Solutions lecture 7

src/7_tight_binding_model_sol.md

@@ -55,7 +55,7 @@ pyplot.tight_layout();
 
 ### Subquestion 4
 
-Hint: The group velocity is given as $v = \frac{d\omega}{dk}$, draw a coordinate system **under** or **above** the dispersion graph with $k$ on the x-axis in which you draw $\frac{d\omega}{dk}$. Draw a coordinate system **next** to the dispersion with *$g(\omega)$ on the y-axis* in which you graph $\big(\frac{d\omega}{dk}\big)^{-1} \approx \frac{1}{\frac{d\omega}{dk}}$ (This is **definitely** not an equality sign, but you can use this 'approximation' to graph the density of states).
+Hint: The group velocity is given as $v = \frac{d\omega}{dk}$, draw a coordinate system **under** or **above** the dispersion graph with $k$ on the x-axis in which you draw $\frac{d\omega}{dk}$. Draw a coordinate system **next** to the dispersion with *$g(\omega)$ on the y-axis* in which you graph $\big(\frac{d\omega}{dk}\big)^{-1} \approx \frac{1}{\frac{d\omega}{dk}}$.
 
 ??? hint "Plots"
 
@@ -75,12 +75,59 @@ For the heat capacity we have: $$C = \frac{\partial U}{\partial T} = \int g(\ome
 
 ### Subquestion 1
 
-The Schrödinger equation is given as: $|\Psi\rangle = \sum_n \phi_n |n\rangle$ such that we find $$ E\phi_n = E_0\phi_n - t\phi_{n-1} - t\phi_{n+1} - t'\phi_{n-2} - t'\phi_{n+2}$$.
+The Schrödinger equation is given as: $|\Psi\rangle = \sum_n \phi_n |n\rangle$ such that we find $$ E\phi_n = E_0\phi_n - t\phi_{n-1} - t\phi_{n+1} - t'\phi_{n-2} - t'\phi_{n+2}$$
 
 ### Subquestion 2
 
 Solving the Schrödinger equation yields dispersion: $$E(k) = E_0 -t\cos(ka) -t'\cos(2ka)$$
 
-### Subquestion 3
+### Subquestiion 3
+
+$$m_{eff} = \frac{\hbar^2}{2a^2}\frac{1}{t\cos(ka)+4t'\cos(2ka)}$$
+
+Plot for t=t':
+
+```python
+k1 = np.linspace(-pi, -pi/2-0.01, 300);
+k2 = np.linspace(-pi/2+0.01, pi/2-0.01, 300);
+k3 = np.linspace(pi/2+0.01, pi, 300);
+
+pyplot.plot(k1, 1/(5*np.cos(k1)),'b');
+pyplot.plot(k2, 1/(5*np.cos(k2)),'b');
+pyplot.plot(k3, 1/(5*np.cos(k3)),'b');
+pyplot.xlabel('$k$'); pyplot.ylabel('$m_{eff}(k)$');
+pyplot.xticks([-pi,0,pi],[r'$-\pi/a$',0,r'$\pi/a$']);
+pyplot.yticks([],[]);
+pyplot.tight_layout();
+```
+
+### Subquestion 4
+
+Plots for 2t'=t, 4t'=t and 10t'=t:
+
+```python
+k1 = np.linspace(-pi, -pi/2-0.01, 300);
+k2 = np.linspace(-pi/2+0.01, pi/2-0.01, 300);
+k3 = np.linspace(pi/2+0.01, pi, 300);
+
+pyplot.plot(k1, 1/(5*np.cos(k1)),'b',label='t=2t\'');
+pyplot.plot(k2, 1/(5*np.cos(k2)),'b');
+pyplot.plot(k3, 1/(5*np.cos(k3)),'b');
+pyplot.xlabel('$k$'); pyplot.ylabel('$m_{eff}(k)$');
+pyplot.xticks([-pi,0,pi],[r'$-\pi/a$',0,r'$\pi/a$']);
+pyplot.yticks([-20,20],[]);
+pyplot.tight_layout();
+
+pyplot.plot(k1, 1/(17*np.cos(k1)),'r');
+pyplot.plot(k2, 1/(17*np.cos(k2)),'r');
+pyplot.plot(k3, 1/(17*np.cos(k3)),'r',label='t=4t\'');
+
+pyplot.plot(k1, 1/(41*np.cos(k1)),'k');
+pyplot.plot(k2, 1/(41*np.cos(k2)),'k');
+pyplot.plot(k3, 1/(41*np.cos(k3)),'k',label='t=10t\'');
+
+pyplot.legend();
+pyplot.show()
+```