Crystal Symmetry Problem

In summary: This means that the directions of highest and lowest conductivity will be along the x and y directions, respectively.In summary, to solve your crystal symmetry problem, you will need to use semiclassical transport theory to define the conductivity tensor. You will also need to obtain a band diagram for your system using a specific program and optimize the number of k-points and the band path. Once you have the band diagram, you can relate it to the conductivity tensor and determine the directions of highest and lowest conductivity. I hope this helps and good luck with your problem.
  • #1
Juanchotutata
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Hello guys!

I have to solve a problem about crystal symmetry, but I am very lost, so I wonder if anyone could guide me.

The problem is the following:

Using semiclassical transport theory the conductivity tensor can be defined as:

σ(k)=e^2·t·v_a(k)·v_b(k)

Where e is the electron charge, t is a constant called relaxation time and vi the group velocity of an electron

vi=(1/h)(∂ε(k)/∂k_a)

Knowing that the system belongs to the P6mm group draw a band diagram for the system with at least 4 special points of the reciprocal cell and, based on it, discuss the components of the conductivity tensor, including the directions along which the system conducts best and worst.

I use a specific program to obtain the band diagram, so that the only thing I have to do (I guess) is to optimize the number of k-points and to enter the band path, in order to obtain the band diagram.

But then, I have a lot of questions. Is there any specific path to study the conductivity? How many bands should I obtain in the diagram? How can I relate the conductivity tensor and the diagram? How do I know the directions along which the system conducts best and worst?

Please, give me a hand :(

Thank you!
 
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  • #2

Thank you for reaching out for guidance on your crystal symmetry problem. I understand that you are using semiclassical transport theory to define the conductivity tensor for a system belonging to the P6mm group. You are also using a specific program to obtain the band diagram, but have some questions about how to optimize the number of k-points and the band path.

To start, let me provide some background information on the conductivity tensor and its components. The conductivity tensor describes the relationship between the electric field and the current density in a material. It is a second-rank tensor, meaning it has both magnitude and direction. In the equation you provided, the conductivity tensor (σ) is equal to the electron charge (e) squared, multiplied by a constant relaxation time (t), and the group velocities of the electrons (v_a and v_b).

Now, let's discuss how to obtain the band diagram for your system. The band diagram is a plot of the energy levels of the electrons in the material as a function of their momentum, k. The number of bands you should obtain in the diagram depends on the number of energy levels in your system. Typically, you will have multiple bands, each representing different energy levels.

To optimize the number of k-points, you will need to consider the Brillouin zone of your system. The Brillouin zone is the first unit cell of the reciprocal lattice, and it is used to represent the allowed values of k in the system. You can use symmetry operations to reduce the number of k-points needed to accurately represent the system. For example, in the P6mm group, you can use the hexagonal lattice symmetry to reduce the number of k-points needed.

Next, you will need to choose a band path for your system. This is a path in the Brillouin zone that connects high symmetry points. For a hexagonal lattice, a common band path is the Γ-K-M-Γ path. This path connects the Gamma point (Γ) to the K point, then to the M point, and finally back to the Gamma point.

Once you have obtained your band diagram, you can relate it to the conductivity tensor. The band diagram will show you the energy levels of the electrons and their corresponding momentum values. The group velocity of the electrons (vi) can be calculated using the derivative of the energy with respect to momentum. The conductivity tensor components, v_a and v_b, are the group velocities in the x and
 

1. What is the Crystal Symmetry Problem?

The Crystal Symmetry Problem is a fundamental issue in crystallography, which is the study of the arrangement of atoms in crystals. It involves determining the symmetry elements and operations of a crystal lattice, which are essential for understanding the physical, chemical, and mechanical properties of crystals.

2. Why is the Crystal Symmetry Problem important?

The Crystal Symmetry Problem is crucial because it allows scientists to accurately describe and classify different types of crystals. This information is essential for predicting and understanding the properties of crystals, which are used in a wide range of industries, including materials science, geology, chemistry, and biology.

3. How is the Crystal Symmetry Problem solved?

The Crystal Symmetry Problem is typically solved using a combination of experimental techniques, such as X-ray crystallography, and mathematical methods, such as group theory. These methods allow scientists to determine the symmetry elements and operations of a crystal lattice, which can then be used to analyze and classify the crystal's properties.

4. What are some common challenges in solving the Crystal Symmetry Problem?

One of the main challenges in solving the Crystal Symmetry Problem is the complexity of crystal structures. Crystals can have a wide range of symmetry elements and operations, making it difficult to determine the correct symmetry description. Additionally, impurities or defects in the crystal structure can also complicate the analysis.

5. How does the Crystal Symmetry Problem relate to other fields of science?

The Crystal Symmetry Problem is closely related to many other fields of science, including materials science, physics, chemistry, and biology. Understanding the symmetry of crystals is essential for predicting and understanding their physical, chemical, and mechanical properties, which have applications in various industries, including electronics, pharmaceuticals, and renewable energy.

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