Relation between eigenvector and atomic displacements

The eigenvalue problem is described as,

$\displaystyle \omega^{2}\mathbf{e}(\mathbf{k},\nu) = \mathbf{D}(\mathbf{k})\cdot\mathbf{e}(\mathbf{k},\nu),$ (3.19)

where $ \omega^{2}$ and $ \mathbf{e}(\mathbf{k},\nu)$ are an eigenvalue and an eigenvector of this problem, respectively. The coordinate system is Cartesian. The $ \mathbf{e}(\mathbf{k},\nu)$ is called polarization vector which corresponds to the normalized atomic displacements weighted by the square root of the atomic mass as written by,

$\displaystyle \mathbf{e}(\mathbf{k},\nu) = \begin{pmatrix}\sqrt{m_1}U_{x}(1,\ma...
...thbf{k},\nu) \\ \vdots \\ \sqrt{m_{n}}U_{z}(n,\mathbf{k},\nu) \\ \end{pmatrix},$ (3.20)

where $ \mathbf{U}$ is the amplitude vector of an atomic displacement. The $ \omega$ corresponds to vibrational frequency. The $ \mathbf{e}(\mathbf{k},\nu)$ is complex number and the phase factor at given instant is expressed by $ \arg(\mathbf{e}(jl,\mathbf{k},\nu))$. The normalization is expressed as,

$\displaystyle (\mathbf{e}(\mathbf{k}))^{\mathrm{T}}\cdot (\mathbf{e}(\mathbf{k}))^{*} = (\mathbf{e}(\mathbf{k}))^{\mathrm{T}}\cdot (\mathbf{e}(\mathbf{-k})) = 1$ (3.21)

The normalized atomic displacement is obtained by,

$\displaystyle \mathbf{u}(jl,t)=\sum_{\mathbf{k},\nu}\mathbf{U}(j,\mathbf{k},\nu)\exp(i[\mathbf{k}\cdot\mathbf{r}(jl)-\omega(\mathbf{k},\nu)t]).$ (3.22)

The origin where the radial vector is measured can be set at an arbitrary point in the real space. An eigenvector calculated at an exact k-point can be represented by the atomic modulations in the calculated supercell at an arbitrary instance $ t$. However the amplitude of the displacements is not described by Eq. (3.24), but is calculated by the thermodynamic properties.

togo 2009-02-12
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