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19.5: Eigenvectors of the Linear Chain

  • Page ID
    29523
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    Let’s get back to our chain, with eigenfunction equation of motion the \(N\) dimensional equivalent of

    \begin{equation}
    -m \Omega^{2}\left(\begin{array}{c}
    A_{1} \\
    A_{2} \\
    A_{3} \\
    A_{4}
    \end{array}\right)=\left(\begin{array}{cccc}
    -2 \kappa & \kappa & 0 & \kappa \\
    \kappa & -2 \kappa & \kappa & 0 \\
    0 & \kappa & -2 \kappa & \kappa \\
    \kappa & 0 & \kappa & -2 \kappa
    \end{array}\right)\left(\begin{array}{c}
    A_{1} \\
    A_{2} \\
    A_{3} \\
    A_{4}
    \end{array}\right)
    \end{equation}

    . We see the matrix is a circulant, so we know the eigenvectors are of the form

    \begin{equation}
    \left(1, \omega^{j},\left(\omega^{j}\right)^{2},\left(\omega^{j}\right)^{3}, \ldots,\left(\omega^{j}\right)^{N-1}\right)^{T}
    \end{equation}

    , which we’ll now write

    \begin{equation}
    \left(1, e^{2 \pi i j / N}, e^{2 \pi i 2 j / N}, e^{2 \pi i 3 j / N}, \ldots, e^{2 \pi i j(N-1) / N}\right)
    \end{equation}

    What does this mean for our chain system? Remember that the \(n^{\mathrm{th}}\) element of the eigenvector represents the displacement of the \(n^{\mathrm{th}}\) atom of the chain from its equilibrium position, that would be proportional to \(e^{2 \pi i j n / N}\)

    The steady phase progression on going around the chain \(n=0,1,2,3, \ldots\) makes clear that this is essentially a (longitudinal) wave. (The actual \(n^{\mathrm{th}}\) particle displacement is the real part of the \(n^{\mathrm{th}}\) element, but there could be an overall complex factor fixing the phase.)


    This page titled 19.5: Eigenvectors of the Linear Chain is shared under a not declared license and was authored, remixed, and/or curated by Michael Fowler.

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