Solutions lecture 7

src/8_many_atoms_sol.md

@@ -17,7 +17,7 @@ pi = np.pi
 
 ### Subquestion 1
 
-Accoustic branch corresponds with (-) in the equation given in the lecture notes. Use the small angle approximation $\sin(k) \approx k$ to ease calculations. For the Taylor polynomial take $\omega^2 = f(x) \approx f(0) + f'(0)k + f''(0)k^2$ (some terms vanish, computation is indeed quite tedious). You should find: $$|v_g| = \sqrt{\frac{\kappa a^2}{2(m_a+m_2)}}$$
+Accoustic branch corresponds with (-) in the equation given in the lecture notes. Use the small angle approximation $\sin(x) \approx x$ to ease calculations. For the Taylor polynomial take $\omega^2 = f(x) \approx f(0) + f'(0)k + f''(0)k^2$ (some terms vanish, computation is indeed quite tedious). You should find: $$|v_g| = \sqrt{\frac{\kappa a^2}{2(m_a+m_2)}}$$
 
 ### Subquestion 2
 
@@ -59,11 +59,11 @@ pyplot.figure()
 k = np.linspace(-2*pi, 2*pi, 400)
 t1 = 1;
 t2 = 1.5;
-pyplot.plot(k, np.sqrt(t1**2 + t2**2+2*t1*t2*np.cos(k)),'b',label='2 atom dispersion')
-pyplot.plot(k, -np.sqrt(t1**2 + t2**2+2*t1*t2*np.cos(k)),'b')
 pyplot.plot(k, -(t1+t2)*np.cos(k/2),'r',label='1 atom dispersion')
 pyplot.plot(k[199:100:-1],-(t1+t2)*np.cos(k[0:99]/2),'r--',label='1 atom dispersion with folded Brillouin zone')
 pyplot.plot(k[299:200:-1],-(t1+t2)*np.cos(k[300:399]/2),'r--')
+pyplot.plot(k, np.sqrt(t1**2 + t2**2+2*t1*t2*np.cos(k)),'b',label='2 atom dispersion')
+pyplot.plot(k, -np.sqrt(t1**2 + t2**2+2*t1*t2*np.cos(k)),'b')
 
 pyplot.xlabel('$ka$'); pyplot.ylabel(r'$E-\epsilon$')
 pyplot.xlim([-2*pi,2*pi])
@@ -75,4 +75,7 @@ pyplot.fill_between([-3*pi,3*pi], -np.abs(t1-t2), np.abs(t1-t2), color='red',alp
 
 pyplot.legend(loc='lower center');
 ```
+(Press the magic wand tool to enable the python code that created the figure to see what happends if you change $t_1$ and $t_2$.)
+
+Notice that the red shaded area is not a part of the *Band structure* anymore!