The tight binding model

Exercises

Preliminary provocations

1. Compare the expression of the effective mass with Newton's second law. Do you observe any similarities?

2. Check the units of the effective mass. Is it what you expect?

   $$[m^*] = \frac{[p^2]}{[E]} = kg$$

3. Calculate the effective mass of the free-electron dispersion relation. Is this what was expected?

   $$m^* = m_e,$$

4. Under what condition is the effective mass the same for each electron?

   If the dispersion relation is parabolic, so in the free electron model.

Exercise 1: The next-nearest neighbour chain

Let's expand our one-dimensional chain model by extending the range of interaction to include the next-nearest neighbours:

$$\langle \phi_n | H | \phi_{n+2}\rangle \equiv -t' ≠ 0.$$

1. Write down the new Schrödinger equation for this system.

   The Schrödinger equation is given by: $|\Psi\rangle = \sum_n \phi_n |n\rangle$ such that we find $$E\phi_n = E_0\phi_n - t\phi_{n-1} - t\phi_{n+1} - t'\phi_{n-2} - t'\phi_{n+2}$$

2. Solve the Schrödinger equation to find the dispersion relation $E(k)$.

   Solving the Schrödinger equation yields dispersion: $$E(k) = E_0 -2t\cos(ka) -2t'\cos(2ka)$$

3. Calculate the effective mass $m^*$.

   $$m^* = \frac{\hbar^2}{2a^2}\frac{1}{t\cos(ka)+4t'\cos(2ka)}$$

   Plot for t=t':

   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(r'$k$'); pyplot.ylabel('$m_{eff}(k)$');
   pyplot.xticks([-pi,0,pi],[r'$-\pi/a$',0,r'$\pi/a$']);
   pyplot.yticks([],[]);
   pyplot.tight_layout();
   

4. Sketch the effective mass as a function of $k$ for the cases $t=2t'$,$t=4t'$ and $t=10t'$.

   Plots for 2t'=t, 4t'=t and 10t'=t:

   def m(k,t):
       return 1/(np.cos(k)+4*t*np.cos(2*k))
   
   k1 = np.linspace(-1.6, -0.83, 300);
   k2 = np.linspace(-0.826, 0.826, 300);
   k3 = np.linspace(0.83, 1.6, 300);
   
   pyplot.plot(k1, m(k1,2),'b');
   pyplot.plot(k2, m(k2,2),'b');
   pyplot.plot(k3, m(k3,2),'b',label='t=2t\'');
   pyplot.xlabel('$k$'); pyplot.ylabel('$m_{eff}(k)$');
   pyplot.xticks([-1.6,0,1.6],[r'$-\pi/a$',0,r'$\pi/a$']);
   pyplot.yticks([0],[]);
   pyplot.tight_layout();
   
   k1 = np.linspace(-1.58, -0.81, 300);
   k2 = np.linspace(-0.804, 0.804, 300);
   k3 = np.linspace(0.81, 1.58, 300);
   
   pyplot.plot(k1, m(k1,4),'r');
   pyplot.plot(k2, m(k2,4),'r');
   pyplot.plot(k3, m(k3,4),'r',label='t=4t\'');
   
   k1 = np.linspace(-1.575, -0.798, 300);
   k2 = np.linspace(-0.790, 0.790, 300);
   k3 = np.linspace(0.798, 1.575, 300);
   
   pyplot.plot(k1, m(k1,10),'k');
   pyplot.plot(k2, m(k2,10),'k');
   pyplot.plot(k3, m(k3,10),'k',label='t=10t\'');
   
   pyplot.legend();
   

---

Last update: October 30, 2023

Comments