Solutions lecture 7

src/8_many_atoms_sol.md

@@ -36,17 +36,33 @@ A unit cell consits of exactly one $t_1$ and exactly $t_2$ hopping.
 ### Subquestion 2
 
 Using the hint we find:
-$$ E \phi_n = \epsilon \phi_n + t_1 \psi_n + t_2 \psi_{n-1} $$
-$$ E \psi_n = \epsilon \psi_n + t_2 \phi_{n+1} + t_2 \phi_n $$
+$$ E \phi_0 = \epsilon \phi_0 + t_1 \psi_0 + t_2 \psi_{n-1} $$
+$$ E \psi_0 = \epsilon \psi_0 + t_2 \phi_{n+1} + t_2 \phi_0 $$
 
 Notice that the hopping, in this case, is without the '-'-sign!
 
+### Subquestion 3
+
 Using the Ansatz and rearranging the equation yields:
 
-$$ E \begin{pmatrix} \phi_n \\ \psi_n \end{pmatrix} = 
+$$ 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_n \\ \psi_n \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='2atom')
+pyplot.plot(k, -np.sqrt(5+2*2*np.cos(k)),'b',label='2atom')
+pyplot.plot(k, -3*np.cos(k/2),'r',label='1atom')
+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([2atom,1atom],['2 atom unit cell dispersion','single atom unit cell dispersion'])
+```