+l$ is odd, diffraction peaks dissapear 2. Solving for $h$, $k$, and $l$ results in
#### Question 3. $$
Let $f_1 \neq f_2$, then S(\mathbf{G}) = \begin{cases}
2f, \: \text{if $h+k+l$ is even}\\
0, \: \text{if $h+k+l$ is odd}.
\end{cases}
$$
$$ Thus if $h+k+l$ is odd, diffraction peaks are absent even though the Laue condition is satisfied. The reason is that the Laue condition is based on the reciprocal lattice vectors constructed from the conventional unit cell instead of a primitive unit cell.
S(\mathbf{G}) = \begin{cases}
f_1 + f_2, \text{if $h+k+l$ is even}\\
f_1 - f_2, \text{if $h+k+l$ is odd}
\end{cases}
$$
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:
$(110),(200),(211), (220), (310)$ $(110),(200),(211), (220), (310)$
#### Question 5. 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.
$a = 2.9100$Å
## 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/extra_exercises.md
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View file @ 90f15ca1
...@@ -219,6 +219,17 @@ The figure below shows a conventional unit cell of one of the possible crystal s ...@@ -219,6 +219,17 @@ The figure below shows a conventional unit cell of one of the possible crystal s
4. Let the atomic form factor of zinc be $f_{Zn}$ and that of sulfur be $f_{S}$. Derive an expression for the structure factor of the primitive unit cell of the crystal. 4. Let the atomic form factor of zinc be $f_{Zn}$ and that of sulfur be $f_{S}$. Derive an expression for the structure factor of the primitive unit cell of the crystal.
5. What will happen to the diffraction pattern if $f_{Zn} = f_S$? 5. What will happen to the diffraction pattern if $f_{Zn} = f_S$?
### Exercise 2: Equivalence of direct and reciprocal lattices
The volume of a primitive cell of a lattice with lattice vectors $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$ [equals](https://en.wikipedia.org/wiki/Parallelepiped#Volume) $V = |\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)|$.
1. Find the volume of a primitive unit cell $V^* = \left| \mathbf{b}_1 \cdot (\mathbf{b}_2 \times \mathbf{b}_3) \right|$ of the corresponding reciprocal lattice.
2. Derive the expressions for the lattice vectors $\mathbf{a}_i$ through the reciprocal lattice $\mathbf{b}_i$.
??? hint
Make use of the vector identity $\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = \mathbf{B}(\mathbf{A}\cdot\mathbf{C}) - \mathbf{C}(\mathbf{A}\cdot\mathbf{B})$
## Nearly free electron model ## Nearly free electron model
### Exercise 1*: nearly free electrons in aluminium ### Exercise 1*: nearly free electrons in aluminium
One material that can be described particularly well with the nearly free electron model is aluminium. Aluminium has a fcc crystal structure, which means that the reciprocal space can be described with an bcc lattice. The first Brillouin zone of a fcc crystal is depicted in the figure below. We further set the edge length of the cubic conventional unit cell of the reciprocal bcc lattice to $2\pi/a$. One material that can be described particularly well with the nearly free electron model is aluminium. Aluminium has a fcc crystal structure, which means that the reciprocal space can be described with an bcc lattice. The first Brillouin zone of a fcc crystal is depicted in the figure below. We further set the edge length of the cubic conventional unit cell of the reciprocal bcc lattice to $2\pi/a$.
......
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