A simple model of a crystal consists of a 1D linear array of sites at positions $x=n a$, for all integer $n$ and separation $a$, each occupied by a similar atom. The potential due to the atom at the origin is $U(x)$, which is symmetric: $U(-x)=U(x)$. The Hamiltonian, $H_{0}$, for the atom at the $n$-th site in isolation has electron eigenfunction $\psi_{n}(x)$ with energy $E_{0}$. Write down $H_{0}$ and state the relationship between $\psi_{n}(x)$ and $\psi_{0}(x)$.

The Hamiltonian $H$ for an electron moving in the crystal is $H=H_{0}+V(x)$. Give an expression for $V(x)$.

In the tight-binding approximation for this model the $\psi_{n}$ are assumed to be orthonormal, $\left(\psi_{n}, \psi_{m}\right)=\delta_{n m}$, and the only non-zero matrix elements of $H_{0}$ and $V$ are

$$\left(\psi_{n}, H_{0} \psi_{n}\right)=E_{0}, \quad\left(\psi_{n}, V \psi_{n}\right)=\alpha, \quad\left(\psi_{n}, V \psi_{n \pm 1}\right)=-A$$

where $A>0$. By considering the trial wavefunction $\Psi(x, t)=\sum_{n} c_{n}(t) \psi_{n}(x)$, show that the time-dependent Schrödinger equation governing the amplitudes $c_{n}(t)$ is

$$i \hbar \dot{c}_{n}=\left(E_{0}+\alpha\right) c_{n}-A\left(c_{n+1}+c_{n-1}\right)$$

By examining a solution of the form

$$c_{n}=e^{i(k n a-E t / \hbar)}$$

show that $E$, the energy of the electron in the crystal, lies in a band given by

$$E=E_{0}+\alpha-2 A \cos k a$$

Using the fact that $\psi_{0}(x)$ is a parity eigenstate show that

$$\left(\psi_{n}, x \psi_{n}\right)=n a .$$

The electron in this model is now subject to an electric field $\mathcal{E}$ in the direction of increasing $x$, so that $V(x)$ is replaced by $V(x)-e \mathcal{E} x$, where $-e$ is the charge on the electron. Assuming that $\left(\psi_{n}, x \psi_{m}\right)=0, n \neq m$, write down the new form of the time-dependent Schrödinger equation for the probability amplitudes $c_{n}$. Verify that it has solutions of the form

$$c_{n}=\exp \left[-\frac{i}{\hbar} \int_{0}^{t} \epsilon\left(t^{\prime}\right) d t^{\prime}+i\left(k+\frac{e \mathcal{E} t}{\hbar}\right) n a\right]$$

where

$$\epsilon(t)=E_{0}+\alpha-2 A \cos \left[\left(k+\frac{e \mathcal{E} t}{\hbar}\right) a\right]$$

Use this result to show that the dynamical behaviour of an electron near the bottom of an energy band is the same as that for a free particle in the presence of an electric field with an effective mass $m^{*}=\hbar^{2} /\left(2 A a^{2}\right)$.