The tight-binding approximation

It is instructive to look at the simple example of a chain composed of hydrogen-like atoms with a single s-orbital. This will serve to illustrate the main concepts in band structure calculations, such as momentum space, and Bloch functions.

Let us first define some identities: The wave function of an isolated atomic orbital centered on atom $j$ is $\phi({\bf r}-{\bf R}_j)$. We are going to use Direc's notation from now one, meaning that:

$$\phi({\bf r}-{\bf R}_j) \equiv \langle {\bf r} \vert j \rangle,$$ (169)

$$\int d^3r \phi^*({\bf r}-{\bf R}_j)\phi({\bf r}-{\bf R}_i) \equiv \langle i\vert j\rangle$$ (170)

We propose a solution of the form:

$$\vert\psi\rangle = \sum_{i=0}^{N-1} c_i \vert i\rangle$$ (171)

We are going to make the following assumptions:

$$\langle i\vert j \rangle = \delta_{i,j}$$ (172)

$$\langle i \vert H\vert j\rangle = -t \delta_{j,i\pm1}$$ (173)

$$\langle i\vert H\vert i \rangle = \epsilon.$$ (174)

The first one implies orthogonality of orbitals sitting on different sites, and this implies, as we are going to see later, that these are ``Wannier orbitals''.The second line means that the Hamiltonian only mizes orbitals sitting on neighboring sites, while the third just defines a ``site energy", which is just a constant shift. We are also neglectning the electron-electron interaction.

As a consequence of the above, the Hamiltonian matrix will be band diagonal:

$$H = \left( \begin{array}{ccccccc} \epsilon & -t & 0 & 0 & & ... ...n & -t \\ 0 & \cdots & & & & -t & \epsilon \end{array}\right)$$ (175)

Here we assumed periodic boundary conditions, meaning:

$$\langle 0 \vert H\vert N-1 \rangle = -t$$ (176)

$$\langle N-1 \vert H\vert 0 \rangle = -t$$ (177)

and obviously, the Hamiltonian is real and symmetric.

We find the solution by writing the wavefunction as a plane wave:

$$c_i = \frac{1}{\sqrt{N}} e^{ikR_i}$$ (178)

Because of the periodic boundary conditions, we have to impose a condition over the allowed values of $k$:

$$e^{ikNa}=1 \Rightarrow k=\frac{2\pi}{Na}m {\mathrm ;m:integer}$$ (179)

The resulting wavefunction is:

$$\vert\psi_k\rangle = \frac{1}{\sqrt{N}} \sum_{i=0}^{N-1} e^{ikR_i} \vert i\rangle.$$ (180)

It is easy to verify by calculating $H\vert\psi_k\rangle$, that it is indeed an eigenstate with an energy

$$\epsilon(k) = \epsilon-2t\cos{(ka)}$$ (181)

Next, we are going to verify that it also is an eigenstate of the displacement operator $T_R$, i.e that is invariant under translations of the lattice: First, we rewrite the wavefunction as:

$$\vert\psi_k\rangle = \frac{1}{\sqrt{N}} \sum_{i=0}^{N-1} e^{ikR_i} T_{-R_i}\vert\rangle.$$ (182)

Then, it is easy to see that

$$T_{R_j} \vert\psi_k\rangle = \frac{1}{\sqrt{N}} \sum_{i=0}^{N-1} e^{ikR_i} T_{-R_i+R_j}\vert\rangle.$$ (183)

$$= \frac{1}{\sqrt{N}} e^{ikR_j} \sum_{i=0}^{N-1} e^{ik(R_i-R_j)} T_{-R_i+R_j}\vert\rangle.$$ (184)

$$= e^{ikR_j} \vert\psi_k\rangle.$$ (185)

Hence, our wavefunction is a Bloch state. Another thing we notice is that the energy band is periodic, with perdio $2\pi/a$. Its is customary to represent it in a region between $-\pi/a$ and $+\pi/a$, which is nothing else, but the 1D Brillouin zone.