Solutions lecture 7

src/7_tight_binding.md

@@ -14,6 +14,14 @@ pi = np.pi
 # Electrons and phonons in 1D
 _(based on chapters 9.1-9.3 & 11.1-11.3 of the book)_  
 
+!!! success "Expected prior knowledge"
+
+    Before the start of this lecture, you should be able to:
+
+    - Formulate Hooke's law
+    - Know your way around complex exponents
+    - Take derivatives of inverse $\sin$ and $\cos$ functions
+
 !!! summary "Learning goals"
 
     After this lecture you will be able to:
@@ -50,6 +58,8 @@ For a system of size $L = Na$ with periodic boundary conditions, we also have $u
 
 ### Electrons
 
+The following figure shows the inter-atomic potential with atoms placed at $ka$ with $k \in \z$:
+
 ![](figures/lattice_potential.svg)
 
 Formulating the equation of motion for electrons is very similar: we consider a LCAO wave function $|\Psi \rangle = \sum_n \phi_n |n \rangle$ and assume a nearest-neighbour hopping $-t$ and on-site energy $E_0$. We thus need to solve the Schrödinger equation
@@ -71,11 +81,11 @@ and for electrons:
 $$\phi_n = Be^{i E t/\hbar - i k x_n},$$
 where $x_n=na$ and where we wrote the time-dependent solution of the Schrödinger equation to emphasize the similarity between the two systems.
 
-As usual, periodicity quantizes k-space by requiring
+As we have seen before, the periodic boundary conditions (\phi_0=\phi_N) quantize k-space by requiring
 
-$$e^{ikx_n} = e^{ik(x_n+L)} \quad ⇒ \quad k = p \frac{2\pi}{L} $$
+$$ e^{ik0 + 2\pi i p} = e^{ikL} \quad ⇒ \quad k = p \frac{2\pi}{L} $$
 
-with $p\in \mathbb{Z}$. As such, $e^{ikx_n} = e^{i p \frac{2\pi}{L} n a} = e^{i \frac{2 \pi n p}{N}}$ and we see that changing $p→p+N$ corresponds to exactly the same solution. Therefore, we have $N$ different solutions in total. Furthermore, solutions with $k$-values that differ by an integer multiple of $N\frac{2\pi}{L} = \frac{2\pi}{a}$ are identical (see figure).
+with $p \in \mathbb{Z}$. As such, $e^{ikx_n} = e^{i p \frac{2\pi}{L} n a} = e^{i \frac{2 \pi n p}{N}}$ and we see that changing $p→p+N$ corresponds to exactly the same solution. Therefore, we have $N$ different solutions in total. Furthermore, solutions with $k$-values that differ by an integer multiple of $N\frac{2\pi}{L} = \frac{2\pi}{a}$ are identical (see figure).
 
 ```python
 x = np.linspace(-.2, 2.8, 500)
@@ -133,8 +143,9 @@ draw_classic_axes(ax)
 ax.annotate(s='', xy=(-pi, -.15), xytext=(pi, -.15),
             arrowprops=dict(arrowstyle='<->', shrinkA=0, shrinkB=0))
 ax.text(0, -.25, '1st Brillouin zone', ha='center')
-ax.set_ylim(bottom=-.7);
+ax.set_ylim(bottom=-.3);
 ```
+
 Here $k_p = 2\pi p / L$ for $0 ≤ p < N$ are *all the normal modes* of the crystal, and therefore we can rederive a better Debye model!
 
 Before we had $\sum_p → \frac{L}{2\pi}\int_{-\omega_D/v_s}^{\omega_D/v_s}dk$, because we introduced the cutoff to fix the problem with the number of modes.
@@ -176,7 +187,7 @@ Comparing this with $E=(\hbar k)^2/2m$, we see that the dispersion is similar to
 In the following lectures we will see that an electron dispersion usually has multiple options for $E(k)$, each called an energy band. The complete dispersion relation is also called a *band structure*.
 
 
-## Group velocity, effective mass, density of states
+### Group velocity, effective mass, density of states
 
 *(here we only discuss electrons; for phonons everything is the same except for replacing $E = \hbar \omega$)*
 
@@ -205,11 +216,11 @@ The connection to quantum mechanics is made by $p = \hbar k$. Analogously, the g
 
 Similarly, the **effective mass** is defined by:
 
-$$m_{eff} \equiv F\left(\frac{dv}{dt}\right)^{-1} $$.
+$$m_{eff} \equiv F\left(\frac{dv}{dt}\right)^{-1} $$
 
 Substituting, we get
 
-$$m_{eff} = \hbar^2\left(\frac{d^2 E(k)}{dk^2}\right)^{-1}$$.
+$$m_{eff} = \hbar^2\left(\frac{d^2 E(k)}{dk^2}\right)^{-1}$$
 
 ```python
 pyplot.figure(figsize=(8, 5))