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513.001 Molecular and Solid State Physics | |
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Tight-binding in 1-DElectrons move in periodic one dimensional potential with periodicity $a=$ 5 Å. Between $\frac{-a}{2}$ and $\frac{-a}{2}$, the potential is defined as, \[ \begin{equation} V(x) = 50\left(\frac{x}{a}\right)^2 \hspace{0.5cm}\text{[eV]}. \end{equation} \]
A harmonic oscillator has a similar parabolic potential. A mass-spring system has a potential energy of $\frac{\kappa x^2}{2}$ and oscillates at a frequency of $\omega=\sqrt{\frac{\kappa}{m}}$, where $m$ is the mass and $\kappa$ is the spring constant. Comparing this with the potential above we find that $\omega=\sqrt{\frac{25e}{a^2m_e}}=4.2\times 10^{15}$ [rad/s]. Here $e$ is the charge of an electron and $m_e$ is the mass of an electron. The energy levels of the harmonic oscillator problem are, \[ \begin{equation} E_n = \hbar\omega \left(n+\frac{1}{2}\right) = 2.75 \left(n+\frac{1}{2}\right) \hspace{0.5cm}\text{[eV]}. \end{equation} \]Use the tight-binding model to solve for the dispersion relation ($E$ vs. $k$) of an electron moving in the periodic potential. Use first two solutions to the harmonic oscillator as the atomic orbitals. The tight-binding wave function is then, \[ \begin{equation} \psi(x) = \sum\limits_{n} e^{ikna}\left(c_1\phi_0(x-na)+c_2\phi_1(x-na)\right). \end{equation} \]Where the first two normalized harmonic oscillator wave functions are, \[ \begin{equation} \phi_0(x) =\left(\frac{m_e\omega}{\pi \hbar}\right)^{\frac{1}{4}}e^{-m_e\omega x^2/2\hbar}\hspace{1cm}\text{and}\hspace{1cm}\phi_1(x) =\left(\frac{m_e\omega}{\pi \hbar}\right)^{\frac{1}{4}}\sqrt{\frac{2m_e\omega}{\hbar}}xe^{-m_e\omega x^2/2\hbar}. \end{equation} \]Using the parameters from above these solutions are, \[ \begin{equation} \phi_0(x) =5.8\times 10^4 e^{-1.8\times 10^{19} x^2}\hspace{1cm}\text{and}\hspace{1cm}\phi_1(x) =5\times 10^{14}xe^{-1.8\times 10^{19} x^2}. \end{equation} \]Compare your result with the bandstructure that is calculated numerically with the app: Band structure in 1-D. |
