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PHY.K02UF Molecular and Solid State Physics | ||||
The eigen value problem that must be solved to determine the fcc phonon dispersion relation is, (show derivation)
\begin{equation} \left[ \begin{array}{c} 4 - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) & - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) & - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) \\ - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) & 4 - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) - \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) & - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) \\ - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) & - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) & 4 - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) - \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) \end{array} \right] \left[ \begin{array}{b} u_{\vec{k}}^x \\ u_{\vec{k}}^y \\ u_{\vec{k}}^z \end{array} \right]=\frac{M\omega^2}{C} \left[ \begin{array}{b} u_{\vec{k}}^x \\ u_{\vec{k}}^y \\ u_{\vec{k}}^z \end{array} \right] \end{equation}The eigen values $\lambda = \frac{M\omega^2}{C}$ can be found by setting the determinant of the matrix to zero.
\begin{equation} \left| \begin{array}{c} 4 - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) -\lambda & - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) & - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) \\ - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) & 4 - \cos\left(\frac{k_xa}{2} + \frac{k_ya}{2}\right) - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_ya}{2}\right) - \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) -\lambda & - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) \\ - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) & - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) + \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) & 4 - \cos\left(\frac{k_xa}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_ya}{2} + \frac{k_za}{2}\right) - \cos\left(\frac{k_xa}{2} - \frac{k_za}{2}\right) - \cos\left(\frac{k_ya}{2} - \frac{k_za}{2}\right) -\lambda \end{array} \right|=0 \end{equation}The matrix is symmetric with six independent elements. It has the form,
\begin{equation} \left| \begin{array}{c} m_{11} -\lambda & m_{12} & m_{13} \\ m_{12} & m_{22} -\lambda & m_{23} \\ m_{13} & m_{23} & m_{33} -\lambda \end{array} \right|=0 \end{equation}This can be written as a cubic equation in $\lambda$.
$$-\lambda^3 +(m_{11}+m_{22}+m_{33})\lambda^2+(m_{12}^2+m_{13}^2+m_{23}^2 - m_{11}m_{22} - m_{11}m_{33} - m_{22}m_{33})\lambda + m_{11}m_{22}m_{33} +2m_{12}m_{13}m_{23} - m_{12}^2m_{33} - m_{13}^2m_{22}-m_{23}^2m_{11} =0$$Cubic equations can be solved using Cardano's formula. The standard form of a cubic equation is,
$$a\lambda^3 + b\lambda^2 + c\lambda + d =0$$Code that will calculate the coefficients $a,b,c,d$ and the matrix elements from the $\vec{k}$-vector is,