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
 
@@ -36,8 +36,8 @@ A unit cell consits of exactly one $t_1$ and exactly $t_2$ hopping.
 ### Subquestion 2
 
 Using the hint we find:
-$$ E \phi_0 = \epsilon \phi_0 + t_1 \psi_0 + t_2 \psi_{n-1} $$
-$$ E \psi_0 = \epsilon \psi_0 + t_2 \phi_{n+1} + t_2 \phi_0 $$
+$$ E \phi_0 = \epsilon \phi_0 + t_1 \psi_0 + t_2 \psi_0 e^{-ika} $$
+$$ E \psi_0 = t_1 \phi_0 e^{ika} + t_2 \phi_0 + \epsilon \psi_0 $$
 
 Notice that the hopping, in this case, is without the '-'-sign!
 
@@ -55,22 +55,67 @@ $$
 The dispersion is given by: $$ E = \epsilon \pm \sqrt{t_1^2 + t_2^2 + 2t_1t_2\cos(ka)} $$
 
 ```python
-ppyplot.figure()
+pyplot.figure()
 k = np.linspace(-2*pi, 2*pi, 400)
 t1 = 1;
 t2 = 1.5;
+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])
+pyplot.ylim([-2*(t1-t2),2*(t1-t2)])
+pyplot.xticks([-2*pi, -pi, 0, pi,2*pi], [r'$-2\pi$',r'$-\pi$', 0, r'$\pi$',r'$2\pi$'])
+pyplot.yticks([-t1-t2, -np.abs(t1-t2), 0, np.abs(t1-t2), t1+t2], [r'$-t_1-t_2$',r'$-|t_1-t_2|$', '0', r'$|t_1-t_2|$', r'$t_1+t_2$']);
+pyplot.vlines([-pi, pi], -2*(t1-t2)*1.1,2*(t1-t2)*1.1, linestyles='dashed');
+pyplot.hlines([-np.abs(t1-t2), np.abs(t1-t2)], -2*pi, 2*pi, linestyles='dashed');
+pyplot.fill_between([-3*pi,3*pi], -np.abs(t1-t2), np.abs(t1-t2), color='red',alpha=0.2);
+
+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!
+
+### Subquestion 5
+
+Both $v_g$ and $m_{eff}$ can be calculated using the respective formulae given in the previous lectures
+
+### Subquestion 6
+
+We know $g(\omega) = \frac{dN}{dk} \frac{dk}{d\omega} = \frac{L}{2\pi} \frac{1}{v_g}$ with $v_g$ from the previous subquestion. Graphically the density of states looks accordingly:
+
+```python
+pyplot.subplot(1,2,1)
+pyplot.subplot(1,2,1)
+k = np.linspace(-2*pi, 2*pi, 400)
+t1 = 1;
+t2 = 1.5;
 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.xticks([-2*pi, -pi, 0, pi,2*pi], [r'$-2ka$',r'$-ka$', 0, r'$ka$',r'$2ka$'])
-pyplot.yticks([-t1-t2, np.abs(t1-t2), 0, -np.abs(t1-t2), t1+t2], [r'$-t_1-t_2$',r'$-|t_1-t_2|$', '0', r'$|t_1-t_2|$', r'$t_1+t_2$']);
-pyplot.vlines([-pi, pi], -3, 3, linestyles='dashed');
-pyplot.hlines([np.abs(t1-t2),-np.abs(t1-t2)],-2*pi,2*pi, linestyles='dashed');
-
+pyplot.xlim([-2*pi,2*pi])
+pyplot.xticks([-2*pi, -pi, 0, pi,2*pi], [r'$-2\pi$',r'$-\pi$', 0, r'$\pi$',r'$2\pi$'])
+pyplot.yticks([-t1-t2, -np.abs(t1-t2), 0, np.abs(t1-t2), t1+t2], [r'$-t_1-t_2$',r'$-|t_1-t_2|$', '0', r'$|t_1-t_2|$', r'$t_1+t_2$']);
 pyplot.legend(loc='lower center');
+
+pyplot.subplot(1,2,2)
+w = np.sqrt(t1**2 + t2**2+2*t1*t2*np.cos(k))
+pyplot.hist(w,30, orientation='horizontal',ec='black',color='b');
+pyplot.hist(-w,30, orientation='horizontal',ec='black',color='b');
+pyplot.xlabel(r'$g(\omega)$')
+pyplot.ylabel(r'$\omega$')
+pyplot.yticks([],[])
+pyplot.xticks([],[])
+pyplot.tight_layout();
 ```
 
+
+