A segment of trans-polyacetylene.
Consider a one-dimensional crystal with two valence orbitals per primitve unit cell. The prototypical example is trans-polyacetylene.
A segment of trans-polyacetylene.
The tight binding wavefunction is written in terms of two atomic orbitals, one centered on each carbon atom in the basis.
\[ \begin{equation} \psi_{k}\left(x\right)=\frac{1}{\sqrt{N}}\sum\limits_{n}e^{inka} \left( c_1\phi_1\left(x-na\right) +c_2\phi_2\left(x-na\right)\right) . \end{equation} \]To determine the dispersion relation, multiply the time independent Schrödinger equation from the left by each of the atomic orbitals and integrate over all space. This results in the following two equations,
\[ \begin{equation} \begin{array}{a} \langle\phi_1|\hat{H}|\psi_{k}\rangle = E\langle\phi_1|\psi_{k}\rangle , \\ \langle\phi_2|\hat{H}|\psi_{k}\rangle = E\langle\phi_2|\psi_{k}\rangle . \end{array} \end{equation} \]In the tight binding approximation, only the on-site and nearest neighbor matrix elements are retained on the left side of these equations and only the on-site term is retained on the right side. It is assumed that all the other terms are small enough that they can be ignored.
\[ \begin{equation} \begin{array}{a} \epsilon_1 c_1 -tc_2(1+e^{-ika}) = Ec_1 ,\\ \epsilon_2 c_2 -tc_1(1+e^{ika}) = Ec_2. \end{array} \end{equation} \]Where $\epsilon_1 = \langle\phi_1(x)|\hat{H}|\phi_1(x)\rangle$, $\epsilon_2 = \langle\phi_2(x)|\hat{H}|\phi_2(x)\rangle$, and $t = - \langle\phi_1(x)|\hat{H}|\phi_2(x)\rangle= - \langle\phi_2(x-a)|\hat{H}|\phi_1(x)\rangle$. Due to symmetry, the matrix element $t$ is the same for both nearest neighbors. This can be written in matrix form.
\[ \begin{equation} \begin{bmatrix} \epsilon_1 -E & -t\left(1+e^{-ika}\right) \\ -t\left(1+e^{ika}\right) & \epsilon_2 -E\end{bmatrix}\left[ \begin{array}{c} c_1 \\ c_2 \end{array} \right] =0. \end{equation} \]This matrix equation will have solutions when the determinant equals zero.
\[ \begin{equation} E^2-(\epsilon_1+\epsilon_2)E + \epsilon_1\epsilon_2 -2t^2(1+\cos(ka))=0. \end{equation} \]Solving for the energy, the dispersion relation is,
\[ \begin{equation} E=\frac{(\epsilon_1+\epsilon_2)\pm \sqrt{(\epsilon_1-\epsilon_2)^2+8t^2(1+\cos(ka))}}{2}. \end{equation} \]The form below generates a table of where the first column is the wavenumber $k$ in 1/m and the second and third columns are the energies of the two bands in units of eV.
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For trans-polyacetylene, the lower band is occupied while the upper band is unoccupied which makes trans-polyacetylene a semiconductor.