Verified Commit cd102caa authored by Anton Akhmerov's avatar Anton Akhmerov
Browse files

add one more lecture video

parent 5113aa77
......@@ -18,6 +18,10 @@ _(based on chapter 16 of the book)_
- examine 1D and 2D band structures and argue if you expect the corresponding material to be an insulator/semiconductor or a conductor.
- describe how the light absorption spectrum of a material relates to its band structure.
??? info "Lecture video"
<iframe width="100%" height="315" src="https://www.youtube-nocookie.com/embed/eTx8FnVQ0pw" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
## Band structure
How are material properties related to the band structure?
......@@ -42,7 +46,7 @@ For a single band:
$$ N_{states} = 2 \frac{L^3}{(2\pi)^3} \int_{BZ} dk_x dk_y dk_z = 2 L^3 / a^3 $$
Here, $L^3/a^3$ is the number of unit cells in the system, so we see that a single band has room for 2 electrons per unit cell (the factor 2 comes from the spin).
Here, $L^3/a^3$ is the number of unit cells in the system, so we see that a single band has room for 2 electrons per unit cell (the factor 2 comes from the spin).
We come to the important rule:
......@@ -225,10 +229,10 @@ _(based on exercise 15.4 of the book)_
Suppose we have a square lattice with lattice constant $a$, with a periodic potential given by $V(x,y)=2V_{10}(\cos(2\pi x/a)+\cos(2\pi y/a))+4V_{11}\cos(2 \pi x/a)\cos(2 \pi y/a)$.
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 $(-\pi/a,0)$; ($\psi_{+}\sim e^{i\pi x /a}$ ; $\psi_{-}\sim e^{-i\pi x /a}$).
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.
\ No newline at end of file
4. Find its eigenvalues.
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment