Solutions lecture 7

src/8_many_atoms_sol.md

@@ -27,3 +27,42 @@ Optical branch corresponds with (+) in the equation given in the lecture notes.
 
 Density of states is given as $g(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega}$. $\frac{dN}{dk} = \frac{L}{2\pi}$ since we have 1D. $\frac{dk}{d\omega}$ can be computed using the group velocity: $$\frac{dk}{d\omega} = \bigg(\frac{d\omega}{dk} \bigg)^{-1} = (v_g)^{-1}$$
 
+## Exercise 2: the Peierls transition
+
+### Subquestion 1
+
+A unit cell consits of exactly one $t_1$ and exactly $t_2$ hopping.
+
+### Subquestion 2
+
+Using the hint we find:
+$$ E \phi_0 = \epsilon \phi_0 + t_1 \psi_0 + t_2 \psi_{n-1} $$
+$$ E \psi_0 = \epsilon \psi_0 + t_2 \phi_{n+1} + t_2 \phi_0 $$
+
+Notice that the hopping, in this case, is without the '-'-sign!
+
+### Subquestion 3
+
+Using the Ansatz and rearranging the equation yields:
+
+$$ E \begin{pmatrix} \phi_0 \\ \psi_0 \end{pmatrix} = 
+\begin{pmatrix} \epsilon & t_1 + t_2 e^{-ika} \\ t_1 + t_2 e^{ika} & \epsilon \end{pmatrix} 
+\begin{pmatrix} \phi_0 \\ \psi_0 \end{pmatrix}
+$$
+
+### Subquestion 4
+
+The dispersion is given by: $$ E = \epsilon \pm \sqrt{t_1^2 + t_2^2 + 2t_1t_2\cos(ka)} $$
+
+```python
+pyplot.figure()
+k = np.linspace(-2*pi, 2*pi, 300)
+pyplot.plot(k, np.sqrt(5+2*2*np.cos(k)),'b',label='2 atom dispersion')
+pyplot.plot(k, -np.sqrt(5+2*2*np.cos(k)),'b',label='2 atom dispersion')
+pyplot.plot(k, -3*np.cos(k/2),'r',label='1 atom dispersion')
+pyplot.xlabel('$ka$'); pyplot.ylabel(r'$E-\epsilon$')
+pyplot.xticks([-2*pi, -pi, 0, pi,2*pi], [r'$-2ka$',r'$-ka$', 0, r'$ka$',r'$2ka$'])
+pyplot.yticks([-3, 0, 3], [r'$-t_1-t_2$', '$E_0$', r'$t_1+t_2$']);
+pyplot.legend()
+```
+